berkley general chemistry 1

GeneralChemistry

PartiSections I-V

Section IStoichiometry

Section IIAtomic Theory

Section IIIEquilibrium

Section IVAcids & Bases

Section VBuffers & Titrations

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Specializing in MCAT Preparation

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Section I

Stoichiometryby Todd Bennett

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Unit Conversion

a) Dimensional Analysisb) Density Determinationc) Typical Conversions

Elemental Analysisa) Mass Percentb) Empirical Formulasc) Molecular Formulasd) Combustion Analysis

Solution Concentrationa) Units and Terminology

i. Molarityii. Molalityiii. Mass Percent (in Solution)iv. Density

b) Dilutionc) Beer's Law

Balancing Reactionsa) Standard Balancingb) Limiting Reagents

Reaction Typesa) Common Reactionsb) Oxidation States

Test-Taking Tipsa) General Adviceb) Mathematical Tricks

i. Addition and Subtractionii. Averagingiii. Multiplicationiv. Division

f?EBKELEYSpecializing in MCAT Preparation

Stoichiometry Section GoalsKnow how to convert one kind of concentration unit into another.The concentration of a solution can be measured in terms of molarity, molality, and density. Youmustknow thedefinitions ofeach unitandhow they differ from one another. Although the testdoes not feature a great deal of math, you should have an idea of how to convert between units.

Understand the difference between empirical and molecular formulas.Know the difference between the molecular formula (actual ratio of atoms in a molecule) and theempiricalformula (simplestwholenumber ratio of the atoms in a molecule). Befamiliarwith theexperimentsand informationneeded to determineboth of the formulas.

Know the effect of standard conditions.Standard conditions are defined as 1 atm. and 298 K for thermodynamics, but STP (standardtemperature andpressure) isdefined as1atm. and273 K. Many calculations ofgasvolume usetheideal gas assumption that at STP, one mole of gas occupies22.4liters.

Understand dilution and its effect on concentration of a solute.Dilution involves a reduction in the concentration of a solute in solution by the addition of solventto the mixture. The addition of solvent therefore dilutes the concentration, but does not change themoles of solute. The equation that you must recallis based on the constant number of solute moles:Minitial-Vinitial = Mfinal-Vfinal.

Recognize standard reactions from general chemistry.Themost commonly recurring reactions in general chemistry that you are expected to know includecombustion, single replacement, double displacement, and proton transfer, to name just a few. Youmust recognize these reactionsand have a basicunderstanding of them.

Recognize the limiting reagent in a reaction and know its effect on the reaction.The limiting reagent dictates theamount ofproduct thatcanbeformed andconsequently thepercentyield for a reaction. Using only starting values andthestoichiometric equation, you mustbe ableto determine which reactant is the limiting reagent in the reaction.

Understand the stoichiometric ratios in combustion reactions.In the combustion of both hydrocarbons and carbohydrates, there is a consistent relation betweenthe number of oxygenmolecules on the reactant side, and the number of water and carbon dioxidemolecules that form on the product side. Know each reaction so that you may easily balance thecoefficients.

General Chemistry Stoichiometry

StoichiometryThe perfect spot to start any review of general chemistry is the basics, whichtraditionally include stoichiometry and chemical equations. The mostfundamental perspective of a chemical reaction, where bonds are broken so thatnew bonds can be formed, is at the atomic and molecular levels. Due to theminute size of atoms, we can never actually view a chemical reaction (so statesHeisenberg's uncertainty principle). We must therefore rely upon developingmodels that can account for changes in all of the atoms and molecules involvedin a chemical reaction or physical process. At the molecular level, we considermolecules. At the macroscopic level, we consider moles. Stoichiometry allowsus to convert one into the other and to shift between these two perspectives. Thenumber of molecules is converted into the number of moles using Avogadro'snumber (6.022 x 1023). The concept of a mole is based upon the amount ofcarbon-12 that is contained in exactly 12.0 grams of carbon, a quantitydetermined by knowing the volume of a 12.0-g carbon sample, the type ofmolecular packing in it, and the dimensions of the carbon atom. This task ofquantifying atoms in a mole is similar to guessing the number of peas that arecontained in an aquarium. It is important that you utilize the mole concept tounderstand, and later to balance and manipulate, chemical equations.

In the stoichiometry section, we focus on those skills needed to solve ratioquestions. Stoichiometry is most commonly thought of as the mathematicalportion of general chemistry. The MCAT, however, has relatively fewcalculations. It is a conceptual test, emphasizing logical thought process ratherthan calculations. For some of you, this is great news. But before celebrating toomuch, consider where the mathematical aspects of general chemistry fit into aconceptual exam. The MCAT does involve some math, but it is not toocomplicated. Math-related calculations required for MCAT questions involvemaking approximations, determining ratios, setting up calculations, andestimating the effect of errors on results. The initial problems presented in thissection involve slightly more calculations than you should expect to see on theMCAT. Some of them may look familiar to you from your general chemistrycourses and should stimulate your recall. As the section proceeds, less emphasisis placed on calculating and estimating, and more emphasis is placed on the artof quickly determining ratios and approximating values.The focus of the stoichiometry section is problem-solving, with special attentionto the idiosyncrasies of each type of problem. Definitions of important terms arepresented with sample questions and their solutions. Answer solutions discusstest strategy and the information needed to obtain the correct answer. Eachproblem in the stoichiometry section represents what we might call the "bookkeeping" of reactions in general chemistry, and it offers an ideal opportunity tobeginwork on fast math skills as well. As you do each of the questions, learn thedefinitions and develop an approach that works well for you. Youmay want toconsider multiple pathways to arrive at the correct answer. It is important thatyou be able to solve questions in several different ways and to get into themindset of the test writers. As you read a passage, think about the questions thatcould be asked about it. If a passage gives values for various masses andvolumes, there will probably be a question about density. If it gives values formoles and solution volume, there will probably be questions on concentrationand dilution. Use your intuition and common sense as much as you can, andmake every effort to develop your test-taking logic.

Introduction

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General Chemistiy Stoichiometry Unit Conversion

Unit ConversionDimensional AnalysisDimensional analysis is a mathematical conversion from one set of units intoanother. It involves multiplyinga given value by a conversion factor or a seriesof conversion factors until the value is finally expressed in the desired units.Converting from one set ofunits to another is a critical skill needed toeliminateincorrect answer choices in the physical sciences section. Be systematicwhenconverting between units. The standard measurements that can be expressed ina variety ofunits are distance (1 m =1.094 yd, 2.54 cm=1.00 in,and 1.609 km=1.00 mile), mass (1.00 kg = 2.205 lb and 453.6 g = 1.00 lb), volume (3.79 L = 1.00gal and 1.00 L= 1.06 qt), and time (3600 s = 1.00 hr). Always convert units asthey appear in a problem into the units indicated in the answer choices (the so-called "target units").

Example 1.1Sprinters can run 100 meters injustunder 10 seconds. Atwhat average speed inmiles perhourmusta runner travel to cover 100 meters in 10.0 seconds?A. 3.7 miles/hourB. 11.2 miles/hourC. 22.4 miles/hourD. 36.0 miles/hour

SolutionThe first task is to determine the given units and the target units. From there,convert the given units into the target units. We are given 100 meters in 10seconds, but the answer choices are expressed in miles per hour. Use the correctconversion factors, as follows:

10.0 s hour

Conversion ofdistance: 10°mxmiles - miles10.0 s rn s

Conversionof time: 10°mx-§- =-^3_10.0 s hr hour

mnm.. 1km x 1mile y 3600s _ 100x3600 miles10.0 s 1000 m 1.609 km 1.00 hr 10 x 1000 x 1.609 hour

100x3600 _ 3600 _ 36 miles10x1000x1.609 10x10x1.609 1.609 hour

36<2

36 ,<36 where 36 _ 18and 36 = 36. So18<^6_ < 361.609 1 2 1 1.609

Only choice C fallswithin the range of 18 to 36.

A frequent task in chemistry is the conversion between various types oftemperature units, volume units, pressure units, and concentration units.Chemists, like most scientists, employ the MKS system, so the conversion fromconventional units less commonly used in science to MKS units is routine.Example 1.2 demonstrates the interconversion between the conventionalFahrenheit unit and the scientific Celsius unit of temperature.

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General ChemiStiy Stoichiometry Unit Conversion

Example 1.2At what temperature is the numerical value the same, whether the units are inCelsius or Fahrenheit?

A. 32°B. 0°C. -40°D. -273°

SolutionThe formulas for conversion between Celsius and Fahrenheit are as follows:

T.F =£t.c +32 .-. T-c = §• (T-f- 32)5 9

Answering the question requires setting T>f= T>(\

T-F = £ T-c +32 becomes: T=£T+32, soT=1.8 T+325 5

T = 1.8T + 32=> -0.8T = 32.\ T = -40°

Density DeterminationThe density of a material or solution is the mass of the sample divided by thevolume of the sample. Density is a measured quantity, determinedexperimentally. Understand the techniques used to measure density. The termspecific gravity refers to the density of a material relative to the density of water,and may be used in a question in lieu of density. For our purposes, specificgravity means the same thing as density, but it has no expressed units.Determining density is a typical example of dimensional analysis.

Example 1.3Exactly 10.07 mL of an unknown non-volatile liquid is poured into an empty25.41-gram open flask. The combined mass of the unknown non-volatile liquidand the flask is 34.12grams. What is the density of the unknown liquid?

A 34.13 g n 8.71 gB.

10.07 mL 10.07 mL

c 10-07 g D 8.71 mL8.71 mL * 10.07 g

SolutionThe density of the liquid is found by dividing the mass of the liquid by thevolume of the liquid. This results in units of grams per milliliter, whicheliminates choice D. The volume of the liquid is 10.07mL, so 10.07 should be inthe denominator. This eliminates choice C. The mass of the liquid is thedifference between the final mass of the flask and liquid combined, and the massof the flask (34.12 - 25.41), which is equal to 8.71 grams. This means that thenumerator should be 8.71. The correct answer is choice B. incidentally, thequestion did not state the reason for using a non-volatile liquid. The liquid mustbe non-volatile, to prevent any loss due to evaporation from the open flask. Inthe event the liquid evaporates away, then the mass you determine is too small,due to the loss of vapor molecules.

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General Chemistry Stoichiometry Unit Conversion

Density questions may onoccasion involve a more intricate conversion ofunits.Forany density question, keep in mind that the targetunits are mass solutiondivided byvolume solution. The mass percent ofa solute ismass solute dividedbymass solution. The product of thedensity andmass percent is mass solutedividedby volume solution. Converting the massof solute intomoles of soluteyields molarity. Example 1.4shows this.

Example 1.4Whatis themolarity ofa 3%NaClsolution with a densityof1.05 grams/mL?A. 0.497 M NaClB. 0.504 M NaClC 0.539 M NaClD. 0.724 M NaCl

SolutionThefirststep is to determine theunits you are looking for, whichin this exampleis moles solute per liter solution. You must find both moles solute and literssolution. The density of the solution is 1.05 grams/mL which means that oneliter of the solutionweighs1050 g. Threepercent (3%) of the solution is sodiumchloride, so the mass of sodium chloride is 0.03 x 1050g. This is the same as 3%of 1000 g + 3% of50g, which is 30g + 1.5 g = 31.5 g ofNaClper liter solution.The grams of sodium chloride are converted to moles by dividing by themolecular weightofNaCl(58.6 grams/mole). Theunit factor method is shownbelow:

1.05 gsolution x1000mLx 3gNaCl x1mole NaClmL solution L 100gsolution 58.6g NaCl

_ 1.05 x 1000 x 3 moles NaCl - 3.15 x lOmoles NaCl100 x 58.6 L solution 58.6 L solution

On the MCAT, you will not have time to solve for values precisely, so you mustmakean approximation. Select the answer that is closest to that approximation.

3L5.>30 _1_ somevalue is greaterthan0.500 M58.6 60 2

315.<33 _11 __55./ somevalue is lessthan 0.550 M58.6 60 20 100

The value falls between 0.500M and 0.550 M, so choices A and D are eliminated.Next youmust choose between 0.504 M and 0.539 M. The value is not closeenough to 0.504 M, so you should choose C, and be a wise student! Wisestudents are a good thing. Some questions on the MCAT may presentmathematical set-ups,without solvingfor an exactnumber.

The physical sciences section of the exam incorporates physics and generalchemistry, so from the beginning of your review, make a conscious effort toconsider physics when working on general chemistry and to consider generalchemistry whenworking on physics. Determining density is a problem commonto both disciplines. Example 1.5 shows an approach to the concept of densitythat is more typical ofwhat is found in a physicsproblem.

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Example 1.5What can be concluded about the density of a metal object which, when placed ina beaker of water at room temperature, sinks to the bottom?A. The density of the metal is less than the density of either water or ice.B. The density of the metal is less than the density of water, but greater than the

density of ice.C. The density of the metal is greater than the density of water, but less than the

density of ice.D. The density of the metal is greater than the density of either water or ice.

SolutionWhen an object floats in a liquid medium, its density is less than that of themedium surrounding it. The fact that it floats means the buoyant force pushingupward against it (pmedium'Vobjecfg) Is greater than gravitation force pushingdownward (weight = mg = PobjecfVobject'g)- Thus, an object floats whenPmedium > Pobject- Because the metal object sinks in water, it must be denserthan water. Ice floats in water, meaning that ice is less dense than water and thusless dense than the metal object. The density of the metal must be greater thanthe density of either water or ice. The correct answer is therefore choice D.

Typical ConversionsIn chemistry, conversions between products and reactants are common, so themole concept is frequently employed. The mole concept is pertinent in theinterconversion between moles and mass, using either atomic mass (forelements) or molecular mass (for compounds). These calculations involve usingthe unit factor method (also known as dimensional analysis.)

Example 1.6How many moles of NaHCC>3 are contained in 33.6grams NaHCC>3?A. 0.20 moles NaHC03B. 0.40 moles NaHC03C. 0.50 moles NaHC03D. 0.60 moles NaHC03

SolutionThe first step in determining the number ofmoles is to determine the molecularmass of NaHC03. The mass is 23 + 1 + 12 + 48 = 84 grams. The number of molesof NaHC03 is found by dividing 33.6 by 84, which is less than 0.50. Thiseliminates choices C and D. The number is greater than 0.25 (21 over 84) andthus greater than 0.20, so choiceA is eliminated. The only value left is choiceB,0.40 moles.

Beyond deteirnining the moles from grams for the same compound are questionswhere the moles of products are determined from the grams of reactants. Thesequestions require converting from grams of a given substance to moles of thegiven substance, and then expressing the quantity of a final substance in terms ofmoles, grams, or liters. Bybalancing the reaction, the mass of a selected productthat is formed in the reaction can be calculated based on the mass of a selectedreactant (which must be the limiting reagent). Examples 1.7 and 1.8 involvedetermining moles, mass, and volume from the given values.

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Example 1.7Based on the following reaction, how many grams of water would form from0.33moles C4H10O reactingwith an unlimited amount of oxygen gas?

C4HioO(g) + 6 02(g) * " 4C02(g) + 5H20(g)A. 18.00 gramsB. 24.00 gramsC. 30.00 gramsD. 36.00 grams

SolutionWith an excess of oxygen, the limiting reagent in this reaction is C4H10O. Theamount of water formed is determined by the 0.33 moles of C4H10O reactant.Using the balanced equation, the ratio of H20 to moles C4H10O is 5 :1, so 1.667moles of water are formed. At 18 grams per mole, this means that fewer than 36grams but more than 27grams are formed. Thismakes choiceC the best answer.

0.33molesC4H10Ox 5molesH2Q x 18gH2Q =ix5x 18gH20=30gH201 mole C4H10O 1 mole H20 3

Example 1.8How many liters ofC02(g) result from the complete decomposition of 10.0gramsof CaC03(s) to carbon dioxide and calcium oxide at STP?

CaC03(s) *~ CaO(s) + C02(g)A. 1.12 litersB. 2.24 litersC. 3.36 litersD. 4.48 liters

SolutionYou are asked to determine the amount of product from a known quantity ofreactant. The first step in problems of this type is to make sure the reaction isbalanced. In this case, it is already balanced. The mole ratio of the twocompounds is 1 : 1. The required conversion involves changing from massreactant, to moles reactant, to moles product, and finally to volume product.This is one variation of unit conversion via mole ratio calculation. In addition,there is the "g - m - m - g" conversion and the "v - m - m - g" conversion. Youneed three steps to go from grams reactant to the target (liters product). Unitsare important here. The units for the mass of reactant is grams. You need tomultiply massbymolesand divide by grams. This is the same as dividing by theMW. The second step is to read the mole ratio from the balanced equation. Inthis reaction, the mole ratio is 1:1 (the units of both numerator and denominatorare moles). The third and final step is to convert from moles product into litersproduct (i.e., multiply by liters and divide by moles.) This is done bymultiplying by the molar volume of the product gas, which at STP (standardtemperature and pressure) is 22.4 liters.

10gramsCaCO3X l™leCaC03 x 1mole CO; x22.4 liters CQ2100gramsCaC03 lmoleCaC03 lmoleCOfc

=10x22.4 = 2.24 liters C02100

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General Chemistry Stoichiometry

Elemental AnalysisElemental analysis determines the atomic composition of an unknown molecule.It is based on the idea that all molecules of the same substance combine atoms ofthat substance in the same way. In other words, water always has two hydrogenatoms and one oxygen atom. Because of this feature of structural uniformity, it ispossible to determine the atomic composition of any molecule. The fundamentalprocess of elemental analysis involves oxidizing an unknown completely andcollecting the products. The amount of each element that was present in theunknown compound can be determined from the amount of oxidized product.These mass values can be converted to mass percent and mole ratio values. Inthe determination of the empirical formula, the mass percent is converted to arelative mass value and then a relative mole value. The mass percent of anelement within a compound must be determined prior to determining theempirical formula for an unknown compound. An empirical formula, you mayrecall, is the simplest whole number ratio of the atoms in a molecule.

Mass Percent (Percent Composition by mass)The mass percent of a particular element within a compound is found bydividing the mass of that element by the mass of the compound and thenconverting the fraction to a percentage. This is shown in Equation 1.1.

mass percent = mass atoms x 100o/omass compound

(1.1)

Mass percent can never exceed 100%for any component element. Determiningthe mass percent of an element from the molecular formula is a straightforwardtask, although the math may be challenging. Mass percent questions can beasked in a conceptual or mathematical manner. Mass percent is independent ofthe total mass of the sample of compound.

Table 1.1shows the relative masses of oxygen and carbon from different samplesof carbon dioxide. This demonstrates the law of multiple proportions. Atomscombine in a fixed ratio in terms of mass and moles. Note that the outcome is thesame in all four trials measuring the ratio of oxygengas that reactswith a knownmass of carbon. The experiment involves oxidizing a known amount of carbonand collecting the product gas. Themass of this product gas is deterrnined, andthe mass of oxygen is assumed to be the difference between the initial and finalweighed masses of the carbon sample.

Mass Carbon Mass Oxygen MassO/'Mass C

1.33 g 3.53 g 3'53/l.33 =2-651.07 g 2.87 g lx 07 =2.68

1.11 g 2.96 g 2.96 / _ 9 fin/l.H-2.67

1.27 g 3.39 g 339/127 =2.67Table 1.1

The mass ratio of oxygen to carbon in the four trials averages out to be 2.67 : 1,which is roughly 8 : 3. This means that for the oxidation product of carbon, theratio of oxygen to carbon is 8 grams to 3 grams, equivalent to 2moles to 1mole.

Elemental Analysis

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General Chemistry Stoichiometry Elemental Analysis

Example 1.9How much calciummetal combineswith one gram of oxygen to form CaO?A. 1.00 gCaB. 1.25 gCaC. 1.67 gCaD. 2.50gCa

SolutionFrom the molecular formula, the mole ratio of calcium to oxygen is 1 : 1. Theatomic mass of calcium (Ca) is 40.08, while the atomic mass of oxygen (O) is16.00. The mass ratio for the compound is 40.08to 16.00, which reduces to 2.505 :1,which rounds to 2.50 to 1. Thismeans that 2.50grams of calcium combine with1.00 grams of oxygento formCaO. ChoiceD is best.

Example 1.10What is the mass ratio of iron to oxygen in Fe203?A. 1.08g Fe to 1.00g OB. 1.63 g Fe to 1.00 g OC. 2.33 gFe to 1.00gOD. 3.49 g Fe to 1.00g O

SolutionFrom the molecular formula, the mole ratio of iron to oxygen is 2: 3. The atomicmass of iron (Fe) is 55.85,while the atomic mass of oxygen (O) is 16.00. The massratio of iron to oxygen for the compound is 2(55.85) to 3(16.00), which equals111.7 : 48.0. This ratio is approximately equal to 116 : 50, or 232 : 100, whichreduces to 2.32 :1. Both numbers must be increased proportionally to keep theratio the same. Choice C is a ratio of 2.33 to 1, which is the closest of the choices.This means that 2.33 grams of iron combine with 1.00grams of oxygen to formFe203- Choice C is best. You should note that iron and oxygen can combine tomake other compounds (with different molecular formulas). One of thesecompoundsis FeO, with a massratioof55.85 to 16.00, which reduces to a ratioof3.49 :1.00. The mass ratio (and mole ratio) of an oxide can be used to identify aspecific compound. Thisprocessis known as combustion analysis.

Examples 1.9 and 1.10 demonstrate how mass percent questions can bemathematical. Masspercent questions can also be asked in a conceptual manner,where the relative masspercentage of a specific element is compared for severalcompounds. Examples 1.11,1.12, and 1.13 demonstrate some different formsofthis type of question, starting with typical examples and graduating to moreabstract ways of asking for mass percent.

Example 1.11What is the mass percent of oxygen in carbon dioxide?A. 27.3%B. 57.1%C. 62.5%D. 72.7%

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SolutionThe mass of carbon in C02 is 12 grams, and the mass of oxygen in C02 is 32grams. The total mass of C02 is 44 grams, so the mass percent of oxygen is theratio of 32 to 44. This ratio reduces to 8 over 11.

Mass percentO = 32g° x100% =22. x 100% =-§- x100%44gC02 44 11

Quick Calculation Technique:Quick calculations require knowing the values of selected fractions. One-eleventh is equal to 0.091; therefore, eight-elevenths is equal to 8(0.091) = 0.728.Thismethod gets an exact value and is very fast, if you know how to do it.

_8_ = 8x J- = 8x 0.091 = 0.728 = 72.8%11 11

Narrowing-Down-Choices Technique:On a multiple-choice exam, you can eliminate answers by narrowing down therange into which the answer fits. 8 over 11is less than 9 over 12,but greater than7 over 10. A range has beenestablished between -2- and -7-. 9 over12is 75%,

12 10and 7 over 10 is 70%, so the correct answer falls between 70% and 75%.

-2_ > _8_ > _7_, where -2- = 75% and-7- = 70%. So: 75% >-§- > 70%12 11 10 12 10 11

Choice D is the best answer.

Example 1.12What is the mass percent of nitrogen in NH4NO3?A. 28%B. 35%C. 42%D. 50%

SolutionThe total mass of the nitrogen in the compound is 28 g/mole, because there aretwo nitrogen atoms in the compound at 14 g/mole each. The mass of thecompound is 28+ 4 + 48= 80g/mole. You must divide 28 by 80quickly. Thecommon denominator ofboth is 4. Reducing by 4yields afraction of -7-.

Mass percent N= ?**£** x 100% =2&x 100% =-7- x 100%80gNH4NO3 80 20

Quick Calculation Technique:Quick calculations may involve getting a denominator to some easy-to-usenumber, such as 10,100, or 1000. It is easy to convert a fraction into decimals orpercents when the denominator is either 10, 100, or 1000. For this question, adenominator of 100 works well. To convert 20 to 100, we must multiply by 5.Multiply both numerator and denominator by 5, to change the fraction -7- to-^2_.

V3 y 6 20 100The percentage is 35%, so choice B is correct.

28 = _7_ = 7x5 = .35. = 0.35 = 35%80 20 20x5 100



Narrowing-Down-Choices Technique:Narrowing down the range into which the answer fits can be applied to anymultiple-choicemath question. 28 over 80 reduces to 7 over 20. The value of 7over 20 is greater than 7 over 21, but less than 8 over 20. A range has beenestablished between-7- and -£-. Thevalue of 7 over 21is 33.3%, and the value of

21 208 over 20 is 40%, so the correct answer falls between 33.3% and 40%. Only choiceB fits in this range, so choice Bmust be the correct answer.

-7_ < -Z. < A, where -7- = 33.3% and-§- = 40%. So33.3% <-7- < 40%21 20 20 21 20 20

Example 1.13Which of the following samples yields the MOSTmoles of sodium cation?A. 1.0 g NaClB. l.OgNaBrC 1.0gNaNO3D. 1.0gNa2CO3

SolutionThis question is a subtle way of asking, "Which salt has the greatest mass percentof sodium?" All choices are 1.0 g of compound, so the most moles of sodium arefound in the compound with the greatest mass percentage of sodium. Choices A,B, and C have the same number of sodium atoms in the compound (one), so theyeach have the same numerator in the mass percent formula. The best choice ofthose three salts is the lightest compound (resulting in the smallest denominatorwhen calculating mass percent). The lightest compound of the three choices A,B, and C is the salt with the lightest anion, which is choice A, NaCl. Now thequestion involves comparing the mass percent of sodium in NaCl to the masspercent of sodium in Na2C03, choice D.

Mass percent Na in NaCl = 23,0gNa x100% =-22_ x 100%58.5 g NaCl 58.5

Mass percent Na inNa2C03 = 46,0gNa— x100% =-^- x100% =23 x100%106gNa2CO3 106 53

22. > -2£_f sochoice D is thebest answer.53 58.5

Empirical FormulaAn empirical formula for molecules uses the smallest whole number ratio of theatoms in a compound. It is the formula that gives the relative numerical valuesfor each element in the molecule in such a way that the numbers in the ratiocannot be reduced without involving fractions. An empirical formula may ormay not be the actual formula of the molecule. It is calculated from the masspercentage of each element within a compound. You may recall from yourgeneral chemistry courses that we start by assuming a 100-gram sample, so thatthe percentages can be changed easily into mass figures. From this point, it is amatter of converting from mass into moles, using the atomic masses for eachelement. The empirical formula is a whole number ratio of these mole values.Empirical formulas must include whole number quantities as subscripts.

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Example 1.14Which of the following is an empirical formula?A. C2H6B. C3H8C. C4H10D. C6H6

SolutionAn empirical formula for a molecule is defined as the formula in which theconstituent atoms are in their smallest possible whole number ratio. Choice A,C2H6, can be reduced to C1H3 (normally written as CH3), so choice A is not anempirical formula. Choice C, C4H10, can be reduced to C2Hs, so choice C is notan empirical formula. Choice D, CgH6, can be reduced to C1H1 (normallywritten as CH), so choice D is not an empirical formula. This eliminates all of thechoices except choice B, C3H8. The ratio of 3 : 8 cannot be reduced any further,so C3H8 is an empirical formula, making choice B the correct answer. In the caseof C3H8, the empirical formula and the molecular formula are the same, becausethe compound is completely saturated with hydrogens. CgHi6 has too manyhydrogens and is not a possible formula. Organic chemistry rules can help tosave time on formula questions.

Example 1.15What is the empirical formula for a compound that is 72% C, 12% H, andcomposed solely of carbon, hydrogen, and oxygen?A. C3H60B. C6Hi20C. C6H14OD. C7H14O

SolutionFor empirical formula calculations, assume a 100-gramsample of the compound.A 100-gramsample in this case would contain 72 grams C, 12 grams H, and 16grams O. The 16 grams of oxygen are determined from the difference betweenthe mass of carbon plus hydrogen and the 100grams of sample. Next, you mustconvert the grams of each element into the corresponding moles of each element.To go from grams to moles, divide the mass of the element by its atomic mass. Inthis case,72 grams of C is equivalent to 6 moles ofC, 12grams ofH is equivalentto 12 moles of H, and 16 grams of O is equivalent to 1 mole of O. This is aparticularly easy example, because the ratios turn out to be whole numbers. Incases where they don't come out whole, you must divide the mole quantity ofeach element in the compound by the lowest mole quantity for any of theelements. However, for this example the best answer is choiceB. Drawn belowis a useful layout of empirical formula calculations. It is often easier just to plugterms into an equation such as this, because you don't have to show your workon the MCAT. Do the questions as quickly and carefully as you can,emphasizing organization in your path to a solution.

Q percentage carbon j-Jpercentage hydrogenQ percentage oxygen = Cl72Hl2Ql6 =CgHl2Olmolarmasscarbon molarmasshydrogen molarmassoxygen 12 1 16

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Example 1.16What is the empirical formulaof an oxide of sulfur that is 60% oxygenbyweight?A. SOB. S02C. s2o5D. SO3

SolutionHere again you should assumea 100-gram sample. A100-gram sample wouldhave 60 grams of oxygen and 40 grams of sulfur. The moles ofO=°-2> and the

16moles ofS=^. $0. ismore than double 40., so there aremore than two oxygen

32 16 32atoms persulfur atom. This eliminates choices AandB. Upon reducing sQ. to20.,

r 32 16we see that the ratio of oxygen to sulfur is 3 :1, making choice D the best answer.

g percentage sulfurQ percentage oxygen = S40_O60 = S20O60 =S \O3molarmasssulfur molarmassoxygen 32 16 16 16

The test emphasizes ratios, so the more numerical intuition you develop, thebetter. To make problem-solving less mathematical, focus on eliminating choicesby comparing relative ratios. An alternative way to ask an empirical formulaquestion with reduced math is shown in Example 1.17.

Example 1.17If a molecule is composed of only two elements (Xand Y), and if X and Ycombine in equal mass quantities, and if Y is less than twice as heavy as X,whichof the following molecular formulas is NOT possible?A. XYB. XY2C X3Y2D. X3Y

SolutionIf Ywere exactly twice as heavy as X, then equal masses of Xand Ywould resultin exactly twice as many moles of Xas Y,a 2 :1 ratio of X to Y. Because Y is lessthan twice as heavy as X, there are fewer than twice as many moles of X as Y.Thus, the ratio of X : Ymust be 2 :1 or smaller. This limiting ratio is true of allthe answers except choice D. The wording of this question allows for thepossibility that the molecular mass of X is equal to or greater than Y.

Molecular FormulaThe molecular formula is the actual mole ratio of the elements within thecompound. The molecular formula is found by multiplying the empiricalformula by the whole number ratio of the molecular mass to the empirical mass(including 1, in some cases). Therefore, conversion from the empirical formula tothe molecular formula requires knowing the molecular mass of the compound. Ifthe molecular mass is double the empirical mass, then all of the elements in theempirical formula are doubled to get the molecular formula.

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Example 1.18What is the molecular formula for a compound that is 82.76%C, has a molecularmass of 58.1 grams per mole, and is composed solely of carbon and hydrogen?A. CH2B. C2H5C. C3H8D. C4H10

SolutionChoice B is eliminated immediately, because a hydrocarbon cannot have an oddnumber of hydrogens. For neutral C2Hs to exist, it would have to be a freeradical. Continuing to use organic chemistry logic, choice A is not physicallypossible. One carbon requires four bonds to hydrogen atoms to form a stablemolecule (methane). CiH2 would be a carbene (:CH2), which is not stable due toits lack of an octet. To decide between choices C and D, you must first find theempirical formula, and then convert that into the molecular formula.

C percentage carbon H percentage hydrogen = C82.76H17.24 = Cg.93H 17.24molarmasscarbon molarmasshydrogen 12 1

empirical formula: CemHlZM.=C\H!Z£. =QH2.5 =QH5r 6.93 6.93 7

An empirical formula of QH5 when multiplied by a whole number cannot yieldC3H8, so choice C is eliminated. This leaves only choiceD. The correct ratio ofmolecular mass to empirical mass confirms that choiceD is the best answer. Themolecular formula is found using the molecular mass of 58.1 grams per mole.The empirical mass is 2(12) + 5 = 29. This value is only half of the molecularmass, so the formula must be doubled to yield C4H10. This question could havebeen solved in seconds by seeing that only choiceD has a molecular mass of 58.

Example 1.19An unknown stable gas is composed of 13.10% H, 52.23% C, and the remainderO. A 0.10-mole sample weighs 4.61 grams. What is the molecular formula forthe compound?A. C2H60B. C3H80C. C3H9OD. C4HgO

SolutionA stable compound made of carbon,hydrogen, and oxygen cannothave an oddnumber of hydrogens, so choiceC is eliminated. Neutral C3H9O would have tobe a free radical. The remaining choices obey the octet rule. The molecular massis 46.1, so choice A is the correct answer. That is the method you should use on amultiple-choice exam. Now let's confirm that by using the molecular formula.First you must assume a 100-gram sample, resulting in 13.10 g H, 52.23 g C, and34.67 g O. Next, the numbers are converted intomoles, and then the values arereduced to a whole number ratio. The calculation of the empirical formula isshown below, where a formula arrangement is used to help keep track of thevalues.

C52.23 H13.10 Q34.67 = C4 4H13 102 2 =CilHlM-O^ =QHaOi12 1 16 ' 2.2 2.2 2.2

The empirical mass is 2(12) + 6 + 16 = 46. This value is equal to the molecularmass, so the empirical formula is the molecular formula.


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Combustion Analysis (an Experimental Procedure)Combustion analysis entails determining the mass percent of each componentelement in an unknown compound. It is accomplished by oxidizing theunknown with excess oxygen (to ensure completecombustion), followed by theseparation and collection ofall ofthe oxidized products. This isanexperimentalprocedure, which makes it a likely topic on a conceptual exam such as theMCAT. When a hydrocarbon is oxidized, carbon dioxide and water are formed.Carbon dioxide and water can be separated using various methods. One methodinvolves passing theC02 gas andH20 vapor across a hygroscopic saltof knownmass. The hygroscopic salt absorbs the water, and thus increases in mass.Thehygroscopic saltmust not react with carbon dioxide. A good choice for thehygroscopic salt is either calcium chloride or magnesium sulfate (both ofwhichyoushould have used as drying agents in yourorganic chemistry lab). Once thewater is absorbed, the remaining gas is passed across a sample of KOH of knownmass. KOH undergoes a combination reaction with carbon dioxide to formpotassium bicarbonate (KHCO3). The potassium hydroxide salt absorbs thecarbon dioxide, and thus increases in mass. Knowing the masses of C02 andH20, we can determine the masses of carbon and hydrogen by multiplying themass percent of each element by the mass of its respective oxide product thatwas collected. Upon dividing these numbers by the mass of the original sample,themass percents ofhydrogen and carbonin the original sampleare determined.The carbon dioxide and water can be separated and collected using a differentmethod than passing the gasesover salts that bind the products. Bylowering thetemperature, carbon dioxide and water can be converted to their solid states.Because solids do not flow, they can be collected easily. The math is the same,once the quantity of each product has been established.Figure 1-1 shows a typical apparatus used in a combustion analysis. The oxygentank serves to provide excess oxygen to the system constantly. The pressurevalve is a one-way valve designed so that oxygen can flow into the samplechamber, but no gases can flow back. The resistor in the base of the samplechamber provides heat to initiate the oxidation. The tube to the right of thesample chamber is connected to a vacuum, to generate a low pressure. Once thereaction is complete, Valve #3 is closed so that no gases are lost to theenvironment, and Valve #1 is opened. Gases flow into the region above themagnesiumsulfate. Magnesium sulfate absorbsmoisture from the gases. Afteratime, Valve #2 is opened so that gases flow into the region above sodiumhydroxide. Sodium hydroxide absorbs carbon dioxide gas. Oxygen gas is thenused to flush any remaining gas in the sample chamber through the system.

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Pressurevalve

Oxygentank

Samplechamber

Resistor

rWWVHn

Adjustable voltage Trap I

Figure 1-1

16

Valve #2 Valve #3

Trap 11

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General ChemiStiy Stoichiometry Solution Concentration

Solution Concentration

Units and TerminolgySolutions are mixtures formed by the addition of a solute to a solvent. A solutionmay contain several different solutes. The amount of solute is measured relativeto the amount of solvent, which results in a certain concentration for the solution.Concentration units include molarity (moles solute per liter of solution), molality(moles solute per kilogram of solvent), mass percent (mass solute per mass ofsolution), and density (mass solution per volume solution). The concentration ofa solute can be changed by changing the amount of solvent. Addition of solventto solution is referred to as dilution and results in a lower concentration by anymeasurement. Paramount to solving problems involving concentrations anddilution is an understanding of the different units.

MolarityMolarity (M) is the concentration of a fluid solution defined as the moles of asolute per volume of solution, where the volume is measured in liters (L). Todetermine the molarity of a solution, the moles of solute are divided by the litersof solution.

Example 1.20What is the molarity of 500.0 mL of solution containing 20.0 grams of CaC03(s)?A. 0.15 M CaC03(aq)B. 0.20 M CaC03(aq)C. 0.33 M CaC03(aq)D. 0.40 M CaC03(aq)

SolutionMolarity is defined as moles of solute per liter of solution. In this question, youmust convertfromgramsCaC03 intomolesCaC03by dividingby themolecularmass of CaC03, and then dividing this value by the liters of solution:

20 grams CaCC^ =Q2Qmoles CaC03^^"lole

0.20 moles CaCQ3 =O20 MCa0o3 =0,40 MCaCo3 =0.40MCa003, choice D0.50 L solution 0.50 1

These questions can be trickier if the units are milligrams, milliliters, ormillimolar. The question uses similar math, but there are more opportunities tomake an error. A common error to avoid is the "factor of a thousand" error. Tobecome more conscious of possible trick questions, ask yourself: "If I werewriting this test question, what would I ask?" Ifyouconsider questions from thetest writer's point of view, the tricks become more apparent.

MolalityMolality (m) is the concentration of a fluid solution defined as the moles of asolute per kilogram of solvent. The molality of a solution does not change withtemperature, so it is often used to determine a change in the solution'stemperature when the change depends on concentration. Notable examples ofthis include boiling-point elevation and freezing-pointdepression. To determinethemolalityof a solution, the moles of solute are divided by the kilogramsof thesolvent.


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General Chemistry Stoichiometry Solution Concentration

Example 1.21What is the molality of a solution made by adding 7.46 g KCl to 250 g of water?A. 0.20mKCl(aq)B. 0.33mKCl(aq)C 0.40mKCl(aq)D. 0.50mKCl(aq)

SolutionMolality is defined as moles of solute per kilogram of solvent. In this question,you must convert from grams KCl into moles KCl by dividing by the molecularmass of KCl, and then dividing this value by the kilograms of solvent:

7.46 grams KCl =010molesKa^"tole

0.10 moles KCl =O10 mKC1=O40 mKCl =0.40 mKCl, choice C0.25kgH2O 0.25 1

Mass Percent (in Solution)Mass percent is the concentration of a fluid solution defined as the mass of soluteper mass of solution multiplied by one hundred percent. The mass percent of asolution remains constant as temperature changes. To determine the masspercent of a solution, the mass of solute is divided by the mass of the solution(where both are measured in grams). Mass percent is a unitless value, becausemass is divided by mass.

Example 1.22What is the mass percent of a 1.0m NaCl(aq) solution?A. 5.53%NaCl by massB. 5.85%NaCl by massC. 6.22%NaCl by massD. 9.50%NaCl by mass

SolutionMass percent is defined as grams of solute per grams of solution. In thisquestion, you must convert from moles NaCl into grams NaCl by multiplying bythe molecular mass of NaCl, and then dividing this value by the total mass ofsolution. The total mass of solution is the sum of the mass of solute and the massof solvent. It is easy to forget to consider the mass of solute in the total mass,which leads to the incorrect answer choice B.

1.0 moles NaCl(aq) x58.5 S/mole =58.5 gNaClTotal mass solution = 58.5 g NaCl +1000 g H2O = 1058.5g solution

Mass%NaCl = 58.5 gNaCl ^^ 58.5 < 585 = 5.85%1058.5g solution 1058.5 1000

Only choice A is less than 5.85%, so choice A is the best answer. Sometimesquestions that would normally require a calculator for a precise answer can bedetermined well enough without one to answer a multiple-choice test question.


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General Chemistiy Stoichiometry Solution Concentration

DensityDensity (p) is the concentration of a fluid solution defined as the mass of solutionper volume of solution. The density of a solution varies with temperature. Thedensity of a solution is uniform throughout, so a small sample of the solution hasthe same density as the entire solution. To determine the density of a solution,the mass of a sample of the solution is divided by the volume of the sample(usually measured in milliliters). Examples 1.3 and 1.5 addressed the topic ofdensity.

DilutionDilution involves the addition of solvent to a solution, thus resulting in anincrease in the volume of the solution and a decrease in the concentration of thesolute in solution. Equation 1.2below describes simple dilution where a solventis added to solution. Determining the concentration when two solutions aremixed requires more work than simple dilution.

MinitiarVinitial = Mfinal'Vfinal ft-2)

Whenworking with dilution questions, be aware of a common twist that thewriterscan employ. Theirquestionmay ask for volume added rather than askingfor the final total volume. Percent dilution may also be discussed. Multiplecontainers are used in standard dilution procedure, so rinsing ensures that theconcentration of solution on the walls of the new containers are equilibrated withthe contents they will hold. Youmay recall filling a volumetric pipette with asolution in general chemistry lab, thendraining thepipette before filling it withthe sample to be transferred. This is done to ensure that anyresidual liquid inthepipette has thesame concentration as thesolution being transferred and thatany water in the pipette is rinsed away.

Example 1.23Whatis the molarity of a solutionmade by mixing 200 mLpure water with 100mL0.75MKCl(aq)?A. 0.25MKCl(aq)B. 0.50MKCl(aq)C. 1.50MKCl(aq)D. 2.25MKCl(aq)

SolutionBecause water has been added to the solution, the concentration must decrease,so choicesC and D are eliminated. Solving this question involvesusing Equation1.2 to determine the effect of dilution on the molarity. The initial molarity(Minitial) &0.75 M, the initialvolume (Vinitial) is 100 mL, and the final volume(Vfinal)is 300 mL- Thequestion requires solving forthe final molarity (Mfinal)-

Minitial-Vinitial = Mnnal-Vfinal - 0.75M-100mL = Mnnal-300 mL

Mnnal -^^ .-. Mnnal •^M-lOOmL, Q75MM =Q25UVfinai 300 mL V3/

Thefinalmolarity is 0.25 M, so choice A is the best answer. Because themolarityis decreased by a factor of three, the dilution process in thisexample is referredto as a threefold dilution. That is, when two parts solvent are added to one partsolution, the volume is tripled and the dilution is threefold. This terminologymaybe unfamiliar at first, but in a short time it should make sense.

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General Chemistiy Stoichiometry Solution Concentration

Example 1.24How many milliliters of water are needed to dilute 80 mL 5.00 M KN03(aq) toLOOM?

A. 160mLH2OB. 320mLH2OC. 400mLH2OD. 480mLH2O

SolutionAgain, Equation 1.2 should be employed to calculate the change in concentrationof a solution after dilution from the addition of solvent. First we must solve forthe final volume. You are provided with values for Minitial' Mfinal/ and Vjnitial/so you can manipulate the equation to solve for Vfinal-

Vfinai =initial'Vinitial =5.00Mx80mL= 5x 80mL =400mLMfinal 1-00 M

The question asks how much water is added, not the final volume. The volumeadded is the difference between the initial volume (80 mL) and the final volume(400 mL). The difference between the two values is 320 mL, so 320 mL of watermust be added to 80 mL of 5.00M KN03(aq) to dilute it from 5.00 M KN03(aq) to1.00 M KN03(aq). Choice B is the best answer.

Example 1.25Which dilution converts 6.00M HCl(aq) to 0.30M HCl(aq)?A. 11 parts water to 1 part 6.00M HCl(aq)B. 19 parts water to 1 part 6.00M HCl(aq)C. 20 parts water to 1 part 6.00M HCl(aq)D. 21 parts water to 1 part 6.00M HCl(aq)

SolutionHydrochloric acid goes from 6.00M to 0.30M, which is a twenty fold dilution.This means that the final volume is twenty (20) times the initial volume. Whendealing with answer choices that present the dilution in terms of parts, the ratiois volume of solvent added to volume of original solution. For the final volumeto be twenty times greater than the initial volume, nineteen parts must be added.

JVfinaL=MinitiaL ^JVfinaL =4QQJM =20/.Vfinal =20 (Vinitial)Vinitial Mfinal Vinitial 0.30 M

Vadded = Vfinal- Vinitial = 20 Vinitial - Vinitial = 19 (Vinitial)

The ratio of the volume added to the volume of solution initially present is 19 :1,so the best answer is choice B. A part can be any set volume. A 20 :1 dilutionwould result in a final volume that is 21 times the initial volume, so the finalconcentration would be less than 0.30 M.

A solution can be diluted by adding solvent or another solution to it. Theaddition of pure solvent is known as a simpledilution. Mixing two solutions ismore complicated than a simple dilution where pure solvent is added, becausesolute from both initial solutions must be considered. The final concentration liessomewhere between the two initial concentration values before mixing. The finalconcentration is a weighted average of the initial concentrations.


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General ChemiStiy Stoichiometry Solution Concentration

Example 1.26What is the final concentration of CI" ions after mixing equal volumes of 0.20MKCl(aq)with 0.40 M CaCl2(aq)?A. 0.20MCl-(aq)B. 0.30MCl"(aq)C. 0.40MCl"(aq)D. 0.50MCl"(aq)

SolutionThe salt KCl yields one chloride ion when it dissociates in water, so the chlorideconcentration is 0.20 M. The salt CaCl2 yields two chloride ions when itdissociates in water, so the chloride concentration is 0.80 M. The finalconcentration equals the total CI" ions from both solutions divided by the newtotal volume. Because equal volumes are mixed, the final concentration will bean average of the two initial concentrations.

0.20Mcr + 0.80Mcr -l.ooMcr = 050MC1-2 2

If the volumes are not equal, then a weighted average yields the finalconcentration. For this example, choice D is the best answer.

Example 1.27What is the concentration of K+ ions in solution after 25.0 mL of 0.10 MK2S04(aq) is added to 50.0mL of 0.40M KOH(aq)?A. 0.25MK+(aq)B. 0.30MK+(aq)C 0.33MK+(aq)D. 0.50MK+(aq)

SolutionThe salt KOH yields one potassium ionwhen it dissociates in water, so the K+concentration is 0.40 M. The salt K2S04 yields two potassium ions when itdissociates in water, so the K+ concentration is 0.20 M. The final concentrationequals the total K+ ions from both solutions divided by the new total volume.Because unequal volumes are mixed, the final concentration is a weightedaverage of the twoinitial concentrations. The mixture involves 25 mL0.20 MK+with 50 mL 0.40 M K+, so the final concentration must fall between 0.20 M and0.40 M. This eliminates choice D. If the two volumes were equal, the finalconcentration would be 0.30M K+, the average of the two concentrations. Butbecause there is more of the more concentrated solution, the final concentration isgreater than 0.30M, so choices A and Bare eliminated. Only choiceC remains.

25 mLx 0.20MK+ +50 mLx 0.40MK+ =1 (0.20MKV- (0-40MK+)75 mL 75 mL 3 3

= 2. + 3. MK+ = I MK+ = 0.33 MK+, choiceC3 3 3

Beer's LawWhen electromagnetic radiation is passed through a solution, the solute mayabsorb some of the light. The fight absorbed is in a specific wavelength range,and the intensity of the absorbance varies with the concentration of solute. Ageneric absorbance spectrum for a hypothetical solute is shown in Figure 1-2.


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General Chemistry Stoichiometry Solution Concentration

^ma Wavelength (nm)

Figure 1-2

Because the absorbance of light varies with concentration, absorbance can beused to determine the concentration of a solute. This is the essence of Beer's law.Beer's law is expressed in Equation 1.3, where e is a constant for the solute atXmax (the wavelength of greatest absorbance), C is the solute concentration (C=[Solute]), and 1is the width of the cuvette (length of the pathway through whichthe light passes).

Absorbance = £C1 (1.3)

The key feature of this equation is its expressionof the principle that absorbanceis proportional to concentration. By knowing the absorbance for solutions ofknown concentration, the concentration of an unknown solution can bedetermined by comparing its absorbance value to the known values.

Example 1.28For 100 mL of a solution with an absorbance of 0.511, what amount of water mustbe added to reduce the absorbance to 0.100?

A. 389mLH20B. 411mLH20C. 488mLH20D. 511mLH20

SolutionFor this question, a hybrid of Equations 1.2 and 1.3 should be employed todetermine the volume that must be added to dilute the solution. Becauseabsorbance is directly proportional to concentration, Equation 1.2 can be rewritten as follows:

Absinitial-Vinitial = Absfinal'VfinalFirst, we must solve for the final volume. You are provided with values forAbsinitial/ Absfinal, and Vinitial, so you can solve for Vfjnal.

Vfi _Absinitial-Vinitial _0.511 x100 mL =5.11 x100 mL _ 511 mLAbsfinal 0.100 1

The question asks for how much water is added, which is the differencebetweenthe initial volume (100 mL) and the final volume (511 mL). The differencebetween the two values is 411 mL; therefore, 411 mL of water must be added.The best answer is choice B.


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General ChemiStry Stoichiometry Balancing Reactions

Standard BalancingLet us briefly address the process of balancing chemical reactions. Reactions arewritten from reactants to products. Because of the law of conservation of matter,the number of atoms must be identical on each side of the reaction. The twosides of the reaction are separated by an arrow drawn from left to right.

C5Hu(l) + Oz(§) • C02(g) + H20(g)There are carbon atoms, hydrogen atoms, and oxygen atoms on both sides of thereaction. To balance the reaction, keep track of the atoms on each side of thereaction. Start with the compound whose atoms are least present in the reaction(carbon and hydrogen are present in only two compounds each, so we start withQ>Hi2). Starting with one C5H12, the atoms must be balanced step by step:

\ CsHi2(D + 02(g) • C02(g) + H20(£)5C ?C12 H ?H

?o ?o

Balance carbon atoms by multiplying C02 by five:

1 C5Hi2(/) + Oz(g) • 5 C02(g) + H20(#)5C 5C12 H ?H?0 10 + ?O

Balancehydrogen atoms by multiplying H20 by six:1C5H12W + Q2^ • 5 C02(g) + 6 H2Ofg)

5C 5C12 H 12 H

?0 16 O

Balance oxygen atoms by multiplying 02 by eight:\C5Hu(l) + 8 02(g) • 5C02(g) + 6H20(g)

5C 5C12 H 12 H16O 16O

ExampleWhat are

1.29the correct coef

Co(OH)3(s) +1 :61 :3

1:31:6

ficients needed to balance the following reaction?

A. 2:3:B. 3:2:C. 2:3:D. 3:2:

1 i2oU4(tuj) *- ^.u^.jw^jim/,/ t ii^vu/

SolutionBalancing equations requires that you keep track of each atom. In this case,because of cobalt, the ratio of Co(OH)3fs) to Co2(S04)3(aq) must be 2 :1, whicheliminates choices Band D. The correct answer is found using water. Two molesof Co(OH)3(s> and three moles of H2SC>4(aq) have a total of in 12 H atoms and 18O atoms, making choice A the best answer.


General ChemiStiy Stoichiometry Balancing Reactions

Balanced equations can be used to determine the amount of a product from agiven amount ofreactant. As wesawinExample 1.7, balanced equations can beused to determine how much product is formed from a given mass of a reactant.Werefer to these questions as gram-to-mole-to-mole-to-gram conversions, wherethe overallconversion processis fromgrams reactant to grams product.

Example 1.30How many grams of water are formed when 25.0 grams of pentane (CsHi2)reacts with oxygen?A. 18.8 gH20B. 25.0 g H20C. 37.5 g H20D. 75.0 g H20

SolutionStep 1:Convertgramsreactantto molesreactant by dividing by molecularmassof the reactant:

grams reactant x mole reactant _moies reactantgrams reactant

Step 2:Convertmoles reactant to moles product using the coefficient ratio fromthe balanced reaction:

, L L moles product , , .moles reactant x *- = moles productmoles reactant

Step 3: Convert moles product to grams product by multiplying by molecularmass of the product:

moles product xsi E =grams productmoles product

The overall conversion is as shown below:

25.0 gC5Hi2 xlmoleC5H12x 6moleH2Q x 18gH2Q =37J.72gC5Hi2 ImoleCsHn lmoleH20

The ratio of water to pentane comes from the balanced oxidation reaction. Theproduct of 6 x 18 is 108, which is greater than 72. This means that the original25.0 grams is multiplied by a number greater than 1, which in turn means thatthe final number is greater than 25.0g. This eliminates choices A and B. 108over72 is less than two, so the final value is less than 50.0 grams, so choice D iseliminated. The only answer that remains is choice C, 37.5 g.

Limiting ReagentsDetermining the limiting reagentin a reactionrequirescomparing the number ofmoles of each of the reactants. The limiting reagent is the reactant that isexhausted first, not necessarily the reactant with the lowest number of moles.Whenthe limiting reagent is completely consumed,the reactionstops, regardlessof the amount of the other reactant. To determine the limiting reagent, theamount of all reactants and the mole ratio of the reactants must be known. If theratio of the moles of Reactant A to Reactant B is greater than the ratio of ReactantA to Reactant B from the balanced equation, then Reactant B is the limitingreagent. If the ratio of themolesofReactantA to ReactantBis less than the ratioof Reactant A to Reactant B from the balanced equation, then Reactant A is thelimiting reagent.

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General ChemiStry Stoichiometry Balancing Reactions

Example 1.31Assuming that the following reaction between oxygen and hydrogen goes tocompletion, which statement is true if 10.0 grams of hydrogen are mixed with64.0grams of oxygen?

2H2(g) + 02(g) • 2 H20(1)A. The limiting reagent is oxygen.B. 74.0 grams of water will form.C. 3.0moles of hydrogen will be left over following the reaction.D. 68.0 grams of water will form.

SolutionIn limiting reagent reactions, you must decide which reactant is depleted first.Limiting reagent questions often look like ordinary stoichiometric questions. Therule is simple: If they give you quantities for all reactants, it is probably a limitingreagent problem. In this question, you are given quantities for both hydrogenand oxygen. 10grams of H is equal to 5 moles of H2, and 64grams of oxygen is 2moles of 02 (remember your diatomic elements!) From the balanced equation,we learn that twice as many moles of hydrogen as oxygen are needed. Thenumber of moles indicates there is a 5 : 2 ratio of hydrogen to oxygen, which isgreater than a 2:1 reaction ratio, so oxygen is depleted first. The correct choice isanswer A. The question gives you opposing choices in A and B. One of thesetwo choices must be true. The correct choice is A.

Example 1.32What is the limiting reagent when 22.0 grams C3H8 are mixed with 48.0 gramso2?

C3H8(1) + 02(g) • C02(g) + H20(g)A. Oxygen is the limiting reagent.B. Propane is the limiting reagent.C. Water is the limiting reagent.D. There is no limiting reagent.

SolutionThis question is more difficult than the previous question, but you are stilldeciding which reactant is depleted first. Because the limiting reagent is areactant, choice C (a product) is eliminated. To solve the question, stick to thissimple rule: Compare the actual ratio of the two reactants to the balancedequation ratio of the two reactants. In this question, you are given unequal massquantities of C3H8 and 02 and a mole ratio that is not 1:1. Good luck.Remember, the first step is to balance the reaction.

1 C3H8(1) + 5 02(g) • 3 C02(g) + 4 H20(g)

22 grams ofC3H8 is22. moles ofC3H8, which is0.50 moles C3H8. 48 grams of0244

is 22- moles of O2: (remember your diatomic elements!), which is 1.50moles of 02.

From the balanced equation, we see that we need 5 moles of 02 for 1 mole ofC3H8. The number of moles indicates there is a 1.50 : 0.50 ratio of 02 to C3H8,which is less than 5:1. This means that oxygen (02) is depleted first. Oxygengas is the limiting reagent, so choice A is the best answer.


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General Chemistiy Stoichiometry Reaction Types

't^m^k^s^E:..•' - "': .1''".:• ..• .• 7.-:.Z-Common ReactionsThereare somereaction typesthat are standard reactions in inorganicchemistry.Included among the common reaction types are the following five: 1)precipitation reactions (also known as double-displacement reactions), 2) acid-base reactions (alsoknown as neutralization reactions),3) composition reactions,4) decomposition reactions, and 5) oxidation-reduction reactions (electron-transfer reactions). Oxidation-reduction reactions can be categorized as eithersingle replacement or combustion. Each reaction typewill be addressed inmoredetail in later sections, so let us consider each type of reaction in minimal detailin this section.

Precipitation ReactionsA reaction that involves two aqueous salts being added together to formspectator ionsand a solidsaltprecipitate that drops out ofsolutionis knownas aprecipitation reaction. It may alsobe referred to as a double-displacement reaction,although that term is not as useful in describingthe chemistry. Drawn below is asample precipitation reaction:

Na2Cr04(aq) + Sr(N03)2(aq) • 2NaN03(aq) + SrCr04(s)AqueousSalt AqueousSalt Ions Precipitate

Precipitationreactions canbe recognizedby the solid salt on the product side ofthe equation. Recognition of the type of reaction is useful for predicting theproduct. Recognizing a precipitate is highly beneficial in identifying a double-displacement reaction. The following solubility rules can be helpful inidentifying the likelihoodof a precipitate's forming:

1. Most salts containing alkali metal cations (Li+, Na+, K+, CS+,Rb+) and ammonium (NH4+) are water-soluble.

2. Most nitrate (NO3") salts are water-soluble.3. Most salts containing halide anions (CI", Br", I") are water-soluble

(with heavy metal exceptions such asAg+ andPb2"*).4. Most salts containing sulfate anions (SO42") are water-soluble

(with exceptions such as Ba2+, Pb2+, Hg2+, and Ca2+).5. Most hydroxide anion (OH") salts are only slightly water-soluble.

KOH and NaOH are substantially soluble, while Ca(OH)2,Sr(OH)2, and Ba(OH)2are fairly soluble in water.

6. Most carbonate anion (CO32"), chromate anion (Cr042"),phosphate anion (P043"), and sulfide anion (S2") salts are onlyslightly water-soluble.

Acid-Base ReactionsA reactionbetweenan acid (a proton donor) and a base (a proton acceptor) formsa neutral salt and water. For now, recognize that proton donors (acids) musthave a proton on an acid (H—Acid). Acids to recognize are HC1, HBr,HI, HNO3,H2S04, and NH4+. Basesto recognize are NaOH, KOH, LiOH, and CaC03-

HC104(aq) + LiOH(aq) • LiC104(aq) + H20(1)Acid Base Salt Water

Acid-base reactions canbe recognized by the formation of a salt and water on theproduct side of the equation. Aqueous acid-base reactionscan be identified bythe transfer of an H from the acid to the hydroxide of the base on the reactantside of the equation.


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General Chemistry Stoichiometry Reaction Types

Composition ReactionsA composition reaction involves the combining of reactants to form a product.The number of reactants exceeds the number of products in a compositionreaction. Entropy decreases and more bonds are formed than are broken incomposition reactions.

PCl3(g) + Cl2(g) • PCl5(g)2 Reactants 1 Product

Composition reactions may fall into other reaction categories as well. In thesample reaction, when PCI3 reacts with Cl2, PCI3 is oxidized and Cl2 is reduced.

Decomposition ReactionsA decomposition reaction is the opposite of a composition reaction. It involvesreactants decomposing to form multiple products. The number of reactants isless than the number of products in a decomposition reaction. Entropy increasesand more bonds are broken than are formed in decomposition reactions.

CaS03(g) • S02(g) + CaO(s)1 Reactant 2 Products

Like composition reactions, decomposition reactions can also fall into otherreaction categories as well.

Oxidation-Reduction Reactions

An oxidation-reduction reaction involves the transfer of electrons from one atomto another. Loss of electrons is defined as oxidation, while gain of electrons isdefined as reduction (LeoGer). The atom (or compound) losing electrons iscausing reduction, so it is referred to as the reductant (reducing agent), while theatom (or compound) gaining electrons is causing oxidation, so it is referred to asthe oxidant (oxidizing agent). The oxidation states must change in an oxidation-reduction reaction. A sample reaction (below) shows how magnesium is losingelectrons (thus being oxidized and having an increase in oxidation state) tobromine (which is being reduced and having a decrease in oxidation state):

Mg(s) + Br2(l) *- MgBr2(s)Reductant Oxidant Salt

Combustion ReactionsCombustion reactions are a special case of oxidation-reduction reactions, wherethe oxidizing agent is oxygen gas, and the products are oxides. Typical examplesof combustion reactions include the oxidation of organic compounds, such ashydrocarbons and carbohydrates, into carbon dioxide and water.

lC3H8(aq) +5 02(g) • 3C02(g) +4 H20(1)Hydrocarbon Oxygen Carbon dioxide Water

Combustion reactions of both hydrocarbons and monosaccharides balance in apredictable manner, as shown below:

Hydrocarbon Combustion

CxHy +(x +y)02(g) • xC02(g) +^H20(g)3 4 2

Monosaccharide Combustion

CxH2xOx + x02(g) • xC02(g) + xH20(g)

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General Chemistry Stoichiometry Reaction Types

Example 1.33Thefollowing reaction is an example ofwhichtypeof reaction?

MgCl2(aq) + AgN03(aq) • Mg(N03)2(aq) + AgCl(s)A. Cation-crossoverB. Oxidation-reductionC. NeutralizationD. Double-displacement

SolutionOf the generic reactions withwhich youare familiar, thereare typical features tonote. In this example, you have two salts undergoing an exchange reaction toyield a precipitate. This makes it a precipitation reaction. No proton wastransferred (eliminating neutralization), and no oxidation states changed(eliminating oxidation-reduction). Cation-crossover is a fabricated name, sochoice A is eliminated. Choice D, double-displacement, is another name for aprecipitation reaction. In double-displacement reactions, you have two saltsundergoing a reaction where they exchange counterions, and one of the newcombinationsformsa precipitate. This is shown in a generic fashion below:

MX(aq) + NY(aq) • MY(aq) + NX(s)

Oxidation StatesAssigning an oxidation state to an atom is a matter of distributing electronswithin a bond based on which atom is more electronegative. The oxidation stateof an atom can be determined by assigning it a value of positive one (+1) foreverybond it formswith a more electronegative atom and assigning it a value ofnegative one (-1) for every bond it forms with a less electronegative atom. Theoxidation state is a sum of all these bonding values. In general chemistry, it isoften easiest to say oxygen is -2 (except in molecular oxygen and peroxides),hydrogen is +1 (exceptin molecular hydrogen and hydrides), and halides are -1(exceptwhen they are a central atom in an oxyacid). The sum of the oxidationstates of the elements in the compound must equal the overall charge, so theoxidation state of any remaining atom can be determined by finding thedifference between its charge and the sum of the known oxidation states.

Example 1.34What is the oxidation state of manganese in KMn04?A. +1B. +3

C. +5D. +7

SolutionTo simplify this example,we will consider that O = -2, and alkali metals = +1. Inthis case, K = +1, and there are 4 Os valued at -2 each, for a net oxidation state of-8. Summingoxygen and potassiumyieldsa total of -7. Thismeans that for themolecule to be neutral, the Mn (the only atom remaining) must cancel out that -7by being +7. In other words, the sum of the oxidation states equals themolecule's formalcharge (zero). So in this case, the oxidation state ofmanganeseis +7. The correct answer is choice D.

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General ChemiStiy Stoichiometry Test-TaldngTips

Test-Taking lipsGeneral AdviceThestoichiometry sectionof this review course is best learnedby trial and errorfrom examples (i.e., practice with many problems.) For the most part, to besuccessful in stoichiometry requires being fast at math and being able to seeimmediately what a questionis askingfor. These are skills that are acquired andnot necessarily memorized. Keep in mind that on a multiple-choice exam, themath has been done for you, so all you need to do is approximate the answer.There is no universal shortcut that works in everysituation, but finding a rangewithinwhichonly one answer choice fits is a goodapproachtomostquestions.

Intuition will also prove useful on the MCAT. Traditional testing at majoruniversities generally rewards memorization over intuitive skills, but preparingfor theMCAT forces you to hone your analyticaland intuitiveskillsas you recallcertain facts from memory. You should try to emphasize this thought processearly and regularly throughout your review studies.

In the examples above, several topics and styles of questions were presented.Before you move on to the practice questions in the passages,make sure that youunderstand the basic principle of each topic and the math typically used toanswer these questions. Math tricks may prove helpful, even for the conceptualquestions in stoichiometry. Keep in mind that you are not graded for showingyour work on the MCAT,so don't solve every problem to the last decimal place.Analyze each question only well enough to eliminate three wrong answers. Beconcise and efficient in your problem-solving, not exhaustive. Generally, thequestions ask you to decide which fraction (or ratio) is larger. This can be doneeasily by converting all the fractions to values over the same denominator, andlooking for an answer choice that falls within a range. Keep it simple.

Mathematical Tips and ShortcutsWhile the MCATdoes not require elaborate calculations, you still must be able todeal with ratios and percentages. Do not use a calculator when practicing for theMCAT. The following strategies are useful ways to calculate a value quickly andto a fair approximation without tables or a calculator:

Addition and SubtractionSplitting numbers and adding common terms is a useful way to make additionand subtraction easier. To split a number, consider how you would round it, andthen split it into the rounded number and the differencebetween the original androunded numbers. 193rounds up to 200,so 193can be thought of as 200- 7. 826rounds down to 800, so think of 826 as 800 + 26. Adding and subtractingrequires linking like terms. Thus, adding 193 to 826can be thought of as:

193 + 826 = 200 - 7 + 800 + 26 = 200 + 800 +26 - 7 = 1000 + 19 = 1019

Subtraction is done in a similar way. 826-193 can be thought of as:

826 - 193 = 800 + 26 - 200 + 7 = 800 - 200 +26 + 7 = 600 + 33 = 633

This approach may seem awkward at first, but it is effective and easy whenadding and/or subtracting several numbers at once. For instance, consideradding 213 to 681, then subtracting 411.

213 + 681 - 411 = 200 + 13 + 700 - 19 - 400 - 11 = 200 + 700 - 400 +13 - 19 - 11

= 500 - 17 = 483

Copyright © by The Berkeley Review 29 Exclusive MCAT Preparation

General Chemistry Stoichiometry Test-Taking Tips

Averaging TermsAveraging terms involves estimating amean value, and then keeping a runningtally of the differences between the actual values and the estimated average. Tofind theaverage difference, therunning tally is divided by thenumber ofvaluesbeing averaged. For instance, the average of25, 33, 21, 28, and30 can be thoughtof as being around 28 (the median value), so the actual average is 28 +/- theaverage difference:

The total difference is -3 + 5 - 7 +0 +2 = -3

When the total difference is divided by 5, it yields an average difference of - 0.6The average of the five values is thus 28 - 0.6= 27.4

MultiplicationMultiplication can also be made easier by splitting numbers asyou would roundthem. For instance, 97 is 100 - 3. Only one number need be split inmultiplication. Thus,multiplying 97by 121 can be thought of as:

97 x 121= (100 - 3) x 121= (100x 121) - (3 x 121) = 12,100 - 363

12,100 - 363 = 11,700 + 400 - 363 = 11,700 + 37 = 11,737

DivisionIf you memorize the following set of fraction-to-decimal conversions, thenproblems involvingdivisionwill be far easier:

1 = 0.200,1 = 0.166,1 = 0.143,1 = 0.125,1 = 0.111, -L = 0.091, -1- = 0.0835 6 7 8 9 11 12Memorizing these decimal values canbeuseful in several ways. For instance, thedecimal equivalent of the fraction 18/66 can be found in the followingmanner:

18=J_=3xi = 3x (0.091) = 0.27366 11 11

Knowing these decimal values is also useful for estimating in decimal termsfractions that are just less than 1. For instance, the decimal equivalent of thefraction 11/12 can be found in the following manner:

li -12-1 =121212 12

These decimal values are also useful in deciding what to multiply a denominatorby to convert it to some number close to 10, 100, or 1000. For instance, thenumerator and denominator of the fraction 47/142 should be multiplied by 7,because 0.143 = 1/7 so 7 x 143 is nearly 1000 (actually, it's 1001). The decimalequivalentof the fraction 47/142canbe found in the following manner:

_4Z = 7x47 =329 = _329_ + a Uttle = o.329 + a little142 7x142 994 1000

12 121 - 0.083 = 0.917

"Better living through chemistry!"


StoichiometryPassages

12 Passages

100 Questions

Suggested Stoichiometry Passage Schedule:I: After reading this section and attending lecture: Passages I - III & VI - VIII

Grade passages immediately after completion and log your mistakes.

II: Following Task I: Passages IV, V, & IX, (20 questions in 26 minutes)Time yourself accurately, grade your answers, and review mistakes.

Ill: Review: Passages X - XII & Questions 87-100Focus on reviewing the concepts. Do not worry about timing.

^SrR-E-V-I.E'W


I. Density Experiment

II. Combustion Analysis

III. Empirical Formula Determination

IV. Molar Volume of a Gas

V. Elemental Analysis

VI. Dilution Experiment

VII. Solution Concentrations and Dilution

VIII. Beers Plot and Light Absorption

IX. Beers Law Experiment

X. Reaction Types

XI. Calcium-Containing Bases

XII. Industrial Chemicals

Questions Not Based on a Descriptive Passage

Stoichiometry Scoring Scale

Raw Score MCAT Score

84 - 100 13 - 15

66 -83 10 - 12

47 -65 7 -9

34-46 4-6

1 -33 1 -3

(1 -7)

(8- 13)

(14- 20)

(21 - 26)

(27 - 33)

(34 - 40)

(41 -47)

(48 - 54)

(55 -61)

(62 - 68)

(69 - 78)

(79 - 86)

(87 - 100)

Passage I (Questions 1 - 7)

A student fills a 50-mL graduated cylinder exactlyhalfway with water, adds a previously weighed sample of anunknown solid, and records the new water level indicated bythe markings on the side of the graduated cylinder. Afterrecording the volume, she removes the unknown solid andadds water to the cylinder to raise the volume back toprecisely 25 mL, replacing any water that may have adheredto the solid. This procedure is repeated for a total of fiveunknown solids, and it is discovered that each of the solidssinks to the bottom of the graduated cylinder. Table 1 showsthe data for all five trials.

Unknown Mass Volume Reading

1 9.63 g 31.42 mL

2 12.38 g 31.19 mL

3 14.85 g 29.95 mL

4 8.22 g 28.00 mL

5 5.64 g 26.41 mL

Table 1

A second experiment is conducted with liquids, using a10.00-mL volumetric cylinder (one that holds exactly 10.00mL of solution) that weighs 42.61 grams when empty. Infour separate trials, unknown liquids are poured into thecylinder exactly to the 10.00-mL mark on the cylinder eachtime, and the combined mass of the cylinder and the liquid isrecorded. Table 2 shows the results of the second experiment.

Unknown Mass of Cylinder with Liquid

6 51.33 g7 58.72 g8 53.21 g9 49.03 g

Table 2

Given that all of the unknown liquids are immiscible(will not dissolve) in water, how many of the unknownliquids can float on water?

A. 1B. 2C. 3D. 4

2. Which of these unknown solids is the DENSEST?

A. Solid#2B. Solid #3C. Solid #4D. Solid #5

Copyright© by The Berkeley Review® 33

3. What would be the volume of a 20.0-gram piece ofunknown Solid #1?

A. 13.3 mLB. 15.0 mLC. 25.0 mLD. 30.0 mL

4. How many of the unknown solids can float on Liquid#7?

A. 0B. 1

C. 2D. 3

5. Which of the following sequences does NOT accuratelyreflect the relative densities of the unknown liquids?A. Liquid #7 > Liquid # 6 > Liquid #8B. Liquid #8 > Liquid # 6 > Liquid #9C. Liquid #7 > Liquid # 8 > Liquid #9D. Liquid #7 > Liquid # 8 > Liquid #6

6. How would the results in Experiment 2 differ from theactual results, if a heavier graduated cylinder had beenused?

A. Both the mass of the cylinder with the liquid andthe density of the liquid would increase.

B. The mass of the cylinder with the liquid wouldincrease, while the density of the liquid woulddecrease.

C. The mass of the cylinder with the liquid wouldincrease, while the density of the liquid wouldremain the same.

D. The mass of the cylinder with the liquid woulddecrease, while the density of the liquid wouldincrease.

7. Which of the following is NOT a unit of density?

g

mLoz.

3cm

kg

A.

B.

in

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Passage II (Questions 8-13)

Elemental analysis is often a preliminary study instructural analysis. A sample compound is placed into achamber with a positive pressure of oxygen gas flowing in.The chamber has an ignition coil that is heated by a current.As the reaction proceeds, the pressure in the chamber buildsup. After a short time, Valve #l is opened to allow theproduct gas mixture from the reaction to flow into anevacuated tube containing some magnesium sulfate, whichabsorbs water vapor from the product gas mixture. ThenValve #2 is opened, allowing the gas to flow into a secondevacuated tube containing some sodium hydroxide, whichabsorbs carbon dioxide from the product gas mixture. Theapparatus is shown in Figure 1. The oxygen tank providesoxygen in excess throughout the process. Valve #3 isconnected to a line that can either evacuate the system orsupply nitrogen to the system.

Pressurevalve

.. . , , ,, MtiS04(s) NaOH(s)Variable voltage e 4Figure 1

Fourdifferent samples are analyzed. The samplemass ofeach unknown substance is approximately two grams. Table1 shows the sample mass placed into the reaction chamber,and the initial and final masses of magnesium sulfate andsodium hydroxide in the side tubes.

Unknown Sample Mass MgS04 Tube NaOH Tube

I 2.011 gInit: 40.00 gFin: 41.21 g

Init: 30.00 gFin: 32.94 g

II 1.995 gInit: 40.01 gFin: 41.26 g

Init: 30.00 gFin: 34.39 g

III 2.003 gInit: 40.00 gFin: 42.00 g

Init: 30.00 gFin: 34.89 g

IV 2.001 gInit: 40.00 gFin: 41.99 g

Init: 30.00 gFin: 36.53 g

Table 1

Which of the following relationships best describes therelative mass percent of carbon in the four unknowncompounds used in the experiment?A. I > II > III > IVB. I>III>II>IV

C. IV > II >III> I

D. IV > III > II > I

Copyright © by The Berkeley Review® 34

Ideally, the properties of sodium hydroxide shouldinclude being:A . hydrophobic and semi-reactive with CCb.B. hydrophobic and highly reactive with CCb.C . hygroscopic and semi-reactive with CCb-D . hygroscopic and highly reactive with CCb.

10. Why is the oxygen tank attached to a pressure valve?A. It absorbs excess oxygen.B . It helps cool the reaction chamber.C . It ensures that oxygen gas is in excess.D . It ensures that oxygen is limiting.

1 1. Which of the four unknown compounds is LEASTlikely to contain oxygen?A. Compound IB. Compound IIC. Compound IIID. Compound IV

12. Why are the U-tubes containing the two salts arrangedin the order that they are?A. To ensure that water is absorbed before the gases

interact with the NaOH chamber

B . To ensure that carbon dioxide is absorbed before thegases interact with the NaOH chamber

C. To enhance the reaction between water and carbondioxide

D. To absorb any excess oxygen gas before it reactswith NaOH

13. What solid is being formed in the second tube after theproduct gas mixture interacts with the salt?A . Magnesium bicarbonateB. Magnesium carbonateC. Sodium bicarbonate

D. Sodium carbonate


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Passage III (Questions 14-20)

Exactly 10.0 grams of an unknown organic compound ispoured into a flask. The compound is then exposed to excessoxygen gas to oxidize it to CO2 gas and H2O gas. Theoxidized vapor flows through a tube filled first with copperoxide to ensure complete oxidation. The vapor then flowsthrough 100.00 grams of powdered anhydrous sodium sulfate,which binds water vapor to form 112.16 grams of hydratedsalt. The vapor continues to flow through 100.00 grams ofpowdered anhydrous sodium hydroxide, which binds carbondioxide vapor to form 123.79 grams of bicarbonate salt.

The unknown compound contains only oxygen, carbon,and hydrogen. The mass percent of carbon in the compoundis determined to be greater than 50%. In a subsequentexperiment, the compound is found to have a molecular masssomewhere between 70 and 80 grams per mole. When thebottle containing the unknown compound is left uncapped,its contents slowly evaporate.

The information from the combustion reaction can beconverted into mass percent for both carbon and hydrogen.By multiplying the grams of CO2 times the mass of onecarbon atom and dividing by the mass of carbon dioxide, themass of the carbon in the original sample can be determined.The mass of hydrogen in the original sample can be found ina similar manner. These two mathematical proceduresconvert the grams of product molecules into the grams ofeach atom. The final numbers are the grams of carbon andhydrogen, respectively, in the unknown compound. To.determine the mass percent, the mass of the atom is dividedby the mass of the sample. The mass percent of oxygen inthe unknown compound is determined by difference.

The information from the mass percents of thecomponent atoms can be used to determine the empiricalformula (formula of the lowest coefficients) for the unknowncompound. To determine the molecular formula from theempirical formula, the compound's molecular weight must beknown. For compounds containing only carbon, hydrogen,and oxygen, the molecular formula must always have an evennumber of hydrogens. Molecular formulas with an oddnumber of carbons and oxygens, however, are possible.

14. If 20.0 grams of the unknown compound described inthe passage were oxidized, what would be observed?A. The moles of CO2 would double, while the

percentage of carbon in the sample would remainthe same.

B. The moles of CO2 would double, and thepercentage of carbon in the sample would alsodouble.

C. The moles of CO2 would remain the same, and thepercentage of carbon in the sample would alsoremain the same.

D. The moles of CO2 would remain the same, whilethe percentage of carbon in the sample woulddouble.


15. The percentage of carbon by mass in the unknowncompound can be calculated as:

A. 23.79 x±2-x 10 x 100%44

B. 23.79 x 44 x 10 x 100%12

C. 23.79 x^xlx 100%44 10

D. 23.79 x4ix^-x 100%12 10

16. What can be said about the boiling point (b.p.) andmelting point (m.p.) of the unknown compound relativeto ambient temperature (Tg)?

A. m.p. > Ta, and b.p. > TaB. m.p. > Ta, and b.p. < TaC. m.p. < Ta, and b.p. < TaD. m.p. < Ta, and b.p. > Ta

17. Which of the following formulas CANNOT be amolecular formula?

A. C2H40B. C2H5OC. C3H60D. C4H802

18. How many moles of water are formed from thecombustion of 10.0 grams of the unknown compound?A. 0.31 moles H2OO)B. 0.69 moles H20(l)C. 1.10 moles H20(l)D. 1.38 moles H20(l)

19. What is the empirical formula for the unknowncompound?A. C3H602B. C4H802C. C4H10OD. CgHirjO

2 0. What is the mass percent of carbon in C5H12O2?

A. 26.4%B. 57.7%C. 60.8%D. 68.2%


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Passage IV (Questions 21 - 26)

A researcher completely oxidizes exactly 1.00 grams ofan unknown liquid hydrocarbon in a containment vessel toyield carbon dioxide and water vapor. The two gases thusformed are collected and analyzed for quantity. The watervapor is collected by passing the gas through a tubecontaining anhydrouscalcium chloride. The carbon dioxidegas is collected by passing the remaining gas through a tubecontaining anhydrous sodium hydroxide. The mass of thecarbon dioxide gas thus collected is 3.045 grams at STP.The carbon dioxide gas is regenerated upon heating thesodium carbonate and this gas is found to occupy a volumeof1.55 liters at STP. The experimental apparatus is shown inFigure 1.

C02,H20, and_excess O2 enter

flU-uft-

CaCl2(s) NaOH(s)

Figure 1In a second experiment, the researcher places a 5.0-mL

aliquot of the unknown liquid into a capped 1.00-liter flask.The cap has a tiny hole in the top, and the empty flask withcap weighsexactly 120.00grams. The compound is heateduntil it reaches a gentle boil. The vapor escapes through thetiny pore in the cap. The liquid continues boiling at 31"C,until none of it remains visible in the flask. The heat sourceis removed from the flask, and the contents are allowed tocool back to ambient temperature. As the flask cools, thevapor in it condenses into a small pool of liquid at the baseof the flask.

The flask and cap are then massed with the condensedliquid present. The entire system is found to have a mass ofexactly 122.32 grams. This means that the mass of theliquid is 2.32 grams. It is assumed that at the moment whenthe heat source was removed, the flask was completely filledwith vapor from the liquid and that all of the air originally inthe flask was displaced. Table 1 lists the molar volume foran ideal gas at selected temperatures.

Temperature (K) Molar Volume

273 22.41 L

288 23.64 L

298 24.46 L

304 24.96 L

313 25.69 L

323 26.51 L

Table 1

Copyright ©by TheBerkeley Review® 36

21. What is the mass percent of carbon in CO2 gas?

A. 25.0%B. 27.3%C. 31.4%D. 35.0%

22. How can the molecular weight of this unknown liquidbe determined?

A 22.41 grams2.32 mole

B 24.96 grams2.32 mole

C. (2.32 x22.41) ^^mole

D. (2.32 x24.96) ^^mole

23. If themasspercentof carbon in the unknown compoundis found to be 82.9%, what is the empirical formula ofthe unknown hydrocarbon?A.B.

C.D.

CH2C2H5CH3C2H7

24. Using the data from the first experiment, how can themass percent of carbon in the unknown compound bedetermined?

A. 1-55 x 12.011 x 100%22.41 x 1.00

B. 1.55 x 22.41 x 12.011 xl00%1.00

c> 22.41 x 12.011 xl00%1.55 x 1.00

D. 22.41 x 1.00 x iq0%1.55 x 12.011

25. How many moles of CO2 gas were formed in the firstexperiment?

A. J^55_ moles CO222.41

B. -L55_ moles CO223.64

C. 22AL moles C021.55

D. LQQ moles CO222.41 x 1.55


26. If in the second experiment the organic vapor had notfully displaced all of the air from the flask by the timethe heat was removed from the flask, how would theresults have been affected?

A . The mass of unknown liquid collected would be toogreat, so the calculated molecular mass would betoo high.

B. The mass of unknown liquid collected would be toosmall, so the calculated molecular mass would betoo high.

C . The mass of unknown liquid collected would be toogreat, so the calculated molecular mass would betoo low.

D. The mass of unknown liquid collected would be toosmall, so the calculated molecular mass would betoo low.


Passage V (Questions 27 - 33)

The empirical formula for a compound can be determinedusing the technique of elemental analysis. For hydrocarbonsand carbohydrates, the process involves trapping andremoving water vapor and carbon dioxide gas and thenquantifying the amounts collected. The products can betrapped in many different ways. In this experiment, thetrapping of the gases is accomplished by passing the productgas through a series of low-temperature gas traps. Bylowering the temperature, the gas can be converted intosolids, which cannot flow and thus are easily collected at thebases of their respective temperature traps. The system isevacuated completely and then flushed with nitrogen gas toremove any remaining air. The vapor from the reactionvessel is then allowed to enter one trap at a time. The gasremains isolated in the region of each trap for a short interval,to allow any gases to form a solid or freeze into a liquid. Theapparatus is shown in Figure 1.

Reactionvessel

\

Gastrap I

irv

Gastrap II

Figure 1

•£Tl

Gastrap III

Vacuuni

Oil bubbler

It is important to allow the excess oxygen gas to flowout of the system. To accomplish this, the line is fitted witha one-way oil bubbler. The oil bubbler maintains the closedsystem by not allowing air to flow into the system, whileallowing the pressure to equilibrate with the environmentthrough venting.

27. At what temperature should the first trap be held inorder to isolate water vapor?

A . 25"C (standard temperature)B. (fC (melting point of ice)C . -33°C (boiling point of Freon refrigerant)D . -I96°C (boiling point of liquid nitrogen)

2 8. The temperatures of the successive traps (i.e., Trap I,Trap II, and Trap III) should be set in what manner?A . The temperatures should gradually increase, so that

each gas is selectively removed one trap at a time.B. The temperatures should gradually increase, so that

each gas can be trapped into all three traps,allowing one to determine the moles by difference.

C. The temperatures should gradually decrease, so thateach gas is selectively removed one trap at a time.

D . The temperatures should gradually decrease, so thateach gas can be trapped into all three traps,allowing one to determine the moles by difference.


29. What additional piece of information is necessary todetermine the molecular formula for the experimentalcompound?A. The volume of CO2 collectedB. The volume of the water collectedC. The volume of the hydrocarbon before the reaction

was carried outD. The molecular mass of the hydrocarbon

30. If an unknown compound were combusted in thepresence of excess oxygen, what by-product of thecombustion would be collected to determine thepercentageof sulfurwithin that compound?A. S02B. CS2C. H2SD. S8

31. The mass percent of oxygen within a compound cannotbe determined directly using elemental analysis. Whichof the following is NOT an explanation for this?A. Oxygen gas does not exist in the solid phase at any

temperature.B. When a carbohydrate is oxidized, the oxygen of the

unknown carbohydrate can be found in both waterand carbon dioxide.

C. The procedure requires adding excess oxygen, so theoxygen atoms from the carbohydrate cannot bedistinguished from the oxygen reactant.

D. Oxygen in the carbohydrate, being fully reduced,does not react with oxygen gas.

3 2. Why is the bubbler filled with mineral oil?A. The oil traps out any unreacted organic vapor.B. The oil can transfer heat to warm the gas rapidly.C. The oil prevents back-flow of gas from the outside

environment.D. The oil filters out any liquid products from the

reaction.

Copyright ©by The Berkeley Review® 38

33. Which of the following is NOT associated with anincreasing mass percent ofcarbon in a hydrocarbon?I. An increase in the mass of carbon per gram of the

compoundn. An increase in the mass of water formed upon

oxidation of one gram of the compoundHI. An increase in the mass of hydrogen per gram of

the compoundA. IlonlyB. DlonlyC. I and II onlyD. II and ffl only


Passage VI (Questions 34 - 40)

Dilution reduces the concentration of a solute by addingmore solvent to the solution. The addition of solventincreases the volume of solution while having no effect onthe moles of solute. Molarity is defined as moles solute perliter solution, so the denominator is increased by the additionof solvent, while the numerator is unaffected. To determinethe concentration, use:

MjVi = MfVf

Equation 1where Mj is the initial molarity, Vj is the initial volume,Mf is the final molarity, and Vf is the final volume.

Dilution can be described by the relative concentration ofthe initial and final solutions. For instance, a fifty percentdilution involves a reduction of the molarity by fifty percent.This would result from doubling the volume of the solution,achieved by mixing one part solvent with one part solution.

A solution is diluted as a solvent is added to it in avolumetric flask, until the desired volume is reached. Toensure complete transfer of the solute, the original flask isoften rinsed with the new solvent, and then the contents arepoured into the volumetric flask. The laboratory instructionsfor a tenfold dilution are:

I: Fill a volumetric pipette with a sample of solutionfrom a beaker and then discard the solution. Repeat thisprocedure two additional times to equilibrate theconcentration of the solution on the walls of the pipettewith the concentration of the solution in the beaker.

II: Using the treated volumetric pipette, transfer tenmilliliters of solution to a 100-mL volumetric flask.

HI: Rinse pure water through the pipette and into thevolumetric flask until the flask is roughly eightypercent full.

IV: Set the pipette aside, and continue to add water to theflask until the base of the meniscus is flush with the100-mL line on the volumetric flask.

34. Which of the following solutions has the GREATESTmolarity?

A. 4% by mass KBr in waterB. 4% by mass KCl in waterC. 4% by mass NaBr in waterD. 4% by mass NaCl in water

35. To convert 300 milliliters of 0.150 M solution to asolution with a concentration of 0.0075 M, how muchwater must be added?

A. 6.00 litersB. 5.70 litersC. 4.30 litersD. 4.20 liters


3 6. What is the final CI" concentration after you mix 50.00mL 0.25 M HC1 with 25.00 mL 0.50 M NaOH?

A. 0.33 MB. 0.25 MC. 0.17 MD. 0.15 M

3 7. Why in step HI is water passed through the volumetricpipette?A. To ensure complete transfer of solutionB. To measure the volume of the water addedC. To cool the volumetric pipetteD. To warm the water prior to mixing

38. Which of the following mixtures results in a 10-folddilution?

A. 9 parts solvent with 1 part solutionB. 10 parts solvent with 1 part solutionC. 10% solvent with 90% solutionD. 91% solvent with 9% solution

3 9. Addition of water to an aqueous salt solution would doall of the following EXCEPT:A. lower the molality.B. lower the molarity.C. increase the density.D. increase the mass percent of solvent.

40. Which of the following would MOST dilute 0.10 MLiCl(aq)?A. The addition of 100 mL H20(1) to 25 mL 0.10 M

LiCl(aq)B. The addition of 200 mL H20(l) to 60 mL 0.10 M

LiCl(aq)C. The addition of 50 mL H20(1) to 15 mL 0.10 M

LiCl(aq)D. The addition of 150 mL H20(1) to 50 mL 0.10 M

LiCl(aq)


Passage VII (Questions 41 - 47)

There are many ways in which the concentration of asolution can be expressed, including:Molarity: The concentration of a solution as measured inmoles solute per liter solution.

M _ moles soluteliters solution

Molality: The concentration of a solution as measured inmoles solute per kilogram solvent.

m _ moles solutekilogram solvent

Percent solution: The percent of solute in a solution bymass or moles.

%Solution by mass = mass solute x 100%mass solution

%Solution by moles = moles solute x 100%total moles in solution

Density: The mass of the solution divided by the volumeof the solution.

p _ mass solutionvolume solution

The concentration of a solution can be expressed in anyof these units, which can be converted into one another aslong as the molecular mass of the solute and solvent areknown. For instance, when the percent solution by mass ismultiplied by the density, the result is mass of solute pervolume of solution. When the mass of solute is convertedinto the moles of solute (which requires knowing themolecular mass), the molarity can be determined. Thepercent solution by mass can be converted into molality bysubtracting the mass of solute from the mass of solution tofind the mass of solvent. That determines the denominator.To get the numerator, the mass of solute is converted intomoles solute, and solving for the molality becomes a simpledivision problem.

Adding solvent to a solution dilutes the solution andthus reduces the concentration of the solute in the solution.Addition of solvent to the solution decreases all of the abovemeasurements of concentration, with the exception of thedensity. The density change of a solution depends on therelative density of the solvent and solution.

41. Adding water to an aqueous solution of knownconcentration always decreases all of the followingEXCEPT:

A. density.B. molarity.C. molality.D. mass percent of the solute.


42. When 1.0 grams of a salt are dissolved into 100 mL ofwater, the volume of the solution is greater than 100mL but less than 101 mL. What can be said about theconcentration of the solution?

A. The molality of the solution is greater than themolarity of the solution; the density of the solutionis greater than that of pure water.

B. The molarity of the solution is greater than themolality of the solution; the density of the solutionis greater than that of pure water.

C. The molality of the solution is greater than themolarity of the solution; the density of the solutionis less than that of pure water.

D. The molarity of the solution is greater than themolality of the solution; the density of the solutionis less than that of pure water.

4 3. An organic compound with a density that is less than1.00 g/mL is added to an organic liquid, also with adensity that is less than 1.00 g/mL. What can be saidabout the concentration of the solution?

A. The molality of the solution is greater than themolarity of the solution; the density of the solutionis greater than that of pure organic liquid.

B. The molarity of the solution is greater than themolality of the solution; the density of the solutionis greater than that of pure organic liquid.

C. The molality of the solution is greater than themolarity of the solution; the relative densities ofthe solution and the organic liquid cannot bedetermined without more information.

D. The molarity of the solution is greater than themolality of the solution; the relative densities ofthe solution and the organic liquid cannot bedetermined without more information.

44. Given that a solute is denser than the solvent intowhich it dissolves, what is TRUE of the concentrationmeasurements of different solutions made up solely ofthe two components?A. The solution with the greatest density also has the

greatest molarity and molality.B. The solution with the greatest density also has the

greatest molarity, but the molality is the same forall of the solutions.

C. The solution with the greatest density also has thelowest molarity and molality.

D. The solution with the greatest density also has thelowest molarity, but the molality is the same forall of the solutions.


45. How many milliliters of pure water must be added to100 mL 0.25 M KBr to dilute it to 0.10 M?

A. 100

B. 150C. 250D. 500

46. To achieve the same chloride ion concentration as 1.0grams NaCl(s) dissolved into 100 mL solution, howmany grams of MgCl2(s) must be added to enoughwater to make 100 mL of solution?

A. 1.0xlx5-MgMgCl2(s)2 94.9

B. 1.0 x2 x5-&4 gMgCl2(s)94.9

C. 1.0x1x24,9 gMgCl2(s)2 58.4

D. 1.0x2x9-^9-gMgCl2(s)58.4

4 7. Given two compounds, Compound A and CompoundB,and the fact that B has a higher molecular mass than A,but A is denser than B, which of the following mixtureswould have the greatest mole fraction of A?A. The mixture of 1.0 grams Compound A with 1.0

grams Compound BB. The mixture of 1.0 moles Compound A with 1.0

moles Compound BC. The mixture of 1.0 mL Compound A with 1.0 mL

Compound BD. Themixture of 1.0x 1023 molecules Compound A

with 1.0 x 1023 molecules Compound B


Passage VIII (Questions 48 - 54)

The absorbance of visible light by colored aqueoussolutions is directly proportional to the concentration ofsolute in the solution. Based on this fact, the concentrationof a solution can be determined by monitoring the absorbanceat one wavelength of light. For best results, the detectorshould be focused on the wavelength of highest absorbance(known as Amax). The relationship between absorbance andsolute concentration is expressed as

Absorbance = e[C]l

Equation 1

where £ is the molar absorbtivity constant of the solute, [CIis the concentration of solute, and 1 is the path length of thelight passing in through the cuvette.

A student measures the absorbance for a series of standardsolutions. Once enough data points are collected, themolarity of another solution using the same solute in anunknown concentration is analyzed by comparing itsproperties with the experimental data. The molarabsorbtivity constant and cuvette path length remain constantthroughout all the trials, so any difference in absorbancebetween the unknown and the reference compounds can beattributed to differences in solute concentration. Figure 1 is agraph of thestudent's datacollected for the standard solutions.

Concentration (in molarity)

Figure 1

Table 1 lists the same data summarized graphically above.

Molarity Absorbance

0.10 0.0930.20 0.188

0.30 0.2780.40 0.3630.50 0.4560.60 0.5600.70 0.636

Table 1

48. If the concentration of a solute were doubled, whatwould happen to the absorbance of the solution?A. It would increase by a factor of four.B. It would double.C. It would be cut in half.D. It would decrease by a factor of four.


49. How could the plateau of the following curve beexplained?

Solute added to solution

A. The solvation catalyst in solution has becomesaturated.

B. The reverse reaction is favored at higher soluteconcentration.

C. As more solute is added to the solution, the solutethat is already dissolved begins to repel the solvent.

D. A maximum solute concentration has been reached,because no more solute molecules can dissolve intosolution.

50. Adding 50 mL of pure water to a 10.0-mL sample ofaqueous salt solution with an absorbance of 0.518would yield a new absorbance of:A. 0.518.B. 0.104.C. 0.086.D. 0.259.

51. The concentration can be found according to which ofthe following equations?

A. [C^^el

B. [C] =

C. [C] =

D. [Cl =

el

AbsAbs-1

8

Abs-e

5 2. According to the data from the experiment, what is theconcentration of an unknown solution, if it has anabsorbance of 0.242?

A. 0.197 MB. 0.240 MC. 0.258 MD. 0.289 M


53. What are the units of e?

A. M-cm

B. M-cm"1C. cm-M-1

1

54.

D.M-cm

Which of the following relationships may be TRUE?I. As the molarity increases, the absorbance increases.II. If Compound X has a lower molar absorbtivity

constant (e) than Compound Y, then to have equalabsorbance readings for separate solutions of X(aq)and Y(aq), the concentration of Y must be greaterthan the concentration ofX.

HI. Absorbance = e[C]l at all A, where absorbance oflight can occur.

A. I onlyB. IlonlyC. I and m onlyD. II and m only


Passage IX (Questions 55-61)

An experiment to ascertain the effects of solutionconcentration on the absorbance of visible light studiessolutions of varying concentration for three differentcompounds. The solutions are analyzed in cuvettes using aUV-VIS spectrometer set at fixed values specific for eachcompound. The goal is to maximize the absorbance, so thewavelength of maximum absorbance is used for eachcompound. For Compound M, the spectrometer was set at561 nm; for Compound Q, the spectrometer was set at 413nm; and for Compound T, the spectrometer was set at 697nm. Table 1 shows concentration and correspondingabsorbance for each solution.

Trial Contents Absorbance

I 0.10 M Compound M 0.362

n 0.10 M Compound Q 0.299

m 0.10 M Compound T 0.511

IV 0.06 M Compound M 0.217

V 0.06 M Compound Q 0.180

VI 0.06 M Compound T 0.307

vn 0.03 M Compound M 0.109

vm 0.03 M Compound Q 0.090

DC 0.03 M Compound T 0.153

Table 1

The absorbance of each solution was compared to theabsorbance of a sample of distilled water, which remained inplace in the spectrometer for the duration of the study.Because all three compounds have absorbance bands in thevisible range, they are all observed to have a distinct colorwhen they are exposed to white light. The absorbed color isthe complementary color of the observed color. The visiblespectrumranges from a wavelength of 400 nm to 700 nm.

55. Why is the spectrophotometer set at 561 nm for thetrial involving Compound M?A. 561 nm is the average wavelength of absorbance for

the complementary color of what is absorbed.B. 561 nm is the average wavelength of absorbance for

the color that is absorbed.C. 561 nm is the wavelength of maximum absorbance

for the complementary color of what is absorbed.D. 561 nm is the wavelength of maximum absorbance

for the color that is absorbed.

56. How can a compound's molar absorbtivity constant beobtained, if absorbance varies with cuvette length?A. e = Abs- [Compound] -1B. £= Abs

[Compound]-1P _ [Compound]-1

Abs

D. 6 = ^-1


5 7. Which solution has the GREATEST concentration?

A. Compound M solution, with an absorbance 0.400B. Compound Q solution, with an absorbance 0.250C. Compound T solution, with an absorbance 0.500D. O.lOMKCl(aq)

5 8. Which graph accurately shows absorbance as a functionof concentration for Compounds M, Q, and T?

Concentration

Concentration

Concentration

D. T

Concentration

5 9. What can be expected for other solutions?I. 0.05 M Compound M has an absorbance of A =

0.181 at X = 561 nm.

n. A solution of Compound Q with an absorbance ofA = 0.225 at X = 413 nm has a concentration of0.075 M.

HI. 0.11 M Compound T has an absorbance of A =0.611 atA. = 710nm.

A. I onlyB. I and II onlyC. I and D3onlyD. II and m only

60. To form a solution of Compound T with an absorbanceof 0.250 at X = 697 nm, what must be done?A. Mix 10.0 mL 0.10 M T with 5.0 mL H2OB. Mix 20.0 mL 0.10 M T with 20.0 mL H2OC. Mix 19.0 mL 0.10 M T with 20.0 mL H2OD. Mix 10.0 mL 0.10 M T with 9.0 mL H2O

61. What are the observed colors for each solution?

A. M: green; Q: violet; T: redB. M: red; Q: violet; T: greenC. M: green; Q: yellow; T: redD. M: red; Q: yellow; T: green


Passage X (Questions 62 - 68)

Stoichiometric reactions can be classified into sixcategories:Combination Reaction: Occurs with the combinationof reactants to form one product.

PBr3(l) + Br2(l) • PBr5(s)

Decomposition Reaction: Occurs with thedecomposition of one reactant to form two products.

CaC03(s) • CaO(s) + C02(g)

Single-Replacement Reaction: Occurs with theexchange of either the cations or the anions in a salt, but notboth. A single-replacement reaction is also referred to as anoxidation-reduction reaction.

3 Mg(l) + 2 ScBr3(g) 2 Sc(s) + 3 MgBr2(s)

Metathesis Reaction: Occurs when two cationsexchange their anions. At least one precipitate falls out ofsolution. A metathesis reaction is also referred to as adouble-displacementreaction.

AgN03(aq) + KCl(aq) • KN03(aq) + AgCl(s)

Combustion Reaction: Occurs with the addition ofoxygen to a reactant to form oxidized products (usuallycarbon dioxide and water, when dealing with hydrocarbons andcarbohydrates).

1 C3H8(g) + 5 02(g) • 3 C02(g) + 4 H20(1)

Neutralization Reaction: Occurs with the reaction of anacid with a base to form water and a salt.

HN03(aq) + KOH(s) KN03(aq) + H20(l)

Each reaction is unique from a stoichiometricperspective. When a solid is formed by the reaction of ions(in the metathesis example above), it is referred to as aprecipitate. Gases can also be formed from reactions. Theneutralization of sodium bicarbonate (NaHC03) yields carbondioxide gas. It is possible to categorize all inorganicchemistry reactions by reaction type.

6 2. How should the following reaction be classified?

Ba(N03)2(aq) + K2S04(aq) •

2 KN03(aq) + BaS04(s)

A. Combination reactionB. Decomposition reactionC. Single-replacement reactionD. Metathesis reaction

Copyright © by TheBerkeley Review® 44

63. A precipitate is LEAST likely to occuras a resultof a:A. combustion reaction.B. decomposition reaction.C. single-replacement reaction.D. metathesis reaction.

64. What is the precipitate formed when aqueous sodiumiodide reacts with aqueous calcium nitrate?A. CalB. NaN03C. Cal2D. Na2N03

65. What is the gas formed when magnesium carbonate istreated with hydrobromic acid?A. Hydrogen gas (H)B. Hydrogen gas (H2)C. Carbon dioxide gas (CO2)D. Magnesium bromide gas (MgBr2)

66. What type of reaction is LEAST likely to form carbondioxide gas?A. Combustion reactionB. Metathesis reactionC. Decomposition reactionD. Neutralization reaction

6 7. How should the following reaction be classified?

CaBr2(l) + Cl2(g) • CaCl2(s) + Br2(D

A. Combination reactionB. Decomposition reactionC. Single-replacement reactionD. Metathesis reaction

68. What type of reaction is MOST likely to have anegative value for the change in entropy (AS)?A. Combination reactionB. Decomposition reactionC. Single-replacement reactionD. Combustion reaction


Passage XI (Questions 69 - 78)

Calcium carbonate is found in many everyday products,such as marble, chalk, and antacids. Commercially, it can beexcavated as either calcium oxide or calcium carbonate.Calcium oxide converts to calcium carbonate upon exposureto carbon dioxide under high pressure. At room temperature,calcium carbonate is relatively insoluble in water. Thesolubility of calcium carbonate increases as the pH of theaqueous solution decreases, because calcium carbonate isbasic. Reaction I can be combined with Reaction II toconvert calcium oxide into calcium carbonate in water.

CaO(s) + H20(1) Ca(OH)2(aq)Reaction I

Ca(OH)2(aq) + C02(g) CaC03(s) + H20(1)Reaction II

Ca(OH)2 and CaC03 both readily form a relativelyinsoluble white solid precipitate in water. Because of thislow solubility, the products of both Reaction I and ReactionII are easy to isolate from solution. Fortunately, calciumhydroxide is more soluble in water than is calciumcarbonate,which allows for the selective precipitation of calciumcarbonate in Reaction II. Because of this, industrial processesfor isolating calcium carbonate are primarily water-based.Calcium carbonate can also be formed according to thefollowing equilibrium reaction:

CaO(s) + C02(g) •« ** CaC03(s)Reaction III

6 9. Ca(OH)2 is considered to be which of the following?

A. An amphoteric saltB. A non-metal hydroxideC. An Arrhenius acidD. An Arrhenius base

70. Which of the following molecules are NOT heldtogether by ionic bonds?I. co32-n. CO2

IE. CaO

A. I onlyB. II onlyC. HI onlyD. I and II only


71. The mass percent of calcium is LEAST in which of thefollowing molecules?

A. CaO(s)B. Ca(OH)2(s)C. CaC03(s)D. CaCl2(g)

72. If 10.00 grams of Ca(OH)2(s) produces 5.00 grams ofCaC03(s), what is the percent yield for the reaction?

A. 37%B. 68%C. 74%D. 100%

73. What is the mass percent of calcium in CaO?(Ca = 40 g/mole O = 16 g/mole)A. 28.6%B. 50.0%C. 66.7%D. 71.4%

74. 28.0grams of CaO(s) whenreactedwith 10.00gramsofH2OO) wouldyield whichof the following?(Ca = 40 g/mole 0 = 16 g/mole H = 1 g/mole)A. 37.00 grams Ca(OH)2 with leftover waterB. 37.50 grams Ca(OH)2 with leftover CaOC. Exactly 38.00 grams Ca(OH)2with no leftoverD. Exactly 42.60 grams Ca(OH)2 with no leftover

75. Which of the following is required to neutralize 5.00mL of 0.20 M CaC03(aq)?

A. 10.0 mL 0.10 NNaOH(aq)B. 10.0 mL 0.10 NHN03(aq)C. 10.0 mL 0.30 N H3P04(aq)D. 10.0 mL 0.20 N HCl(aq)

76. What is the final concentration of Ca(OH)2(aq) after50.0mLof pure water are added to 5.0 mL of 0.50MCa(OH)2(aq)?A. 0.055 MCa(OH)2(aq)B. 0.050 M Ca(OH)2(aq)C. 0.046 M Ca(OH)2(aq)D. 0.025 M Ca(OH)2(aq)


77. If 20.0 mL of 0.20 M CaCl2(aq) were mixed with 30.0mL of 0.30 MCaC03(aq), what would the final Ca2+concentration be?

A. 0.233 MCa2+(aq)B. 0.250 MCa2+(aq)C. 0.260 MCa2+(aq)D. 0.267 MCa2+(aq)

78. If 50.0 grams of CaC03(s) are completely neutralizedwith HC1, how many liters of C02(g) form at STP,knowing that one CaC03 yields one CO2?

A. 4.48 LB. 10.00 LC. 11.20 LD. 13.56 L


Passage XII (Questions 79 - 86)

Every year, a substantial amount of the countless tons ofchemicals produced worldwide is used in the manufacture offertilizers and plastics. Described below, according to acommon element each one contains, are some of the typicalchemical fertilizers used in America.

Potassium:

Salts containing potassium are referred to as potash. Themost common forms are K2SO4, 2 MgS04-K.2S04, andKCl. The amount of potassium per gram of salt is importantto know when determining the quantity of fertilizer needed fora job. The compound richest in potassium (by mass percent)is potassium oxide, K2O. The potassium content of anypotash is expressed as a fraction of the potassium inpotassium oxide (K2O).

Nitrogen:Salts containing nitrogen are very useful as fertilizers. Themost common forms are NH4N03, (NH4)2S04, andNH4H2PO4. A very common organic fertilizer containingnitrogen is urea, H2NCONH2, which is made from ammoniaand carbon dioxide. Ammonium sulfate is also made fromammonia by combining Reaction I with Reaction II:

2 NH3(aq) + C02(g) + H20(1) -• (NH4)2C03(aq)

Reaction I

(NH4)2C03(aq) + CaS04(aq)(NH4)2S04(aq) + CaC03(s)

Reaction II

Phosphorus:Salts containing phosphorus are also very useful asfertilizers. The most common form is Ca(H2P04)2-Calcium bisdihydrogenphosphate is produced by Reaction HI:

2 Cas(P04)3F(s) + 7 H2S04(aq) •3 Ca(H2P04)2(aq) + 7 CaS04(aq) + 2 HF(g)

Reaction III

Fluoroapetite (Ca5(P04)3F) is added as the limitingreagent in Reaction HI. This is done to maximize the percentyield of phosphorus in the reaction. Industrially, the percentyield of a reaction and the mass percent of its product arecritical in terms of profit margin for the fertilizer producer.

79. Which of the following compounds has theGREATEST amount of potassium per gram?A. K20B. K2S04C. KClD. KN03


80. Which of the following compounds has the LOWESTmass percent of nitrogen?A. H2NCONH2B. NH4N03C. (NH4)2S04D. NH4H2PO4

81. What is the mass percent of nitrogen in urea?

A. 23.3% NB. 31.9% NC. 46.6% ND. 66.7% N

82. Why is Cas(P04)3F the limiting reagent in thesynthesis of Ca(H2P04)2?A. It prevents leftover Cas(P04)3F from being

wasted, so that P-containing compounds areconserved.

B. It allows leftover Ca5(P04)3F to be wasted, so thatP-containing compounds are conserved.

C. It prevents leftover Cas(P04)3F from beingwasted, so that P-containing compounds are notconserved.

D. It allows leftover Cas(P04)3F to be wasted, so thatP-containing compounds are not conserved.

83. If 10.0 grams of (NH4)2C03 are used in Reaction II toobtain 10.0 grams of (NH4)2S04, then what is thepercentyield for the reaction?A. Less than 50%B. Greater than 50%, but less than 75%C. Greater than 75%, but less than 100%D. Greater than 100%

84. The molarity of potassium is GREATEST for which ofthe following solutions?

A. 10.0 g KCl in enough water to form 100 mL ofsolution

B. 10.0 g K2SO4 in enough water to form 100 mL ofsolution

C. 10.0 g K2C03 in enough water to form 100 mL ofsolution

D. 10.0 g KN03 in enough water to form 100 mL ofsolution


85. From which of the following reactions is it EASIESTto isolate the desired product?

A. A reaction yielding the product as a precipitateB. A reaction yielding the product as a liquidC. A reaction yielding the product as a gasD. A reaction yielding the product as an aqueous solute

8 6. Which of the following relationships must be TRUE?I- pKal(H3PO4) is less than pKa(HF)H. pKa2(H3P04) is less than pKa2(H2S04)D3. pKa(HF) is less than pKai(H2S04)A. I onlyB. IlonlyC. I and II onlyD. II and m only


Questions 87 throughdescriptive passage.

100 are NOT based on a

87. If 25 grams of oxygen are combined with 20 grams ofpropane gas, then which of the following statementswould be TRUE after the mixture is ignited?

A. 45.0 grams of carbon dioxide forms.B. 38.0 grams of water vapor forms.C . Oxygen is the limiting reagent.D . Propane is the limiting reagent.

88. The hemoglobin in red blood corpuscles of mostmammals contains approximately 0.33% iron byweight. If osmotic pressure measurements show thatthe molecular weight is 68,000 for hemoglobin, howmany iron atoms must be present in each molecule ofhemoglobin?

A. 1

B. 4

C. 224D. 400

89. 9.00 grams of a sugar are burned in a containmentvessel, and all the CO2 is collected. The volumeoccupied by the CO2 at STP is 6.72 liters. If themolecular weight of the sugar is 180 g/mole, what isthe ratio of O2 to CO2 in the balanced equation?

A. 3 :3B. 3 :6C. 6:3D. 6:6

90. What is the molecular formula for an unknown gas withthe empirical formula C2H30, if 1.00 grams of theunknown gas occupies 260 mL at STP?A. C2H30B. C4H602C. C6H903D. CgHi204

91. What volume of 02(g) is produced from 1.0 g BaOupon its decomposition to Ba(s) and 02(g) at STP?

A. 0.074 L

B. 0.100 L

C. 0.148 LD. 0.166 L


92. For the following reaction, calculate the mass ofMg2P207(s) that is formed from the decomposition of2.0 grams MgNH4P04(s).

2MgNH4P04(s) —• Mg2P207(s) + 2 NH3(g) + H20(g)

A. 0.8 gramsB. 1.7 gramsC. 2.2 gramsD. 2.8 grams

93. A compound containing 50% by weight of Element X(atomic weight = 40) and 50% by weight of Element Z(atomic weight = 80) is one in which:

A . the molecular formula is XZ or ZX.B. the simplest formula is XZ or ZX.C. the simplest formula is XZ2 or Z2X.D. the simplest formula is X2Z or ZX2.

94. 11.89 grams of hot iron are exposed to a continuousstream of pure oxygen for ten minutes. At the end ofthis time, the completely oxidized sample weighs 16.99grams. The empirical formula for the compound thusformed is MOST accurately written as:A. Fe302-B. FeO.C. Fe203.D. Fe03.

95. A stable compound consisting of 53.4% C, 11.0% H,and the remainder O has a molecular weight of 90grams/mole. The molecular formula for the compoundis:

A. C5H602.B. C3H603.C. C4H26O.D. C4H,o02.

96. In reducing Cr042"(aq) to Cr203(s), how does theoxidation state of chromium change?A . From +6 to +3

B. From +4 to +3C. From +3 to +4D. From +3 to +6


97. An unknown metal is found to combine with oxygen ina 2 : 3 ratio in the molecular formula. The metal oxideis approximately 53%metal by mass, and the remainderis oxygen. What is the MOST probable identity of themetal?

A. CalciumB. IronC. ChromiumD. Aluminum

98. Which of the following organic compounds has theGREATEST mass percent of carbon?A. Acetic acid (CH3C02H)B. Ethanol (CH3CH2OH)C . Methyl acetate (CH3C02CH3)D. Glucose (C6H12O6)

99. When one gram of each of the following organiccompounds is burned (oxidized), which one yields theGREATEST amount of carbon dioxide (by mass ormoles)?

A. Acetic acid (CH3CO2H)B. Ethanol (CH3CH2OH)C . Methyl acetate (CH3CO2CH3)D. Glucose (C6H12O6)

100. What is the mass percentage of chlorine inMg(C104)2?

A. 71.0%B. 40.6%C. 31.8%D. 15.9%

Copyright ©byThe Berkeley Review® 49

"Count on molecules to make your day!"

1. B 2. D 3. A 4. B 5. A 6. C

7. D 8 D 9. B 10. C 11. D 12. A

13. C 14. A 15. C 16. D 17. B 18. B

19. C 20. B 21. B 22. D 23. B 24. A

25. A 26. D 27. C 28. C 29. D 30. A

31. A 32. C 33. D 34. D 35. B 36. C

37. A 38. A 39. C 40. A 41. A 42. A

43. C 44. A 45. B 46. C 47. C 48. B

49. D 50. C 51. A 52. C 53. D 54. C

55. D 56. B 57. A 58. D 59. B 60. C

61. D 62. D 63. A 64. C 65. C 66. B

67. C 68. A 69. D 70. D 71. D 72. A

73. D 74. A 75. D 76. C 77. C 78. C

79. A 80. D 81. C 82. A 83. B 84. C

85. A 86. A 87. C 88. B 89. D 90. B

91. A 92. B 93. D 94. C 95. D 96. A

97. D 98. B 99. B 100. C

YOU ARE DONE.

Stoichiometry Passage AnswersPassage I (Questions 1-7) Density Experiment

Choice B is correct. In order for a liquid to float on water, the liquid must be both immiscible in water and lessdense than water. All of the liquids are assumed to be immiscible in water, according to the question, so theonly stipulation that remains to be considered is the density of each liquid. The density of water is defined as1.00 grams per milliliter. We know that the mass of the empty volumetric cylinder is 42.61 grams and itscapacity is 10.00 mL. For the density of any liquid to be less than 1.00, a 10.00-mL sample of the liquid musthave a mass of less than 10.00 grams. This means that when 10.00 mL of a liquid is added to the volumetriccylinder, the liquid and cylinder must have a combined mass of less than 52.61 grams. If the combined mass isless than that, then the density of the liquid must be less than 1.00. According to Table 1, unknown Liquid #6(with a combined mass with the volumetric cylinder of 51.33 grams) and unknown Liquid #9 (with a combinedmass with the volumetric cylinder of 49.03 grams) are the only two unknown liquids with combined masses ofless than 52.61 grams. Only Liquid #6 and Liquid #9 can float on water. The best answer is therefore choice B.

Choice D is correct. This question requires evaluating the density for unknown Solids #2, #3, #4, and #5.Density is defined as mass per volume. In this case, the volume of the solid is obtained by taking the volumereading from the chart for each unknown and subtracting 25.00 mL for the volume of the water already in thecylinder. This method of measurement is known as the "volume by displacement technique." The followingtable shows the values of mass, volume, and density obtained for each solid:

Unknown Mass (g)

12.38

14.85

8.22

5.64

Volume (ml.)

31.19-25.00 = 6.19

29.95 - 25.00 = 4.95

28.00 - 25.00 = 3.00

26.41 -25.00=1.41

Density

2=*£-zoo-S-6.19 mL mL

2i^g_=3.ooJL4.95 mL mL

^iL=2.74JL3.00 mL mL

5-64g =4.00-g-1.41 mL mL

The unknown solid with the greatest density is unknown Solid #5. The correct answer is thus choice D. Thenumbers could have been compared to one another to obtain the relative values.

ChoiceAis correct. Thedensityof unknownSolid #1 must be determined first. It is found by dividing its massby its volume (found by difference):

p= ?^g =J^g.= 1J0_g.31.42 - 25.00 mL 6.42 mL mL

Because the density is 1.50grams per milliliter, a 20.0-gram piece of the solid must have a volume of 13.3 mL,because 20.0 grams divided by 13.3 mL is equal to 1.50 grams per milliliter. The correct answer is choice A.Choices C and Dshould have been eliminated immediately, because when the volume of an object is greaterthan its mass, then the density of the object is less than 1.00. Solid #1 sinks when placed into water, indicatingthat its density is greater than that of water, which is 1.00 grams per milliliter. Distinguishing betweenchoice A and choice B requires looking more closely at the ratios.

Choice B is correct. In order for a solid to float onunknown Liquid #7, it must have a density less than thatliquid. The density for unknown Liquid #7 is:

58.72-42.61 g 16.11P =

10.00 mL= 1.611-2-

0.00 mL mL

Only Solid #1, with a density of only 1.50 grams per milliliter, has a density less than 1.611 grams permilliliter. The best answer is choice B. The relative densities of Liquid #7 and Solid #1 are shown below:

16.11 g 32.22g 9.63g 28.89g 32.22g 28.89gPLiquid#7 = , = . ; PSolid#l =

10.00 mL 20.00 mL 6.42 mL 19.26 mL 20.00 mL 19.26 mL

Copyright © by The Berkeley Review® 50 Section I Detailed Explanations

Choice A is correct. The following table shows how to solve for the densities of the four liquids:

Density (-%-)mL

872 g =Q.872S-10.00 mL mL

1611 g =1.611-§-10.00 mL mL

m6°g =1.Q60 JL10.00 mL mL

642 g =Q.642 JL10.00 mL mL

From the calculations, the relative densities in descending order are: Liquid #7 > Liquid #8 > Liquid #6 >Liquid #9. The answer choice that doesnot follow this pattern is choice A: Liquid #7 > Liquid #6 > Liquid #8.This question could also have been solved by comparing the masses in the chart for the cylinder and liquidcombined. All of the liquids have the same volume (10.00 mL) and were in the same cylinder, so the same massis subtracted from each in determining the liquid's mass. This would have saved much time.

Choice C is correct. If the mass of the cylinder were heavier than 42.61 grams, then the reading for the mass ofthe liquid and cylinder combined would begreater thanitwas in Experiment 2. However, the mass of liquid ina fixed volume is the same, so its density does not change. No matter what container is chosen to hold it, thedensity of a liquid is an invariant propertyof that liquid. This is best reflected in answerchoice C.

Choice D is correct. Density is defined asameasure ofmass per volume. The units for density should thereforereflect a massunit divided by a volume unit. Inanswerchoice A, themass ismeasured in grams and the volumeis measured in milliliters, which makes choice A acceptable. In answer choice B, the mass is measured inounces and the volume is measured in centimeters cubed, which makes choice B acceptable. You should recallthat a milliliter is a centimeter cubed. In answer choice C, the mass is measured in kilograms and the volume ismeasured in liters,which makes choice C acceptable. In answer choice D, the weight (and not necessarily themass) ismeasured inpounds and isdivided by anarea (dimension squared) andnotbyvolume. Choice D is thusameasureofpressure,not density. Thecorrect answer is choice D.

Unknown Mass (g) Volume (mL)

6 51.33 - 42.61 = 8.72 10.00

7 58.72-42.61 = 16.11 10.00

8 53.21 - 42.61 = 10.60 10.00

9 49.03 - 42.61 = 6.42 10.00

Passage II (Questions 8 - 13) Combustion Analysis

8. Choice D is correct. Given that all of the samples were of nearly equal mass (between 1.995 g and 2.011 g), thegreatest mass percent of carbon is in the compound that has the greatest mass of carbon. The compound with thegreatest mass of carbon produces the greatest mass of carbon dioxide gas. This means that the easiest way tosolve this question is to compare the amount of carbon dioxide collected for each sample, as listed in column 4(the NaOH tube column) inTable 1. From the data in the NaOH tube column, the greatest mass of CO2 iscollected from Compound IV (36.53 -30.00 =6.53), eliminating choices Aand B. Because a greater mass of CO2 isproduced by oxidizing Compound III than by oxidizing Compound II, choice Disthe best answer.

9. Choice Bis correct. The roleof thesodium hydroxide salt is tobindCO2, not to bind H2O. This means that thesalt should be both hydrophobic (non-water-binding) and reactive with carbon dioxide, making choice B thebest answer. The term "hygroscopic" refers toa compound with a high affinity for binding moisture in the air.

10. Choice C is correct. The pressure valve is designed to allow oxygen gas to flow from the tank into the systemwhen a threshold pressure is maintained. The oxygen partial pressure can be controlled and maintained at ahigh level. The oxygen tank is left open with a positive pressure of oxygen to ensure that oxygen gas iscontinually flowing into the system, so choice Ais eliminated. Nothing was mentioned about the temperature ofthe oxygen gas, so choice B is eliminated. Oxygen gas is always present, so it is in excess and is not a limitingreagent. This eliminates choice D and makes choice C the correct answer.

11. Choice D is correct. The greatest mass percent of carbon is found in the compound with the smallest number ofoxygen atoms in its formula. Thus, the compound least likely to contain oxygen is the one that produces the mostCO2 upon combustion. That is Compound IV. In addition, pure carbon when oxidized yields 7.33 g CO2.Compound IV yields 6.53 grams CO2, so itis close enough to pure carbon to assume no oxygen is in it, choice D.

Copyright ©by The Berkeley Review® 51 Section I Detailed Explanations

12. Choice A is correct. When the first valve is opened, the gas is exposed to magnesium sulfate, which binds water,but not carbon dioxide. This means that water is bound first, leaving an atmosphere of excess oxygen gas andcarbon dioxide. This is important, because the sodium hydroxide compound can bind both carbon dioxide andwater. Using magnesium sulfate first to remove the water vapor ensures that all water is removed from the gaswhen the second valve is opened, exposing the gas to the sodium hydroxide salt. Any increase in the mass ofmagnesium sulfate is due to the binding of water. Any increase in the mass of sodium hydroxide is due to thebinding of carbon dioxide. If the gases were first exposed to sodium hydroxide, the increase in mass would be dueto both water and carbon dioxide. Choice A is the best answer. Excess oxygen leaves the system as a free gas.

13. Choice C is correct. The solid formed in the second tube results from the reaction of sodium hydroxide (NaOH)with carbon dioxide (CO2),so it must be a sodium salt. This eliminates choice A and B. The following reactionconfirms that the best answer of the given choices is sodium bicarbonate, choice C:

NaOH(s) + C02(g) • NaHC03(s)

Passage III (Questions 14 - 20) Empirical Formula Determination

14.

15.

Choice A is correct. If 20.0 grams of the unknown were oxidized, instead of 10.0 grams, then the amount ofcarbon dioxide and water formed as products would double. This eliminates choicesC and D. The mass percentof carbon should remain the same, because the mass of carbon dioxide formed and the mass of the compound bothdoubled. The best answer is choice A. The mass percent of carbon is constant, because the molecular formula ofthe compound is constant.

Choice C is correct. To determine the mass percent of carbon in the unknown compound, the grams of carbon incarbon dioxide (CO2) are divided by the total number of grams of original compound. The mass of carbondioxide formed is found by subtracting 100.00 grams of original sodium hydroxide from the 123.79 grams ofbicarbonate salt. The difference is the mass of carbon dioxide that binds the salt.

12gC23.79grams CO2 x44gC°2 x 100% = 23.79 x 12. x J_ x 100o/o

10.0grams unknown 44 10This makes choice C the best answer.

16. ChoiceD is correct. In the first sentence of the passage, we read that the unknown organic compound is pouredinto a flask. The term "poured" implies that the compound flows, which makes it a fluid. The fact that it canbe poured into a flask means that it is flowing down, which defines it more specifically as a liquid. Thecompound is a liquid at ambient temperature (room temperature). This means that the melting point is lessthan ambient temperature, because at room temperature it has already melted into a liquid. The boiling pointis greater than ambient temperature, because at room temperature it has not yet boiled into a gas. Because i treadily evaporates, the boiling pointmay be close to ambient temperature; but because it is a liquid at ambienttemperature, the boiling point must be greater than ambient temperature. The best answer is choice D.

17. Choice B is correct. As stated at the end of the passage, the molecular formula for a compound with justhydrogen, oxygen, and carbon cannot have an oddnumber of hydrogens. An odd number of hydrogens results inan odd number of bonding electrons (electrons present in bonds). Considering that there are two electrons perbond, an odd number of bonding electrons results in only half of a bond somewhere in the compound (whichequates to a single electron, or free radical.) A half-bond is not stable (possible to isolate physically), so choiceB, with five hydrogens in the formula, is not possible. Knowledge from organic chemistry can prove useful insolvinggeneralchemistryquestions. You must incorporate information frommany sources to excel at this exam.

18. ChoiceBis correct. There are 12.16 grams of water produced from the oxidation of 10.0 grams of the unknown.Thiscanbe determined by subtracting100.00 grams for the mass of the anhydrous sodium sulfate from the finalmassof 112.16 grams for the hydrated sodiumsulfatesalt. Themoles ofwater formed are greater than 0.5 moles,because12.16 grams divided by 18 is greater than 9 grams divided by 18. The moles of water formed is less than1.0moles, because 12.16 grams divided by 18 is less than 18grams divided by 18. The best answer is thereforechoice B.

H 12.16 g< . ° <

18g0.5 moles =

18g/mole 18g/mole H mole

= 1.0 moles

Copyright © by The BerkeleyReview® 52 Section I Detailed Explanations

19.

20.

Choice C is correct. The question asks for the empirical formula, but because information is given about themolecular mass, you may wish to consider the molecular formula first, then eliminate choices based on thatinformation. According the passage, the molecular mass has a value between 70 and 80 grams per mole, sochoice B (MW = 48 + 8 + 32 = 88) and choice D (MW= 96 + 10 + 16 = 122)are eliminated. The only choices with amass in the range of 70 grams to 80 grams are choice A (which has a mass of 74 grams per mole) and choice C(which also has a mass of 74 grams per mole). Choice A is — percent carbon, while choice C is — percent

carbon. The mass percent of carbon is greater than fifty percent, so choice A is eliminated, leaving choice C asthe best answer. The exact numerical value for the mass percent of carbon can be approximated as follows:

12gC23.79 grams CO2 x44gC°2 x 100% = 23.79 x 12. x J_ x 10n% = 2379 x 12 x 100% = 23'79 X3 x 100%

10.0grams unknown 44 10 44 x 10 11 x 10

= 7L3Z x 100%, whereZL3Z x 100% > 50%110 110

Choice C is both the molecular formula and the empirical formula.

Choice B is correct. The mass percent of carbon in C5H12O2 is found by dividing the mass of five carbon atoms(60 amu) by the molecular mass of C5H12O2 (60 + 12 + 32 = 104amu). The value determined when 60 is dividedby 104 is less than 60%, because it is less than 60 divided by 100. The best choice for a value close to, but lessthan, 60% is choice B, 57.7%. You really should choose answer choice B, if you know what's best for you.

mass percent of carbon = 5 x 12

60 + 12 + 32

_ 60

104-6Q_ = 60%100

Passage IV (Questions 21 - 26) Molar Volume of a Gas

21. Choice B is correct. The mass percent of carbon in carbondioxide is the mass of carbon (12.011) divided by themass ofcarbon dioxide (44.009) x100%. The fraction I2- reduces to-2- =3x^ =3x 0.091 =0.273 x100% =27.3%.

44 11 11To make math easier, you should memorize the following set of fraction-to-decimal conversions. Ademonstration of how these fractions are useful is found on page 30.

I = 0.333,1 = 0.250,1 = 0.200,1 = 0.166,1 - 0.143,1 = 0.125,1 = 0.100, -1- = 0.091,-1- - 0.091,-1- = 0.0833 4 5 6 7 8 9 10 11 12This trickworkswell, once learned. It may seem awkward at first, but try it. Thecorrect answer is choice B.

22. Choice D is correct. The second experiment was conducted to determine the molecular mass of the unknownliquid. At 304 K, the boiling point of the liquid (and thus, the temperature of the gas), the volume of gas isexactly 1.00 liters, assuming that the flask is completely filled with vapor from the liquid. Molar volume atthis temperature is 24.96 liters per mole. At 304 K, the gas has a density of 2.32 grams per liter. When thisvalue is multiplied bymolar volume at 304 K, the liters cancel outand the remaining units are grams permole.The molecular mass is 2.32 g/L x24.96 L/mole =2.32 x24.96 g/mole=58 g/mole. The best answer ischoice D.You should always consider units when looking at math questions. The question asks for a molecular weight,which has units ofgrams per mole. The numbers in the answer choices are2.32 (which is in grams per liter),22.41 (which is in liters per mole), and 24.96 (which is in liters per mole.) To get the target units, grams perliter are multiplied by liters permole, choices CorD. Choice D has the correct molar volume of the unknownliquid at 304 K.

23. Choice B is correct. You know the molecular mass of the unknown compound (58grams per mole) and the masspercent (82.9%), soyou can make an intuitive determination of the molecular formula. The mass ofcarbon in thecompound must equal either 12, 24, 36, or 48 (multiples of 12). Given that 82.9% of 58 is closest to 48 of themultiples of 12, it can beassumed that the molecule has four carbons. This leaves ten hydrogens (to round outthe 10 grams in 58 grams per mole not accounted for by the four carbons); so the molecular formula is C4H10,which has an empirical formula of C2H5. All of this intuition should lead you to pick B.If you want to solve for the empirical formula exactly, here is the math using the compact formula, where a100-gram sample is assumed. Do lengthy math calculations only if they are absolutely necessary.

C82.9H17.1 = C69H171 = C&MU2A = C1H25 = C.2Hn;12 1 " 6.9 6.9

®Copyright © by The Berkeley Review 53 Section I Detailed Explanations

24. Choice A is correct. The mass percent of carbon in the unknown! compound must be less than 100%. Thisautomatically eliminates choices B, C, and D. If you didn't notice that fact, then the problem can be solvedknowing that the massof carbon in theCO2 isequal to themassof carbon in the unknown compound. Themass ofcarbon in CO2 will be the moles of CO2 (1.55/22.41) multiplied by the mass of carbon (12.011). When this isdivided by the mass of the sample (1.00), the result is answer choice A.

25. Choice A is correct. The key fact in the passage is that the 3.045 grams of carbon dioxide gas formed occupies1.55 L at STP. The question asks for moles of carbon dioxide, so the liters must be converted to moles. Theconversion is given in the chart as 22.4 liters per mole. To convert to moles, either the 3.045 CO2 grams must bedivided by the molecular mass of 44 grams per mole, or the 1.55 liters CO2 must be divided by the 22.41 litersper mole. The second option is answer choiceA, so pick that. Ifyou wish to get an exact value for the moles, youmust get the denominator close to 100. This can be done by scaling the numerator and the denominator (i.e.,adding a value to each that is equivalent to multiplying both by the same value).

155 x 4 = 620 = 6.20 + 0.775 =jx9Z5 s 0.069 moles COz22.4 x 4 89.6 89.6 + 11.2 100.8

This is fairly precise, but unnecessary on the MCAT. The addition of 0.775 to the numerator and addition of 11.2to the denominator accomplish the goal of making the denominator close to 100 while scaling proportionally.

26. Choice D is correct. If the organic vapor had not displaced all of the air in the flask at the time the heat wasremoved, then the flask would not have been filled with pure organic vapor at that time (as was assumed).The actual amount of organic vapor in the flask would be less than the assumed value (100%), so the actualamount of liquid collected would be too low. If the measured mass of the liquid is too small, then the value inthe numerator is too small for the calculation of the molecular mass, so the calculated molecular mass is too lowas well. Pick D to feel that happy sensation of correctness.

Passage V (Questions 27 - 33) Elemental Analysis

27. Choice C is correct. Water must be frozen out of solution, so the temperature of the trap should be less than thefreezing point of water (0°C). This eliminates choices A and B. The temperature of the trap cannot be lowerthan the sublimation point of carbon dioxide, however; otherwise, the carbon dioxide gas would solidify in thetrap along with the water vapor. The trap's temperature must be greater than the sublimation point of carbondioxide, but that temperature is neither given in the passage nor is it common knowledge, unless you happen toown the Jeopardy® Chemistry game. To play it safe, choose a temperature just less than the freezing point ofwater. The value closest to 0°C without exceeding 0°C is -33°C The best answer is therefore choice C.

28. Choice C is correct. The successive traps should be aligned in a manner to isolate each gas separately. Themoles are not determined by difference, so choices B and D are eliminated. The trick here is determining therelative temperature sequence. If the traps were set first to isolate carbon dioxide by deposition (convertingfrom gas into solid), then water vapor would freeze out along with the carbon dioxide gas, and thus thecompounds would not be separated. By freezing water vapor first, the carbon dioxide can pass to a later trapwhere it can in turn be isolated free of water. Because the freezing point of water is 0°C and the sublimationpoint of carbon dioxide is -78°C, the temperatures of successive traps should gradually decrease. ChooseC.

29. Choice D is correct. To determine the molecular formula from the empirical formula, the molecular mass of thecompound must be known. This fact makeschoice D the best answer. The volume of the products and reactantsis dependent on the mole ratio in the reaction, but not exclusively cm the molecular formula. The volume maychange depending on other conditions, such as temperature. This eliminates choices A, B, and C Pick D.Determination of the molecular formula from the empirical formula involves comparing the empirical mass tothe molecular mass. If the empirical mass is equal to the molecular mass, then the empirical formula is themolecular formula. Choose D for the satisfaction of knowing you got this correct.

30. Choice A is correct. Combustion analysis involvesoxidizingcompounds and collecting their oxide gases. Sulfurgas can be oxidized like carbon (given that carbon and sulfur are roughly equivalent in electronegativity). Thismeans that sulfur oxide can be collected and analyzed like the oxide of carbon. The oxide of sulfur that isformed is uncertain, but the only answer choice showing oxidized sulfur is choice A, where sulfur carries apositive four (+4) oxidation state. Sulfur can also be oxidized into sulfur trioxide (SO3), but that is not listed asa choice.


31. Choice A is correct. If oxygen gas were trapped out from the gas mixture, there would be no way to knowwhether the oxygen gas collected was from the original hydrocarbon, or from the excess oxygen gas needed toensure the complete combustion. This makes choice C valid. It is true that oxygen from the carbohydrate isfound in both the carbon dioxide product and the water product; thus, the oxygen from the compound cannot beisolated. This makes choice B valid.

Oxygen in the carbohydrate is fully reduced (has a negative two (-2) oxidation state when bonded to carbon andhydrogen). When looking at complete combustion reactions, remember that oxygen atoms from the compound donot react with oxygen gas at all. This makes choice D valid. Oxygen gas has a sublimation point around -180°C(so oxygen gas undergoes deposition when the temperature drops below -180°C); thus, oxygen gas can becollected out of the air using a liquid nitrogen trap. Nitrogen liquid has a boiling point of -196°C, so thetemperature of the trap must be -196°C. The best choice is therefore choice A, because oxygen can exist as asolid. The correct answer was a double negative.

32. Choice C is correct. As stated in the passage, the bubbler vents any pressure buildup within the system whilekeeping the system closed. The bubbler is a one-way valve that allows gas to effuse from the system, but doesnot let gas infuse from the environment. The oil in the bubbler is intended to interact only with the gas, so nounreacted organic vapor should dissolve into the oil. Choice A is a valid possibility in the reaction (a truestatement), but it is not relevant to the goals of the experiment. Answer choices like choice A are difficult toeliminate sometimes. The oil in the bubbler is not serving as an oil bath, so choice B is eliminated. The bubblerkeeps the system closed by preventing infusion ofgaseous compounds from the outside. The best answer is choiceC. Liquids cannot flow into the oil in the bubbler, so choiceD is eliminated.

33. Choice D is correct. An increase in the mass percent of carbon is defined as an increase in the amount of carbonper gram of the compound, so statement I is valid.The amount of water formed depends on the number of hydrogens in the compound, which depends on the masspercent of hydrogen. As the mass percent of carbon within a hydrocarbon increases, the mass percent ofhydrogen in the hydrocarbon decreases. This reduces the amount ofwater formed from one gram ofcompound, soit is not true that the mass of water formed increases upon oxidation. The decrease in mass percent of hydrogenreduces the mass of hydrogen formed per gram of compound. Thismeans that neither statements II nor III isassociated with an increasingmass percent of carbonwithin a hydrocarbon. ChoiceD is the best answer.

Passage VI (Questions 34 - 40) Dilution Experiment

34. Choice D is correct. The four salts listed as choices are all present in the samemass quantity in solution (4% oftheir respective solution.) All of the solutions have 96% water solvent, so the salt with the greatest number ofmoles has the highest molarity. Given that all of the salts have an equal mass, the greatest number ofmolesbelongs to the salt with the lowest molecular weight. Sodium (Na) is lighter than potassium (K), and chlorine(CI) is lighter than bromine (Br), soNaCl is the lightest salt of the choices. Pick D, and feel fresh.

35. Choice B is correct. The trick here lies in the wording: "...how much water must be added?" This questionrequires that you use the dilution equation, MinitialVinitial =MfjnaiVfirial, where Mis molarity and Vis volume.

0.15Mx300 mL =0.0075MxVfinal .-. Vfina, =0.15Mx300mL =J5_ xmmL =l^ii x300 mL0.0075 M 0.75 3

= 15 x 4 x 100mL = 6000 mL = 6.0 L

This value is Vfjnai, not the answer to the question (volume of water added)! To finish with 6.0 liters ofsolution from an original volumeof0.3 liters, 5.7 liters must be added. Choice B is the correct answer.

36. Choice C is correct. As far as chloride ion concentration is concerned, it doesn't matter whether you add purewater or NaOH(aq) to theHCl(aq) solution, because CI" is justa spectator ion in the acid-base reaction. NeitherNaOH(aq) nor pure water increases the moles of chloride anion in the solution. Solve the question by using thedilution equation,MinjtiaiVinitiai = MfinaiVfjnai, whereM is molarity and V is volume:

0.25 Mx 50 mL =Mfinai x 75 mL .-. Mfina, =a25Mx50mL =SI x 0.25M=2 x 0.25 M= 0.167M75 mL 75 3

The final molarity is 0.167 M, which makes choice C, 0.17M, the best answer.


37. Choice A is correct. Pure water is passed through the volumetric pipette to flush any residue out of thevolumetric pipette. Not allof thecontents of thepipette may come free because of the adhesion ofwater to theglass, so rinsing ensures that the residual solution in the pipette is forced from the pipette into the newsolution. The rinsing is not quantitative, so it does not measure a volume. This eliminates choice B. Notemperature is mentioned, so the water used to rinse the pipette is not responsible for heating or coolinganything. This eliminateschoices C and D. Thebest answer is choice A.

38. Choice A is correct. A ten-fold dilution is defined as a dilution that results in a final concentration of a solutionthat is ten percent of its original concentration. For this to occur, the final volume of the solutionmust be tentimes the initial volume. The final volume is the sum of the initial volume and the added volume, so thefollowing math can be applied:

Vfinal = Vinitial + Vadded and Vfinal = lOVjnitial/ so 10Vinitial = Vinitial + Vadded9Vjnitial = Vadded/ so 1 Part solution is mixed with 9 parts solvent.

Ten parts solventmixedwith one part solution yields a final volume that is eleven times larger; thus, the finalconcentration would be one-eleventh, which is less than ten percent. This eliminates choice B. Choice C is thereverse of a ten-fold dilution. Choice C is a dilution to ninety percent of the original concentration. Choice D isthe same as choice B. Pick choice A to feel that jovial tingle of correctness.

39. Choice C is correct. The addition of water to an aqueous salt solution increases the mass of the solvent withoutchanging the moles of solute, so the molality decreases (given that the denominator increases.) This makeschoice A valid. Addition of water to an aqueous salt solution increases the volume of the solution withoutchanging the moles of solute, so the molarity decreases (given that the denominator increases.) This makeschoice B valid. The salt solution is denser than the pure water, so the addition of water (a less dense solution)to the salt solution decreases the density of the solution, making choice C invalid. Addition of water to anaqueous salt solution increases the mass of the solvent without changing the mass of solute, so the mass percentof solvent increases. This makes choice D valid. The correct answer is choice C.

40. Choice A is correct. The greatest dilution results from the greatest relative addition of solvent. The greatestdilution involves the greatest ratio of solvent added to initial solution present, and it does not depend on thetotal volume of the initial or final solution. The ratio of solvent added to the initial volume of solution inchoice A is 100 : 25, which reduces to 4 :1. The ratio of solvent added to the initial volume of solution in choiceB is 200 : 60, which reduces to 10 : 3. The ratio of solvent added to the initial volume of solution in choice C is 50: 15, which reduces to 10 : 3. The ratio of solvent added to the initial volume of solution in choice D is 150 : 50,which reduces to 3 :1. The greatest ratio of solvent added to initial volume of solution is found in choice A, thebest answer.

Passage VII (Questions 41 - 47) Solution Concentrations and Dilution

41. Choice A is correct. If the density of the solution is less than the density of water, then the addition of waterto the solution may actually increase the density of the solution. The molarity and the molality of the solutionalways decrease as solvent is added, because the denominator in both molarity and molarity increase whilethe numerator remains the same. As solvent is added to the solution, the mass percent of solute in solutiondecreases, because the mass of solute in solution remains constant while the mass of solution is increasing. Onlythe density does not always decrease upon the addition of water to an aqueous solution; thus, the best answer ischoice A. In rare cases where the aqueous solution has a density of less than 1.00, the density increases whenwater is added.

42. Choice A is correct. Calculating the molarity and the molality of the solution involves the moles of salt in thenumerator. The numerator is the same in both the molarity and molality, so the difference between the twovalues depends solely on the denominator. The mass of water is 0.100 kg, while the volume of solution isgreater than 0.100 liters. This means that the denominator in the molarity calculation is greater than thedenominator in the molality calculation. The larger the denominator, the smaller the value. This means thatthe molality is greater than the molarity, which eliminates choices B and D. The density of pure water is 1.00grams per mL. The density of the aqueous salt solution is the mass (101 grams due to salt and water) divided bythe volume. The volume of the solution is less than 101 mL, so the density of the solution is greater than 1.00(given that the numerator is 101 and the denominator is less than 101). The density of the solution is greaterthan the density of water, so the best answer is choice A.


43.

44.

45.

46.

47.

Choice C is correct. Because the density of the solvent is less than 1.0 grams per mL, the volume in litersexceeds themass in kilograms for the solvent. When comparing molality and molarity, both values have thesame numerator. You are looking for the value with the smaller denominator. The mass of solvent is less thanthe volume of solution, somolality is greater thanmolarity. This eliminates choices Band D. The density isthe mass of the solution divided by the volume of the solution. The density values are less than 1, but are notknown specifically. In addition, there is no information about the mixing process. The density cannot bedetermined without information about the density of the components. The best answer is choice C.

Choice A is correct. All of the solutions are made up of one solute and solvent, but in different ratios. The soluteis denser than the solvent, so it can be concluded that the solution with the greatest density has the highestconcentration of solute. As solute concentration is reduced, the density is also reduced. Increasing the amount ofsolute relative to solvent increases both molality and molarity. The best answer is choice A.

Choice B is correct. This questions requires the use of the dilution equation, Mjn;tiaiVjnjtjai = MfjnaiVfjnai,where M is the molarity and V is the volume. Plugging the given values into the equation yields:

0.25 Mx 100 mL = 0.10Mx Vfinai .-. Vfiriai = °'25M x 100mL =-Q^. x 100 mL = 2.5 x 100 mL = 250 mL0.10M 0.10

The final volume (Vfjnai) is 250 mL, but Vnnai is not the answer. The question asks for the volume of wateradded. To start with 100 mL of solution and finish with 250 mL of solution, 150 mL of water must be added.

Choice C is correct. The units of molarity are moles solute per liter solution. Because both solutions have thesame final volume (100mL), the question reduces to asking how many grams of MgCl2 yield the same number ofmoles of CI" as 1.0 grams NaCl. There are two chlorides per magnesium chloride molecule, so only half thenumber of moles of MgCl2 as moles of NaCl is required. This accounts for the factor of one-half in thecalculation. The grams of NaCl must be converted into moles of NaCl, which in turn are converted into grams ofMgCl2 after accounting for the mole ratio of the two salts This calculation requires knowing the molecular massfor both salts. Because the magnesium chloride is heavier than sodium chloride, intuition tells us that moregrams of the magnesium chloride are needed to form a mole quantity equal to that for the sodium chloride.This means that in the calculation, the molecular mass ratio term should be greater than 1, making choice Cthe best answer. The unit factor method solution is as follows:

1.0 grams NaCl x 1 mole NaCl x lMgCl2x 94.9 grams MgCl2 =^ x1 x 9jL£ grams MgCl258.4 grams NaCl 2 NaCl 1 mole MgCl2 2 58.4

The final answer is less than 1.0 grams MgCl2-

Choice C is correct. The mole fraction of a compound in a solution is found by dividing the moles of the givencompound by the total moles of the solution. Choice B has a mole fraction of 0.50 for both Compound A andCompound B, because there is one mole of each compound, and the total number of moles is two. In an equal-mass mixture (one gram each), the greater number ofmoles is present in the compound with the lower molecularmass. Compound A is lighter than Compound B; therefore, in equal masses of Compound A and Compound B,there are more moles of Compound A than Compound B. The mole fraction of Compound A is therefore greaterthan 0.50 in choice A. This eliminates choice B. For an equal volume solution (in choice C there is one mL each),the relative masses can be determined by the densities. Because Compound A is denser than Compound B, equalvolumes of Compound A and Compound B result in the mass of Compound A being greater than the mass ofCompound B. The mass percent of Compound A is greater than 50%, as it is in answer choice A; thus, choice Cresults in an even larger mole fraction of Compound A than in the equal mass solution of choice A. The equal-molecules solution (choice D) has the same mole fraction as the equal-mole solution (0.50 for each). Becausechoice Band choice D are the same answer, they are both eliminated. The greatest mole fraction of CompoundA is found in the equal-volume solution, choice C.

Beer's Plot and Light AbsorptionPassage VIII (Questions 48 - 54)

48. Choice B is correct. According to the absorbance equation (and the data and the graph), when the concentrationof solute is doubled, the absorbance of the solution containing it doubles. This occurs because twice as manymolecules are present to absorb light. The best answer is choice B. Choices C and D should have beeneliminated immediately, because the absorbance increases as the concentration of solute increases. Choice A iseliminated, because the absorbance does not increase as the square of an increase in molarity.


49. Choice D is correct. At low concentrations, the relationship between concentration and absorbance is linear, aspredicted from the absorbance equation. At higher concentrations, deviation begins to occur, until theabsorbance reaches a maximum. There is no reaction occurring, so choice Bcan be eliminated. No solute woulddissolve into the solvent if the two repelled one another, so choice C can also be eliminated. Choice A maysound tempting, if you are blindly choosing anamount on the basis ofwords you recall from biochemistry. Thesaturation of the solvation catalyst sounds like a good answer (especially considering that Michaelis-Mentenkinetics shows a similar graph when saturation ofsome sort occurs). The problem here is that "saturation ofsolvation catalyst" means nothing, as there is no solvation catalyst. Also, catalysts affect the rate (not theconcentration) of reactants, products, solutes, orother components ofa solution. Choice Ashould be eliminated.Once water (the solvent) has interacted with asmuchsoluteas it can, no more solutecan dissolve into solution.The molarity reaches a maximum, so the absorbance reaches a maximum. The graph would be a line that justends at the saturated concentration. The best answer is choice D.

50. Choice C is correct. The addition of 50 mL of water to 10 mL of solution results in a dilution to one-sixth of theoriginal concentration value. The absorbance of light should also be reduced by a factor ofone-sixth. One-sixthof 0.518 is a value less than 0.100,so the only possible answer choice is choice C.

51. Choice A is correct. Dividing the equation Absorbance = e[C]l by e and 1 isolates the value of soluteconcentration ([C]). Theresult is that choice A is thebestanswer. Thisquestion may have seemed toosimple toyou. Onoccasion, there are some simple questions about theMCAT, sodon't try to find tricks thataren't there.

52. Choice C is correct. From the data in Table 1, we know that the absorbance of a solution with a concentration of0.20M is 0.188, while the absorbance of a solution with a concentration of 0.30 M is 0.278. If the absorbance ofthe unknown solution is 0.242, then the molarity of the solution must lie between 0.20 M and 0.30 M. Theaverage of0.20 Mand 0.30 Mis0.25 M, and theaverage of0.188 and 0.278 is 0.233, so theabsorbance fora 0.250M solution should be roughly 0.233. Because 0.242 is slightlygreater than 0.233, an absorbance value of 0.242 isassociated with a concentration value that is slightly greater than 0.250 M. The best answer is choice C.

53. Choice D is correct. Isolating the value for e in the absorbance equation yields absorbance divided by(concentration-pathlength). Absorbance is unitless, while concentration is in molarity, and pathlength is insome fraction of meters (like centimeters). The units of e therefore must involve inverse molarity times inversecentimeters. The best answer is choice D.

54. Choice C is correct. As solute concentration (molarity) increases, the absorbance increases. This makesstatement I valid. At this point, the answerchoices narrow to either choice A or choice C. If you are strappedfor time, just look at statement III. Beer's law applies at all wavelengths (Xmax is chosen because it is thegreatest value, and thus is the easiest wavelength at which to obtain an accurate measure of absorbance).Because statement III is valid, choice C is the best answer. The absorbance depends on both the molarabsorbtivity constant and solute concentration, so ifCompound Xhas a lower constant than Compound Y, thenthe solutionwith Compound Xmust be moreconcentrated than the solutionwith Compound Y, in order to havean equal absorbance. This makes statement II invalid. The best answer is choice C.

Passage IX (Questions 55 - 61) Beer's Law Experiment

55. Choice D is correct. The spectrophotometer measures the absorbance of light, so the wavelength settingcorresponds to an absorbed color, not a complementary color. This eliminates choices A and C, both of whichrefer to the reflected (or complementary) color. To obtain the most accurate value, the spectrophotometer is setat the wavelength corresponding to maximum absorbance. This makes choice D the best answer. Choice Biscoincidentally correct in some cases, where the absorbance band is symmetric. This does not make choice Babetter answer than choice D, but it does raise an important point: Remember that the test writers reward youfor choosing the best answer, not just a correct answer.

56. Choice B is correct. The question states that the absorbance varies with cuvette length. From the passage, weknow that absorbance varies with the concentration of solute in the solution. It makes sense that theabsorbancedepends on the molar absorbtivity constant (e), so the Beer's law relationship of Abs. = e-[C]-l can bededuced. Dividing both sides of the equation by [Compound]-l yields choice B. From the term "molarabsorbtivity," it can be inferred that absorbance is in the numerator and molarity in the denominator, whichmakes choice B the only possible answer. Units may not always be the best route to the correct answer.


57. ChoiceA is correct. This question requires converting absorbance to concentration. A 0.10 M solution ofCompound Mhas anabsorbance of0.362, so if the absorbance is0.400, the molarity is greater than 0.10 M. Thiseliminates choice D. Compound Q with an absorbance of 0.250 is less that 0.10 M (where the absorbance is0.299), so choice B is eliminated. Compound T with an absorbance of 0.500 is less that 0.10 M (where theabsorbance is 0.511), meaningchoice C is eliminated, leaving choice A as the best answer.

58. Choice D is correct. From thedata inTable 1,we see that Compound Thas the greatest absorbance of the threecompounds, sochoices AandBareeliminated. The absorbance values for Compound Mat a concentration equalto that of theothercompounds are closer to thevalues ofCompound Q thanCompound T. Thatmeans that thebetter graph is choice D. This is a question where it easy to make a mistake by choosing answer choice Cwithout examining theotherchoices more closely. Abrief survey offormer students showed that roughly one-third of you will make a carelessmistake on this question and chooseC.

59. Choice B is correct. Compound M has an absorbance of 0.217 at 561 nm when the concentration is 0.06 M.Therefore, when the concentration is reduced to 0.05 M (a value that is about 17%less), the absorbance shouldbe reduced to a value that is five-sixths of 0.217, which is roughly 0.181. Statement I is valid. A solution ofCompound Q with an absorbance of 0.225 at 413 nm must have a concentration between 0.06 M (where theabsorbance is 0.180) and 0.10 M (where the absorbance is 0.299). Because 0.225 is a little closer to 0.180 than0.299, the concentration should be a little closer to 0.06 M than 0.10 M, which makes statement II valid, andmakeschoice B the only possiblechoice. Statement IIIis invalid, because at 697nm, 0.10 MCompound T has anabsorbance of 0.511. This means that at the same wavelength, with a molarity of 0.11 M, the absorbance isabout 0.56+. Given that the wavelength is not 697 nm (the wavelength where maximum absorbance isobserved), the absorbance is actually less. This makes statement III invalid, agreeing with the selection ofchoiceB. PickB to gain the prestige, honor, and pride that goes with knowing you got another question right.

60. Choice C is correct. Compound T has an absorbance of 0.511 when it has a concentration of 0.10M. An absorbanceof 0.250 is just less than half of the 0.511 value, so the concentration must be just less than 0.05M (half of 0.10M). The question is then, "Which solution results in a concentration of Compound T that is just less than0.05M?" The starting solutions are all 0.10M Compound T; so to achieve a solution that is just under 0.05M, theamount of water added should be just a tiny amount greater than the amount of Compound T solution. This isobserved with choice C. The absorbance for each choice is listed below:

Choice A: 10/i5 (0.511) >*/2 (0.511) >0.25; therefore, choice Ahas too high an absorbance.Choice B: 20/4fj (0.511) =1/2 (0.511) >0.25; therefore, choice Bhas too high an absorbance.Choice C: 19/39 (0.511) <1/2 (0.511) .-. 19/39 (0.511) may equal 0.25; therefore, choice Cis good.Choice D: 10/i9 (0.511) >*/2 (0.511) >0.25; therefore, choice Dhas too high an absorbance.

The value of 19/39 (0.511) should be close enough to 0.250 to convince you to choose C.

61. Choice D is correct. Although colors were not reviewed in this section, it is a topic that frequently is discussedalong with the absorbance of light. One of the goals of these review passages is to expose you to this kind ofintegrated material. In each section, there are some questions on topics that have yet to be discussed. Such isthe case here. The color observed for the solution is the complementary color of what is absorbed. The visiblespectrum is from roughly 700+nm to 400nm. Red light is the least energetic of visible light, so red light has abroad wavelength of approximately 700 nm. Violet light is the most energetic of visible light, so violet lighthas a narrow wavelength of approximately 400nm. Compound Q absorbs light with a wavelength of maximumabsorbance (A,max) of 413 nm. This corresponds to violet light, so the color observed for the solution is yellow(the complementary color of violet). This eliminates choices A and B. Compound T absorbs light with awavelength of maximum absorbance (Xmax) of 697nm. This corresponds to red light, so the color observed forthe solution is green (the complimentary color of red). This eliminates choice C, and makes choice D the bestanswer. You should be able to answer this with the background knowledge that 400 nm is at the violet end ofthe visible spectrum, 700 nm is at the red end of the visible spectrum, and a working knowledge ofcomplementary colors. Compound M is irrelevant in solving this question. Appearing red (reflecting red light)means that green light (the complementary color of red) was absorbed. Green light is in the middle of thevisible spectrum, so it could have a wavelength of 561nm. There is no reason for you to know the exact Xvaluesfor each color, but you must be able to deduce the best answer to this question. Absorbed light, reflected light,complementary colors, the color wheel, and the visible spectrum will all be addressed in the next chapter.


Passage X (Questions 62 -68) ReactionTypes62. Choice D is correct. It is nota combination reaction, because nocompounds combined toform a single compound,

so choice Ais eliminated. It is not a decomposition reaction, because no compound broke apart to form multiplecompounds, so choice Bis eliminated. It is not asingle-replacement reaction, because no metal compound wasformed orconsumed, so choice Cis eliminated. Because the barium cation (Ba2+) and the potassium cation (K+)have exchanged their respective anions in this reaction, the reaction is classified as a metathesis reaction(known also as adouble-displacement reaction). The best answer is choice D. By picking choice D, you can feelsatisfied knowing you picked a correct answer. Go ahead, feel good!

63. Choice Ais correct. Aprecipitate is asolid that forms from solution and deposits on the bottom of the flask asa reaction proceeds. The reaction least likely to form a precipitate is the reaction that is least likely to form asolid product. In choice A, the combustion reaction forms water liquid and carbon dioxide gas, but does not forma solid of any kind. For this reason, choice A is the best choice. In a decomposition reaction, a solid can beformed, as in the example offered in the passage. In the single-replacement and metathesis reactions, salts areformed as products, so precipitates areboth likely and probable.

64. Choice C is correct. The reaction of aqueous sodium iodide (Nal(aq)) with aqueous calcium nitrate(Ca(N03)2(aq)) is a double-displacement (metathesis) reaction, so the precipitate formed in the reaction mustbe one of the two possible product salts. Calcium is a dication (carries a +2 charge), so choice Aiseliminated,because the calcium salt is not neutral when bonded to only one iodide anion. Sodium and nitrate are bothhighly soluble in water, so neither sodium nor nitrate is found in the precipitates. Both choice Band choice Dare therefore eliminated. The best choice is answer C. The reaction is given below:

2 Nal(aq) + 1 Ca(N03)2(aq) • 2 NaN03(aq) + 1 Cal2(s)

65. Choice C is correct. The reaction ofaqueous hydrobromic acid (HBr) withmagnesium carbonate (MgCOs) is aneutralization reaction. It is stated in the passage that carbonates, when neutralized, yield carbon dioxidegas. The best choice is answerC. Thereaction isdrawnbelow:

2 HBr(aq) + 1MgC03(s) • 1 H2C03(aq) + 1MgBr2(s)

lH2C03(aq) • 1 H20(aq) + 1 C02(g)

2 HBr(aq) + 1MgC03(s) • 1 H20(aq) + 1 C02(g) + 1MgBr2(s)

You should recall that baking soda (sodium bicarbonate) effervesces when an acid is added. The bubbles thatform are carbon dioxide gas. Hydrogen gas cannot exist as H (molecular hydrogen is a diatomic gas), andmagnesium bromide is an ionic salt (which isdefinitely not a gas at room temperature), meaning that choices Aand D should be discarded immediately. To eliminate choice B, you had to recognize that magnesiumcarbonate is a base, and not a reducing agent. Magnesium metal is rich in electrons, so it would make a strongreducing agent, but not magnesium cation (Mg2+). The addition of HBr to Mg(s) would yield H2(g) by way of anoxidation-reduction reaction.

66. Choice Bis correct. Carbon dioxide gas is a common product in combustion reactions, sochoice A is eliminated.The example reaction for decomposition shows calcium carbonate (CaC03) decomposing to form calcium oxideand carbon dioxide. Choice C is thus eliminated. The same reaction is possible with calcium sulfite (CaS03),so sulfur dioxide can be formed in the decomposition reaction of calcium sulfite. The passage states that theneutralization of sodium bicarbonate (NaHC03) forms carbon dioxide gas; therefore, choice D is eliminated.The only choice left is answer choice B, the metathesis reaction, where cations are interchanged, but no carbondioxide gas is formed. If the anion begins ascarbonate, it finishes ascarbonate, andnot carbon dioxide.

67. Choice C is correct. In the reaction, the chlorine gas is reduced to chloride anion andexchanged for the bromideanion, which is oxidized into bromine liquid. The product is not the result of coupling (combining) of thereactants, so it is not a combination reaction. This eliminates choice A. The products are not the result of thebreakdown (decomposition) of a reactant, so it is not a decomposition reaction. That eliminates choice B. Theanion in the salt is exchanged, while the cation is not. This describes a single-replacement reaction, meaningchoice C is valid. Because the cation remains the same, the reaction cannot be a metathesis reaction, whicheliminates choice D. The best answer is choice C.


68. Choice A is correct. A negative value for AS results when the reaction gains order (or loses randomness). In acombination reaction, the number of molecules decreases from reactant to product, which is an increase in order.As such, a combination reaction is likely to have a negative value for AS. This makes choice A the best answer.In a decomposition reaction, the number of molecules increases from reactant to product, which is an increase indisorder. A decomposition reaction is likely to have a positive value for AS. This eliminates choice B.In a single-replacement reaction, the number of molecules remains the same on each side of the reaction. Bothsides of the equation have a precipitate and a solute, which leads to no conclusion about the change in order forthe system. A single-replacement reaction can have either a positive or negative value for AS. Although asingle-replacement reaction may have a negative AS, choice C is not as good an answer as choice A. In acombustion reaction, there is an increased number of molecules from reactants to products. The products aregases, so there is a large positive value for AS. This eliminates choice D.

Passage XI (Questions 69 - 78) Calcium-Containing Bases

69. Choice D is correct. Ca(OH)2 yields a hydroxide anion (OH") when added to water, which makes Ca(OH)2 anArrhenius base. Ca(OH)2 cannot donate a proton, so it is not an Arrhenius acid. This eliminates choice C.Choice A is also eliminated, because it cannot be an amphoteric species, unless it can act as both an acid and abase. Calcium is a metal, so Ca(OH)2 is a metal hydroxide, not a non-metal hydroxide. Choice B iseliminated. The correct answer is choice D.

70. Choice D is correct. Although C032" is charged and thus an ion, the anion itself is held together by covalentbonds (three a-bonds and a resonating 7i-bond). Carbon dioxide is held together by covalent bonds between thecentral carbon and the two adjoining oxygens atoms. Only CaO (III) has an ionic bond, and it is considered to bepartially ionic. Calcium carries a +2 charge, and oxygen carries a -2 charge. This makes D the correct choice.The structures and bonds are shown below:

Oa

O a a O"

a ao=c=o

Ionic bond2+

Ca O2-

71. Choice D is correct. All of the compounds contain only one calcium per molecule, so the smallest mass percent ofcalcium is found in the compound with the greatest molecular mass. The mass percent is found by dividing themass of calcium by the mass of the compound. The molecular mass for CaO is 56 grams per mole; for Ca(OH)2 itis 74 grams per mole; for CaC03 it is 100grams per mole; and for CaCl2 it is 111grams per mole. The heaviestcompound is CaCl2 (choiceD), meaning choiceD has the greatest denominator and therefore the smallest masspercent of calcium. This makes choice D the best answer.

72. Choice A is correct. To calculate the percent yield, the actual yield (in moles) is divided by the theoreticalyield (in moles). The actual yield in moles is5 g/100g-moles"1, while the theoretical yield in moles is 10 g/74g-moles"1. Thepercentyield is found as follows:

5g

Percent yield =_ 100 g-mole-1

10 gx 100% = 5 x 74 x 100% = —Z4— x 100% = -2Z_ X 100% = 37%

100 x 10 100 x 2 100

74 g-mole"1The percent yield is 37%, which makes choice A the correct answer.

73. Choice D is correct. The mass percent of calcium in calcium oxide is the mass of calcium (40 grams per mole)divided by the mass of calcium oxide (56 grams per mole) and multiplied by 100%. 40/56 x 100% is greater than50%, which eliminates choices A and B, and is greater than 66.7%, which eliminates choice C. Only choice Dis greater than 66.7%, so choice D must be the correct answer.

Range method: ^ x 100% >4Q. x 100% = 2. x 100% = 66.7%56 g 60 3

Calculation method: ^1 =1 =5 x ±- ~ 5 x 0.143 = 0.715 = 71.5%56g 7 7


74.

75.

76.

77.

78.

Choice A is correct. Because matter can be neither creatednor destroyed, the total mass of reactants must equalthe total mass of the products. There are 38.0 grams of reactants, so choice D is eliminated for having morethan 38.0 grams of products. The question from here becomes a limiting reagent problem. We must decidewhether there is leftover water (is CaO the limiting reagent?), leftover calcium oxide (is water the limitingreagent?), or an exact mixture (there is no leftover and thus no limiting reagent.) The moles of CaO equal 28/56(which is one-half) and the moles of H2O equal 10/18 (which is greater than one-half.) Because the calciumoxide reacts with the water in a one-to-one ratio, there is leftover water. This means CaO is the limitingreagent in the reaction, so the reaction does result in leftover H2O. The only choice with leftover water is A.

Choice D is correct. The N in the answer choices stands for normality, which is the molarity of the protonsthat can be generated by the acid. Normality is the molarity of an acid times the equivalents ofhydroniumper molecule. CaC03(aq) reacts according to the equations shown below. A2:1ratio is needed to reach fullneutralization.

CaC03(aq) + 2H+(aq) H2C03(aq) + Ca2+(aq) H2C03(aq) H20(1) + C02(g)

5 mL x 0.2 moles/liter = 1 mmole CaC03, so 2 mmoles of a strong acid (which yield H+) are needed to reachequivalence. Choice Ais a strong base, so eliminate it. Choice Byields only 1mmole of H+, and choice Cyields3 mmole of H+,so eliminate both choices Band C Choice D yields the 2 mmole of H+ that are needed. Thebestanswer is choice D. For a monoprotic acid, normality and molarity are the same. However, for a diprotic acid,the normality is double the molarity; and for a triprotic acid, the normality is triple the molarity.

Choice C is correct. This isa case whereEquation 1.2 in the text, MjnitialVinitial =MfinaiVnnai, must be used.0.500 Mx5.0 mL =Mfinal *55.0 mL .-. Mfinal =5/55 x0.50 .-. Mfinal =Vll x°-50 <VlO x°-50 =°-050

The final molarity is therefore less than 0.05 M. Without pounding numbers, you should see that choice C isthe best choice (the choice just less than 0.050 M). Choice D is too far below 0.050 M to be reasonable. Donotsolve mathquestions exactly, but find a range into which only one answer choice fits. In this case, the range isfrom just less than 0.050 (maybe 0.040 or so) to0.050. Only choice C fits into that range.

Choice C is correct. The final concentration can be determined by taking a weighted average of theconcentrations for the two initial solutionsbefore they are mixed. A quick observation you should apply to thisquestion is that if the mixture were a fifty-fifty mixture by volume (the mixture of equal volume solutions),then the final concentration would lie exactly between (be the mean value of) the two values. In thisparticular question, the mean value would be 0.250 M. Because there is an excess of the higher-concentrationsolution (30 mL at 0.30 M, withonly 20 mL at 0.20 M), theanswer should becloser to 0.30 M than0.20 M(makingit greater than0.250 M). This eliminates choices Aand B. The exact value canbe solved for as follows:

Total moles = (20 mL x 0.20 M) +(30 mL x 0.30 M) = 4+9 = 13 = 0.26, choice C.Total volume 50 mL 50 50

Choice C is correct. 50 grams CaC03(s) is equal to0.50 moles CaC03(s), because themolecular mass ofCaC03(s)is 100 grams per mole. Because there is only one carbon in CaC03(s), 0.50 moles CaC03(s) yield 0.50 moles ofCO2. 0.50 moles of CO2 at STP occupies avolume of: 0.50 moles x22.4 liters/mole =!/2x 22-4 =1L2 L- Pick c

Passage XII (Questions 79 - 86) Industrial Chemicals

79. Choice Ais correct. The greatest amount ofpotassium pergram ofcompound is found in thecompound with thegreatest mass percent ofpotassium. The greatest mass percent of potassium isfound inK2O, aswas mentioned inthepassage. This means that thebestchoice is answer A. The mass percents for the four saltsare as follows:

2 x 39.1K20: 2 x 39.1 _ 78.2 _ 39.1

(2 x 39.1) + 16 94.2 47.1

39.1 _39.1KCl:39.1 + 35.5 74.6

K2SO4:

KN03:

The relative mass percents when comparing the four values are: ^^- > *"•'The greatest mass percent of the choices is found with K2O (choice A).


39.1

47.1

78.2 -39.1(2 x 39.1) + 32 + (4 x 16) 174.2 87.1

39.1 - 39.139.1 + 14 + (3 x 16) 101.1

39.1 ^ 39.1 39.174.6 97.1 101.1

Section I Detailed Explanations

80. Choice Discorrect. The compounds inchoices A, B, and Call have two nitrogen atoms percompound, while thecompound in choice Dhas only one nitrogen atom in the compound. The lowest mass percent ofnitrogen is inchoice D,because thefour compounds areroughly comparable inmolecular mass, but they have different massesof nitrogen. The mass percents for the four choices are shown below:

A. H2NCONH2 => ^ =28 =_7_ <160gH2NCONH2 60 15 2

B. NH4NO3 => ?^? =&=-7_ =35%80g NH4NO3 80 20

C. (NH4)2S04 => ?M^ =_28_ =_7_ = 21+%132g (NH4)2S04 132 33

D. NH4H2P04 => l^N =J4_ < 14%115g NH4H2P04 115

28 > 28 > _28_ > _28_ (= _14\60 80 132 230 115

81. Choice C is correct. The mass percent of nitrogen in urea is just less than 50%:

Mass percent urea = 2 28 _ 7_ < 1_60gH2NCONH2 60 15 2

The only answer choice that is just less than 50% is choice C.

82. Choice A is correct. According to the passage, the phosphorus content of the compound is critical, so nophosphorus should be wasted in unreacted reactant. This can be prevented by adding the reactants in a ratiothat avoids the presence of any leftover phosphorus-containing reactant. The goal is to conserve thephosphoms-containing compound, which makes choice A the best answer.

83. Choice B is correct. The percent yield for a reaction is defined as the actual quantity of product isolated fromthe reaction mixture, divided by the theoretical quantity of product that should form, as determined from thelimiting reagent of the reaction. According to the balanced equation, for every one mole of (NH4)2C03 thatreacts, one mole of (NH4)2S04 forms. This means that 10 grams of (NH4)2C03 (which is 10/96 moles(NH4)2C03) should produce 10/96 moles of (NH4)2S04, which is (10/96 x 132) grams (NH4)2S04. Dividingthe mass of (NH4)2S04 obtained (10 grams) by the theoretical mass of (NH4)2S04 that should have formed(13.75 grams) determines the percent yield. The calculation of the yield is shown below:

Theoretical mass: ±0- x 132 = -1320- = 330=n0>104+6 = 133 = 13.7596 96 24 8 8 8 4

Range for thepercent yield: 50% < 5 = 10. < 1OQ0 < 3 = _J0_ = 75*%7 14 13.75 4 13.33

The value for the percent yield lies somewhere between 50% and 75%,which makes choice B the best answer.

84. Choice C is correct. All of the choices have the same volume (100mL) and the same mass of salt, so the greatestmolarity of potassium is present in the solution with the greatest number of moles of potassium in solution.Hence, the greatest molarity results from using the compound with the greatest mass percent of potassium,which corresponds to the salt with the lowest molecular mass. Choice D is eliminated, because NO3 is heavierthan CI, so KCl has a lower molecular mass than KN03, and thus a greater mass percent of potassium than inKNO3. Choice B is eliminated, because S04 is heavier than CO3, so K2CO3 has a lower molecular mass thanK2S04/ and thus a greater mass percent of potassium than K2S04. The question boils down to determining therelative mass percents of potassium in KCl and K2CO3. The molecular mass of KCl is 74.5 grams per mole, andthe molecular mass of K2C03 is 138.2 grams per mole. Potassium carbonate (K2CO3) yields two potassium ionsper molecule, so to have the same mass percent of potassium as KCl, its molecular mass would have to be doublethat of KCl (which equals 149 grams per mole.) Because K2CO3 is less than twice as massive as KCl, therelative denominator is smaller when calculating the mass percent of K2CO3 than when calculating the masspercent of KCl. The greatest mass percent of potassium, and thus the most grams of potassium in a 10.0-gramsalt sample, is found in K2CO3. This means that the highest concentration of potassium is present in theaqueous K2CO3 solution, so choose answer choice C for best results.

Copyright© by The Berkeley Review® 63 Section I Detailed Explanations

85. Choice A is correct. It is easiest to isolate a solid (precipitate) from solution, because the solid canbe filteredfrom solution rather easily. Equally, a liquid and gas can flow, while a solid cannot, so removing liquid and gasto leave solid behind is easy. This is the crux ofmany lab techniques. A liquid must be distilled away fromsolution, a solute must be either distilled, extracted, or removed by chromatography, and a gas must becollected in a gas-trap free of air. Thebest answer is thus choice A.

86. Choice A is correct. This is another casewhere the questionfocuses on a topicwe have yet to discuss. Tliere area few things you should recall. First, the lower the pKa value is, the stronger the acid is. Second, when the pHof the solution exceeds the pKa of an acid, thatsite exists predominantly in the deprotonated state. StatementIII can immediately be eliminated, because the first proton of sulfuric acid is strong, meaning it has a very low(negative) pKa value associated with it. From the products of the last equation in the passage, it can be seenthat H2PO4-, SO42", and HF all simultaneously exist in solution. Sulfuric acid has lost both of its protons,phosphoric acid has lost its first proton, and hydrofluoric acid has yet to lose a proton. The first proton ofphosphoric acid is lost more readily than the proton of HF, which means that pKai of H3P04 is lower thanthe pKa value of HF. Statement I is thus true. Because the second proton of sulfuric acid has been lost, whilethe second proton of phosphoric acid has not been lost, the second proton of sulfuric acid is more acidic than thesecond proton of phosphoric acid. The result numerically is that pKa2 of H2SO4 is less than pKa2 of H3P04.Statement II is thus false. Only statement I is true, so the best answer is choice A.

Questions 87-100 Not Based on a Descriptive Passage

87. Choice C is correct. This is a limiting reagent question. The balanced equation for the reaction is as follows:lC3H8(g) + 5 02(g) • 3C02(g) + 4H20(g)

There are 25/32 moles 02(g) and 20/44 moles C3Hs(g) as reactants. According to the mole ratio from thebalanced equation, the moles of 02(g) must be five times thatofC3H3(g). The moles of02(g) needed to reactwith20/44moles ofC3Hg(g) is 100/44. The value of25/32 is less than1, so there is less than onemole of02(g)present. The value of 100/44 is greater than 2, so more than two moles of 02(g) are needed. This means thatoxygen (02(g)) is depleted before C3Hs(g), making 02(g) the limiting reagent andchoice C the best answer.

88. Choice Bis correct. If you know this answer from your biology classes, trust your knowledge and don't wastetime doing the calculation. You should know that hemoglobin contains four iron atoms. That is the point ofthis question. The math is time-consuming, sosave time wherever you can. Ifyou didn't know that fact abouthemoglobin, then doing the math was necessary. The mass of iron in hemoglobin is the mass percent of irontimes the total mass ofhemoglobin. 0.33% x68,000 = .0033 x68,000 =3.3 x68 = 204 + 20.4 = 224.4 grams Fe. Thenumber of irons is therefore 224.4/55.8, which makes 4 (choice B) the best answer choice.

89. Choice D is correct. The first method which probably comes toyourmind is themethod learned from generalchemistry classes. In this method, the 9grams ofcarbohydrate are converted to 0.050 moles ofcarbohydrate bydividing by the molecular mass of 180. The 6.72 liters of carbon dioxide at STP are then divided by 22.4 litersper mole to yield 0.30 moles CO2. The ratio of CO2 to O2 in the combustion of a monosaccharide is 1 : 1.Therefore, 0.30 moles of oxygen reacted aswell, and thus the mole ratio of carbohydrate to O2 to CO2 is0.05 :0.30 :0.30, which equals 1:6:6. This is a long-winded, butvalid, solution. Ashorter method is as follows:In the combustion of a monosaccharide, the mole ratio of oxygen reactant to carbon dioxide to water is always 1:1 : 1. A monosaccharide has the molecular formula CnH2nOn, so it has a molecular mass of 30n (12n+ 2n + 16n)grams per mole. For a molecular mass of 180 grams per mole, the value of n is 6, so the formula for themonosaccharide is C6H12O6. Themole ratioofO2 toCO2 is therefore 6 :6. Pick D and feel relaxed.

90. Choice B is correct. Choices A and C are eliminated, because they cannot be a molecular formula. For acompound with justcarbon, oxygen, and hydrogen, the number ofhydrogens must be aneven number. With anoddnumber ofhydrogens, the bonding does not work out. To decide between choices BandD, you must determinethe molecular mass of the compound, by dividing the mass by the volume and then multiplying this value bythe molar volume:

Igram y 22.4 liters = 22.4 grams B22.4 grams =4 x ^ grams = ^ grams0.26 liters 1 mole 0.26 moles 0.25 moles mole mole

OnlyC4H6O2 has a molecular massjust under90 g/mole, sochoice Bis the bestanswer.

Copyright © by The Berkeley Review*3 64 Section I Detailed Explanations

91. ChoiceA is correct. This is a question testingyour knowledge of the unit factormethod. Barium oxide (BaO)canform bariummetal and oxygen gas(O2) upondecomposition. These are theonlytwoproducts possible fromthe decomposition of barium oxide. Hence, for every two moles of BaO, one mole of O2 forms. To solve thisproblem, the 1.0gramsof BaOmust be converted tomoles, and themoles of BaO in the 1.0-gram samplemust bemultiplied by one-half. This gives the molesofO2formed. Multiplying the moles of O2by 22.4 liters per mole(the volume for one mole of gas at standard temperature and pressure), gives liters of oxygen gas produced bydecomposition:

1.0 grams BaO x 1™kBaO x 1mole Q2 x22.4 liters - 22.4 = 2Z4 < _1_153 grams BaO 2 moles BaO lmole 153x2 306 10

Only choiceA is less than 0.1 liters, so pick choiceA and make yourself one point wiser.

92. ChoiceB is correct To determine the grams of product, the conversionmust go through moles. This question is astandard grams reactant-to-moles reactant-to-moles product-to-grams product conversion.

2.0 gramsMgNH4PQ4 y 1Mg2P2Q7 y222.6 gMg2P2Q7 _ 2x 222.6 _ 222.6 - hfeP(y137 3&amsI 2MgNH4P04 1mole Mg2P207 137.3 x 2 137.3

/mole

2> 222A >i137.3

The mass of Mg2P2C>7(s) is greater than 1, but less than 2. The only answer that falls between 1 and 2 is answerchoice B.

93. Choice D is correct. We know nothing of the molecular mass, so any formula we determine is the empiricalformula (simplest formula). This eliminates choice A. Because the elemental mass of Z is twice the elementalmass of X, a compound with equal mass quantities of X and Z contains twice as many moles of X as moles of Z.The mole ratio of X to Z is therefore 2:1; thus, the empirical formula is either X2Z1 or Z1X2. Pick choice D tofeel correct.

94. Choice C is correct. The math here is beyond the MCAT level, but the point of such a question is to encourageyou to use intuition and simplify the question. You can always eliminate some of the answer choices, and atleast narrow your guess from random choice to a fifty-fifty chance of success. This question is simplified byknowing the possible oxidation states of iron. From your experience with hemoglobin, you know that iron has aoxidation state of either +2 or +3, and that oxygen has an oxidation state of -2 when it is coupled with a metal.If oxygen has an oxidation state of -2, then iron cannot have an oxidation state of +2 or +3 in either Fe302 orFe03. Choices A and D are eliminated based on outside knowledge. This problem can now be solved using yourmathematical intuition. In the iron oxide sample, there are 11.89 grams of iron and 5.10 grams of oxygen (foundby subtracting 11.89from 16.99). The molecular weight of iron (Fe) is 55.85 grams per mole, and the molecularweight of oxygen (O) is 16.0 grams per mole. An empirical formula is based on mole ratios, so the elementalmasses must be converted into mole quantities. 5.1 over 16 is just less than one-third, so there are roughly 0.30moles of oxygen in the iron oxide. 11.89 over 55.85 is just over one-fifth,so there is roughly 0.20moles of iron inthe iron oxide. There are more moles of oxygen than iron, which eliminates choice B (FeOhas equal moles ofiron and oxygen) and leaves choice C as the correct answer by default. Minimize doing math by using yourbackground knowledge and logic.

Feii.89 O5.10 = Fei+Or = Feo2+O0 33_ = Fe20355.85 16.0 5 3

95. Choice D is correct. It is not possible to have 26hydrogens and only 4 carbons in a stable compound. Youmayrecall from organic chemistry that the maximum number of hydrogens in a hydrocarbon or carbohydrate is 2n +2, where n is the number of carbons in the compound. This eliminates choice C. Perhaps it is easiest to solvethis question by first determining the number of carbons in the molecule. The mass percent of carbon in themolecule is 53.4%. When this percentage is multiplied by 90 grams/mole, it tells us that just over half the massis carbon. Carbon has a mass of 12grams per mole,so the valuemust be a multiple of 12. Themass due to carbonis 48 grams, which is the mass of four carbon atoms. The molecular formula therefore contains four carbons,eliminating choices A and B. The only choice left is answer D, which does in fact have roughly 11% hydrogen.Again, the goal in preparing for your exam is to be able to solve these questions quickly and using as muchintuition as possible. You are not rewarded only for being thorough; you are also rewarded for being fast. Takelogical short cuts whenever they present themselves. Do not blindly repeat the calculation techniques youlearned in your general chemistry courses.


96. Choice A is correct. Because this is a reduction reaction, the oxidation statemust decrease, which eliminateschoices Cand D. By assuming oxygen has a -2 oxidation state, it is possible to reduce this problem to simplealgebra. In the reactant, the sum of the oxidation states for the four oxygen atoms is -8. In order for the overallcharge on Cr042"(aq) to be -2, the chromiummust have an oxidation state of +6. The sum of +6 and -8 is the ioniccharge of -2. In the product Cr203(s), each chromium must be +3 to cancel out the three oxygen atoms (at -2each) tomake theoverall molecule neutral. For this reason, choice A is thebestanswer.

97. Choice D is correct. Themetal combines in a 2:3 ratio withoxygen to form an oxide that is 53% metal bymass.Assuming a 100-gram sample implies that there are 53 grams of metal and 47 grams of oxygen in the 100-gramsample of the metal oxide. This means that the ratio of (53 divided by the molecular mass of the unknownmetal) to (47 divided by the molecular mass of oxygen, which is 16) isequal to 2 : 3. This can be solved longhand or by intuition. Rather than solve the math exactly, it is a good idea to plug the values for all fouranswer choices into the setup, and get a rough estimate:

53gmetal._47gO_=2.3^Giyen ^4Zferougmy3/53gI^L mustbeabout2MWmetal 16»/ . 16 MWmetal

/moleThe molecular masses of the four answer choices are 40.0, 55.9, 52.0, and 27.0 respectively. 47 divided by 16 isslightly less than 3, so 53 divided by the molecular mass of the metal must be slightly less than 2. Thiseliminates the first threechoices (40.0, 55.9, and 52.0), leaving 27.0 (aluminum) as the best answer. The correctanswer is choice D. You could also have eliminated choiceA from knowing that calcium cannot achieve a +3oxidation state (orcharge), so calcium cannot combine withoxygen in a 2: 3 ratio.

98. Choice Bis correct. The mass percent of carbon is the mass of carbon ineach compound, divided by the totalmass of each compound, multiplied by 100%. For each compound, the mass percent can be determined as follows:Acetic acid: 2i x 100% Ethanol: 2! x 100% Methyl acetate: 26 x 10o% Glucose: -T-2- x 100%

60 46 74 18024<30=1 24>23=1 36 < 37 =1 -Z2_ < -2Q_ = 160 60 2 46 46 2 74 74 2 180 180 2

Of the choices, only choice B (ethanol) has a fraction greater than one-half. This implies that the greatestmass percent ofcarbon is found inethanol. The best answer is therefore choice B.

99. Choice B is correct. TheMCATtest writers can ask about the same concept in many ways. It is more importantthat you walk away from this question knowing the concept than it is getting the question correct. The mostcarbon dioxide results from the compound with the greatest mass percent ofcarbon, as long as there are equalmasses ofeach sample present initially. There was one gram ofeach compound initially, before oxidation, soour only concern iswith the mass percent of carbon. The greatest mass percent of carbon isfound inethanol, sothe greatest amount ofcarbon dioxide results from the oxidation ofethanol. The best answer ischoice B.

100. Choice C is correct. Because there are twoCI in theMg(C104)2 molecule, the mass percent of chlorine is themass of two CI over the mass of Mg(C104)2-

Mass chlorine _ 2 x 35.5 71 _ 71Mass compound 24.3 + (2 x 35.5) + (8 x 16) 24.3 + 71 + 128 223.3

Jl_ > _Z1_ > _ZL, where 3±- =X = 33.3%, and: -ZL =i- = 25%. So 33.3% > -7-1- > 25%213 223.3 284 213 3 284 4 223.3

The mass percent is a little less than 33.3%, so choices Aand Bare eliminated. The value is just a little lessthan 33.3%, because thedenominator (223.3) isjust a little greater than 213. The best answer is thechoice thatisslightly less than 33.3%, which is 31.8%, choice C. Choice Dcan be eliminated, because it is less than 25%.


Section II

AtomicTheoryby Todd Bennett

Voltagesource Filter

Photographicplate

Atomic Structure

a) Subatomic Particlesb) Isotopesc) Average Atomic Massd) Classical Experiments and Machinery

i. Thomson Experimentii. Mass Spectroscopyiii. Millikan Oil Drop Experimentiv. Rutherford Experiment

e) Heisenberg's Uncertainty Principlef) Atomic Modelg) Hydrogen Energy Levels

Electronic Structure

a) Electronic Theoryb) Effective nuclear Chargec) Electronic Spin Pairingd) Spin Pairing and Magnetisme) Electronic Density and Orbitalsf) Collective Orbital View of Energy Levelsg) Electronic Configurationh) Quantum numbers

Periodic Trends

a) The Periodic Tableb) General Elemental Periodic Trends

i. Atomic Radiusii. Ionization Energyiii. Electron Affinityiv. Electronegativity

c) Periodic Families (Groups)

Light Absorption and Emissiona) Excitation and Relaxationb) Atomic Spectrum of Hydrogenc) Electromagetic Spectrumd) Visible Spectrum and Colors

iii. Emitted Coloriv. Reflected Color and the Color Wheel

e) Fluorescencef) Photoelectric Effect

Nuclear Chemistrya) Muclear Particlesb) nuclear Decay and Capturec) Half-Life

BerkeleyUr.E.V.KE'W®


Atomic Structure Section Goals

©*

Be familiar with the location, mass, and charge of sub-atomic particles,Anatom, from thechemist's perspective, iscomposed ofa nucleus madeup ofprotonsand neutrons,surrounded byorbiting electrons. You mustknow theBohrmodel and theproperties ofeachparticlewithregard to itscharge, mass, andeffect ontheatom. Forinstance, additional electrons result inthe formation of an anion, while a decrease in electrons results in the formation of a cation.

Be familiar with energy levels and energy transitions.The electrons orbitin distinct, quantized levels. There is a base (lowest-energy) level, and excitedstates above that to which electrons jump when enerey is added to the atomic system. Energy isabsorbedwhen an electronis elevatedfromthe ground state to an excitedstate. Energy, in the formofa photon,is released whenan electron drops to theground state froman excited state.

Understand electronic configurations.Electronsfill the orbitals according to a defined sequence,described by the Aufbau principle. Thekeys facts to recall about the principle are that there are two electronsper orbital, and there is ones-orbital per level, three p-orbitals per level, five d-orbitals per level, and seven f-orbitals per level.The filling order has exceptions m the transition metals, lanthanides, and actinides. There areexceptionsthat allow for half-filled shell stability, as is seenwith chromium.

a^j* Understand quantum numbers.fit LQuantum numbers are used to define the orientation and location of an electron. There are fourquantum numbers: n, 1m/, and ms. Youmust be familiar with the quantum numbers and how todetermine them for agiven electron within anatom.

§^j* Be familiar with periodic trends.

* Knowing theperiodic trendsforthemain-group elements is essential. Thetrendsthat youmustbefamiliarwith include ionizationenergy, atomicradius, electronaffinity, and electronegativity(althoughelectronegativity of the noble gases need not be committed to memory). It is important that you befamiliar with each basic trend and the reason for that trend. Beable to explain any deviations fromstandard periodic behavior.

Understand isotopes, average atomic mass, and isotopic labeling.Isotopes are atoms with the same number of protons, but a different number of neutrons in theirnuclei. For some elements, there are several isotopes. The elemental mass (average atomic mass)found inthe periodic table isaweighted average ofall ofthe isotopes ofthatgiven element. Isotopesare chemically equivalent, so they are hard to detect without using a mass spectrometer. Isotopes,because of their chemical similarity, are substituted for one another in some reactions to serve aslabels.

Understand nuclear decay and half-life associated with first-order decay.It isessential thatyouunderstand thebreakdown andbuildupofnuclei through thegainand lossof nuclear particles. It is also important to be able to determine the concentration ofa species bycombining the initial concentration with half-life information. The half-life is the period of timerequiredforhalfofa givensampletodecayto someproduct Mostexampleswill involvedeterminingthe concentration at a given time for a reaction (or process) that follows first-order decay kinetics.

General Chemistry Atomic Theory

Atomic TheoryFrom the work of Thomson, Millikan, Rutherford, Bohr, Pauling, and others, wehave a modern view of the atom and its fundamental structure. The core of theatom by this view is a nucleus composed of protons and neutrons held togetherby an unbelievably strong force. The details of the nucleus are not wellunderstood, but using a simplistic model, the majority of the mass and all of thepositive charge of the atom is found at the core. The nucleus is surrounded byorbiting electrons that stay in distinct orbits, no two of which are exactly thesame (Hund's rule). Coulomb's law explains the mutual attraction of the orbitingelectrons and the nucleus. Electron energy levels are based on Coulomb's law,although quantum mechanics is invoked to explain the overall behavior of theorbiting electron. The closer an electron is on average to the nucleus, the moretightly it is held, and the greater the energy required to remove that electronfrom the atom. The space in which an electron is believed to orbit is referred toas an orbital. Each orbital is distinct from all other orbitals, and electrons have theoption of spinning clockwise or counterclockwise as they occupy the orbital.Only two electrons may occupy each orbital at the same time, and they musthave opposite spins to do so. These distinct orbitals are quantized energy levelswhere the electrons are said to reside. This fundamental idea is the basis of allatomic theory.

Atomic behavior and electronic configurations are explained by the concept ofelectronic bookkeeping. The ease or difficulty of gaining, removing, or sharingan electron are determined by the location of the electron. An electronicconfiguration is an account of all of the electrons in an atom. Quantum numbersare specific for each individual electron. Questions about electronicconfiguration and quantum numbers on the MCAT should be some of thesimpler questions you see, so be sure to get them correct, as they are worth thesame amount of points as the more difficult topics. Understanding theexperiments and applications associated with the absorption and emission oflight is critical to performing well on the MCAT. Absorption spectroscopy,lasers, and fluorescent tubes are just a few examples of devices that utilize lightthat you are expected to understand for the MCAT.

Equally important as understanding the structure of an atom and the energeticsof its orbiting electrons is seeing the effect the structure has on the reactivity ofthe atom. There are distinct trends in atomic radius, ionization energy, electronaffinity, and electronegativity that can be traced back to the filling of electrons inthe atom. Knowing the reasoning behind the periodic trends is more importantthan memorizing the direction of each trend and notable exceptions to the trend.

The last of the topics that fits under the heading of atomic structure is nuclearchemistry. This section should be one of the easier sections, as scientists do notunderstand the topic in enough detail to expect you to have a deepunderstanding of it. Nuclear chemistry is made easy by knowing the particlesand the processes of decay and capture, and your ability to do algebra. Know thedefinitions of isotopes, nuclear decay, nuclear capture, nuclear particles, andhalf-life. This chapter starts with an elementary look at subatomic particles andfinishes with an overview of the atom, the energy states of its particles, andtransitions between energy levels.

Introduction


General Chemistry Atomic Theory Atomic Structure

Atomic Structurej^X&£&&U^'uf£2iX \

Subatomic ParticlesOur most fundamental view of matter is that atoms are composed of three subatomic particles. These particles are the proton, neutron, andelectron. Each subatomic particle isunique inits properties andposition. Aproton andneutron arecomparable inmass, while an electron is roughly Vl800 °*^emass °* aProton-The proton andelectron carry the same magnitude ofcharge, but withoppositesign. Physical properties associated with each particle arelisted inTable 2.1

Particle Mass (kg) Mass (amu) Charge (C) Charge (e)

Electron 9.11 x 10"31 5.49 xlO-4 -1.602 xlO"19 -1

Neutron 1.67 x 10"27 1.0087 0 0

Proton 1.67 xlO"27 1.0073 1.602 xlO"19 +1

Table 2.1

Thenucleusis made up of neutrons and protons. Mass and charge are the twomeasurements repeatedly used to describe subatomic particles and whole atoms.By convention, the mass of an atom is said to be due only to the protons andneutrons, because the electrons are essentially massless. An atom normallycarries a neutralcharge, unless it has lostor gained electrons. In a neutral atom,the number ofprotons equals the number ofelectrons. Aconvenient shorthandnotation isused todescribe every element. There is themassnumber(A) used todescribe the number of protons and neutrons in an atom. There is also theatomicnumber (Z) used to describe the number of protons in an atom (and thusthe number of electrons in a neutral atom). The notation used to represent eachatom is shown in Figure 2-1.

A number Elemental Symbol

197^PtZ number Platinum-195

The A number tells us that the combinednumber of neutrons and protons is 195.The Z number tells us that the number ofprotons is 78. Combining these factstells us that there are 117 neutrons and78 protons in this atom.

Figure 2-1

IsotopesIsotopes are atoms of the same element that contain a different number ofneutrons but the samenumber of protonswithin their nuclei. Isotopes, havingthe same number of protons, are chemically similar but have different atomicmasses. Isotopes react the same way chemically andthus can bedistinguishedonlyby mass separation techniques (such as mass spectroscopy). Isotopes areoften used as markers in chemical labeling experiments and as tags in nuclearmagnetic spectroscopy studies. An isotope can betraced from the source (whereit is added) to theendpoint in a physiological system, a biochemical pathway, ora reaction mechanism. Typical examples of isotopes include 1H (standardhydrogen), 2H (deuterium), and 3H (tritium). Common isotopes used inlabelingstudies include deuterium, tritium, carbon-13, carbon-14, phosphorus-32, andiodine-121. These isotopes canbemonitored either by theradioactive decay theyemit orbynuclear magnetic resonance imaging (known asNMR spectroscopy).

Copyright © by The BerkeleyReview 70 The Berkeley Review


Example 2.1How do thesubatomic particles in26A1 and27A1 differ?A. The two are isotopes with a different number of neutrons.B. The two are isotopes with a different number of protons.C. The two are ions with a different number of electrons.D. The two are ions with a different number of protons.

SolutionBecause the symbol is Al, the atomic number is always 13, implying that thereare 13 protons. This eliminates choices Band D. In either case there is no charge,so there are thirteen electrons present and the species is not an ion. Thiseliminates choiceC and leaves choiceA as the correct answer. In 2*>A1, there are13 neutrons (because 13 and 13 sum to 26), while in 27A1, there are 14neutrons(because 14 and 13 sumto27). In26A1, there are13 neutrons, 13 protons, and 13electrons, while in 27A1, there are14neutrons, 13 protons, and13electrons. Thetwo differ by one neutron, making them isotopes. Choice A is good.

Example 2.2What symbol represents the neutral atom with fifteen electrons and sixteenneutrons?

A. i5GaB. IIPC. 31p

15A

D. 31c165

SolutionBecause there are fifteen electrons, there are also fifteen protons in the neutralatom. This makes the atomic number 15, which is associated with the elementphosphorus, and eliminates choices A and D. The mass is 31 (15protons and 16neutrons), so the symbol is 15P, choiceC.

Example 2.3Which of the following is an isotope of element 35 containing 44neutrons?A. ^Br"B. 79BrC. 81BrD. 79Kr

SolutionThe atomic number of bromine is 35, regardless of which isotope is beingobserved. Because the element in this question is bromine, choiceD is eliminatedimmediately. When the 35 protons of bromine are coupled with 44 neutrons, themass of the bromine isotope is 79 amu. The answer which shows a mass of 79amu is choice B. Choice A is an anion, which is formed upon the addition of anelectron.


Zia Siddiqui


Average Atomic MassThe average atomic mass ofanelement is aweighted average of themasses ofallof the isotopes, an average that takes intoaccount abundance. Thereference forall isotopic masses is 12C, which is assigned a mass of12.000 amu. All isotopicmasses are measured relative to 12C

Example 2.4Whena sample ofmagnesium is subjected to mass spectroscopy, it is found thatthere are three detectable isotopes. What is the average atomic mass ofmagnesium, given that the relative isotopic abundance is 79% 24Mg, 10% ^Mg,andll%26Mg?A. 23.71 g/moleB. 24.31 g/moleC 25.83 g/moleD. 26.92 g/mole

SolutionThemath associatedwith this question is time-consuming, so before diving intoit, review the answers to see what can be eliminated. The average mass has tofallwithin the rangebetweenthe lightestand heaviest isotopes. In other words,the averageis in themiddle somewhere. This eliminateschoices A and D. Themost abundant isotope is24Mg, so the average atomic mass should beclose to24grams per mole. Because the heavier isotopes are 21% the total mass of thesample, the average atomic mass is a littlemore than 0.21 grams above 24. Thebest answer is choice B. The mathematical solution is shown below, but inpreparation for thisexam, use rigorous math only to confirm intuitivelyobviousanswers.

Averageatomicmass = 79%(24) +10%(25) + 11%(26)= (0.79 x 24) + (0.10x 25) + (0.11 x 26)

= (0.79 x 24) + (0.10x (24+ 1)) + (0.11 x (24 + 2)= (0.79 x 24)+ (0.10 x24)+ (0.10 x1) + (0.11 x24) + (0.11 x 2)= (0.79 x24)+ (0.10 x24)+ (0.11 x24)+ (0.10 x1) + (0.11 x 2)= (1.00 x 24)+ (0.10 x 1) + (0.11 x2) = 24 + 0.10+ 0.22= 24.32

The math is shown in an intuitive, step-wise fashion that does not require acalculator. The number used as the center point, 24, was chosen knowing thatthe average value was around 24.

Classical Experiments and MachineryClassical experiments inchemistry arestudies that determined the fundamentalfeatures of matter. These experiments are important, because they define thenature of matter and the smallest units of matter. Of interest when looking atmatter are charge, mass, location, and composition. There are three classicalexperiments: the Thomson experiment (used to determine thesignof charges),theMillikan oil drop experiment (used to determine the magnitude of charge),and the Rutherford experiment (used to determine the location of denseparticles). The experiments as listed herearemodified from their original formto emphasize the rationale, rather than the procedure. Of equal significancetoday is the mass spectrometer (used todetermine thecharge tomass ratio for aparticle), which wasdeveloped to support the classical experiments.


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Thomson ExperimentThe Thomson experiment demonstrated the existence of opposite charges in anatom and that charge is a fixed quantity. Thomson deflected a stream of chargedparticles (electrons) using an external electric field (the plates of a capacitor).Because the stream of particles bent in a uniform fashion, Thomson concludedthat there was a consistent charge-to-mass ratio for the particles.Procedure andApparatusIn the Thomson experiment, a beam of electrons was generated traveling left toright, as shown in Figure 2-2. Thomson observed that when he applied anelectric field (a positively charged plate on one side and a negatively chargedplate on the other) perpendicular to the electron beam, he could deflect it by anexact amount each time. The magnitude of deflection depends on the strength ofthe field (charge on the plates) and the mass of the electron. Reversing the platesof the external field gets the opposite deflection.

v l\Circuit open .*. No electron beam Circuit closed/. Electron beam

*

Electron field on .*. Electron beam bends Electron field on .-. Electron beam bends

Figure 2-2

Results and ConclusionsBecause the direction of the deflection changed when the orientation of the fieldchanged, Thomson concluded that there must be two types ofcharge that opposeone another. Because the arc of the deflection was constant, Thomson concludedthat electrons have a fixed charge-to-mass ratio, measured to be 1.76 x108 C/g.Adapted Thomson Experiment (in MCAT testing style)What is presented here is a spin-off of theThomson experiment. The Thomsonexperiment concept canbe applied to various particle beams besides theelectronbeam. A simplified version of the apparatus has an accelerating field (notshown), which accelerates particles to the left, a double filter to ensure uniformlinear trajectory, and an electric field perpendicular to the vector of entry.Charged particles are deflected according to theirsign ofcharge. Four particlesare considered. Their pathways are shown in Figure 2-3.

Particle acceleratedfrom left to right

Doublefilter

+ + + + + + + + + + + + + + + +

III

Figure 2-3

Atomic Structure


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Pathway I is attributed to an electronbeam, while Pathways n, HI, and IVare notidentified. The MCAT test writers are more interested in ascertaining yourintuitive understanding of problems than your superficial knowledge of facts.Questions are likely to introduce obscure particles and test you on theirproperties, which mustbe derivedfrom the experiment. Consider the followingquestion based on the data in Figure 2-3.

Example 2.5Pathway II is taken by which of the following?A. A positronB. A neutrinoC. A gamma rayD. Amuon

SolutionThe particle following Pathway II must be negatively charged, based on thedirection of its deflection. It deflects in the same direction as the electron.Although some of the particles in the answer choices may seem unfamiliar, thequestion still can be answered. Based on the name, you should deduce that apositron is positively charged. This eliminates choice A. A positron is an anti-electron, in that it has the same mass as an electron, but the opposite charge.Based on the name, you should deduce that a neutrino is neutrally charged. Thiseliminates choice B. A neutrino is essentially massless and carries no charge.They are difficult to detect. Detection of neutrinos is done through collision andscintillation. ChoiceC is eliminated, because a gamma ray is a photon. A photondoes not bend in an electric field. Photons may be refracted, but that requires achange in medium. The only choice left is a muon, choice D. In all likelihood,you do not know what a muon is. This question is not testing your knowledge ofparticles; it is testing your reasoning abilities. A muon traces a different pathwaythan the electron, because it is about 200 times as massive as an electron.



Mass SpectrometryA mass spectrometer is designed to measure the charge-to-mass ratio for acharged particle. This is accomplished by sending a particle into a perpendicularmagnetic field and observing the degree to which it curves. The degree of arcing(radius of curvature) for a particle can vary with mass, initial velocity, magnitudeof charge, and the strength of the magnetic field. As momentum increases (eithermass or initial velocity), the particle deflects less, so the radius of curvatureincreases. As the charge magnitude increases, the force causing deflectionincreases, so the particle deflects more, causing the radius of curvature todecrease. Bycomparing the curvature for an atomic or molecular ion to a knownstandard, the mass of the unknown ion can be determined. The massspectrometer is used in general chemistry to determine isotopic abundance.Recall that isotopes are the same element with a different number of neutrons. Inorganic chemistry, the mass spectrometer is used to determine molecular massand fragmentation behavior to help elucidate the structure of an unknowncompound. Figure 2-4 shows a basic schematic design for a mass spectrometer.Note that some mass spectrometers may also have a velocity selector situatedbetween the accelerating region and the deflecting region.

it-

Particle path

H_ X X X X X X X

o

5 • X X_vx X X X X

X X X\ X\ X X X

ri\ ]X X X K ,' X X X

X X x//x XXX

X X X X X X X

r2>rl

m vr oc

qB

Magnetic field into page

Double filter to ensure uniform particlebeam perpendicular to Bfield

Figure 2-4

Procedure andApparatusThemass spectrometer embodies a simple concept. Force depends on mass, sowhen an equal force is applied to different masses, they accelerate at differentrates. The mass spectrometer takes advantage of this by accelerating chargedparticles inmotion using amagnetic field. The procedure isas follows:

O An element or molecule is ionized using high-energy electron impactor incident electromagnetic radiation. The ionized particle is thenaccelerated to the cathode plate. The strength of the acceleratingregion (voltage of the electric field) is adjustable, so the velocity ofthe particle can be set to any desired value.

© The particle passes through the double filter to ensure a uniformperpendicular beam.

© The particle leaves the accelerating region and enters aperpendicular magnetic field (oriented into the page as drawn inFigure 2-4) whereit is deflected in a counterclockwise, radial fashionby the perpendicular magnetic force.

© The radius is ascertained from the strike point against a collisiondetector. The mass-to-charge ratio is calculated from the radius ofthe arc.

Atomic Structure


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Example 2.6Which of thechanges to thesystem wouldincrease the radius ofcurvaturein themass spectrometer?A. Usinga doublyionizedrather than singlyionizedelementB. Using an isotopewith fewer neutronsC. Increasing the acceleratingvoltageD. Increasing the magnetic field

SolutionThe trick here is translating what the answer choices mean in terms of thephysics of theapparatus. Using a doubly ionized element (+2 cation) rather thana singly ionized element (+1 cation) results in a greater q value, so a greaterdegree of deflection is observed. This reduces the radius of curvature (r), sochoice A is invalid. Usingan isotopewith fewer neutrons results in a reducedvalue for mass, so a greater degree of deflection is observed. This reduces theradius of curvature (r), so choiceBis invalid. Increasing the accelerating voltageresults in a greater velocity, so a lessened degreeof deflection is observed. Thisincreases the radius of curvature (r), so choice C is the best answer. Increasingthe magnetic field results in a greater force, so a greater degree of deflection isobserved. This reduces the radius of curvature (r), so choice D is invalid. Thisquestion can be answered from a conceptual perspective or based on therelationship of variables described in a formula, with equal success. Whetheryou use equationsor intuition is a matter of personal preference and timing.

Themass spectrometer canbe used to determine the charge-to-massratio for theelectron and the proton. The mass of a neutron is obtained by looking at themass difference between known isotopes. For instance, the mass differencebetween 1H+ (aproton) and2H+ (deuterium ion) is themass ofone neutron. Themass spectrometer can also be used to determine the isotopic abundance for thecomponentatomsofeachelement. Common isotopes that should be memorizedinclude: 12C (the most abundant isotope ofcarbon), 13C (used in carbon NMR),14C (used incarbon dating, because it undergoes decay), 1H (the most abundantisotope ofhydrogen), 2H (deuterium, used inproton NMR solvents), 3H (tritium,used inradio-labeling experiments), 235U (used innuclear fission), and 238U (themost abundant isotope of uranium).

Millikan Oil Drop ExperimentTheMillikan oil drop experiment is a difficult experiment to perform. Its aim isto suspenda charged oildrop in an electric field. Todo this,an electron must beadded to the oil drop or the oil drop must be ionized by impact, before it isplaced into the electric field. Wewill consider adding the electronhere. Thiscannot be accomplished easily, given that the oil drop is neutral and has noaffinity for the negatively charged electron. Falling oil drops pass through abeam of electrons where some oil drops are penetrated at random by an electron.Ideally, the electron penetrates the core of the oil drop and comes to rest, due tothe viscosity of the oil. Thesuspensionof an electronin the oil drop produces acharged oil drop. Enough charged oil drops continue to fall, one of whicheventually passes through a pore in the upper plate of a capacitor. The twoplates of the capacitor are separated, but both lie within the walls of a glasscylinder. Figure 2-5 is a basic schematic of the apparatusused in theMillikan oildrop experiment.


General Chemistry

Oil can

Figure 2-5

Atomic Theory

If suspended or fallingat constant speed,

qE=-mgA

qE|q =

mg

E

Charged oil drop

Procedure andApparatusIn the Millikan experiment, a fine mist of oil droplets is allowed to fall through atiny pore in the upper capacitor plate into a regionwhere a uniform electric fieldexists. The oil droplets fall because of gravity, so Millikanset out to apply a forcethat could stop the droplets from falling. If the droplets are suspended, then theforce applied equals the gravitational force (a known quantity that is dependenton the mass of the falling object). The applied force is formed by charging the oildropletsby exposure to either an electron beamor an x-raybeam (onlyone of thebeams is applied in different trialsof the experiment). Anelectric field is appliedto suspend the charged droplets. The uncharged oil drops fall unaffected by thefield. Despite the difficulty of this experiment, Millikan obtained enough validdata so that an average measurement was put forth and accepted by the scientificcommunity at large.

O The falling oil drop gains a charge by either losing or engulfing anelectron as it falls (depending on the version of the experiment). Onrare occasions, the falling oil drop both gains an electron afterpassing through the pore in the upper plate. The electric fieldstrength is adjustable, so that the oil droplet can be suspended.

© If the particle is suspended (or falling at a constant velocity), then thenet force is zero, so mg = - qE. Because we know g and we can setthe electric field strength (thus we know E), by plugging in theaverage mass of an oil drop (we know an average m), we can solvefor q, the charge of the electron.

Results and ConclusionsThe charge of an electron has a fixed numerical value that is the same for allelectrons. The value for this fundamental unit of charge is 1.6 x 10"19 C. Thecharge of a proton is found to have the samemagnitude, but opposite signof theelectron. When a proton is combined with an electron, there is no net charge.The Millikan oil drop experiment was difficult to carry out, so even after manytrials, there were only two significant figures in the final number.

The Thomson experiment, the Millikan oil drop experiment, and the use of themass spectrometer generally address questions about the subatomic particles interms of what they are. The where they are questions are answered by interactionswith incident light and particle beams. The Rutherford experiment is the mostsignificant location experiment we will consider.

Atomic Structure


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Rutherford ExperimentTheRutherford experiment determined that atoms havedensenuclei withnearlyall of the atomic mass centrally concentrated, and that metals have uniformlyspaced atoms in their microscopic composition. The Rutherford experimentrelies upona technique thatcanbe employed to find a ball in the bushes in thedark. "What does finding a ball in the bushes at night have to do with theMCAT?" youmayask. Well, when a ball is lost in the bushes, it is easierto findits location by looking for its shadow than by hunting for the ball itself. Byshining light through the bushes and observing the shadow cast against thebackground of a wall, a circular shadow can give hints as to the location of thespherical ball. Depth in thebushes canbe obtained by first moving the flashlightcloser to the bushes, then moving it farther away. The size of the shadow varieswith position of the light source, so the relative position of the ball to the lightsource is determined from shadow dimensions. In essence, the ball can belocated and its dimensions can be found without ever seeing the ball. This sameprinciple was used by Rutherford to find subatomic particles. A subatomicparticle is smaller than a ball, so a light source of significantly shorterwavelength must be employed. In this example, we use x-rays as the lightsource. In the analogy, the bushes had to be thin for the light to pass throughthem, so a thin strip of gold foil is used for the study. Rather than looking for ashadow on the wall, the experiment uses photographic paper to collect the x-raysthat pass through. Figure2-6shows the basic design of the experiment.

Lead reaction vessel

X-ray beam

X-ray source Lead filter

Figure 2-6

Thin gold foil strip

Photographic plateor

Luminescing screen

Procedure andApparatusAn incident beam (x-rays, alpha particles or electrons may be used) is focusedand aimed at a thin slice of gold metal, thin enough that the beam is able topenetrate and pass through the gold foil. Gold is chosen, because it has a largenucleus and its atoms pack in such a manner where light can pass through itslattice more easily than other metals. If alpha particles are used, a luminescentscreen is placed around the gold foil to detect where the particles pass throughthe foil and strike the luminescent screen (which glows when struck by an alphaparticle). If x-rays are used, a photographic plate is placed around the gold foilto detect where the photons pass through the foil and strike the film. In thealpha-particle version of the experiment, some particles are deflected by the goldsample, resulting in parts of the luminescing screen never illuminating.

Residts and ConclusionsBecause the incident beam mostly passes straight through the sample, withdeflection (ricocheting particles) being observed in only a few cases, it isconcluded that the atom is made up predominantly of empty space. The massassociated with the atom occupies very little space and is not spread uniformlythrough the material. Atoms are composed of a nucleus holding the mass (theprotons and neutrons.) Thisdense nucleus carries all of the mass besides that ofthe electrons. But given that the electrons are of such low mass, it is not possibleto discern their whereabouts from the Rutherford experiment.



The Rutherford experiment disproved the diffuse particle model (referred to asthe plum pudding model.) In the version of this experiment using the x-ray beam,the photographic screen displayed the output shown in Figure 2-7.

Figure 2-7

Dark spots represent areas where no x-rays struck the film. In essence, the spotsare the shadows of the sub-atomic particles. Based on the distribution of the darkspots, it is concluded that the mass of gold is not evenly distributed in the goldfoil, but is in fact found in concentric, dense nuclei. Because the nuclei (spots) areevenly spaced, the gold atoms must be arranged in a lattice structure. This isbecause the shadow pattern on the filmmimics the distribution of particles in thematerial. Gold is chosen because it is malleable and it has a massive nucleus,thus it diffracts x-rays more readily than lighter elements, like aluminum. This iswhy x-rays in medical imaging show bones and teeth (rich in calcium nuclei)rather than tissue (rich in the light carbon, oxygen, and nitrogen nuclei). In orderto analyze blood using x-rays, a heavy salt must be added to the solution, oftenbarium iodide.

Example 2.7What hypothesis did the Rutherford experiment support?A. Atoms combine in definite proportions.B. Atoms contain subatomic particles.C. Protons and electrons carry opposite charges.D. Solids are made of atoms with a dense nucleus and vast empty space

between nuclei.

SolutionChoice A is Dalton's law of definite proportions. Choice B is particle theory.ChoiceC is the Thomson experiment, conducted by observing the deflection ofthe beam in a cathode ray tube when an external electric field is applied. TheRutherford experiment involves the bombardment of a thin piece of metal foilwith either high-energy photons (x-rays) or a beam of electrons (Rutherford didboth in separate experiments.) The photons pass through and strike anilluminating screen, which forms a shadow pattern indicative of the material'sshape. Most of the beam passes through, with a minimal amount being eitherreflected or diffracted. The conclusion is that the material is essentially emptyspace, with a few dense nuclei scattered throughout the material. The spots onthe screen are uniformly spaced, so the nuclei must also be evenly spaced withinthe element. This makes choice D the best answer.

The location of the electrons is not determined by Rutherford experiment.Electrons are too small and moving too fast to locate precisely. This is describedby Heisenberg's uncertainty principle.

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Heisenberg's Uncertainty PrincipleThe Heisenberg uncertainty principle quantifies the idea that it is not possiblesimultaneously to identifya particle'sposition (where something is) and velocity(how fast and where it's going). Equation2.1 is the mathematical versionof theHeisenberg uncertainty principle, where Ax is the uncertainty in position andA(mv) is the uncertainty in momentum.

Ax.A(mv)>h/4jl (2.1)

The basic premise here is that you can know either where something is, or howfast it is going,but not both at the same time. Think of using a camera to focuson a movingball. If your aperture is smalland the shutter speed is fast, then thepictureof theballshowsyouwhereit is, but you don't knowwhere and how fastit's going. If your aperture is large and the shutter speed is slow, then the pictureof the ball is a streak that shows you where and how fast it's going, but you don'tknow exactly where the ball is. Becausewe cannot locate an electron's preciseposition, we settle for a view where the electron is observed over time. Thisresults in orbitals as a model for the orbiting electron.

Atomic ModelThe Bohrmodel presents a simplified picture that explains the quantization oflight and the reproducibility of spectra. The basic premise is that electronsoccupy specific circular orbits about the nucleus, and thus the electrons havespecific energy levels (associated with each orbit). Electrons can exist only inspecified orbits (electronic shells), so eachenergylevelof an atom is quantized.Figure 2-8 shows this:

n = 4

n = 3

n = 2

n = l

ywWW A.4_,

w w w

Electronic energy levels Electronic shells

Figure 2-8

Theenergy levels arespaced according to theenergetics of transition between thelevels. More energy is required to carry out transitions when the electron isnearest to the nucleus. The electrons are situated in various energy levels(known more accurately as orbitals). These are quantized states that electronsoccupy. Principle energy levels arenumbered 1 to °°, where n = 1 is the lowestelectronic energy level. Energy mustbe absorbed by the atomfor an electron toelevate to a higher energy level. This is referred to as both excitation of anelectron and absorption of energy. Conversely, energy is emitted when anelectron drops from a higher energy level (excited state) to the lowest energylevel (ground state). A good analogy to electrons climbing energy levels is a


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rocket ship escaping Earth's gravitational pull. It takes less energy to increasedistance as the rocket ship pulls farther away from the Earth. Likewise, it takesless energy for an electron to increase energy levels as the electrons increase tohigher energy levels (farther from the nucleus). For instance, it requires moreenergy for an electron to go from the n = 2 level to the n = 3 level, than for theelectron to go from the n = 3 level to the n = 4 level in any atom. This is why theenergy levels for electron states are drawn closer and closer as the principlequantumnumber (n) increases in Figure2-8. Thispicturebecomes a littlemorecomplicated if the rotational energylevels associated withan atomare combinedwith the electronic energy levels. At the level of understanding needed foranswering MCAT questions, we ignore the rotational energy levels when welookat electronic energy levels. Weshall consideronly the principalenergy levelwhen considering electrons. Equation2.2 is used to determine the energyof anelectron in its principal energy level.

E=2*2mZ2e4 (2.2)n2h2

E=energy (principal energy level) m=mass of an electron (9.11 x10"31 kg)Z=nuclear charge e=the charge of an electron (1.6 x10"*9C)n=the electronic energy level h=Plank's constant (6.63 x10"34 J-sec)

Themass of an electron,m, the charge of an electron, e, and Planck'sconstant, h,are all constants, so when Equation 2.2 is considered as a proportionality, itbecomes Equation 2.3.

EocZ?_ (2.3)n2

Example 2.8According to Figure 2-8, howwould the photonfrom ann =4ton =2 transitioncompare to the photon from an =2ton = l transition?A. The n = 4 to n = 2 transition is twice as energetic as the n = 2 to n = 1

transition.B. The n = 4 to n = 2 transition is more than twice as energetic as the n = 2 to n =

1 transition.C. The n = 2 to n = 1 transition is twice as energetic as the n = 4 to n = 2

transition.D. The n = 2 to n = 1 transition is more than twice as energetic as the n = 4 to n =

2 transition.

SolutionThe distance between the n = 4 and n = 2 levels is less than the distance betweenn = 2 and n = 1, so transition energy is greater from the n = 2 level to the n = 1level. Thephoton released froman n = 2 level ton = 1 level transition hasmoreenergy thanthephoton released from ann =4 level ton =2level transition. Thiseliminates choices A and B. The transition energy from the n = 2 level to the n =1 level is more than twice the transition energy from n = «> to n = 2, so it isdefinitely more than twice the energy of the n = 4 level to n = 2 level transition.ChoiceD is the best answer. This energy difference is shown in Figure 2-8. Onthe MCAT, you should assume that the diagrams are drawn to scale, unlessotherwise noted in the question or passage.

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General ChemiStry Atomic Theory Atomic Structure

The energy required to excite an electronfrom a lower level (orbital) to a higherlevel is often in the ultraviolet (UV) to visible range of electromagnetic radiation,so the energy givenoffas the electrondrops back down to the lower level (orbit)is emitted as light energy. This is the basic principle behind spectroscopy.Lower energy levelshave less absolute energy and thus are more stable states inwhich an electron can exist. The smaller the gap between energy levels, the lessenergy that is given off, and therefore the longer the wavelength of light that isemitted. One formula is important for understanding the relationship betweenthe speed of light, the frequency and wavelengthof light, and light energy. Theenergy of a photon and its wavelength of light are inversely proportional.Equation 2.4sums this up, where E is the energy of the photon, v is frequency, cis speed of the wave, and Xis wavelength.

E = hv = he. (2.4)X

Hydrogen Energy LevelsHydrogen is the simplestatom to study, because it has only one electronand oneproton. Much of our atomic theory is extrapolated from what we know abouthydrogen. Because energy is quantized and the energy of the electron dependson features of the hydrogen atom, the energy levels can be calculated. Equation2.5represents the energy of the different levelsof hydrogen.

E=-2.178x lO'18^2-] (2.5)Energy levels are defined as being negative relative to a free electron. If theelectron is in the n = oo energy level, then E = 0, and the electron is free from anucleus. Considering that photons are absorbed and emitted when electronschange energy levels, themore useful application of the energy equation involvestransition energy. Equations 2.6and 2.7show the relationship between transitionenergy and the corresponding wavelength of the photon involved.

AE =Efina! - Einitiai /. AE =-2.178x lO'18U ±-\ (2.6)18/ 1 _ 1 \

\nfinal ninitial/

X= h£- (2.7)AE

The ionization energy of hydrogen from its ground state (from the n = 1 level) is1312 kj/mole. Because of the squaring of the principle energy level, theionization of an electron in hydrogen from the n = 2 level is one-fourth of thatvalue (328 kj/mole). The transition energy from the n = 1 level to n = 2 level isthe difference between the two values, 984 kj/mole.


General ChemiStry Atomic Theory Electronic Structure

Electronic Structure r-r;...;.;;; ^Yl/^I^L;^Electronic TheoryIn atoms, the electrons orbit in distinct shells. Not all shells can hold the samenumber of electrons. Shells farther from the nucleus have a greater radius, andthus a greater capacity to hold electrons. Equation 2.8 gives the maximumoccupancy of electrons in a shell, where n is the principle quantum number.

Numberofelectrons in shell=2 (n)2 (2.8)

Figure 2-9 shows electrons of the lithium atom in their respective shells. The firstshell holds two electrons, so the third electron must occupy the second shell. Thefirst shell is the core shell, while the outermost shell is the valence shell.

Valence shell 2"d •""»'level(electron occupancy up to 8)

Core shell 1$'"*<&level ,(electron occupancy of 2)

Nucleus: made up of bothneutrons and protons

Figure 2-9

Effective Nuclear Charge (Nuclear Attraction)Orbitingelectronsare held in their orbitsby an attractive electrostatic force to thenucleus. In addition to nuclear attraction, electrons are also repelled by otherelectrons. The net force is responsible for holding the valenceelectrons in place.The net charge exerted upon the valence electrons is referred to as the effectivenuclear charge. Theeffective nuclear chargeaccounts for attraction to the nucleus,repulsion from core electrons, and minimal repulsion by other valenceelectrons.When approximating the effective nuclear charge (Zeff), the nuclear charge isadded to the core electron charge (a negative term). Figure2-10 shows effectivenuclear charge increasing whilemoving leftto rightacross theperiodic table.

7Li /> v\ 9Be

nucleus: +3; core electrons -2 nucleus: +4; core electrons -2.-.Zeff = +1 •••Zeff =+2

Figure 2-10

When we move from left to right in the periodic table, the nucleus of eachsucceeding atomadds a proton and the valence shelladds an electron. Theeffectof the extra valence electron is not as significant as the effect of the additionalproton. As a result, the effective nuclear charge increases as the periodic table isscanned from left to right.

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General Chemistry Atomic Theory Electronic Structure

Each electron in an element travels a unique pathway, dictated by severalprinciples. Electrons are difficult to track, so we have theories to explainelectronic behavior, but they are just models to explain observed behavior.According to the Heisenberguncertainty principle we cannot see an electron,butwe can study its pathway over time. Charged particles in motion createmagnetic fields, so by studying the magnetic field generated by a movingelectron, it is possible to learn about the pathway and position of the movingelectron. This is why two of the four quantum numbers associated with anelectron refer to magnetism that results from a moving electron. We shall blendthe many ideas about the electron that have evolved over time, starting with themost simplistic model, the Bohr model.

Electron Spin PairingElectrons fill orbitals in a pre-determined sequence, filling evenly into orbitals ofequal energy with like spin (all orbitals get a single electron, said to be "spin up"),before placing a second electron with opposite spin into each orbital. Thephysical reality is that electrons may spin either clockwise or counterclockwiseabout their axis. Spinning charged particles generate magnetic moments, so thetwo opposite spins produce opposite magnetic fields. The magnetic fieldsgenerated by electrons revolving about their axis are referred to as either spin up(implying that the spin produces a magnetic field vector oriented upward) orspin down. Byconvention, electrons are said to fill orbitals spin up first, beforefilling spin down. Figure 2-11 shows the electron filling of lithium-7 andberyllium-9, where arrows represent electrons, and the orientation of the arrowimplies spin.

n = l

(Core shell: Is2)•n = 2

(Valence shell: 2s1)

With unpaired electrons, it is paramagnetic.

n = l2

(Core shell: Is )* n = 2

(Valence shell: 2s2)

ii„

ilii

2s

Is

With all electrons paired, it is diamagnetic.

Figure 2-11

The shells represent energy levels an electron can occupy, while orbitalsrepresent the region in which the electron is likely to be found. An s-orbital hasspherical electron density. The difference between the Is and 2s orbitals lies intheir dimensions. The Is has a smaller radius and has no nodal shells (regionswhere the electron has zero probability of existing.)


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Spin Pairing and MagnetismA paramagnetic species is defined as an atom or molecule that contains at least oneunpaired electron. In organic chemistry, paramagnetic compounds are referredto as radicals. An unpaired electron is an electron that has no second electronspin paired with it. By convention, the first electron into an orbital is said toenter in a spin up fashion, thus an unpaired electron is a spin up electron in anorbital that has no spin down electron. Because the electron is unpaired, it issusceptible to magnetic fields. If an external magnetic field is applied to aparamagnetic species, the electron spins align with the field. This induces amagnetic moment into the compound, thus making it magnetic. This is to saythat paramagnetic species can have magnetism induced into them.

A diamagnetic species is defined as an atom or molecule that contains no unpairedelectrons. All electrons in the atom or molecule are spin-paired, meaning thatevery electron that is spin up will have a spin down electron sharing its orbital.By convention, the first electron into an orbital is said to enter in a spin upfashion, so the second electron is a spin down electron. Because all of theelectrons are spin-paired, diamagnetic compounds are not susceptible tomagnetic fields. If a magnetic field is applied to a diamagnetic species,half of theelectron spins align with the field, forcing the other half to align against the field.No magnetic moment is induced into the compound. This is to say thatdiamagnetic species cannot have magnetism induced into them.

Electron Density and OrbitalsAtomic orbitals are three-dimensional pictorial representations of the regionwhere an electron is likely to be found. Because we observe electrons over time,we look at where the electron usually is, and draw a probability map of theelectron distribution. It's like look at a spinning fan. You cannot see each bladeas they turn, but you can see over time the area where they spin. Figure 2-12represents the electrondensity of an electron in an s-orbital over time, the orbitalrepresentation, and the probability map based on distancefrom thenucleus.

*** v * .

Electrondensity map

Orbitalrepresentation

from nucleus

Figure 2-12

The electron density map shows that electrons are found most often near thenucleus. This is also represented by the graph of the probability of finding anelectron as a function of its distance from the nucleus. The orbital representationis typically used by chemists to depict the s-orbital. The size of the sphere varieswith the electron density map, depending on the atom. The shape of an orbital isdefined by the distribution of electrons about the nucleus. Ninety-five percent ofthe time, the electron can be found within the boundaries of the orbital. We willlook at the s-, p-, and d-orbitals in substantial detail, while f-orbitals will beconsidered, but in minimal detail. Most common elements do not have electronsoccupying the f-orbitals.



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S-orbitalsS-orbitals result from spherical distribution of the electrons about the nucleus.Figure 2-13 shows three different s-orbitals, where the principle quantumnumber represents the energy level and the average distance from the nucleus.

Is 2s

Figure 2-13

P-orbitalsP-orbitals result from barbell-like distribution of the electrons about the nucleus.Figure 2-14 shows the three different p-orbitals, each oriented about a differentaxis. Electrons are not found at the nucleus in p-orbitals. Absence of electrondensity at any point is referred to as a node. P-orbitals have one node at thenucleus that is part of a nodal plane between the two lobes.

u'-y

D-orbitalsD-orbitals result from double barbell-like distribution of the electrons about thenucleus. Figure 2-15 shows the five different d-orbitals, each oriented differently.D-orbitals have two nodal planes, and electrons are not found at the nucleus.


Indxz, dxy, anddyz, lobes lie between theaxes

In dv2 ,,2and d _2, lobes lie on the axisx - y t.

Figure 2-15

86 The Berkeley Review

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F-orbitalsF-orbitals result from triple barbell-like distribution of the electrons about thenucleus. There are the seven different f-orbitals, each oriented about a differentplane or axis. Electrons are not found at the nucleus. F-orbitals have three nodalplanes. Little chemistry is carried out with the f-orbitals, so they are uncommon.

Collective Orbital View of Energy LevelsOrbitals result from probability calculations, where energetics is considered.Different orbitals are associated with different energies. Conceptually, we useorbitals to show the energy and most frequent location of an electron. Figure 2-16 shows orbitals with relative size emphasized, from lowest energy levels to thehigher energy levels. Levels are spaced according to energetics. Arrowsrepresent electrons and their spin orientation. Because there are twelve electronsshown, the element represented is magnesium.

2pxOO 2?yCf>

The upward single-headedarrow represents an electron inthe Is orbital with its magneticspin orientation up.

2s

Is

Figure 2-16

Sd^

2P,

The downward single-headedarrow represents an electron inthe Is orbital with its magneticspin orientation down.



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Electronswithin an element fill the energy levels starting from the lowest energy.This means that the electrons within an element follow a specific filling order.There are some rules to consider when looking at electronic configurations.Pauli's exclusion principle: No two electrons can have the same set of quantumnumbers (n, /, n\, ms).

Hund's rule: Electrons completely fill lower energy levels before starting to fillhigher energy levels. Ina degenerate setoforbitals, electrons singly occupy eachorbital before a second electron pairs up within the same orbital. Figure 2-17demonstrates Hund's rule.

Not allowed, because electrons filldifferent orbitals before pairing up.

Allowed, because electrons have filledeach orbital singly without pairing.

Figure 2-17

Aufbau principle: Electrons are added oneby one to the shells, startingwith thelowest energy level, and then into sequentially increasing energy levels. Thenumbers in Figure2-18 represent the sequenceof addition for the electrons.

©

2p®

2s

Is

i

©

o

i

0

©

r

9

Figure 2-18

Electronic ConfigurationElectronicconfigurations are shorthand notation for the electrons present in anatom and their energy levels. Electrons fill according to a set pattern, one that isderived from the Aufbau principle chart shown in Figure 2-19. Bydrawing thetable and then sequentially following the arrows, the orbital filling sequence isgenerated. Couple this information with the orbital occupancy, and electronicconfigurations are seen to be systematic. For instance, the first line shows thatthe Is level fills first, to an occupancy of two electrons. The next arrow crossesthrough the 2s level, so the 2s orbital is filled next. From here the third arrowshows that the 2p levelfollowed thenby the 3s levelare filled. It continuesdownthe chart. The first break from numerical sequencing comes when the 4s level isfilled before the 3d level, despite the fact that the perimeter of the 3d level iscloser to the nucleus than the perimeter of the 4s orbital. The reason for theapparent discrepancy is that the energy of the level is based on an averageposition of the electron, not the extreme position. Ionizing electrons are notremoved from the atom in reverse order, however. Outer shell electrons arealways removedfirstwhen forming cations. Figure2.8shows only the first fivearrows, but the pattern continues. You should also be able to deduce theelectronic configurations for neutral atoms, cations, anions, excited states, andany exceptions to the rules.


General ChemiStry Atomic Theory Electronic Structure

1: Fills the Is orbital to: Is22: Fills the 2s orbital to: 2s23: Fills the 2p then 3s orbitals to: 2p63s24: Fills the 3p then 4s orbitals to: 3p64s2

5g 5: Fills the 3d, 4p then 5s orbitals to: 3d104p65s2

Figure 2-19

Figure 2-19 shows that the electrons fill according the orbitals listed by thesequential arrows. It works by following the arrows sequentially. Arrow 1 goesthrough Is, so the Is orbital is filled first. Arrow 2 goes through 2s, so the 2sorbital is filled next. Arrow 3 goes first through 2p, then through 3s, so the 2porbital is filled after the 2s orbital, followed by the filling of the 3s orbital. Theprocess is repeated arrow after arrow until all of the electrons have beenaccounted for. Although g-, h-, and j-orbitals exist in theory, the periodic tablecontains no elements that have electrons in either g-, h-, or j-orbitals.

Example 2.9The electronic configuration for manganese is which of the following?A. ls22s22p63s23p63d7B. ls22s22p63s23p63d5C. ls22s22p63s23p64s23d5D. ls22s22p63s23p64s23d7

SolutionManganese (Mn) is element number 25, so a neutral manganese atom mustcontain 25 electrons. This eliminates choice B (only 23 electrons) and choice D(containing 27 electrons). Because the 4s orbital is filled before the 3d orbital,choice A is eliminated. This leaves only choice C.

Example 2.10An element in which column of the periodic table is diamagnetic?A. Column 1 (alkali metals)B. Column 2 (alkaline earth metals)C. Column 6 (chalcogens)D. Column 7 (halogens)

SolutionA diamagnetic compound has all of its electrons spin-paired. This means thatthere must be an even number of electrons in the element. Based on the evennumber constraint, choices A and D are eliminated. Column 6 elements (thechalcogens) have a valence electronic configuration ofns2np4, which results intwo p-orbitals having only one electron each. This means that chalcogens areparamagnetic, eliminating choice C. This means that the alkaline earth metals incolumn 2 arediamagnetic, with a valance electronic configuration ofns2. Thealkaline earth metals and the noble gases are diamagnetic. The best answer ischoice B.

Copyright©by TheBerkeley Review 89 Exclusive MCATPreparation

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Electronic configurations may include abbreviations based on filled coreshells.A filled coreis represented by the noble gas that contains those sameelectrons.For example, aluminum (Al) is ls22s22p63s23p1, which isequivalent todrawingit as [Ne]3s23p1. This shorthand is typical. In addition, youmust be aware ofsomecommon exceptions to theAufbau principle. Half-filled d-shelland filledd-shell stability results when a singleelectronis elevated from a lower energylevel that ispaired (usually thes-orbital) toyield evendistribution ofelectrons inthe d-level. Half-filled d-shell stability is seen with chromium, molybdenum,and tungsten. Filled d-shell stability is seen withcopper, silver, gold, and somesayplatinum. Figure 2-20 shows theelectronic configurations for chromium andcopper, exceptions to the Aufbau filling order.

Half-filled d-shell stability inchromium: [Ar]4s13d5 rather than [Arl4s23d4Filled d-shell stability in copper: [Ar]4s13d10 rather than [Ar}4s23d9

Figure 2-20

Elements in the samecolumn of the periodic tablehave similarvalenceshellsandelectronic configurations, with thenotabledifference being the shell number. Forinstance, Na is{Ne^s1 and potassium is [Ar^s1. This means thatalkali metalsare s1 metals, and exhibit similar chemical behavior, given their commontendency to lose one electron. Blocks in the periodic table are named after thelast electron in the electronic configuration. Alkalimetals fall into the s-block byvirtue of their last electron in an s-orbital. So far, we have viewed ground stateelectronic configurations. Groundstateelectronic configurations occurwhen theelectrons occupy the orbitals in the exact predicted order, starting from leastenergetic and filling orbitals that areprogressively ofhigherenergy. An excitedstate electronic configuration occurs when any electron absorbs energy andmoves to a higher energy level than it normally occupies in the ground state.The absorption and emission of energy, usually in the form of a photon, isassociated with the excitation and relaxation of an electron, as it moves betweenthe ground and excited states.

Example 2.11Which electronic configuration represents an excited state?A. F:ls22s22p6B. N:ls22s22p3C. He: Is2D. Liils^p1

SolutionAn excited state electronicconfiguration does not follow energetic sequence. Anexcited state has at least one electron in an energy level higher than what isdrawn as standard for the ground state. Besure not to confuse an ion (eithercation or anion) with an excited state. A cation is an atom that has a deficit of atleast one electron and thus carries a positive charge. An anion is an atom thathas an excess of at least one electron and thus carries a negative charge. In thisquestion, choice Ais a fluorine anion(itcontains an extraelectron), and choices BandCarenormal. For Li, it should havels22s1 as a groundstate. The electronicconfiguration given in the answer choicehas the last electron in a 2p-orbital,which is of higher energy than the ground state 2s. This makes choiceD thecorrect answer, because it is an excited state.


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Quantum NumbersQuantum numbers are a set of four numbers that uniquely describe an electronwithin an atom. Four factors describe an electron: the shell, the orbital, theorientationof the orbital, and the alignment of themagnetic field resulting fromthe precessing electron. These terms translate into the quantum numbers. Afundamental rule with quantum numbers is that no two electrons can have thesame set of quantum numbers (this is the Pauli exclusion principle). Quantumnumbers are used to describe the motion and location of each electron in anelement. Quantum numbers describe the shape of an electron's cloud, themomentum of the electron, the orientation of the electron density, and therotation of the electron about its axis. There are four quantum numbers used todescribe an electron; n, /, m/, and ms. There are also rules that must be followedwhen assigning quantum numbers to an element. Each number has its specificguidelines, which often depend on the other quantum numbers. Table 2.2 liststhe rules for assigning quantum numbers.

Quantum # Rules

n

Principle (n): Describes the shell (average radius of theelectron from the nucleus and its energy level) in which theelectron resides. It can be any integer greater than zero.

/

Angular Momentum (/): Describes the orbital (shape of theelectron cloud formed by the orbiting electron) in which theelectron resides. It must be less than the value of n. It can bea positive value or zero.

i*VMagnetic (mi): Describes the orientation of the orbital about aplane or axis. It can be any value in the range from negative 1to positive 1, including zero.

msSpin (ms): Describes the rotation (counterclockwise orclockwise) of the electron about its axis. It can be eitherpositive or negative one-half (spin up or spin down).

Table 2.2

Table 2.3 shows the correlation of quantum numbers to electrons within an atom.

n / ™l ms orbital totalelectrons

Description

1 0 0 + 12

Is 2 1st lsi

2 0 0 + 12

2s 2 2st 2si

2 1 ±1,0 + 12

2p 6 2pxt 2pvT 2pzT 2px42pvi 2pz!

3 0 0 + 12

3s 2 3st 3si

3 1 ±1,0 + 12

3p 6 3pxT 3pvt 3pzt 3pxl 3pv43pz4

4 0 0 + 12

4s 2 4sT 4s!

3 2 ±2, ±1, 0 + 12

3d 103dxyt 3dxzT 3dyzt 3dx2. y2t 3dz2t3dxv! 3dxzl 3dvz! 3dx2. v24 3dz2i

Table 2.3




When assigning quantum numbers to an electron, you must keep inmind that"no two electrons orbiting the nucleus of the same element have the same set ofquantum numbers." To assign quantum numbers, you must first describe theelectron in words. Consider an electron in the third shell. There are eighteenelectrons held in the third shell, so we need to be more specific. An electron inthe third shellcanbe in eitheran s-orbital, a p-orbital, or a d-orbital. Forsakeofargument, let's consider a p-orbital. There are three p-orbitals, each capable ofholding two electrons, sothere are six electrons that can befound in the 3plevel.Weneed to bemorespecific. Each p-orbital has a differentorientationin space(px along the x-axis, py along the y-axis, and pz along the z-axis). For sake ofargument, let's consider the p-orbital aligned onthe x-axis. This is the px-orbital.Two electrons can be found within a 3px-orbital, one with a magnetic spinmoment upward and the otherwith a magnetic spin moment downward. Bydescribing the electron as spin up, the electron is unique. There is only oneelectron that canbe spinup withina 3px-orbital. It tookfour terms to narrow itdown to a unique electron, hence there are four quantum numbers. Let'sconsider that same electron:

A 3px-spin up electron has the following description in words and thereforethese corresponding quantum numbers:

therefore, the principle quantum number (n) is 3therefore, the angularmomentum quantum number (/)is 1therefore, the magnetic quantum number (ny) is -1

Third shell

P-orbital

X-axis orientation

Spin up therefore, the magnetic spin quantum number (ms) is +^

The quantum numbers for the electron are: n=3,/ =1, ny =-1, and ms =+A. Theskill youmust (re)develop is getting the four numbers quickly. Determining nand / is relatively easy. The n-value is the shell number, so it is prettymuchagiven. The /-value is theorbital. Orbitals increase from s top to d tof, andsoon,and the /-values increase from 0 to 1 to 2 to 3 and so on. An /-value of 0corresponds to an s-orbital, an /-value of 1 corresponds to a p-orbital, and soforth. Thechallenging part is finding the ny and ms values. In all likelihood,you have no idea why x-axis orientation leads to the conclusion that ny = -1.That's just thewaythey do it. Justas x comes before y and z in the alphabet, -1comes before 0 and +1 numerically. Toget the ny and ms values,youmust drawout theenergy levels. The assignment ofny values for the p-orbitals is showninFigure 2-21.

Pxny = -l

Pym/ = 0

Figure 2-21

Pzny = +l

Chemistry convention tells us to number the different orbitals from -/ to +/sequentially. The middle orbital always has annyvalue of0. This is true for allorbitals. The assignment ofnyvalues for thed-orbitals is shownin Figure 2-22.

dxym/ = -2


dxzny = -l

92

dyzm/ = 0

Figure 2-22

dx2.y2ny = +l

dz2ny = +2

The Berkeley Review

Ueneral ChemiStry Atomic Theory Electronic Structure

Electrons are filled into orbitals one at a time from left to right, with spin upgoing in first, followed by spin down once each degenerate orbital has oneelectron. Filling spinup first is a convention anddoes not represent thephysicalreality of electrons. When an electron is spin up in an orbital, it has a spinquantum number (ms) of+1/2- When anelectron isspin down in anorbital, ithas ms =-1/2. For the 3px spin upelectron, the filling isshown inFigure 2-23.

Px Py pzny = -1 ny = 0 ny = +1

Figure 2-23

Theelectron falls into the first p-orbital, so ny is -1. Theelectronis spin up, so msis+V*

To be able to apply these quantum numbers, keep in mind that each electronwithin an element has a unique set of quantum numbers. An electron can bedescribed in terms of words (such as an electron in the second energy level in ap-orbitalwith x-orientationand spin up is a 2px T) or in terms ofnumbers (n = 2,/ = 1, ny = -1, mg = +V2)- Quantum numbers followarbitrary guidelines. Forinstance, the electron in a p-orbital oriented on the z-axis in the third shell withspin up has the quantum numbers n = 3 (for the third shell), / = 1 (because theelectron is inap-orbital), ny =+1 (for the z-axis), and ms =+I/2 (for spin up).

Example 2.12All of the following are true of an electron EXCEPT:A. electrons in a lower energy level can absorb energy and elevate to a higher

energy.B. exactly the same amount of energy is emitted when an electron relaxes

between the same states.C. there are many energy absorptions possible, but they are always of a

precisely known energy.D. an electron in the n = 1 energy level can be found at an infinite number of

distances from the nucleus.

SolutionOur model of atoms is a positive concentric nucleus surrounded by orbitingelectrons. These electrons may occupy only specific orbits, which have distinctenergies and pathways. According to this description, electrons in a lowerenergy level can absorb energy and elevate to a higher energy, so choice A isvalid and thus eliminated. According to this description, exactly the sameamount of energy is emitted when an electron relaxes between the same states, sochoice B is valid and thus eliminated. According to this description, there aremany possible energy absorptions possible, but they are always of an exactenergy, so choice C is valid and thus eliminated. According to this description,an electron in the n = 1 energy level is found at only one distance from thenucleus, not at an infinite number of distances. This makes choice D invalid, andthus the best answer.


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General Chemistry Atomic Theory Periodic Trends

PenodH

The Periodic TableThe periodic table isorganized according to valence electrons. Figure 2-24 showstheshell of the periodic table. The blocks arenamed after the lastelectron thatfills each respective atom in that section of the periodic chart. For instance, anelement in the D-block has its last electron (a valence electron) in a d-orbital.Understanding the blocks helps tounderstand theperiodic trends. The S-blockcomprises only metals. The D-block houses the transition metals, where periodicbehavior is not necessarily an obvious trend. The P-block includes metals andmetalloids in its left side and non-metals in its right side.

11

i i i i •

...§...blnck

•

: :p block :• i i i i• i i i •

'• '• '• Deblock! _i i i i _^^^^r^

• •

1̂ ^^^^

» • • • •

i 1i i

• i i i• i i i

t-block-

Figure 2-24

General Elemental Periodic TrendsPeriodic trends refer to any chemical behavior that can be matched to a trendwithin the periodic table. All chemical properties depend on the valenceelectrons, so periodic trendsultimately growout of valence electrontrends. Anyfeature of an atom that affects how tightly a valence electron is held contributestoperiodic trends. The two major factors arethe effective nuclear charge andthevalence shell, bothofwhichsupport periodic trends. Theyare listedbelow:

O As you move from left to right across a period in the periodic table, theeffective nuclear charge increases.

© Asyou descend a family in the periodic table, the valence shell increases, sothe distance of the valence electron from the nucleus of the atom increases.

The effective nuclearcharge(Zeff) is the net charge exerted upon the outermostelectrons (valenceelectrons). This value is empirically determined and takes intoaccount attraction due to the protons, shielding due to the neutrons, andrepulsion due to the core electrons. It is generally approximated as the protoncharge minus the electron repulsion. The effective nuclear charge affects howtightly the electrons are held, which affects the ionization energy, the electronaffinity, and the atomic radius. The effective nuclear charge increasesacross arow in the periodic table. Although we generally approximate the effectivenuclear charge, it canbe derived from the ionization energy. This procedureassumes that ionization energyis purely related to the effective nuclear charge,and fails to account forshell stability, particularlyfilled-octet stability.



Example 2.13Which ofthefollowing atoms has theGREATEST effective nuclear charge?A. CarbonB. FluorineC. SodiumD. Sulfur

SolutionEffective nuclear charge increases from left to right in the periodic table, so theelement in the column that is farthest to the right has the greatest effectivenuclear charge. Fluorine is to the right of carbonwithin the same period, so ithas a greater effective nuclear charge (Zeff). ChoiceA is eliminated. Sodium is inthe first column of the periodic table, so it has the smallest effective nuclearcharge. Choice C is eliminated. The correct answer is choice B.

Periodic trends depend on both the effective nuclear charge (affecting thestrength with which valence electrons are held) and the valence shell (affectingthe distance between electrons and the nucleus.) Periodic trends as we movefrom left to right across a row of the periodic table (period) are attributed toincreasing effective nuclear charge. Periodic trends as we move up through acolumn of the periodic table (family) are attributed to decreasing valence shells.The net result of these two effects is represented by the bold arrow shown inFigure 2-24. As we move along the pathway of the bold arrow, the followinggeneral atomic trends are observed:

O The atomic size decreases (the radius of the atom is defined as the distancefrom the center of the nucleus to the exterior of the valence electron cloud).

@ The ionization energy increases (the energy required to remove theoutermost electron from the atom).

© The electron affinity increases (the energetics associated with an atomgaining an electron).

0 The electronegativity increases (the tendency to share an electron withanother atom within a bond).

Example 2.14Which sequence accurately lists increasing ionization energy of the atoms?A. Br>F>Cl>TeB. 0>S>P>BrC. Br<F<Cl<TeD. 0<S<P<Br

SolutionThe appearance of fluorine in the middle of a trend should get your attention as awrong answer, eliminating choices A and C. Oxygen is directly above sulfur inthe periodic table, so oxygen has a greater ionization energy than sulfur. Thecorrect answer is choice B.


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None of the trends is uniform or perfect. The effective nuclear charge does notuniformly increase when we scan across a period. There are.deviations insometrends. One of the more common deviations is seen with electron affinity andionization energy, due to half-filled stability and filled-shell stability. Forinstance, nitrogen has a greater ionization energy than oxygen, because uponionization, nitrogen loses its half-filled p-shell. On the contrary, oxygen gainshalf-filled stability upon being ionized. The test writers may not prey on theseexceptions, but they certainly can emphasize the conceptual aspects by looking atthe factors that affect periodicity. For instance, rather than ask about atomicradius, theymay askabout ionic radius.

As a general rule, cations are smaller than neutral atoms, because the loss ofelectrons allows the atom to compact more tightly, given the diminishedrepulsion associated with the missing electrons. As a general rule, anions arelarger than neutral atoms, because the gain of electrons causes the atom toexpand, given the enhanced repulsion associated with the additional electrons.Valance electrons account for the size of anions, neutral atoms, and cations.Extra electrons repel and thus increase the atomic size ofananion, while a loss ofelectrons results in less repulsionand a smaller radius for a cation.

Example 2.15When strontium (Sr) becomes an ion, what is observed?A. It forms a +1 cation that is smaller than Sr.B. It forms a +1 cation that is larger than Sr.C. It forms a +2 cation that is smaller than Sr.D. It forms a +2 cation that is larger than Sr.

SolutionChoiceC is correct Strontium (Sr) is found in the secondcolumnof the periodictable. Alkaline earth metals lose two electrons to gain octet stability. As such,strontium carries a +2 charge, so choices A and Bare eliminated. Cations aresmaller than neutral species, because there are fewer electrons and thus lessrepulsion. Thismakeschoice C the best answer.

Example 2.16Which of the following ions is the LARGEST?A. Cl-B. Na+C. K+D. Br"

SolutionWithin a period, anions are larger than cations, so chloride (CI") is larger thansodium cation,and bromide (Br") is larger than potassium cation. This eliminateschoices Band C. Because Bris lower in the periodic table than CI,Br is larger asa neutral atom than CI. This same trend holds true, if both Br and CI pick up thesame number of electrons. In this case, both bromide and chloride picked up oneelectron each, so bromide, with its electrons in a higher valence shell, is largerthan chloride. The answer is choice D.


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Atomic RadiusThe atomic radius is thedistance from thecenter of thenucleus to theedgeof thevalence cloud of electrons. However, most of the empirical measurements ofatomic radii that exist are not from electron density maps, but instead aredetermined by dividing bond distances between like atoms in half. Because ofoverlapping electron clouds, this method does not generate a true atomic radius,but rather a covalent bonding radius. However, it does lead to internallyconsistent values. Atomic radii are measured in units of picometers. The radiusof an atom decreases as a family in the periodic table is ascended, because thenumber of electronic shells decreases. The radius of an atom decreases as aperiod in the periodic table is scanned from left to right, because the effectivenuclear charge increases. The trend is fairly uniform from left to right, with nodistinct exceptions due to half-filled stability.

Figure 2-25 lists the atomic radii of the first twenty elements. As a generalobservation, within a period, atomic radius decreases as the atomic numberincreases.

2.01

1.9

1.8

1.7-

1.6-

1.5"

-< 1-4'

2

•y i.nS•2 1.01<

0.91

0.8

0.7-

0.6-

0.5

0.4-

0.3-

0.2-

0.1

1.2-

H

Li

Be

N

He

Na

Mg

O

Ne

Figure 2-25

Al

Si

T 1 1 1 1 1 1 1 1 1 1—T—T—T—T—T—i 1 r1 2 3 4 5 6 7 8 7 10 11 12 13 14 15 16 17 18 19 20

Atomic Number

K

Ca

CI

Ar

Periodic Trends



The sudden increase in size from He to Li, Ne to Na, and Ar to K is attributed tothe expanded valance shell associated with the additional electron. For instance,the electronic configuration of Ar is ls22s22p63s23p6, while K isls22s22p63s23p64s1. The fourth shell (n =4quantum level) has a larger radius,so thepotassium atom is larger than the argon atom. Table 2-25 terminates atelement 20,becausefromelement 21 to 30 the radius stays roughly equal, sincetheelectrons are being added to the third quantum level (3d orbitals). This doesnot affect the radius of the electronshell drastically. The transition metals havevery similar atomic radii, although they arenotexactly equal.

The trend isconsistent through theelements listed, with theexception ofheliumand hydrogen. The larger atomic radius ofhelium when compared tohydrogengoes against the discussed trend ineffective nuclear charge. The best explanationfor this deviation involves both the shielding effect of the two neutrons in thehelium nucleus and the electron repulsion experiencedby electrons in the firstquantum level where they are closer together than in any other quantum shell.In other words, the electrons in the n = 1 level repel one another more thanelectrons in the n = 2 level, because they have the smallest interelectronicdistance. This repulsion forces the electrons away from one another, resulting ina greater area being occupied by the orbiting electrons. Keep inmind that theelectrons, notprotons orneutrons, define the radius ofanatom. The atomic radiiofatoms may beused topredict the bond length within molecules. The smallerthe atomic radius of the atom, the shorter the bond it forms when sharingelectrons with another atom. Shorter bonds are stronger bonds, so there exists acorrelation between an element's location in the periodictableand the strengthofthe bonds that element can form.

Example 2.17Which of the following elements has theLARGEST atomicradius?A. OB. FC. NeD. Na

SolutionThis is justa simple case of reading from the periodic table. The element in thelowest and furthest leftposition is sodium, Na, so choiceD is the best answer.


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Ionization EnergyIonization is the process of losing an electron from the valence shell. When anatom is ionized, it becomes a cation. The energy required to remove the outermostelectron from the valence shell is knownas the ionization energy. A genericreaction for ionization is shownbelow, whereErepresents anyelement.

E(g) E+(g) + e"The energy required to carry out ionization depends on the attraction of theelectron to the nucleus, its distance from the nucleus, and the stability of itselectronic configuration. Because severalfactors influence ionization energy, it istoo difficult to calculate,and thus it is generally evaluated in a qualitative sense.

Figure 2-26 lists the ionization energies of the first twenty elements in theperiodic table. Within a row in the periodic table, ionization energy increases asthe atomicnumber increases. This is a general trend, but with some exceptions.

2500

2400

2300

2200

2100

2000

1900

1800

^ 1700o 1600 \a^ 1500

^1400oS 1300c

" 12005•J3 1100

.a 1000

900

800

700

600

500

400

300

200

100

He

H

Be

Li

12 3 4 5 6

Ne

Ar

N

O CI

SiMg

Al CaNa

K

7 8 9 10 11 12 13 14 15 16 17 18 19 20

Atomic Number

Figure 2-26

Periodic Trends



Ionization energy for anelement generally increases asyou move left toright inthe periodic table. Notable exceptions occur when there ishalf-filled stability ofthe energy level and when there is an s2-shell. Ionization energy for an elementincreases asyou ascend a column in the periodic table, with the element higherup the column having the greater ionization energy. This is because as thenumberofelectronic shells decreases, the proximity ofan electron to the nucleusincreases, and thus the attraction to the nucleus increases. The sudden decreasein ionization energy from He to Li, Ne to Na, andAr to Kis attributed to theexpanded valance shell (and thus reduced attraction) associated with theadditional electron. For instance, the electronic configuration of Ne is ls22s22p6,while for Nait is ls22s22p63s1. The third shell (n=3quantum level) hasa largerradius, so the sodium atom canmore easily lose an electron than the neon atom(with its outermost electron being more attracted by the nucleus). As withatomic radius, from element 21 to element 30 the ionization energy remainsroughly equal, because the electrons are being removed from the same 4s-orbital.The effective nuclear charge on the4s-electrons does not change drastically. Theexceptions in the transition metals are also due to half-filled and filled d-shellstability.

Ionization energy may beused topredict the oxidation and reduction potentialsof an atom. The easier it is to ionize an atom, the easier it is to oxidize that atomby one electron. This leads to a larger (more positive) value for the oxidationpotential. Alow ionization energy for an atom correlates to a smaller (or morenegative) value for the reduction potential ofthecation thatis formed.

Example 2.18Why is the ionization energy of beryllium greater than the ionization energy oflithium?

A. Behas a larger principalquantum number than Li.B. Lihas a greater density than Be.C. Be has a largereffective nuclearcharge than Li.D. Li has a bigger proton count than Be.

SolutionBoth beryllium and lithium have their last electron (the electron lost uponionization) in a 2s-orbital. This eliminates choice A, because the principlequantum number (valence shell) is the same for both. Lithium is less massiveandlarger than beryllium, soit is less dense. This eliminates choice B. Lithiumhas three protons, while beryllium has four, so choice D is a false statement.Only choice Cremains. The difference between lithium andberyllium lies in theeffective nuclear charge. The beryllium nucleus has four protons, while thelithiumnucleus has onlythreeprotons. Thegreaternumber of protons increasesthe attractive pull on the electron. Because the pull is greater, the effectivenuclear charge isgreater. The greater the effective nuclear charge, thegreater theionization energy; therefore beryllium has a greater ionization energy thanlithium,because it has a greatereffective nuclearcharge than lithium.

TheMCAT testwriters design questions to encompass logical analysis. Totestaconcept andgenerate the illusion ofunfamiliarity, they can make subtle changesto a question. The second ionization energy can be tested in a question.Consider the periodic trends for elements, but remember thatwhen youevaluateions, reading from theperiodic table doesnot always givethe rightanswer.


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General ChemiStry Atomic Theory Periodic Trends

Example 2.19How does the ionization energy of sodium compare to the ionization energy ofmagnesium?A. Na has a greater ionization energy, because it has the greater effective

nuclear charge.B. Mg has a greater ionization energy, because it has the greater effective

nuclear charge.C. Na has a greater ionization energy, because it has the greater electronic shell.D. Mg has a greater ionization energy, because it has the greater electronic shell.

SolutionThis is just a simple case of reading from the periodic table. Both magnesiumand sodium are in the same period (third), so they have the same valence shell.This eliminates choices C and D. The element that is furthest to the right in theperiodic table is magnesium, so magnesium has the greater ionization energy.Choice B is the best answer, because Na and Mg are in the same row of theperiodic table, where effective nuclear charge is the reasoning behind periodicdifferences.

Example 2.20How does the second ionization energy of sodium compare to the secondionization energy of magnesium?A. Na has a greater second ionization energy, because it has the greater effective

nuclear charge.B. Mg has a greater second ionization energy, because it has the greater

effective nuclear charge.C. Na has a greater second ionization energy, because it has the smaller

electronic shell.D. Mg has a greater second ionization energy, because it has the smaller

electronic shell.

SolutionThe second ionization energy is the energy associated with losing the secondelectron, which takes the element from +1 to +2. For sodium, an octet is obtainedby losing the first electron, thus the second electron lost drastically destabilizesthe electron cloud. This makes the second ionization energy very high. Formagnesium, an octet is obtained by losing the first and second electrons, thus thesecond electron lost stabilizes the electron cloud. This makes the secondionization energy very low for magnesium, eliminating choicesBand D. Becausewe are talking about shell stability, the best (albeit not perfect) answer is choiceC.

Na+ • Na2+ Mg+ • Mg2*ls22s22p6 ls22s22p5 ls22s22p63s1 ls22s22p6

The electronic configurations show that sodium loses octet stability upon itssecond ionization, while magnesium gains octet stability upon its secondionization.

By asking about the second ionization energy, the concept rather than thememorization of periodic trends is tested.

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Electron AffinityElectron affinity measures the tendency of an element to gain an electron. It is ameasurement of theenergy absorbed or released when an electron is added intothe valenceshell. Electron affinity can be either negative or positive, meaningthat gaining an electron canbe either exothermic or endothermic. The genericreaction for electron affinity is shownbelow, whereE representsany element:

E(g) + e- E"(g)

Figure 2-27 lists the electron affinity for the first twenty elements in the periodictable. Within a rowin the periodic table, electron affinity correlates with atomicnumber, but there are some extremespikes in the trend.

250

225

200

175

150

125

^ 10°o 75

,*

fr

50

25

I °< -25

Jj "50u

jaj -75w

-100

-125

-150

-175

-200

-225

-250

-275

-300

-325

-350


Be

He

N

Li

H

12 3 4 5 6 7

Mg

Ca

Ne Ar

Na Al K

Si

O

CI

8 9 10 11 12 13 14 15 16 17 18 19 20

Atomic Number

Figure 2-27



The biggest deviations are attributed to the stability associated with a filled s-shell. The graph may seem confusing at first, because the lowest numbers havethe greatest electron affinity. The numbers listed are energy released upongaining an electron, so a negative number refers to an element with a highelectron affinity. Thesudden increase in electron affinity (energy released upongaining an electron) from Be to B, Mg to Al, and Ca to Ga is attributed to theinstability ofone electron in the p-level. For instance, upon gaining anelectron,the electronic configuration ofMg goes from ls22s22p°3s2 to ls22s22p63s23p1,which creates a new energy level, and is unfavorable. Upon gaining an electron,the electronic configuration ofNagoes from ls22s22p^3s1 to ls22s22p^3s2, whichfills the s-shell and generates stability. From element 21 to element 30, electronaffinity is erratic, because the d-shell stability is changing. No trend for electronaffinity is evident in the transition metals.

Like ionization energy, the energy associated with electron affinity depends onthe attraction of an electron to the nucleus, its distance from the nucleus, and thestability of its electronic configuration. Because several factors influence electronaffinity, the trend across a period is erratic. In general, an element releases moreenergy upon gaining an electron as you move left to right in the periodic table.Drastic exceptions occur when there is half-filled stability of the energy level andwhen there is an s2-shell. In general, an element also releases more energy upongaining an electron as you ascend a column in the periodic table. This is becauseas the number of electronic shells decreases, the new electron is closer to thenucleus, and thus the attraction to the nucleus increases.

Example 2.21The electron affinity of an element is MOST similar to which of the followingproperties?A. ElectronegativityB. Ionization energyC. Oxidation potentialD. Reduction potential

SolutionThe electron affinity for an element measures the energy associated with the gainof one electron. Choices B and C are out, because both of them deal with losingan electron. Electronegativity is not the best choice, because it deals with thesharing of electrons in a bond, not the gaining of an electron. The best answer ischoice D, reduction potential, because reduction is the gain of an electron. Anelement with a high reduction potential has a high electron affinity.


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ElectronegativityElectronegativity is a measure of an atom's tendency to gain and retain anelectron from a neighboring atom within a bond. It is formally defined as theability of an atom to attract towards itself the electrons in a chemical bond.Electronegativity is related toboth ionization energy andelectron affinity. Thisis to say that the electronegativity of an atom depends on both the electronaffinity andionization energy ofthat atom. Linus Pauling generated amethod tomeasure electronegativity, and created a scale, referred to as the Pauling scale.Electronegativity ismeasured ona relative scale, with thevalues measured fromthe electron distribution within a bond. The standards are 0.9 for sodium and 4.0for fluorine, and all other values are based on dipole moments associated withbonds to these atoms.

The electronegativity of an atom increases as the periodic table is ascended,because as the number of electronic shells decreases, causing the attraction to thenucleus to increase. The electronegativity of an atom increases as the periodictable is scanned from left to right, because the effectivenuclear charge increases.The trend in electronegativity is very clean, showingno exceptions. Figure2-28reflects these trends.

(860O)c2u

W

4.0

3.8

3.6

3.4

3.2

3.0

2.8

2.6

2.4

2.2

2.0

1.8

1.6

1.4

1.2

1.01

0.8

0.6

0.4

0.2


O

N CI

H

Si

Be Al

Mg

Li CaNa

K

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Atomic Number

Figure 2-28



When the electronegativity values of the two atoms within a bond are close, thebond is covalent. When the electronegativity difference exceeds 2.0, then thebond is ionic. Figure 2-29 represents the electron clouds and shows the relativeelectronegativity of the atoms involved in an ionic bond and a covalent bond.

Unequal sharing Relatively equal sharing(AEneg = 2.1, .\ the bond is ionic) (AEneg = 1.0, .*. the bond is covalent)

Figure 2-29

Example 2.22Electronegativity difference between bonded atoms is BEST determined by:A. measuring the bond length.B. measuring the dipole moment.C. calculating the difference in electron affinity between the two elements.D. calculating the difference in ionization energy between the two elements.

SolutionBecause electronegativity measures the tendency to share an electron, and thedipolemoment represents the degree of sharing between two atoms in a bond,the best answer is choice B. Electronegativity is related to both electron affinityand ionization energy, so electronegativity can be estimated knowing bothionizationenergy and electron affinity, but not just one of them. Choices C andD are eliminated, because you need both to approximate the electronegativitydifference. Choice A is eliminated, because bond length dictates bond strength,but not necessarily the relative electronegativity. This question may seemchallenging, because the terms are interconnected.

Example 2.23The trend in electronegativity increases with which of the following?A. Ionization energyB. Atomic radiusC. Atomic numberD. Number of valence electrons

SolutionThisquestion is close to verbatim in reproducing a question from a recentMCATexam. Electronegativity follows a cleanly predictable trend, so choice A iseliminated, because ionization energy follows an erratic trend. Electronegativityincreases as atomic radius decreases, so choice B is eliminated. Increasing atomicnumber sounds tempting; but when a new shell is formed, electronegativitydrops, while atomic number increases. This can be seen in going from fluorine toneon to sodium. This eliminates choice C. As the number of valence electronsincreases, we are moving from left to right across a period of the periodic table.Increasing valence electrons does not affect shells. From left to right in a period,electronegativity increases. This makes choice D the best answer.


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why not Ionization Energy?!?!?!

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Periodic Families (Groups)Columns in the periodic table (families) contain elements that have the samevalence electron count and thus show similar chemical reactivity. They arereferred to as groups of elements. Someof the groups you should know are thealkali metals (first column of H through Fr), the alkaline earth metals (secondcolumn of Be through Ra), the chalcogens (sixth column of O through Po), thehalogens (theseventh column ofF throughAt), and thenoble gases (last columnofHe through Rn). The lastcolumn of elements is observed to be generally non-reactive. They are the inert gases. The MCAT test-writers have been known toinclude a great deal of excess information in the passage. It is up to you torecognizewhat is pertinent and what is not.

Alkali Metals (Group I)Alkali metals are in the first column of the periodic table. Included are lithium,sodium, potassium, rubidium, cesium, francium, and to some extent hydrogen.Hydrogen can act as both a halogen and an alkali metal. The common feature isthat their valence shell isns1, where n isany integer greater thanone. Asneutralelements, they are strong reducing agents, because they readily lose an electronto become a +1 cation (which gives them a complete octet). They are some of thestrongest reducing agents (most favorably oxidized). They react with anycompound or element that has even a slight electron affinity. Their reactivityincreases as you descend the column, mostly because it is easier to lose an s-electron from a shell that is further out (greater n quantum number). Theircation form is very soluble in water with almost any anion.

All alkali metals react favorably with water to form the metal hydroxide andhydrogen gas. Reaction 2.1 is the generic reaction:

2 Mfe) + 2 H20(g; »~ 2 MOH(aq) + H2(g)

Reaction 2.1

The oxides they form are variable with the metal. Lithium forms an oxide(M2O), sodium forms a peroxide (M2O2), and potassium, rubidium, and cesiumform superoxides (MO2). Reactions 2.2, 2.3, 2.4, 2.5, and 2.6 show the oxidationreactions of the alkali metals:

4 U(s) + 02(g) • 2 Li20(s)

Reaction 2.2

2Na(s) + 02(g) • Na2C»2(s;

Reaction 2.3

K(s) + 02(g) • K02(s)

Reaction 2.4

Rb(s) + 02(g) • Rb02fs)

Reaction 2.5

Cs(s) + 02(g) • Cs02(s)

Reaction 2.6


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Besides reacting with oxygen, alkali metals are oxidized by halogens, nitrogen,and hydrogen. Reactions 2.7,2.8, and 2.9 are a sampling of these reactions.

6Li(s) + N2(g) • 2Li3N(s)Reaction 2.7

2Na(s) + H2(g) • 2NaH(s)Reaction 2.8

2Cs(s) + Br2(l) • 2CsBr(s)Reaction 2.9

Example 2.24What happens to sodium metal when it is added to water?A. It is oxidized to yield sodium hydroxide, which is insoluble in water.B. It is oxidized to yield sodium hydroxide, which is soluble in water.C. It is reduced to yield sodium hydride, which is insoluble in water.D. It is reduced to yield sodium hydride, which is soluble in water.

SolutionAs shown generically in Reaction2.1,a metal hydroxide is formed upon additionof an alkali metal to water. Because sodium is going from neutral to +1 whenlosing an electron to oxygen, sodium is oxidized. This eliminates choices C andD. You may recall from your acid-base chemistry experience that sodiumhydroxideis a strong base, and it readily dissociates in water. ThismakesNaOHvery soluble in water. The best answer is choiceB.

Alkaline Earth Metals (Group II)Alkaline earth metals are metals from the second column of the periodic table.Included are beryllium, magnesium, calcium, strontium, and barium. Mostberyllium complexes are covalent in nature. The commonfeature is that theirvalence shellis ns2,wheren is any integer greater thanone. Asneutralelements,they are strong reducing agents, because they readily lose two electrons tobecome a +2cation with a filled octet. However, they are not as reactive as alkalimetals. Like alkali metals, they are strong reducing agents. They too react withany compound or element that has a high electron affinity. Their reactivityincreases as you descend the column, because the first and second ionizationenergies both decrease. Their cation form is not as soluble in water as are thealkali metals, primarily due to their +2 charge and smaller radius.

Alkaline earth metals, except beryllium, react favorably with water to form ametal hydroxide and hydrogen gas. Reaction2.10 is the generic reaction:

M(s) + 2H20(g) *- M(OH)2(aq) + H2(g)Reaction 2.10

The alkaline earth metals all form oxides (MO) when oxidized by oxygen gas.The generic reaction is shown in Reaction 2.11.

2M(s) + 02(g) • 2MO(s)Reaction 2.11


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Besides reacting with oxygen, alkaline earth metals can also be oxidized byhalogens, nitrogen, and hydrogen. Reactions 2.12, 2.13, 2.14, and 2.15 are arandom sampling of these reactions:

Mg(s) + 2HCl(aq) • MgCl2(aq) + H2(g)

Reaction 2.12

3Sr(s) + 2N2(g) • 2Sr3N2(s)

Reaction 2.13

Be(s) + H2(g) • BeH2(s)

Reaction 2.14

Ba(s) + I2(g) • Bal2(s)

Reaction 2.15

Chalcogens (Group VI)Chalcogens aremetalloids and non-metals from the sixth columnof the periodictable. Included are oxygen,sulfur, selenium, tellurium, and polonium. Oxygen,sulfur, and selenium are non-metals, while tellurium and polonium aremetalloids. The common feature is that their valence shell is ns2np4. They formseveral covalent molecules with non-metals. As neutral elements, they areoxidizing agents, because they gain two electrons to become a -2 anion with afilled octet. However, their reactivity decreases as you descend the column,because the first and second electron affinities are not as great. They are ofteninsoluble, although it varies with their counterion (cation). Oxygen exists as adiatomic molecule (C^), sulfur and selenium exist as octatomic molecules (Ss andSes), and tellurium and polonium exist in vast molecular matrices.

Halogens (Group VII)Halogens are non-metals from the seventh column of the periodic table.Included are fluorine, chlorine, bromine, iodine, and astatine. The commonfeature is that their valence shell is ns2np5. They form covalent molecules withnon-metals and ionic compounds with metals. As neutral elements, they arestrong oxidizing agents, because they readily gain an electron to become a -1anion with a filled octet. However, their reactivity decreases as you descend thecolumn, because the electron affinity is not as great. They are often soluble,although it varies with their counterion (cation). They all exist as a diatomicmolecules (X2), although little is known of astatine due to its radioactivity.

Noble Gases (Group VIII)Noble gases are non-metals from the eighth (and last) column of the periodictable. Included are helium, neon, argon, krypton, xenon, and radon. Thecommon feature is that their valence shell is complete at ns2np6. For the mostpart, they form no bondsand exist asmonatomic atoms. Thanks to the workofNeilBartlettofU.C. Berkeley, xenonand krypton are known to form compoundswith halogens. The compounds with fluorine show more electron densityaround fluorine, implying that fluorine ismoreelectronegative than either xenonor krypton.


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General Chemistry Atomic Theory Light Absorption and Emission

Light Absorption and EmissionExcitation and RelaxationIt requires energy to ionize an atom, because one of its electrons is removed froma stable environment and placed into a less stable environment. However, whena quantum of energy less than what is required to ionize the element is absorbedby the element, then an electron can be excited to a higher energy state (known asan excited state). Different forms of energy may be absorbed to excite an electronup to a higher energy level. For instance, heat energy can be absorbed to excitethe electron. When the electron relaxes back toward its ground state, the energy isreleased as a photon. This is seen with fireworks. Some different forms ofenergy include light energy, thermal energy (usually vibrational kinetic energy),mechanical energy (usually translational kinetic energy), and electrical energy. Itis important to consider different energy forms, because there are many thingscapable of absorbing one form of energy and emitting a different form.

Absorption (Excitation): The gain of energy by an element or molecule, resultingin the excitation of an electron from a lower energy state (often the ground state)to a higher energy state (an excited state). The form of energy absorbed can vary.Emission (Relaxation): The loss of energy by an element or molecule, resulting inthe relaxation of an electron from a higher energy state (whichmust be an excitedstate) to a lower energy state (which can be the ground state or another excitedstate, but of lower energy). The form of energy emitted can vary.

Absorption of energy is defined as any process in which a photon is absorbed bya compound to excite an electron in the compound to an elevated electronicenergy level (excited state). Thismeans that the electrongoes to a higher energystate. Theperiod of time that the electronremains in thiselevatedenergylevel isreferred to as the lifetime of the excited state. A compound can store energy bymaintaininga high population of electrons in elevatedenergy states. When theelectronrelaxesback down to the ground state, energy is emitted in the form of aphoton. Absorption and emission are therefore oppositeprocesses. Figure 2-30shows the absorption and emission processes for a theoretical pair of energylevels. Because the energy levelsare equal, theenergyof thephotonabsorbedforexcitationis equal in energy to the photon released upon relaxation.

Excited state

hv absorbed

Ground state-

Absorption (Excitation)

Electron excitesup to a higherenergy level

Excited state

Electron relaxesdown to a lowerenergy level

Ground state-

hv emitted

Emission (Relaxation)

Figure 2-30


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In this generic example, the energy of the photon absorbed is equal to the energyof the photon emitted. In actuality, there ismore than one singular energy levelfor the ground state and the excited state due to the coupling of electrical energylevelsand rotationalenergy levels associated with the atom. Transitions involvea change in electronic state as well as a change in the rotational energy of theatom. In molecules, there are vibrational energy levels to consider, in addition tothe rotationaland electronic energylevels. The result is that the ground state andexcited state exists as a band of energy levels, not a single level. This means thatmultiple possibilities exist for the energy of transition. Rather than a single lineabsorption or emission spectrum being formed, a range is formed. This is whyFigure1-2 is a broad peakand not a sharp line. Figure2-31 shows different typesof emission and absorption spectra where the color range is shown. Absorptionspectra show all light exceptwhat was absorbed, which appears as a black line,due to the absence of light. They are black lines in a rainbow. Emission spectrashow only the emitted light. They are colored lines against a dark background(due to only selected frequencies being emitted). They are stripes of color.

Absorption spectra

Wavelength

Emission spectra

Wavelength

Figure 2-31

The lifetime of an excited state is often in the picosecond to nanosecond range, sothe interval of combined excitation and relaxation is very fast. The conversionbetween energy forms is exploited in many devices. Incandescent bulbs absorbthermal energy resulting from the resistance associated with electrical flow andemit light energy (although some energy is dissipated in the form of heatthrough conduction and convection). The operation of tube lighting involves thisconversion process. Plate charges build up at each end of the tube, creating anelectric field. A cation in the gas tube is accelerated by the electric force. As itcollides with other gas molecules, its kinetic energy is transferred to the othermolecule, which absorbs the energy to excite an electron. Upon relaxing, theenergy is released as light. In reality, the tubes must run on alternating current(AC), so that the cation never reaches a plate. This is why tube lights, such asfluorescent tubes, are actually high-frequency strobe lights.


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Atomic Spectrum of HydrogenHydrogen is highly studied in terms of its absorption of light. Separate samplesof hydrogen gas exhibit exactly the same emission and absorption spectra. Thismeans that all hydrogen molecules absorb the same frequency of photons. Therepeatability of the hydrogen spectrum is attributed to the quantization of theenergy transitions. What is meant by "quantized energy transitions" is that theabsorption and emission of light by hydrogen occurs in distinct increments (orquantities). Because hydrogen has distinct transitions between energy levels, itmust also have distinct energy levels. However, with the large number of energylevels, there are a large number of transitions that are possible. For hydrogen,groups of transitions are named for the energy level to which they relax. Forinstance, all transitions down to the n = 1 level are categorized into the Lymanseries and emit photons in the ultraviolet region of the EM spectrum. Selectedenergy levels and transition series for hydrogen are shown in Figure 2-32.

n = l

Figure 2-32

n = co

_ymanseries

These electronic transition series names are specific to hydrogen. Of the fourseries shown in Figure 2-32, only the Balmer series emits photons in the visiblerange. When a sample of hydrogen gas glows, it is because electrons are relaxingdown to the n = 2 level, which is still an excited state. Figure 2-33 shows the linespectrum associated with the Balmer series:

Balmer series emission spectrum

^8-2 h-2 V2Wavelength

Figure 2-33

The Balmerseries emits photons in the visible range. In the Balmerseries, A.3-2 is656 nm, A.4.2 is 486 nm, A5.2 is 380 nm and A^-2 is 410 nm, so they fall in thevisible range of the spectrum. This also means that A4-2 and A5.2 fall in the

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visible range. The Paschen series involves photons in the infrared region of theEM spectrum. The Brackett series involves photons in the low infrared andmicrowave regions.

Example 2.25Given that A.3.2 is 656nm, which of the following is NOT true?A. All Balmer series transitions are less than 800 nm.B. All Lymanseries transitionsare less than 400nm.C. The lowest energy Paschen transition is greater than 800nm.D. The highest energy Brackett transition is less than 800nm.

SolutionThis requires looking closelyat Figure 2-32.Choice A: Transitions to the n=2 level fall into the Balmer series. The lowestenergy transition in the Balmer series is from n=3 to n=2 and is given as 656nm.All other transitions in the Balmer series are more energetic, so these transitionsemit photons with a wavelength less than 656nm. ChoiceA is a valid statement.Choice B: Transitions to the n=l level fall into the Lyman series. The lowestenergy transition in the Lyman series is fromn=2 to n=l, and it is of significantlyhigher energy than the n=°° to n=2 transition. This means that all photons in theLyman series are of higher energy and shorter wavelength than the transitions inthe Balmer series. According to Figure 2-33, the Balmer series emits photonsnear 400nm, so transitions in the Lyman series emit photons with a wavelengthless than 400nm. Choice B is a valid statement.

Choice C: Transitions to the n=3 level fall into the Paschen series. The lowestenergy transition in the Paschen series is from n=4 to n=3, and it is ofsignificantly lower energy than the n=3 to n=2 transition. This means that thelowest energy photon in the Paschen series is of lower energy and higherfrequency than the transitions in the Balmer series. The Balmer series emits aphoton at 656nm, so the lowest energy transition in the Paschen series emits aphoton with a wavelength greater than 656nm, and according to Figure 2-32,higher than 800nm. Choice C is a valid statement.

Choice D: Transitions to the n=4 level fall into the Brackett series. The highestenergy transition in the Brackett series is from n=°° to n=4, and it is of muchlower energy than the n=3 to n=2 transition. This means that the highest energyphoton in the Brackett series is of lower energy and higher frequency than thetransitions in the Balmer series. The Balmer series emits a photon at 656nm, sothe lowest energy transition in the Brackett series emits a photon with awavelength greater than 656nm, and according to Figure 2-32, higher than800nm. Choice D is an invalid statement, and thus is the correct answer choice.

Electromagnetic SpectrumPhotons have associated with them a distinctive frequency, energy amount, andwavelength. The relative energetics of various radiation must be known. If youever have difficulty recalling the electromagnetic radiation spectrum, think aboutthe fact that you wear sunglasses to protect your eyes from ultraviolet radiation,not visible radiation. This means that ultraviolet radiation is of higher energythan visible light. Also, lead shields are used to filter out x-rays, so they must beof even higher energy than ultraviolet photons. Conversely, we are fairly safefrom radio waves considering the fact that millions of radio waves pass throughus every second. Common sense can be applied to recall the relative spectrum.Table 2.4 lists the common EMranges in terms of frequency and wavelength.

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EM Radiation Wavelength (m) Frequency (Hz) Common Usage

AM 103 -104 105 to 106 Cheaper radiocommunication

FM 1-102 107 to 108 Expensive radiocommunication

Microwave 10-4 -1 108 to 1012 Satellite, cell phones,radar, heating water

Infrared 10-6 _10-3 10H to 1014 Line-of-sight com,molecular ID, heat

Visible 4xl0-7-7xl0-7 4.3xl014-7.5xl014 Vision, fiber opticcommunication

Ultraviolet 10-8 . io-7 1015 to 1016 Bond-breaking,exciting electrons

X-ray 10"12-10-8 10l6 to 1020 Nucleus detection,core e" ionization

Gamma 10-15_10-IO 1018to 1023Nuclear excitation,world destruction

Table 2.4

Common practical uses of EM radiation are something that should be learned,because of the tradition of test writers asking about such devices. It should becommonknowledge that radio waves and low-energy microwaves are used incommunications. Microwave communications include satellite transmissions,cable television, cellular phone networks, and airport landing systems.Microwaves are also used for cooking and heating things that contain water,because the frequency required to rotate a water molecule is found in themicrowave region. Remote control units (also known as line-of-sightcommunicators) oftenuse infraredsignalsto send information. Thefrequency ofthe EM radiation used in these common tools and appliances to which we aredirectly exposed cannot be near 1700 cm"l, 3000 cm'l, or 3500 cm'l, because thoseare the bond-stretching frequenciesassociatedwith molecularbonds common toall animal life. Ultraviolet light is used to breakbonds,and x-rays are used forimaging atomic structure (including bones and teeth).

Visible Spectrum and ColorsThe visible spectrum runs from red (around 700 nm), orange, yellow, green, blue,throughviolet(around400 nm). The relative wavelength, frequency, and energyofphotonsin the visible spectrumof light (bydecreasing wavelength, increasingfrequency, and increasing energy) is given in Figure2-34.

Red < Orange < Yellow < Green < Blue < Violet700 to 627 run 627 to 594 nm 594 to 561 nm 561 to 477 nm 477 to 438 nm 438 to 400 nm

Figure 2-34

The order of the colors of the visible spectrum is often recalled using themnemonic ROYG. BiV, an acronym for the spectrum of visible light in sequenceof increasing energy. It is also important to know that the visible range of light isfromapproximately 700 nm for red light to approximately 400 nm forvioletlight.Ultraviolet spectroscopy ranges from 20 nm to 400 nm.



Example 2.26For some substance X, AEi yields yellow light, while AE3 yields blue light. Whatmust be true of the photon associated with AE2?A. It is probably orange.B. It has a wavelength of 700nm.C. It is probably green.D. It has a wavelength of 400 nm.

SolutionAE2 falls between AEi and AE3, so the photon emitted must be between yellowand blue light in the visible spectrum. According to the spectrum mnemonic(ROYGBiV), green light falls between yellow light and blue light. This eliminateschoice A and makes choice C the best answer. A wavelength of 700 nmcorresponds to red light, and a wavelength of 400nm corresponds to violet light,eliminating both choice Band choice D.

Color is a phenomenon associated with vision. Light in the 400 nm to 700 nmrange can be detected by the eye and processed in the brain. Color is the result oflight within this range being absorbed by the cones of the eye. Rods also detectlight (in the green range), but this is for the purpose of intensity analysis ofdetails in the image. Consider the eye to function like a visible lightspectrophotometer, analyzing stimuli for intensity and wavelength. There arethree kinds of cones, each one responsible for detecting different frequencyranges of light. Color, in a photon sense, can result from one of two phenomena.Color is perceived from either the emission of light at a specific frequency or thereflection of light at a specific frequency. Hence, there are two types of color toconsider: emitted and reflected.

Emitted ColorColors is emitted from a source that radiates visible photons. Atoms emit lightenergy when an electron relaxes from an excited state to a lower level. Figure 2-35 illustrates the concept of emitted color.

^\/\/W» r v ; -zam!\/\/\Ar^

Figure 2-35

An electron relaxes from an excited state and releases a photon of somewavelength. All of the photons in Figure 2-35 are of essentially the samewavelength. The color that is observed is the color emitted. Simply put, withemitted color, "What you get is what you see." Emitted colors can be seen atnight, because their source is emitting light. A good test to determine whether acolor is an emitted color is to ask whether it can be seen in the dark. If theanswer is yes, then it is an emitted color. Examples of emitted light sourcesinclude: neon light tubes, television screens, fireworks, and glow-in-the-dark ink.



Reflected Color and the Color WheelReflected colors are produced when incident light (we'll consider it to be whitelight, for the sake of simplicity) strikes a surface, and the surface absorbs certainfrequencies, thus reflecting only a portion of the incident light to the eye. Thereflected radiation appears as a color that is a combination of the reflectedphotons. The color we observe is the complementary color of the frequency thathad the highest intensity of absorption. This is where the color wheel can help.Complementary colors are on opposite sides in the color wheel. Given that whitelight is a combination of all colors, it is a combination of all complementary pairs.This is a over-simplification, because the primary colors of light are differentfrom the primary pigment colors, but we shall use the two interchangably to aidin answering chemistry questions. Complementary pairs include red and green,blue and orange, and yellow and violet. A color wheel is shown in Figure 2-36.

The color wheel is used to determine complementarycolors. Red and green are opposing on the color wheel,so they are complementary colors. Moving around thewheel, violet is the complementary color of yellowand blue is the complementary color of orange.

Figure 2-36

Chlorophyll appears green in the presenceofwhite light. Theconclusion is thatchlorophyll containsa pigment that has a maximum intensity absorbance of redlight, the complementary color of green. One type of chlorophyll has anabsorbance maximum at 740 nm, while another has an absorbance maximum at680 nm. Both of these values correspond to red light. Figure 2-37 illustrates theconcept of emitted color as it relates to chlorophyll.

Incidentwhite light

Reflected color(appears green)

( Chlorophyll sample )(absorbs red) \ = 700± nm

Figure 2-37

Reflected colors can be seen only in the presence of an external light source. If acolor cannot be seen in the dark, then it is a reflected color. Examples of reflectedlight sources include paint pigments, pen inks, and fabric dyes.



Example 2.27Which of the following statements is NOT true?A. A yellowpaint pigmentabsorbs lightwith a Xmax of 411 nm.B. A green lightbulb emits lightwith a Xmax of 522nm.C. A blue shirt has a dye that absorbs light with a A,max of 611 nm.D. A television screenglowingviolet emits light with a A.max of 573 nm.

SolutionThis question entails comparing reflected and emitted colors. The first thing youshould ask yourself is whether the color can be seen at night. If the answer isyes, then it is an emitted color and what you see is the color involved in theelectronic transition (emission). If the answer is no, then it is a reflected colorand what you see is the complementary color of the light involved in theelectronic transition (absorption). Choice A is a paint pigment, which cannot beseen at night. The yellow color is a reflected color, so according to the colorwheel, violet light has been absorbed. Violet light has a wavelength near 400 nm,so choice A is viable. Choice B is a light bulb, which can be seen at night. Thegreen color is an emitted color, so green light has been emitted. Green light has awavelength between 561 nm and 477 nm, so choice B is valid. Choice C is afabric dye, which cannot be seen at night. The blue color is a reflected color, soaccording to the color wheel, orange light has been absorbed. Orange light has awavelength between 627 nm and 594 nm, so choice C is valid. Choice D is aradiating screen, which can be seen at night. The violet color is an emitted color,so violet light has been emitted. Violet light has a wavelength near 400 nm, sochoice D is invalid, and thus the best answer to this question.

The absorption of energy followed by the emission of a photon of light is a usefulprinciple that can be applied to many forms of spectroscopy. Infraredspectroscopy can be used in organic chemistry to identify bond types andfunctional groups. It is also involved in UV-visible spectroscopy, where acompound is subjected to incident light at a specific frequency, and the relativeintensity of a transmitted beam compared to a reference beam is measured.Conclusions about the concentration and the kind of compound itself can beinferred from the results. Classical experiments that involve measuringquantities of components in a gaseous mixture often focus on analysis of theemission and absorbance spectra of the gases. This is at the heart of atomicabsorption spectroscopy, used to analyze amongst other things planetary gases.

The MCAT is not printed in color, so questions involving color will presentnumerical data for wavelengths or graphs of absorbance spectrum. Figure 2-38shows an absorbance spectrum and its analysis.

Abs = e[C]l

Copyright © by The Berkeley Review 116

\nax is the wavelength at the greatestabsorbance. A Xmax at570nm means thatyellow light is absorbed, so the compoundappears violet to the human eye.

570 Wavelength (in nanometers)

Figure 2-38

The Berkeley Review

General ChemiStry Atomic Theory Light Absorption and Emission

FluorescenceFluorescence at the simplest level is the conversion of ultraviolet light into visiblelight. A material with a semi-stable excited state can absorb an ultravioletphoton, relax a slight amount giving off infrared photons, and then relaxcompletely back to its original state. The visible photon emitted is of slightly lessenergy than the ultraviolet photon absorbed, because of the small loss of energywhen the infrared photon was lost. The next energy range down from ultravioletis violet light in the visible range. This is why fluorescing materials appearpurple. The term "black light" refers to an ultraviolet light (it emits light outsideof the visible range) that when shined upon fluorescing material appears purple.You may have seen this in organic chemistry lab when you analyzed tic plates orat social gatherings with fluorescent hand stamps. This concept does not need tobe interpreted at any higher level, so this rough explanation is adequate.Fluorescence will be addressed in a passage.

Photoelectric EffectThe photoelectric effect is exactly what the name implies. An incident photoncauses the release of an electron. It is a fancy way of saying that a compound canbe ionized with a photon, as long as the photon has energy greater than theionization limit of the material. The energy required to remove an electron fromthe surface of the solid material is referred to as the work function, (J). Excessenergy (energy in excess of the work function) becomes kinetic energy for theelectron that is emitted. The photoelectric effect is really quite simple, butconfusion sometimes arises from the different terminology used in chemistry andphysics. Don't be fooled; just say no to physics (just kidding, but I'm a chemistrynerd and just had to say that). Equation 2.9 is the equation for the photoelectriceffect.

hv= <t> + i/2mv2 (2.9)

hv = energy to ejectelectron + kineticenergy of ejected electron (2.10)

The photoelectric effect is the principle at work in solar panels. Photons of alldifferent wavelengths and energy are emitted by the Sun. This energy can becollected at the Earth's surface. Some solar photons are sufficiently energetic toeject an electron when they strike a metal surface. Todo so, the photonmust beof an energy high enough to remove the electron from the valence shell of themetal. The loss of an electron from the metal creates electrical flow (electricity),which can be converted to either potential energy (stored in an electrochemicalcell) or mechanical energy (used to power a device).

As the energy of the incident photon increases, the kinetic energy of thedischarged electron increases. Figure 2-39 shows an apparatus devised tomeasure the photoelectric effect, while Figure 2-40 is a graph of the electron'skineticenergy as a function of the frequency of the incident photon.



Incident beam of light

Detector of KE31

c

co1-

o

w

Ejected electrons

Stopping Voltage

Figure 2-39

hv = hv0 + KEe.

v0: threshold frequency

Frequency of incident photon (Hz)

Figure 2-40

The threshold frequency corresponds to the photon equal in energy to theionizing energy. Any photon with a frequency less than the threshold frequencydoes not have enough energy to ionize the material, so no electron is ejected.

Example 2.28Which material probably has the LOWESTthreshold frequency for thephotoelectric effect?A. BoronB. CarbonC. SodiumD. Sulfur

Solution

The threshold frequency corresponds to the work function of the material (<J> =hv0). Each atom holds onto its valence electrons with a different work function,therefore each atom requires a different amount of energy to undergo ionization.The lowest threshold frequency for the photoelectric effect corresponds to thematerial that is easiest to ionize. This question is asking for the material with thelowest work function. Of the choices, only one is a metal, so only one has a lowwork function. The best answer is choice C, sodium. This answer correspondswith lowest ionization energy as well.



Nuclear ChemistryNuclear ParticlesSimply put, nuclear chemistry is the chemistry that the nucleus of an atom canundergo. Instead of referring to it as reactivity, it is referred to as instability of thenucleus. The science community knows little of nuclear behavior at this point inhistory, but there are theories. From our perspective, however, all we willconcern ourselves with is nuclear decay and nuclear capture. The process ofparticle loss from the nucleus that results in a different nucleus is referred to asnuclear decay. In physics, it may also be referred to as fission. The process ofparticle gain by the nucleus which results in a different nucleus is referred to asnuclear capture. In physics, it may also be referred to as fusion. As a generalrule, particles with mass less than 56 amu undergo fusion, while particles withmass greater than 56 amu undergo fission. Particlesthat we consider to be lost orgained are alpha particles (helium nucleus), beta particles (an electron), positrons(a positively charged particle with the mass of an electron), neutrons, andneutrinos (an uncharged particle with the mass of an electron). The nucleus alsohas ground states and excited states associated with it (just as the orbitingelectrons have energy levels), so nuclei can also undergo photon absorption andemission. The high-energy photon is a gamma ray. Table 2.5 lists some of thecommon nuclear processes. You should treat these reactions as exercises insimple algebra. The total mass and number of protons should be equal on bothsides of the reaction. This is because mass and charge are conserved (along withmomentum) in each process.

Process

a-Decay

fi-Decay

fi+-Decay

y-Emission

oc-Capture

fi-Capture

fi+-Capture

y-Absorption

Reaction Tracking

8* -> fa +!J|Y

w -> -ae + }?z

W -> ¥* +!$Q

'§8x* -> hv +i§8x

l^X+^lgA

I8x -fe -> ^Q

gX + }e -> igz

igjX + hv -» !§8X*

Notes

The mass works: 120 = 4 +116.The proton number works: 50 = 2 + 48.The mass works: 120 = 0 + 120.The proton number works: 50 = -1 + 51.The mass works: 120 = 0 + 120.The proton number works: 50 = 1 + 49.Mass and proton number do not change.The nucleus relaxes to emit a photon.The mass works: 120 + 4 = 124.The proton number works: 50 = 2 + 48.The mass works: 120 + 0 = 120.The proton number works: 50 + -1 = 49.The mass works: 120 + 0 = 120.The proton number works: 50 + 1 = 51.Mass and proton number do not change.The nucleus absorbs a photon to excite.

Table 2.5

Nuclear Decay and CaptureA heavy element undergoes a decay process to increase its nuclear stability. Alight element undergoes a capture process when struck by a high-energy particleto increase its nuclear stability. There are a few points you should note for theseprocesses. With C-decay, the electron may or may not be added to the electricalshell of the atom. If it is, the atom remains neutral. If the electron escapes, the

Nuclear Chemistry


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General Chemistry Atomic Theory Nuclear Chemistry

compound becomesa cation. Usually the particle escapes, and an ion is formed,althoughwe typically ignore this fact in nuclearchemistry. The representationofthe beta particle as having a -1 for its Z term is not accurate, but it works for thealgebra of the equation. Beta capture and positron decayboth result in the sameelementbeing formed. Likewise, beta decay and positron capture both result inthe same element being formed. The positron behaves just as a beta particledoes, only it carries a positive charge and is an anti-electron. This makes it antimatter, but for our purpose that doesn't matter.

The reactions in Table 2.5 are simplified. We have ignored conservation ofenergy and conservation of momentum. To balance these more accurately, themass of the electron (beta particle) and positron (anti-electron) cannot be treatedas zero. A neutrino is required in some cases to balance the equation. Solvingdecay and capture questions in the manner that these examples are presentedwill work fine on test questions you may see. The questions you may see willinvolve deterrnining the particle that was lost in a process, or determining thefinal elemental product after a nuclear reaction has taken place. It is sort of oddthat if you reverse the first two letters in nuclear chemistry the phrase becomesunclear chemistry. Given the sciencecommunity's level of understanding at thepresent time,unclear is an accurate depiction of the subject.

Example 2.29Whichof the following elements results from two consecutive alpha decays of21°At?A. 210FrB. 206BiC# 202jiD. 2<>2Bi

SolutionAlpha particles are helium nuclei (alpha-helium sounds like alpha-helix, soremember the phrase "alpha helium"). An alpha particlehas a mass of 4 amu,thus losing two alpha particles results in the loss of 8 amu. The element thatremains after two consecutive alpha decays has four fewer protons than theoriginal element (from 85 to 81) and has a mass eight less than the originalelement (from 210 to 202 amu). At is element number 85, so the final elementmust have atomicnumber 81. Consulting a periodic table shows that element 81isTl. 202jj ismebeS£ answer, sopick C.

Example 2.30Electron captureby a nucleus resultsin whichof the following?A. An increase in atomicnumber by one, and a mass increase of one amu.B. An increase in atomic number by one, and no change in atomic mass.C. A decrease in atomicnumber by one, and a mass decrease of one amu.D. A decrease in atomic number by one, and no change in atomic mass.

SolutionElectron capture involves thegainofan electron by thenucleus. Anelectron hasa -1 charge and no mass. This means that the element should experience adecrease in its positive charge by one, and no change in its mass. This is bestdescribed in choice D.


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iieneral ChemiStry Atomic Theory Nuclear Chemistry

Example 2.31Which of the following processes would NOT result in the formation of tritium?A. Alpha decay of7LiB. Positron emission by3HeC. Beta decay of4HeD. Gamma emission from 3H

SolutionAnalphaparticle is a heliumnucleus(4He), sowhen7Li loses a helium nucleus,the mass decreases by four to three and the number of protons decreases by twotoone. The final product is tritium (3H). Choice Aisvalid. Apositron isamass-less positively charged particle that when lost, converts a proton into a neutron.The mass remains the same, but the 3Heconvert into tritium (3H). Choice Bisvalid. Beta decay involves the loss of an electron from the nucleus. This convertsa neutron into a proton. The mass does not change and thus remains at four.The number of protons increases by one to three. The product is 4Li, which isNOT tritium. The best answer is choice C. Gamma emission does not change theparticle, thus choice D is valid.

Example 2.32Inthe conversion from 165Ho to157Gd, at least two alpha particles were emitted.What else was emitted?

A. A third alpha particleB. One neutronC. One beta particleD. One positron

SolutionInconverting from 16^Ho to ^Gd, themass decreases byeight, which equatesto the mass of two alpha particles. This means that any other particles emittedare massless. This eliminates choices A and B. The atomic number has decreasedby three. Loss of two alpha particles decreases the atomic number by four, so theother particle emitted must increase the atomic number by one. Loss of apositron decreases the atomic number by one, so choice D is eliminated. A betaparticle must also have been emitted in addition to the two alpha particles.Choice C is correct.


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Half-LifeThe half-life ofa sample ofmaterial is theperiod of time required for50% oftheconcentration ofmaterial to decay to a different (possibly stabler) form. It ismostcommon to see half-lives associatedwith first-order decay processes. The graphsshown inFigures 2-41 and 2-42 show the concentration ofa component over timefor first-order decay (exponential decay) and zero-order decay (linear decay).The graphs shown inFigures 2-43 and 2-44 are the derivative graphs ofFigures2-41 and2-42, which represent thechange in rateas a function ofchange in time.

d[X]dt

First-order decayi(exponential decay)

[X]Zero-order decay

lear decay)

Time • *• Tune *•

Figure 2-41 Figure 2-42

\. First-order decay\fexponential decay)

rate =d[X]dt

d[X]dt

n

Zero-order decay(linear decay)

T" te.Tune •

Figure 2-43

i nne ^

Figure 2-44

In Figures 2-45 and 2-46, the graphs aremarked at 50% and 25% of the initialconcentration values. These are the various concentrations after consecutive half-lives. It takes one half-life to cut the concentration in half, and a second half-lifeto cut that concentration in halfagain, whichresultsin one-quarter of the originalconcentration. Figure 2-45 shows half-life as a function of time for a first-orderdecay. Figure 2-46 shows half-life as a function of time fora zero-order decay.

25%- •


First-Order Decay(exponential decay)Half-life is constant

time

Figure 2-45

122

25%"

Zero-Order Decay(linear decay)

Half-life decreases

time

Figure 2-46

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Ueneral ChemiStry Atomic Theory Nuclear Chemistry

Figure2-45 demonstrates that for a first-order process, the half-life is constant,regardless of concentration. As the concentration decreases, the duration of thehalf-life remains the same. Figure 2-46 demonstrates that for a zero-orderprocess, the half-life is directly proportional to the concentration of thecompound. As the concentration decreases, so does the duration of the half-life.The second half-life is half as long as the first half-life,because the concentrationchange is half as large. As a point of interest, for a second-order process, thehalf-life is indirectly proportional to the concentration of the compound. As theconcentration decreases, the length of the half-life increases.

These graphs should look familiar from biology and physics. In biology,bacterialgrowth is exponential, and enzyme kinetics can be either exponential orlinear. In physics, displacement, velocity and acceleration follow similarpatterns. The point is that you must understand what these graphs tell youabout some values increasing or decreasing as a function of time. Half-lifeanalysis is most commonly evaluated using the upper graphs.

On the MCAT,you can expect to have to solve some half-lifequestions. As far ashalf-life problems are concerned, there are two methods by which they aresolved. One method is to use Equation 2.11, the first-order decay equation:

Q =C0e-kt (2.11)

where Q is the concentration at a given time, C0 is the initial concentration, k isthe rate constant for the decay process, and t is the elapsed time. This works butrequires the use of natural logs, which can be time-consuming.

The second method involves the sequential summing of half-lives required toreach a specific concentration. The test traditionally uses numbers that work wellwith this method. It is the most time-efficient method. When a problem askshow much time passes until a certain percentage of the original quantity remains,it is easiest to figure how many half-lives are required to reach that percentageand then convert the quantity of half-lives to total time. When using thismethod, use reasonable approximations. For instance, if the half-life is 110years,then 335 years is three half-lives plus a little bit. There is 50% remaining after thefirst half-life, 25% remaining after the second half-life, and 12.5%remaining afterthe third half-life. Thus, there is just under 12.5%remaining after 335 years.

Example 2.33How many half-lives are required to decompose 93.75%of some material?A. 2B. 3

C. 4D. 5

SolutionDecomposing (decaying) 93.75% of the initial material results in 6.25%remaining. The quick and easy method is to cut the percentage in halfcontinually until 6.25 is reached.

100% -> 50% -> 25% -> 12.5% -> 6.25%

It requires a total of four half-lives to reach a point where 6.25% remains, whichis a decomposition of 93.75% of the material. The best answer is choice C.


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Example 2.34If the initial concentration of a toxic material is 1500parts per million and thehalf-life is4.5 days, how long will it take for thelevel to reach a concentration of100 ppm?A. 13.6 daysB. 17.8 daysC. 18.2 daysD. 22.3 days

SolutionWe mustsequentially halve 1500 untila number close to100 is reached.

1500 ppm -> 750 ppm -> 375 ppm -> 187.5 ppm -» 93.75 ppmAfter four half-lives, there is a little less than the target value of 100 ppmremaining. This means that just under four half-lives are needed to reach 100ppm. Four half-lives areeighteen days, so thebestanswer is slightly less than18days. Thebest answer is choiceB.

Sequential halving oftheconcentration in first-order decay questions is far easierthan usingEquation 2.11 or other rate equations. The test-writers may giveyouthe appropriate equations needed to solve the question in longhand, but thismethodshouldbe yourmethod of choice. Justbecause the test writersprovideyouwitha piece ofinformation, doesnotmeanthat youneed to use it.

This technique can alsobe applied to exponential growth problems, as seen inbiology withpopulation genetics that obey exponential growthbehavior. Withgrowth, you continually double the concentration rather thancut it in half. Forinstance, if a bacterial population doubles in 12 seconds, then in one minute itincreases 32times (25).

X -* 2X -> 4X -> 8X -> 16X -» 32X

Example 2.35If the half-life ofa particular atomis 20.4 minutesand the initialconcentration is1200 partspermillion, whatwilltheconcentration beafterexactly onehour?A. 582 ppmB. 291 ppmC. 155 ppmD. 145 ppm

SolutionOnehour is not exactly equalto threehalf-lives (three half-lives = 1hour and 1.2minutes). After three half-lives, we have the following:

1200 ppm -> 600 ppm -> 300 ppm -» 150ppmWe haven't quite used a full three half-lives, so the concentration should nothave decreased all the way down to 150 ppm. This means that the correctanswershouldbe justover150 ppm (150+ ppm). Theonlyanswer choice that isslightly greater than 150ppm is 155ppm, choiceC.


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AtomicStructure

Passages14 Passages

100 Questions

Suggested Atomic Structure Passage Schedule:I: After reading this section and attending lecture: Passages I, IV - VI, & VIII


U: Following Task I: Passages II, III, VII, IX, & XIV, (34 questions in 44 minutes)Time yourself accurately, grade your answers, and review mistakes.

m*: Review: Passages X - XIII & Questions 96-100Focus on reviewing the concepts. Do not worry about timing.

R-E-V-I-E»WSpecializing in MCAT Preparation

Atomic Theory Study Passages

I. Classical Experiments

II. Isotopic Abundance and Average Atomic Mass

III. Bohr Model of Hydrogen

IV. Paramagnetism and Liquid Crystal Displays

V. Migration through a Membrane

VI. Ionization Energy

VII. Transition Metal Trends

VIII. Lasers

IX. Paint Pigments

X. Fluorescence and Phosphorescence

XI. Flame Test

XII. Glyphosate

XIII. Technetium Decay

XIV. Cold Fusion

Questions Mot Based on a Descriptive Passage

Atomic Structure Scoring Scale


84 - 100 13- 15

66 -83 10 - 12

47 -65 7 -9

34-46 4-6

1 -33 1 -3

(1 -8)

(9- 14)

(15 - 21)

(22 - 28)

(29 - 35)

(36 - 42)

(43 - 49)

(50 - 56)

(57 - 63)

(64 - 69)

(70 - 76)

(77-81)

(82 - 88)

(89 - 95)

(96 - 100)


Two of the more famous classical experiments inestablishing the properties of subatomic particles are theThomson experiment and the Millikan oil drop experiment.TheThomson experiment determined that the mass-to-chargeratio for an electron is 5.686 x 10*12 kg/C. The Thomsonexperiment involved the application of a perpendicularelectric field to a beam of electrons produced by a cathode raytube. The Thomson apparatus is shown in Figure 1.

Double filterUpper cathode plate

ITm

Lower anode plate

Electron beam

Figure 1The double filter ensures that the electron beam is in a

single direction. The electron beam, when exposed to theelectric field generated by the two plates, is bent towards thepositively charged plate. The curvature is constant as long asthe field strength is constant. The bending of the electronbeam is attributed to both the attractive force of theoppositely charged plate and the repulsive force of the likecharged plate. A magnetic field may also be used to bend theelectron beam. When using a magnetic field to deflect theelectrons, the radius of the arc of the electron beam can beemployed to determine the mass-to-charge ratio usingEquation 1.

m _Brq v

Equation 1In Equation 1, the mass of the electron is m, the charge ofthe electron is q, the strength of the magnetic field is B, theradius of the arc of the electron pathway is r, and the velocityof the electron is v.

In the Millikan oil drop experiment, after ten years ofdata collection, it was determined that the charge of anelectron is 1.6x 10"19 C. The experiment involved chargingan oil drop by bombarding it with an ionizing electron beam.Several oil drops passed through the ionizing beam, but onlya few gained a charge. The oil drops then fell into an electricfield generated by two charged plates in a glass cylinder. Thefield was aligned so that the lower plate repelled the oil dropupward, while the upper plate pulled the oil drop upward.

The force upward on the oil drop is thus qV, where q isthe charge of the oil drop and V is the voltage of the field.The force pulling the oil drop downward is mg, where m isthe mass of the oil drop, and g is the gravitational forceconstant. If the oil drop is suspended, then the two forces areequal, and it is possible to determine the charge of the oildrop. The charge of the oil drop must be a whole numbermultiple of the charge of an electron. The apparatus used inthe Millikan oil drop apparatus is shown in Figure 2.


Oil can

XSuspendedcharged oil drop

Figure 2The hole in the top plate allows for oil droplets to fall

into the chamber. The field between the two plates isassumed to be uniform, so that the force on the oil droplet isequal at any position between the two charged plates. Thefield is adjustable, so that droplets of variable mass can besuspended. Not all oil droplets are of equal mass, so anaverage value is assumed.

1. An assumption of the Millikan oil drop experiment iswhich of the following?A. Oil drops are naturally charged.B. The pull of gravity is equal in all directions at the

microscopic level.C. The mass of electrons is negligible compared to the

mass of the oil drop.D. The magnitude of charge for an electron can vary.

2. The electron beam bends downward as it passes betweenthe two charged plates in the Thomson experiment.What can we expect for a beam of neutrons andprotons?A. The proton beam would arc upwards as it travels,

while the neutron beam would not arc at all.

B. The neutron beam would arc upwards as it travelswhile the proton beam would not arc at all.

C. Both the neutron and the proton beams would arcupwards as they travel between the plates.

D. Neither the neutron nor the proton beam would arcas they travel between the plates.

3. How could a value of 3.2 x 10"19 C in an oil drop inthe Millikan oil drop experiment be explained?A. The oil drop was doubly ionized.B. The oil drop gained an electron rather than lost an

electron.

C. The oil drop lost a proton rather than lost anelectron.

D. The oil drop gained a proton rather than gained anelectron.


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Saad Azam

Saad Azam

4. The Thomson experiment determined which of thefollowing?A. The mass of an electronB. The mass of a protonC. The mass-to-charge ratio of an electronD. The mass-to-chargeratio of a proton

5 . The Millikan oil drop experiment requires knowing allof the following EXCEPT:A. the mass of the oil droplet.B. the voltageof the electric field.C. the gravitational force constant.D. the temperature of the chamber.

6. In a neutral atom, what can be said about the number ofelectrons relative to the number of protons andneutrons?

A. The number of electrons must equal both thenumberof protons and the number of neutrons.

B. The number of electrons must equal the number ofprotons only, and the number of neutrons isirrelevant.

C. The number of electrons must equal the number ofneutrons only, and the number of protons isirrelevant.

D . The number of electrons must equal the sum of theprotonsand neutrons combined.

7. The STRONGEST attractive electrostatic force is feltbetween:

A. a proton and a neutron.B. a neutron and an electron.C. an electron and an electron.D . an electron and a proton.

8. Ifanoildropwere slowly descending once it was in theelectric field betweenthe twochargedplates, then whichof the following changes would NOT stop it fromcontinuing to fall?A . Increase the number of electrons on the drop.B. Decrease the number of electrons on the drop.C. Increase the voltage difference between the two

plates.D . Offset the gravitational pull withan electric force.



To determine the relative abundance of the isotopes foran element, a mass spectrometer is employed. The elementis first vaporized into its gaseous state, if it is not already ingas form. Because this is carried out in a vacuum, a hightemperature may not be required to vaporize the material.The cationic compound is generated by high-energy electronimpact with the neutral atoms. A neutral atom loses avalence electron to ionization upon impact. The cationicspecies formed is then accelerated in an electric field until itreaches a set velocity (v). The particle beam produced isprojected through a double filter to ensure that no deflectedparticles pass through to the detector.

Once particles are passed through the double filter, theyaresubjected to a magnetic field perpendicular to the originalelectric field. This curves the pathway of the particles. Theforce on charged particles moving through a magnetic field isfound using Equation 1.

F = qvBEquation 1

where q is thecharge of theparticle (assumed to be +1) and Bis themagnetic field strength. According to Newton's secondlaw, the force on the particle is F = ma, where m is mass anda is acceleration. The particle is moving along a curved path,so it has angularacceleration, which is found using Equation2.

a =

R

Equation 2where R is the radius of the arc. Substituting ma for F inEquation 1andmanipulating thevariables yields Equation 3.

qB=mvR

Equation 3The equations support the intuitive perception of how

the path is affected by mass, charge, velocity, and externalfield strength. Figure 1 shows a basic schematic of a massspectrometer.

Detector

High energyelectron beam

Double

Gaseousatoms

Atomicions x x x x x

Magnetic field

Figure 1Different isotopes have different atomic masses, so each

isotope has a unique radius (R). The detector determines therelative quantities of each isotope. This information is thenused to determine the average atomic mass, a value that takesinto account all of the isotopes for a particular element.


Saad Azam

Table 1 lists the isotopic abundance for the common isotopesof boron, magnesium, and silver.

Element Atomic Mass PercentageBoron-10 10.013 amu 19.78%

Boron-11 11.009 amu 80.22%

Magnesium-24 23.985 amu 78.70%



Silver-107 106.91 amu 51.80%

Silver-109 108.90 amu 48.20%

Table 1

9. Why is carbon added to a sample before it is vaporizedin a mass spectrometer?A. Carbon-12 is used as a standard for mass reference.B. Carbon-13 is used as a standard for mass reference.

C. Carbon helps to lubricate the atoms at an atomiclevel.

D. Carbon absorbs heat from the atoms and preventsthe sample from overheating.

10. Which of the following plots represents the outputfrom the mass spectrometer for magnesium?A. B.

11. Which of the following calculations accuratelydetermines the average atomic mass for silver?A. 0.787 (23.985) + 0.1013 (24.986) + 0.1117 (25.983)B. 0.1117 (23.985) + 0.1013 (24.986) + 0.787 (25.983)C. 0.482 (106.91) + 0.518 (108.90)D. 0.518 (106.91) + 0.482 (108.90)

Copyright © by The BerkeleyReview® 129

12. When a mass spectrum is to be obtained for a solidcompound, the first step in the process is which of thefollowing?A. VaporizationB. Condensation

C. Ionization

D. Sublimation

13. Identify the TRUE statement(s) from the following:I. The average atomic mass of an element is always

greater than the mass of the heaviest isotope forthat element.

n.

ffl.

A.

B.C.D.

The average atomic mass of an element is alwaysgreater than the mass of the lightest isotope forthat element.

When there are two isotopes for an element, theaverage atomic mass is closer to the more abundantisotope than the less abundant isotope.I onlyII onlyI and III onlyII and III only

14. The difference between the two isotopes of boron is:A. the isotope of boron with mass 11.009 amu has

one proton less than the isotope of boron withmass 10.013 amu.

B. the isotope of boron with mass 11.009 amu hasone proton more than the isotope of boron withmass 10.013 amu.

C. the isotope of boron with mass 11.009 amu hasone neutron less than the isotope of boron withmass 10.013 amu.

D. the isotope of boron with mass 11.009 amu hasone neutron more than the isotope of boron withmass 10.013 amu.



The Bohr model for hydrogen proposes that there arequantized energy levels in which an orbiting electron may befound. The theory is derived from the observation that exactfrequencies of light are absorbed by hydrogen gas, no matterwhat the concentration of the sample or intensity of theincident light source. Given that light frequency directlyrelates to light energy, the idea is that quantized amounts ofenergy are absorbed and emitted asanelectron moves betweenenergy levels. Figure 1 shows the Bohr representation ofhydrogen.

n = 1

Energy levels

Orbital levels

Figure 1

Due to Coulombic attraction, the most stable energylevels for the electron correspond to the closest proximity tothe nucleus. This means that the energetics of the orbitingelectron depend on the charge nucleus and the distance fromthe nucleus (or quantum level). The energy level for theelectron of hydrogen is found using Equation 1.

E=-2.178 x 10"I8|^|J

Equation 1

15. What is the ionization energy for hydrogen?

A. 1.634 x 1018 JB. 2.178 x 1018 JC. 4.356 x 10",8JD. 8.712 x 10*,8J

16. ThedeBroglie equation relates theenergyof a photon toits equivalent mass according to E=mc2. How can thedeBroglie wavelength of an electron be found?

A. x =-h-mc

B. X = mch

C. X=^->2

D

mc'

2h


17. Which of the following transitions relates to theGREATEST energy of absorption?A. From the n = 2 level to the n = 4 levelB. From the n = 2 level to the n = 1 levelC. From the n = 1 level to the n = 2 levelD. From the n = 2 level to the n = 5 level

18. Which statements are valid regarding the energy levelsof hydrogen?I. As the value of n increases, the electron is farther

from the nucleus, therefore the electron is presentin a higher energy state.

II. As the value of n increases, the difference betweenadjacent energy levels (with n values differing byone), gets smaller.

III. A larger nuclear charge has no effect on transitionenergies, although it lower all of the energy levels.

A. I onlyB. II onlyC. I and II onlyD. II and III only

19. For some compound X, which emits orange light for atransition from the n = 4 level to the n = 3 level, whatmust be TRUE?

A. A transition from the n = 5 level to the n = 4 levelemits green light.

B. A transition from the n = 4 level to the n = 2 levelemits red light.

C. A transition from the n = 6 level to the n = 4 levelemits violet light.

D . A transition from the n = 5 level to the n = 3 levelemits yellow light.

20. Which of the following observation leads to theconclusion that energy levels are quantized?A. A sample always absorbs the same amount of

light.B. A sample always absorbs the same color of light.C. A sample always emits different frequencies at

different temperatures.D. A sample always emits different frequencies at

different concentrations.



Liquid crystal displays (LCDs) utilize the properties ofparamagnetic species. When plane-polarized light passesthrough a solution containing a paramagnetic species, itrotates the plane-polarized light. If the paramagnetic speciesis subjected to a magnetic field, then the particles align in thefield, and the plane-polarized light is not rotated. The displaycommonly seen in calculators and digital watches uses thisprinciple to filter or pass light through templates.

The basic design of the display board involves passinglight through a polarizing plate into a cell containing theparamagnetic species and then having the light pass througha second polarizing plate that is rotated 90' relative to thefirst plate. If the field is off, the light will rotate and thus bepassed through the second polarizer. This results in a brightspot on the display board. If the field is on, the light willnot rotate and thus will not pass through the secondpolarizer. This results in a dark spot on the display board.Figure 1 shows a simple display circuit.

Circuit on (Solution is aligned)

Vertical polarizer(only vertical waves pass)

Vertical polarizer(only vertical waves pass)

Horizontal polarizer(only horizontal waves pass)

light passes

Aligned paramagnetic solution does NOT rotate light plane

Circuit off (Solution is random)

Horizontal polarizer(only horizontal waves pass)

Light passes

Unaligned paramagnetic solution rotates light plane

Figure 1

A paramagnetic species is defined as a compound thathas unpaired electrons. Many common examples involvetransition metals with a d-shell count of d* tod9. Organiccompounds that are paramagnetic are referred to as radicals.Oxygen gas has two unpaired electrons; thus, when exposedto a magnetic field, oxygen molecules will with the field.


21. Which of the following can be used in an LCD?

A. CdCl2B. F2C. CoCl2-6H20D. H2

22. What are the quantum numbers associated with theLAST electron in the nitrogen atom?

A.n =2 1= 1 mj =+l ms =+J-

B.n =2 1= 1 mj =+l ms =-I

C.n =2 1= 1 mi = -l ms =+J-

D. n=2 1=1 mi =-l ms =-^

23. What is the shape ofFe(CN)63"?A. Octahedral

B. Square planarC. Tetrahedral

D. Trigonal bipyramidal

2 4. Which of the following compounds is polar?A. FeCl63-B. Ni(CN)53-C. cis-Pt(NH3)2Cl2D. trans-Pt(NH3)2Cl2

2 5. Which ion has the MOST unpaired electrons?A. Cr3+B. Mn2+C. Cu+

D. Ni2+


26. Which of the following statements is NOT true?A. Cr has a higher formal charge in Cr(CN)6 than

Cr(NH3)63+.B. Cu(H20)6+ has ten d-electrons.C. Fe(NH3)63+ has the same shape as FeCl63".D. Scandium is more likely to have +4 formal charge

than titanium.

27. Which of the following pairs of molecules does NOThave the central atom (transition metal) with the samenumber of d-electrons in both compounds?

A. Fe(NH3)63+andFeCl63-B. Co(H20)63+ and MnCl64-C. Cr(NH3)63+andMoCl63"D. Os(NH3)42+ and RhCl3(PR3)3



In vivo, semi-permeable membranes are used by thebody to control diffusion andosmosis (entropically favorableoccurrences). In the kidney, Bowman's capsule functions insuch a manner. Semipermeable membranes havemicroscopic pores that allow the passage of somemoleculesor, in certain cases such as the sodium-potassium pump,ions. They function by allowing only selected ions to passfrom the solution on one side of the membrane to thesolution on the other side of the membrane. Semi-permeablemembranescan distinguish by size, charge, or both.

Semipermeable membranes can be applied to bench topchemistry (in vitro) as well. This application can increase areaction's selectivity. Semipermeable membranes used invitro, that separate by size, are often referred to as molecularand atomic sieves. While some molecular sieves differentiatebyatomic radius, others differentiate bycharge. Tostudy theeffectiveness of a semipermeable membrane, twosolutions ofdifferentsoluteconcentrations are separatedby themembrane.Figure 1 shows a U-tube with a semipermeable membraneseparating the two solutions.

v

Sidel

Side 1 starts with solute thatcannot cross the membrane.

Side 2 starts with solute thatcan cross the membrane.

-Semipermeablemembrane

Side 2= c=

V:j:::i:i:-:j::&ii| /

After the solute migrates across themembrane, there is a concentrationdifference. Flow stops when osmoticpressure equals hydrostatic pressure.

Figure 1Table 1 the rates of effusion from Solution I through a

synthetic membrane into Solution II for a generic study.

Cation Rate (M/s)Li+ 4.7 x 10'2Na+ 1.1 x 10-2Mg2+ 2.9 x 10"2K+ 3.5 x 10'3Ca2+ 8.3 x 10'3

Table 1


Note: The synthetic membrane used in the experimentrepresented in Table 1 is believed to be aprotic anddiscriminates by atomic size. It is modeled after a proteinthat is found in the lipid bilayer.

28. Based on the data in Table 1, which of the followingconclusions is valid?

A. As cationic size increases, the rate of migrationacross the membrane decreases.

B. As cationic size increases, the rate of migrationacross the membrane increases.

C. The cationic size has no direct effect on the rate ofmigration across the membrane.

D. Because cations are larger than neutral atoms, theeffect of cationic size on the rate of migrationcannot be determined.

29. The BEST explanation for why potassium (K) has alower ionization potential than sodium (Na) is that:A. potassium is more metallic than sodium.B. potassium has a larger enthalpy of reaction with

water than does sodium.

C. the 4s-orbital is larger than the 3s-orbital.D. potassium has a larger electron affinity than does

sodium.

3 0. The BEST explanation for why fluorine is smaller thancarbon is which of the following?A. Fluorine's electrons are at a lower quantum level

than those of carbon.

B. Filled p-orbitals contract more than half-filled p-orbitals.

C. Fluorine has a larger effective nuclear charge thancarbon.

D. Fluorine is heavier than carbon.

31. Which of the following is the SMALLEST?A. F-

B. Ne

C. Na+

d. cr


32. What is the charge on the MOST stable ion ofmagnesium?A. +2

B. +1

C -1D. -2

3 3. Which of the following atoms is LARGEST?A. F

B. CI

C. Br

D. I

34. If the pore of the membrane used in the experimentassociated with Table 1 were to distinguish by both sizeand charge, the SLOWEST rate would occur with whichcation?

A. Li+

B. K+

C. Mg2+D. Ca2+


Saad Azam


2500-

2400-

2300-

2200-

2100-

2000-

j» 1900-

1 1800-3 1700-^ 1600-£1500-£ 1400-c 1300-

ts 120°-jjj 1100-.2 1000-

900-

800"

700-

600-

500-

400-

8 9 10 11 12 13 14 15 16 17 18 19 20Atomic Number

Figure 1

Figure 1 shows the first ionization energy for the firsttwenty elements in the periodic table. The first ionizationenergy is defined as the energy required to remove the firstelectron from the valence shell of an element to form amonocation. The general reaction, where E symbolizes anyelement from the periodic table, is shown in Reaction 1.

E(g)—>E+(g) + e-Reaction 1

To negate solvent effects, ionization reactions are donein the gas phase. The first ionization energy is proportionalto the square of the effective nuclear charge of the element.The effective nuclear charge is the charge exerted upon thevalence electrons by all other charged species in the atom.The effective nuclear charge takes into account the numberofprotons in the nucleus and the number of core electrons(electrons in principalquantum shells lower than the valanceelectrons). Ionization energy depends on the valence shellprincipal quantum number. Ionization energy decreases asyou descend a family in the periodic table. A larger principlequantum number corresponds to a lower ionization energy.

Figure 1 shows deviations from the behavior predictedfrom effective nuclear charge data. Deviations are seen withelements having valance electronic configurations of ns2np1and ns2np4, which is attributed to the filled s-level and thehalf-filled p-level associatedwith the cations that are formed.There is additional stability when the p-orbitals have equaloccupancy of electrons. The symmetry associated with thehalf-filled state results in cation stability.


35. Which of the following BEST explains the highionization energy of helium?A. Helium has no valance electrons to remove.

B. Helium has an effective nuclear charge greater thantwo.

C. The electron must be removed from the firstquantum level, which experiences the greatestnuclear charge.

D. Helium requires a great amount of energy to beconverted first into the gas phase.

36. What can be expected for the second ionization energyof lithium, sodium, and potassium?A. The second ionization energies are only slightly

larger thanthefirst ionization energies, because theZeffhas not changed drastically.

B. The second ionization energies are only slightlylarger than thefirst ionization energies, because thevalance shell has changed.

C. The second ionization energies are substantiallylarger than thefirst ionization energies, because theZeffhas not changeddrastically.

D. The second ionization energies are substantiallylarger than thefirstionization energies, because thevalance shell has changed.


3 7. How can the lower ionization energy for oxygen thannitrogen BEST be explained?A. Nitrogen is more electronegative than oxygen, so it

does not share electrons as readily.B. Nitrogen has more protons than oxygen, so it

carries a greater effective nuclear charge.C. Nitrogen is larger than oxygen, so it is harder to

remove a valance electron.D. Nitrogen has a half-filled p-level, so it does not

lose an electron as readily. Oxygen, when it losesone electron, attains half-filled p-level stability.

38. How can the trend in first ionization energy betweenaluminum, silicon, and phosphorus BEST beexplained?A. The effective nuclear charge increases as you move

from Al to P in the periodic table.B. The effective nuclear charge decreases as you move

from Al to P in the periodic table.C. The number of core electrons increases as you

move from Al to P in the periodic table.D. The number of core electrons decreases as you

move from Al to P in the periodic table.

39. Why is the first ionization energy of fluorine greaterthan the first ionization energy of chlorine?A. Fluorine is less electronegative than chlorine, so it

shares electrons more readily than chlorine.B. Chlorine has more protons than fluorine, so it

carries a greater effective nuclear charge.C. Chlorine is losing an electron from a shell that is

farther from the nucleus than fluorine, so chlorinecan lose an electron more easily.

D. Chlorine has no electrons in the p-level, so it doesnot lose an electron. Fluorine, when it loses oneelectron, completes its valance shell.

40. Which of the following elements can be oxidizedMOST easily?A. Sulfur

B. MagnesiumC. Boron

D. Argon


41. What can be expected for the first ionization energies ofbromine, krypton, and selenium?

A. LE.Br > IE.Kr > IE.SeB. I.E.Kr > LE.Br > IE.SeC. I.E.Se > I.E.Br > LE.KrD. I.E.Se > LE.Kr > LE.Br

4 2. The first ionization energy for sodium is 495 kJ/moleand the second ionization energy is 4558 kJ/mole. Thefirst ionization energy of magnesium is 732 kJ/mole,and the second ionization energy is 1451 kJ/mole.How can the change in second ionization energy beexplained?A. The first and second ionizations of sodium are from

different quantum levels, while the first and secondionizations of magnesium are from the samequantum level.

B. Sodium is less electronegative than magnesium,which implies that the second ionization energymust be greater.

C. Once one electron has been lost from magnesium,the second electron is repelled by the positivecharge of the magnesium cation.

D. Sodium metal gains half-filled stability by losingone electron, while magnesium requires losing twoelectrons to attain half-filled stability



Transition metals do not follow the same periodic trendsassociated with the main-group elements. The reasoningbehind their separate trend is rooted in their valenceelectrons.The d-electrons of transition metals are not the outermostelectrons, despite the fact that the d-level is being filled asyou move left to right through any row of the transitionmetals. When we look at first-row transition metals, we seeelectrons fill the levels according to the Aufbau principle, sothey fill the 4s-level prior to the 3d-level. However, whenelectrons are lost (when the element undergoes ionization),the electrons are lost from the 4s-level before the 3d-level.Table 1 shows the ionization energy data associated with thefirst row of transition metals. The first three ionizationenergy valuesare listed for eachof the transition metals.

M M2+ M3+ 1st IE 2nd IE 3rdIE

Sc 4s23d' — [Ar] 631 1235 2389

Ti 4s23d2 3d2 3d* 658 1309 2650

V 4s23d3 3d3 3d2 650 1413 2828

Cr 4sJ3d5 3d4 3d3 652 1591 2986

Mn 4s23d5 3d5 3d4 717 1509 3250

Fe 4s23d6 3d6 3d5 759 1561 2956

Co 4s23d7 3d? 3d6 760 1645 3231

Ni 4s23d8 3d8 3d? 736 1751 3393

Cu 4s13d10 3d9 3d8 745 1958 3578

Zn 4s23d10 3d*0 —922 1772 —

Table 1

Periodic trends also include atomic radius, ionic radius,and atomic density at 25°C. Transition metals also show adeviation from main group elements in terms of trends inatomic size. Table 2 lists radius data for selected metals.

Radius Physical Properties

M M2+ M3+ M.P. B.P. Density

K 227 — — 63.7 760 0.86

Ca 197 99 — 838 1441 1.54

Sc 162 — 81 1539 2738 3.0

Ti 147 90 77 1668 3259 4.51

V 134 88 74 1903 3447 6.1

Cr 130 85 64 1874 2667 7.19

Mn 135 80 66 1241 2152 7.43

Fe 126 77 60 1536 3001 7.86

Co 125 75 64 1495 2908 8.9

Ni 124 69 — 1453 2731 8.92

Cu 128 72 — 1083 2597 8.96

Zn 138 74 —419.5 906 7.14

Table 2


Trends are predictable when based on the effective nuclearcharge and the outermost electron shell (valence shell).However, the effective nuclear charge and valence cloud of thedifferent transition metals do not follow the same pattern asmain block elements. Similar reasons can be employed toexplain deviations in trends. Half-filled d-shell stability isassociated with chromium, and filled d-shell stability isassociatedwith copper, which accounts for deviations in theirelectronic configurations from what is expected according tothe Aufbau principle.

43. What is the density of manganese at 1000°C?A. Between 1.00 and 2.00 g percm3.B. Between 2.00 and 7.43 gpercm3.C. Between 7.43 and 9.00 gper cm3.D. Between 9.00 and 12.00 g per cm3.

44. What is true of the relative second ionization energyvalues of Mo, Tc, Pd, and Ag?A. Mo>Tc>Pd>AgB. Ag>Pd>Tc>MoC. Ag>Tc>Mo>PdD. Ag>Pd>Mo>Tc

45. How can the lower melting point and boiling point forzinccompared to other transition metalsbe explained?A. Increased interatomic forces with the filled 3d-shell.B. Reduced interatomic forces with the filled 3d-shell.C. Increased interatomic forces with the half-filled 3d-

shell.D. Reduced interatomic forces with the half-filled 3d-

shell.

4 6. What can be concluded about the ease of removing a 3d-electroncompared to a 4s-electron?A. The 4s is more easily removed because it is closer

to the nucleus.B. The 4s is less easily removed because it is closer to

the nucleus.C. The 4s is more easily removed because it is farther

from the nucleus.D. The 4s is less easily removed because it is farther

from the nucleus.


47. What is TRUE about changes in density and atomicradius as one scans across the 3d row of the periodictable?

A. As atomic radius increases, density increases.B. As atomic radius increases, density decreases.C. As atomic radius increases, density increases,

except for elements with filled d-shell stability.D. As atomic radius increases, the density does not

vary in a predictable fashion.

4 8. How can the reduced radii of transition metal cations beexplained?A. Transition metals lose their outermost electrons

from the 4s-orbital, when becoming a cation.B. Transition metals lose their outermost electrons

from the 3d-orbital, when becoming a cation.C. Transition metals lose their highest-energy

electrons from the 4s-orbital, when becoming acation.

D. Transition metals lose their highest-energyelectrons from the 3d-orbital, when becoming acation.

49. If the reduction potential ofMn2+ is-1.18 Vand Co2+is -0.28 V, what can be concluded about the reductionpotential ofFe2+?A. E'reductjon ofFe2+ is less than -1.18V.B. E0reduction ofFe2+ is between -1.18 and -0.73 V.C. Eduction °f Ee2+ is between -0.73 and -0.28 V.D. E0reduction °f Ee2+ 1S greater than -0.28 V.



The basic principle behind the operation of a laser is theexcitation of electrons from the ground state of a compoundto some meta-stable excited state. The term meta-stablerefers to an excited state where an electron can exist for asustained period of time before it falls back to its groundstate. By constantly supplying energy to the compound, theelectrons can be "held" in the excited state. This preferencefor a higher energy level is in accordance with Boltzmann'sdistribution law of energy for any chemical or physicalsystem. Boltzmann's distribution law predicts that the energyof a system will be distributed throughout the different levelsaccording to a standard probability function that follows abell curve. At higher total energy for the system, moreelectrons are present in an excited state than at lower totalenergy for the system. Figure 1 shows a molecular systembefore and after energy has been added. The excited andground states are represented as levels capable of holdingmultiple electrons.

hvin

Before energy is added After energy is added

II i i i ii in i ii

Meta-stable excited state

ii tiilllilL IIGround state energy level

Figure 1Energy is most easily introduced into the system by

incident light (photons). A system analogous to the laser isthe electrical capacitor, which can be filled with electricalenergy and then instantly drained, releasing a pulse of energy.Lasers can also store energy (light energy) and then release apulse of photons. A laser emits light energy correspondingto the transition-level changes. Once the crystal or gas in thelaser releases photons, the light is collected and reflectedbetween mirrors, before it is released through an apertureemitting the laser beam. Light emitted from a laser is closeto beingmonochromatic (that is, light of one wavelength),but not precisely monochromatic. To date, absolutelymonochromatic light has yet to be observed. This isattributed to the fact that the ground state and excited statesare not single energy levels, but are in fact a band ofquantized levels with identical electronic energy, butvaryingvibrational and rotational energies.

5 0. Which of the following electronic configurations is theground stateelectronic configuration foraluminum?A. ls22s22p63p3B. ls22s22p63s3C. ls^s^p^s^p1D. ls22s22p63s!3p2


51. Monochromatic light is NOTobserved for which of thefollowing reasons?A. Lasers are not precise in their emissions.B. Crystals always have some imperfections.C. The laser re-absorbs some of the energy emitted.D. Each electronic energy level has many rotational

sub-levels.

5 2. What is the energy in joules associated with a photon oflight with a wavelengthof 330 nanometers?

(E =be., where h=6.6 x 10"34, X, =wavelength in nm)X

A. 6.0 x 10"28 JB. 6.0 x 10*,9JC. 6.0 x 10'17 JD. 6.0 x 1015 J

53. Which of the following sets of quantum numbers areassociated with the last electron in neutral vanadium(element # 23)?

--J.2

__L2

A. n = 3; 1= 2;

B. n = 3; 1= 2;

C. n = 4; 1= 0;

D. n = 4; 1= 0;

m| = 0;

mi = 0;

mi = 0;

ms = +J

ms =

mc = +-i-

mj = +l; ms = -L2

54. Which of the following elements would have anelectronic configuration of ns^(n-l)d^ due to half-filledstability?A. CopperB. Silicon

C. ManganeseD. Molybdenum


55. According to Figure 1, the photons emitted compare inwhat way to the photons absorbed by the transitionbetween energy levels in the laser system?A. The light emitted is of higher wavelength.B. The light emitted is of equal wavelength.C. The light emitted is of lesser wavelength.D. The light emitted is incomparable.

5 6. For a laser to emit visible light, the material used musthave an emission in what energy range, knowing thatthe visible range is approximately 400 nm to 750 nm?

1.24 x 10"6eV(E =

A.

B.

C.

D.

-, where X is measured in meters)

6.25 eV to 9.90 eV

3.10 eV to 6.25 eV

1.65 eV to 3.10 eV

1.24 eV to 1.65 eV


Passage IX (Questions 57 - 63)

Long before the advent of modern technology, coloredpaints existed. The pigments for these paints were extractedfrom natural sources, such as plants and mineral deposits. InEurope and the Middle East, the majority of the pigmentsused for painting were metal oxides. In the Orient, moreorganic pigments were used. By tracing the origins ofpigments, art historians have been able to determine thesource not only of paintings, but of stylistic influences aswell. Listed in Table 1 are some of the more commonsources of colors used in the early centuries A.D.

Pigment Chemical Formula Color

Alizarin 1,2-dihydroxyanthraquinone Yellow

Azurite 2CuC03Cu(OH)2 Blue

Cinnabar HgS Red

Massicot PbO Yellow

Minium Pb304 Scarlet

Orpiment As2S3 Yellow

Realgar AS2S2 Orange redRust Fe203«H20 Brick red

Ultramarine blue SiS2/Al2S3 Blue

Verdigris Cu(C2H302)2-Cu(OH)2 Greenblue

White lead 2PbC03Pb(OH)2 White

Table 1

As a rule, inorganic paint pigments last for a far longerperiod of time than the organic paint pigments. For thisreason, inorganic pigments are a more reliable source used toidentify the time and region from which many paintingsoriginated. The colors in inorganic paint pigments are causedby the absorption of light in the visible range of the lightspectrum. The absorption is due to transitions between thed-orbitals, which because of the asymmetry of the ligands arenot degenerate. The splitting of the d-orbitals results indifferent energy levels within the d-level. Figure 1 shows theenergetic splitting of the d-orbitals in an octahedral complex.

dx2. y2 d22

dxy yz

Figure 1

The color we see is the reflected color, thecomplementary color of the color that is absorbed. If a paintappears blue, it is because of a paint pigment is present in itthat is absorbing orange light (the complementary color ofblue). Figure 2 shows an artist's color wheel, which can beused to determine complementary colors. Complementarycolors oppose one another on the color wheel. Blue opposesorange, so they are complementary colors of one another.Red and green are also a complementary pair.


Figure 2

White light is a combination of all colors, so when onecolor is absorbed from white light by a pigment, the othercolors reflect. All the colors are no longer present, so atleast one of the colors has no complementary color present.The result is that the reflected light takes on a hue of thecolor that has no complementary color to cancel it. Whenwhite light reflects off of a pigment, the reflected light can beanalyzed for intensity versus wavelength after a prism splitsthe light via refraction. The output is an absorptionspectrum, which appears as a rainbow band with verticalblack lines, due to the absence of the light absorbed.Absorption spectra are the opposite of emission spectra,althoughboth require a prism to refract (grate) the light.

5 7. Which of the following pigments absorbs light of theSHORTEST wavelength?A. Alizarin

B. Azurite

C. Cinnabar

D. Verdigris

58. When a sample is heated by flame, and the emitted lightis put througha prism, what result is produced?A. An emission spectrum of dark background with

some colored bands.

B. An absorption spectrum of dark background withsome colored bands.

C. An emission spectrum of colored background withsome black bands.

D. An absorption spectrum of colored backgroundwith some black bands.

59. In the absorption spectrum of realgar, what color bandwould contain black lines?

A. Orange and redB. Orange and greenC. Blue and redD. Blue and green


60. Given that the wavelength of visible light has a rangefrom about 740 nm to 390 nm, what is theapproximate wavelength of light absorbed by the copperdication in Cu(C2H302)2'Cu(OH)2?A. 450 nm

B. 550 nm

C. 650 nm

D. 750 nm

61. White lead absorbs which colors?

A. All colors

B. Green, blue, and violetC. Red, orange, and yellowD. No colors in the visible range

62. All of the following exhibit reflected color EXCEPT:A. Candle wax

B. Fabric dyesC. Pen ink

D. Gas-filled light tubes

63. Which of the following pigments will decompose first?A. Alizarin

B. Azurite

C. RustD. Verdigris



Fluorescence occurs with the absorption of a photon bya molecule, exciting one of its electrons to a higher energystate. Because the material is molecular and not atomic,there are also vibrational and rotational energy levelsassociated with it. The molecule can change its vibrationalenergy level (by changing its bond-stretching frequency) andgive off energy in the form of IR photons or heat. Heat isreleased during collisions with less energetic molecules.When the electron falls back to the ground state, a photon ofless energy than was initially absorbed is given off. Afluorescing compound can thus absorb high-energy light(such as ultraviolet) and give off lesser-energy light (such asvisible light).

Phosphorescence also begins with the absorption of aphoton by a molecule, exciting one of its electrons to ahigher energy state. The electron can flip its spin in thisexcited state, dissipating some of the extra energy. When theelectron relaxes, because it shares the same spin as theground state electron, it cannot fall back to its ground state.It must relax to an intermediate state, so a lower-energyphoton is emitted than was initially absorbed. Figure 1represents what goes on during both fluorescence andphosphorescence. The Xq potential well represents theground electronic state, and Xi represents the excited state.The lines within the well represent the different stretchingmodes (vibrational energy levels) of the molecule.

u

I

Bond radius (A)Figure 1


The wiggly arrows in Figure 1 represent energy that isdissipated as an excited electron drops in vibrational energy,,which is released in the form of an IR photon. Thevibrational energy thereby released can be measured as heatgiven off from the system. The amount of heat emittedvaries with different molecules, because the electronic energylevels are not singular levels (they have both vibrational androtational energy levels associated with them). Thetransition energy is thus the result of a random combinationof energy levels. Because the electronic energy levels are notsingular levels and because the transition is random, it isimpossible for a molecule to absorb or emit light in such amanner that all of the photons simultaneously have the samefrequency (or wavelength). For this reason, monochromaticlight is not physically possible.

64. A spin flip in the excited state is associated with whichof the following processes?A. Fluorescence onlyB. Phosphorescence onlyC. Both fluorescence and phosphorescenceD. Neither fluorescence and phosphorescence

65. What can be said about the photon absorbed relative tothe photon emitted by a fluorescing diatomic molecule?A. The energy of the photon absorbed is greater than

the energy of the photon emitted, while thewavelength of the photon absorbed is less than thewavelength of the photon emitted.

B. The energy of the photon absorbed is less than theenergy of the photon emitted, while the wavelengthof the photon absorbed is greater than thewavelength of the photon emitted.

C. Both the energy and wavelength of the photonabsorbed are greater than the energy and wavelengthof the photon emitted.

D. Both the energy and wavelength of the photonabsorbed are less than the energy and wavelength ofthe photon emitted.

66. What is TRUE about monochromatic light?

A. Monochromatic light is not possible, because lightis not quantized.

B. Monochromatic light is possible, because light isquantized.

C. Monochromatic light is not possible, because thereare vibrational energy levels of lower energydifference than the electronic energy levels withwhich they are associated.

D. Monochromatic light is possible, because there arevibrational energy levels of greater energydifference than the electronic energy levels withwhich they are associated.


67. According to the graph, how do the photons emittedfrom fluorescence and phosphorescence compare?

A. The photons emitted from fluorescence have alonger wavelength than the photons emitted fromphosphorescence.

B. The photons emitted from fluorescence have ashorter wavelength than the photons emitted fromphosphorescence.

C. The photons emitted from fluorescence have thesame wavelength as the photons emitted fromphosphorescence.

D. The photons emitted from phosphorescence are ofhigher energy than the photons emitted fromfluorescence.

68. Which of the following transitions emits theSHORTEST wavelength of light?A. Xi V = 1 to Xrj V = 0, where V is the vibrational

energy level.B. XiV = 0toXrjV = 0, where V is the vibrational

energy level.C. XrjV = 1 to Xi V = 0, where V is the vibrational

energy level.D. Xq. V = 0 to Xi V = 0, where V is the vibrational

energy level.

69. Which of the following statements about fluorescenceis FALSE?

A. Fluorescence is possible with molecules.B. Fluorescence converts visible light into ultraviolet

light.C. Fluorescence can dissipate energy in the form of

heat.

D. There are some atoms for which fluorescence is notpossible.



A student conducts an experiment where a series ofaqueous solutions of group I cations (alkali metals) or groupII cations (alkaline earth metals) are each heated by a flame.The heat of the flame, if hot enough, can stimulate electronsin the cations to leave their ground state. When they relax tothe ground state, light is emitted. Because the light is anarrowband of photons with predominantly one wavelength,it appears colored. The color is used to identify the cation insolution. Table 1 shows the student's result.

Barium Lime-green ring around flameLithium Brilliant crimson flame

Potassium Light purple ring around flameSodium Orange-yellow flameStrontium Red flame

Table 1

The student then conducts the same experiment usingtransition metals rather than main-group metals. The trendin color is predictable with main-group cations of the samecolumn (family) in the periodic table. The color of thetransition metal cations depends more on the ligands attachedto the metal cation. Table 2 also shows the colors obtained.

Vanadium (II) Violet flame

Copper (I) Orange flameNickel (H) Red flame

Chromium (DT) Red flame

Cobalt (H) Green flame

Table 2

With transition metals, the color emitted by the flame isoften the color of the light absorbed when the cation insolution is exposed to white light. The color of the aqueoussolution that is detected by the eye is the complementarycolor of the color absorbed. The color that is associated withtransition metals is due to an electronic transition within thed-orbitals (between the d-levels, which split due to thepresence of ligands). These electronic transitions are referredto as d-d transitions, and they occur only with transitionmetals that are not d^ord^ in their electronic configuration.

7 0. Which cation is paramagnetic?A. Na+

B. Sr2+C. Cu+D. Cr3+

71. What is the electronic configuration of chromium?A. [Ar]4s23d4B. [ArHs^d5C. [Arl3d6D. [Ar]3d3


72. Which of the following transition metals has theGREATEST d-d transition energy, given that the colorobserved in the flame test is due to a d-d electronictransition?

A. Ni2+B. Cr3+C. Co2+D. V2+

7 3. Identify the TRUE statement(s) from the following.I. As ionic size increases; the energy of the transition

increases.

II. The transition energy of K+ is greater than thetransition energy ofCr3+.

DX Cu+ and Ni2+ have the same electronicconfiguration.

A. I onlyB. II onlyC. I and H onlyD. II and IE only

74. What are the quantum numbers associated of the lastelectron of nickel dication?

ms=- -!•2

m<t = - i-

A.n = 3 1= 2 mi = -l

B. n = 3 1= 2 mi = 0

C. n=3 1=2 mi =+l ms = -i-

D.n =3 1=2 mj = +l ms =+±-

75. Given that lithium cation emits a red flame, sodiumcation emits an orange-yellow flame, and potassiumcation emits a violet flame, what color flame will beemitted by rubidium cation?A. Yellow

B. Green

C. Blue

D. Colorless (and thus the normal orange)

76. An emission spectrum of lithium would appear in whatmanner?

A. It would be a rainbow with a black line in the redregion of the spectrum.

B. It would be a rainbow with a black line in thegreen region of the spectrum.

C. It would have a black background with a red line.D. It wouldhave a black backgroundwith a green line.


Passage XII (Questions 77-81)

Glyphosate, shown in Figure 1, is a synthetic compoundthat is currently the world's largest selling commercialherbicide. Glyphosate kills vegetation by binding metalswithin a plant, thus starving the organism of essentialmetallic nutrients needed for the transport of other nutrientsand waste, such as carbon dioxide and oxygen. WhenGlyphosate leeches metals from the plant, its respiration isshut down. Glyphosate is shown in Figure 1 in its saltform, as extracted from pH = 7 aqueous solution.

O HII I

H0 A AH H H H

Figure 1

After the plant has died, Glyphosate decomposes in thepresence of moisture into phosphate, carbon dioxide, andammonia. The decomposition products are environmentallysafe, and some are useful nutrients for the soil. In a study ofthe decomposition of Glyphosate in vivo, isotopically labeledGlyphosate was applied to vegetation, and the isotopicabundance was monitored at regular intervals over a twenty-day period. Based on kinetic data associated with hydrolysisand decomposition, the conclusion was reached that the decaytime of Glyphosate to its inorganic products is approximatelyfourteen days. The compound was radio-labeled in threeseparate experiments, using a different radio-label in eachexperiment. The first experiment employed tritium (3H), thesecond experiment employed 14C, and the third experimentemployed 32P to monitor the half-life and decay rates.Carbon-14 was used as a marker, but not for kineticpurposes. The half-life of carbon-14 is far greater than thelifetime of a typical lab experiment. The isotopic markerhelps to study the migration of Glyphosate through theenvironment.

77. Which of the following is an isotope of32P?A. 32SB. 32P-C. 3IPD. 3,Si

78. The molecular shape about the nitrogen in Glyphosateis BEST described as:

A. trigonal planar.B. trigonal pyramidal.C. square planar.D. bent.


79. The half-life of 32P is approximately 14 days. If theconcentration of a 32P radio-labeled Glyphosate sample(C3HgN06P) is found to be 188 parts per millioninitially, how long will it take until the concentrationis 25 ppm?A. 27 daysB. 33 daysC. 41 daysD. 43days

8 0. Which of the following molecules have ionic bonding?I

II

III

A.

B.

NH3

C02

Na2P04H

I onlyII only

C. Ill onlyD. I and II only

81. As 32P decays, it changes into a product that is stillphosphorus. This is necessary in a marker so that thechemical behavior of the labeled species does notchange. What is the MOST likely particle given off inthe decay of phosphorus-32?A. A neutron

B. An alpha particleC. A beta particleD. A positron


Passage XIII (Questions 82 - 88)

In order to image vital organs by photon emission,technetium-99 is ingested into the body so that its low-levelgamma radiation (140-keV) may be detected. For brainimaging, 99Tc is introduced via the bloodstream, and itsubsequently diffuses throughout the body, including to thebrain. Technetium-99 has a half-life of approximately sixhours, so scanning and analysis must closely followingestion. For 99Tc tomigrate toall of the target organs ina concentration high enough to be detectable requires roughlytwo hours. After two hours, the concentration of activetechnetium-99 is approximately eighty percent (80%) of itsoriginal value. Technetium-99 decays through gammaemission, meaning that its nucleons simply drop in one stepfrom an excited nuclear state to a ground nuclear state whenthey emit photons. The nucleus of technetium does notchange upon the emission of the gamma photon.

For brain imaging, the skull is encompassed by adetection device that collects and records the gamma radiationat arbitrary sites. Radiation escapes more readily from theareas of the brain that have a lower tissue density and fluidthan those that have a greater density, but a complication inthe analysis of brain images is that the radiation dissipates inliquid. This means that slight variations in the texture andcomposition of brain matter can produce different imagingpatterns. The display pattern collected can be converted intoa contour map of the brain. This technique work well withorgans that have a relatively low water content.

82. Beta decay of 210Po results inwhich of the following?A. 210AtB. 210BiC. 206PbD. 2<>9At

83. Alpha decay is observedtransformations EXCEPT:

A. 254Esto250BkB. 238uto234ThC. 223Frto221BiD. 247Cmto243Pu

in all of the following

84. If the initial dosage of99Tc gave a reading of 120 p:Ci,what will the reading be after 12 hours?A. 60 uCiB. 40 uCiC. 30 uCi

D. 20 |iCi


85. Electron capture by carbon produces the same elementas which of the following processes?A. Positron capture by boronB. Positron decay of carbonC. Beta capture by nitrogenD. Alpha decay of oxygen

8 6. Gamma radiation is considered to:

A. weigh4 amu and have a nuclear charge of +2.B. be massless with a nuclear charge of -1.C. weigh4 amu and have no nuclear charge.D. be massless and have no nuclear charge.

87. The electronic configuration for 99Tc is which of thefollowing?

A. [Ar]5s23d5B. [Krl5s24d5C. [Kr]5sUd6D. [Kr]5s24d7

88. When 99Tc undergoes gamma decay, the final productis which of the following?A. "TcB. "RuC. 95NbD. "Mo


Passage XIV (Questions 89 - 95)

As of March 1989 physicists and chemists had yet tocarry out a nuclear fusion reaction that did not require moreenergy input than output. Fusion reactions require hightemperature and strong magnetic fields to be carried out.This is the reason that nuclear power generators are fissionreactors. Reaction 1 is an example of a fusion reaction,while Reaction 2 is an example of a fission reaction.

^Li +2H -> 2^HeReaction 1

292U +0n^32He+286Rn+0nReaction 2

Then in March of 1989, B. Stanley Pons and MartinFleischmann announced that they has achieved a nuclearfusion reaction in a test tube at room temperature. Pons andFleischmann set up a simple electrochemical cell with apalladium cathode in 0.1 M LiOD(D20) and a platinumanode. The design of the cell was to form D2 gas by passingan electron flow from the platinum anode to the palladiumcathode. The cell generated more heat than expected, so itwas proposed that a nuclearreaction musthave transpired.

The researchers believed that the deuterium migrated intothe palladium metal of the cathode and gathered in thepockets of the lattice, so they hypothesized that thedeuteriumfused together in the palladium electrode by eitherReaction 3 or Reaction 4.

2H +^H Jh +^hReaction 3

2H +2H -> lQn +\UeReaction 4

Pons and Fleischmann in their original paper stated thatboth tritium and neutrons were observed in the cathode oftheir electrochemical cell, but the concentrations were toolow to account for the extra heat generated. Since that time,the hypothesis of a "cold fusion" nuclear reaction takingplace under these experimental conditions hasbeen dismissedas improbable by most scientists in the field.

89. What is NOT a product when two deuterium atomsundergo a fusion reaction?A. Tritium

B. Helium-4

C. Helium-3

D. Hydrogen

90. What evidence did Pons and Fleischmann point to thatwould indicate that a nuclear reaction had transpired inthe electrochemical cell?

A. Less heat was released than expected.B. More heat was released than expected.C. That electrons were emitted by the cell.D. That electrons were absorbed by the cell.


91. Cold fusion is BEST described as:

A. an endothermic nuclear reaction capable of rapidlycooling the environment.

B. an exothermic nuclear reaction capable of rapidlycooling the environment.

C. an endothermic nuclear reaction capable of beingcarried out at room temperature.

D. an exothermic nuclear reaction capable of beingcarried out at room temperature.

9 2. Nuclear power plants employ what reaction to generatepower?A. Fission, because fusion requires such a great energy

input that it is inefficient, and the net energychange is unfavorable.

B. Fusion, because fusion requires such a great energyinput that it is inefficient, and the net energychange is unfavorable.

C. Fission, because fission requires such a greatenergy input that it is inefficient, and the netenergy change is unfavorable.

D. Fusion, because fission requires such a great energyinput that it is inefficient, and the net energychange is unfavorable.

9 3. Which of the following is a fusion reaction?

A. 23952U +in^232Th +4HeB. 239>-*292U +2Hec 23<£u +11B-> 2$CfD.f3Tc*-*99Tc +h/

94. What is formed when 238U emits an alpha particle anda deuterium nucleus?

A. 234AcB. 232AcC. 236UD. 232Ra

95. Capture ofwhich of the following particles would NOTresult in a change in the atomic number?A. AlphaB. Beta

C. Neutron

D. Tritium nucleus


Questions 96 - 100 are NOT basedon a descriptive passage.

9 6. Which of the following supports the conclusion thatelectrons have quantized energy levels?

A. The existence of a nucleusB. Scattering ofX-rays by a thin sheet of a materialC. The bending of nuclear radiation particles by a

magnetic fieldD. Distinct lines in an electromagnetic radiation

spectra

97. The LONGEST wavelength of light would beassociated with:

A. X-rays.B. violet light.C. green light.D. infrared light.

98. The ionization energy of H(g) is 1312 ^ . A goodmole

approximation for the second ionization potential ofhelium (g) is:

A.

B.

C.

D.

5248-^-mole

2624 -&-mole

656-kJLmole

328^-mole

99. Which of the following is the electronic configurationfor an excited state ofNa+?

A. ls22s22p53s2B. ls^s^p^s1C. ls22s22p6D. ls^s^p^s1


100. What is the MOST common shape for a transitionmetal with five ligands attached?

A. Square planarB. Trigonal bipyramidalC. Pentagonal planarD. Hexahedral

\M'.y'^,x

"I know where my nucleus is. do vou?"

1. C 2. A 3. A 4. C 5. D6. B 7. D 8. B 9. A 10. B11. D 12. D 13. D 14. D 15. B16. A 17. C 18. C 19. D 20. B21. C 22. A 23. A 24. C 25. B26. D 27. B 28. A 29. C 30. C31. C 32. A 33. D 34. D 35. C36. D 37. D 38. A 39. C 40. B41. B 42. A 43. B 44. D 45. B46. C 47. B 48. A 49. C 50. C51. D 52. B 53. A 54. D 55. B56. C 57. A 58. A 59. D 60. C61. D 62. D 63. A 64. B 65. A66. C 67. B 68. A 69. B 70. D71. B 72. D 73. C 74. B 75. D76. C 77. C 78. B 79. C 80. C81. A 82. A 83. C 84. C 85. B86. D 87. B 88. A 89. B 90. B91. D 92. A 93. C 94. B 95. C96. D 97. D 98. A 99. D 100. B

STOP, YOU'VE HAD ENOUGH!

Atomic Theory Passage AnswersPassage I (Questions 1-8) Classical Experiments

1. Choice C is correct. If oil drops were naturally charged, then the experiment could not work, because the totalcharge on the droplet would not be generated by the electron beam alone. The charge determined from theexperiment would not necessarily reflect the charge of an electron. This eliminates choice A, because it isassumed in the experiment that the oil droplet is initially uncharged before exposure to the electron beam.Because the gravitational pull on the oil droplet is calculated as opposing the electric field, it must be assumedto be in one direction (downward). If the gravitational pull were in all directions, the experiment would not bepossible. This eliminates choice B. The mass used in the calculation is the average mass of an oil droplet, sothe massof the electrons is in fact ignored. This can be assumed knowing that protons and neutrons are far moremassive than electrons. The change in mass by gaining or losing electrons is negligible compared to the rest ofthe atomic mass. This makes choice C the best answer; an answer to be chosen by many, like you. The charge ofan electron is assumed to be constant, if you are solving for an exact value. The experiment could be solved for anaverage value, but there is no reasoning behind one electron being of a different charge than another electron.This is to say that charge is quantized (has an exact value).

2. ChoiceA is correct. In the tube in Figure 1, the electron beam is negatively charged, so it bends downwards dueto its attraction to the positively charged plate below and its repulsion from the negatively charged plateabove. It may seem odd that the cathode plate (plate above) is negatively charged and the anode plate (platebelow) is positively charged, but the plates form a charged capacitor, not a discharging battery. For adischarging battery and charging capacitor, the plate charges are opposite. Because a proton has the chargeopposite from an electron (protons carry a positive charge), protons exhibit behavior opposite from an electron.Proton beams therefore bend upwards. Aneutron isneutral, so its pathway would not arc at all between the twocharged plates. The neutron does not arc, because it is unaffected by the charged plates. The best answer ischoice A.

3. ChoiceA is correct. The value of the charge is twice as large as expected. This can be explained in terms ofmagnitude ofcharge rather than sign of charge. Avalue that is twice as large as expected is attributed to adoubly charged oil droplet. Choice A, doubly ionized, would explain this. Choices B, C, and Dall address signof charge and thus are eliminated.

4. Choice C is correct. TheThomson experiment, as stated in the passage, determined the mass-to-charge ratio ofthe electron. It can be used to determine the mass-to-charge ratio for any charged particle, but in thisparticular experiment, the particle was an electron. The best answer is choice C. A mass spectrometer is amechanical variation of the Thomson experiment theme.

5. Choice D is correct. The charge of an electron is determined by equating the mass of the droplet and thegravitational force constant (mg) with the voltage of the electric field and charge of the oil droplet (qV). Tosolve for q, the other three variables must be known. This means that choices A, B, and Care all values thatmust be known. There isno temperature factor in the equation (mass, charge, gravity, and voltage do not varywith temperature), so choice D represents the factor that is least important in the calculations.

6. Choice B is correct. An atom is neutral when it carries no net charge. Klectrons carry a negative charge, whileprotons carry a positive charge. Neutrons are uncharged. In order for an atom to be neutral, the number ofelectrons must be equal to the number of protons. The best answer is choice B. Choice A is not false per se, butbecause there exist isotopes whose number of electrons does not equal the number ofneutrons, no conclusion can bedrawn about the number of electrons and neutrons within a neutral atom.

7. Choice D is correct. The strongest electrostatic attractive force is felt between particles of opposite charge.This is known as Coulomb's law. An electron and proton carry opposite charges, so the best answer is choice D.Aproton and neutron exhibit no Coulombic attraction, but because they are held tightly in the core of thenucleus, there must be some attractive force between them. Without the electrostatic caveat, this questioncould not have a proton and a proton listed as an answer choice because of the strong force associated with thenucleus that holds protons together. This is a more advanced topic that you may see in courses in nuclearchemistry (atomic physics) but as it is presented in the MCAT, the ambiguity presented by nuclear attractions(the uncertaintyof the source of the strong force) is not a probable topic forquestions.

Copyright © by The Berkeley Review® 147 Section II Detailed Explanations

Choice Biscorrect. If an electron is descending, a net upward force must be applied to accelerate the particle inthe opposite direction. To increase the net upward force, the force upward must be increased or the forcedownward reduced. This would slow the particle and eventually stop it from dropping. This means that eitherthe mass must be decreased, the gravity must be decreased, the charge must be increased, or the voltage of thefield mustbe increased. The easiest thing todo is increase the voltage of the field. In the actual experiment,this is what is done. Increasing the number ofelectrons in the drop increases the charge, which should stop (orat least slow) the descent orchange the direction of its path to an ascent. Decreasing the gravitational pull,which requires placing the apparatus in an anti-gravity environment, would also alleviate the descendingdifficulty. The only choice that would definitely not stop the descent, but would in fact increase the descent, isto increase the mass of the oil droplet. The best answer is choiceB.

Passage II (Questions 9 - 14) Isotopic Abundance and Average Atomic Mass

9. Choice A is correct. All atomic masses in theperiodic table are referenced against carbon-12. Carbon-I2 is thestandard traditionally used. The carbon-12 isotope is defined as having amass ofexactly 12.0000 amu. Carbonis added toa sample to standardize the mass ofthe peaks given in the spectrum. Not all elements have exactwhole number masses, so a reference is necessary. Youare wise to pickA.

Choice Bis correct. According toTable 1, magnesium has three major isotopes. The isotope with the lowestatomic mass is the most abundant (78.70%). This means that the lowest peak by mass (the first peak from leftto right) mustbe the largest peak. This is trueonly in the spectrum ofchoice B.

Choice Dis correct. The average atomic mass for any element is the weighted average ofall of the isotopes ofthe given element. Choices Aand Bare eliminated, because Table 1 lists only two isotopes for silver, and thecalculation shows three values being summed. Choice A is the calculation of average atomic mass formagnesium. The correct answer is choice D, because that has the correct percentages multiplied by the correctisotopic masses. You would be a very wisdom-laden soul, ifyou were toselect Dasyouranswer.

12. Choice D is correct. In order to send the isotopes of the element through the mass spectrometer, the elementmust be in the gas phase. This means that any sample not in the gas phase initially must be converted to thegas phase to be analyzed using mass spectroscopy. The process of converting a solid into a gas is referred to assublimation, thus making choice D correct.

13. Choice D is correct. Because the average atomic mass is an average of the masses of all of the isotopes, theaverage atomic mass must fall within the range of the isotopic masses, meaning that the average atomic massis greater than the lightest isotope but less than the most massive isotope. When there are two isotopes only,then the average atomic mass is the weighted average of the two. In a fifty-fifty mixture of two isotopes, theaverage atomic mass lies perfectly between the two masses of the isotopes. If the mixture favors one of theisotopes, then the average atomic mass would be closer to the more abundant isotope than the less abundantisotope. The true statements are II and III, making the bestanswerchoice D.

14. Choice D is correct. The difference between isotopes of the same element is found in the number of neutrons. Iftwo atoms have a different number of protons, then they are not the same element. Choices A and B areeliminated, because they allude toa different number of protons. The greater mass associated with the boron-11 isotope is attributed to an extra neutron being present in the Boron-11 isotope as compared to the boron-10isotope. The best answer is choice D.

10.

11.

Passage III (Questions 15 - 20) Bohr Model of Hydrogen

15. Choice Bis correct. The ionization ofhydrogen involves an electronic transition from the n = 1 energy level tothe n = oo energy level. Using Equation 1 todetermine the transition energy yields the following:

2.L78xl0-18(Z2)[4r--Uj\nr nf2/

where Z=1, nj =1, and nf =<*>. Having «. in the denominator makes the number zero, which results in a value of2.178 xlO'18 (Z2)J. With a value of Z=1, the ionization energy is 2.178 x10'18 J. The ionization energy ofhydrogen from its valence level is 2.178 x10"18J, which makes the best answer choice B.


16. Choice A is correct. This solution requires relating the formula for the energy of a photon to the energyequation given in the question. The relationship is as follows:

E = mc2 and E =&- .-. mc2 = M => mc =-k => X= ^-X X X mc

which is choice A.

17. Choice C is correct. Choice B is eliminated, because from the n = 2 energy level to the n = 1 energy level, thereis a drop in energy, which corresponds to emissionof light, not absorption. The best answer is choiceC, becausethe energy levels get closer together as the n value increases, so the transition from the n = 1 energy level to then = 2 energy level is always of greater energy than any transition starting at an energy state where n > 1.However, to be certain, the following mathematical relationship can be derived:

Given that the energy level of an electron is E=-2.178 x10~18|2_ J,the transition energy is AE =-2.178 x10-18 (Z2)U-- -U J=2.178 x10"18 (Z2)U-- -U J-

Inr2 ni2/ Um2 nf2/

The relative energies are therefore comparable by the value of the 1-1---M term.W2 ni2/

\a2 22J \22 52J \22 42j \i 4/ l4 25/ U 16/f1-1| >(-25 4_j >|_4_. _1_| => 3>_21_ >JL, therefore choice Cis the best answer.\ 4/ VlOO 100/ U6 16/ 4 100 16

18. Choice C is correct. Statement I is valid, because as an electron's principle quantum number (n) increases, theelectron is in an energy level (orbit) that is farther from the nucleus. Statement II can be viewed from thepictorial representation of the energy levels in Figure 1. As the value of n is increasing, the energy levelsbecome closer to one another. This makes statement II a valid statement. According to the answer choices, thismakes choice C the best answer. A larger nuclear charge (Z) affects the energy at each level, and thus affectsthe transition energies as well. This makes statement III incorrect, and confirms that choice C is the bestanswer. Choose C and feel a little giddy.

19. ChoiceD is correct. The electronic transition from the n = 5 energy level to the n = 4 energy level is of lowerenergy than the electronic transition from the n =4energy level to the n =3energy level. Therefore, the photonemitted is oflower energy than a photon corresponding toorange light. Because green light is ofhigher energythan orange light, choice A is eliminated. The electronic transition from the n = 4 energy level to the n =2energy level is ofhigher energy than the electronic transition from the n =4 energy level to the n =3 energylevel. Therefore, the photon emitted is ofhigher energy than a photon corresponding to orange light. Becausered light is of lower energy than orange light, choice Bis eliminated. The electronic transition from the n =6energy level to the n =4energy level is of lower energy than the electronic transition from the n =4energy levelto the n = 3 energy level. Therefore, the photon emitted is of lower energy than a photon corresponding toorange light. Because violet light isofhigher energy than orange light, choice C iseliminated. The electronictransition from the n =5 energy level to the n =3energy level isofhigher energy than theelectronic transitionfrom the n =4energy level to the n =3energy level. Therefore, the photon emitted is ofhigher energy than aphoton corresponding to orange light. Because yellow light isofhigher energy than orange light, choice D isthe best answer.

20. Choice B is correct. Choice A is an invalid statement, because according to Beer's law, the absorbance of lightis proportional to sample concentration. Because samples can have varying concentrations, they can in factabsorb different amounts of light. The color of light (or frequency) corresponds to transition energy. Becausethe sample, independent of concentration, absorbs the samefrequency of light each time, the transition energymust be a fixed value. This leads to the conclusion that energy levels are fixed, and thus quantized. Choice Bis the best answer. The sample emits light of the same frequency, regardless of solution's temperature andconcentration. Temperature can affect the intensity of the light that is emitted, but not the frequency. Thefrequency of the light remains constant. This eliminates choice C (temperature effects) and choice D(concentration effects). Choice B is your answer choice.

Copyright ©byThe Berkeley Review® 149 Section II Detailed Explanations

Paramagnetism and Liquid Crystal DisplaysPassage IV (Questions 21 - 27)

21. Choice C is correct. A liquid crystal display uses a paramagnetic compound in its polarizing cells to rotate (ornot rotate) light. A paramagnetic species has at least one unpaired electron. Because chlorides are -1 each,cadmium in CdCl2 carries a +2 charge. This is the result of losing two 5s electrons from elemental cadmium,which results in an electronic configuration of ls22s22p63s23p64s23d104p64d10 for Cd2+. Each level is full, sothere are no unpaired electrons. Choice A is eliminated. In elemental fluorine, the two fluorine atoms sharetheirunpaired electrons in the form of a bond. This means that the fluorine molecule has no unpaired electrons.Choice B is eliminated. Because chlorides are -1 each, cobalt in CoCl2-6 H2O carries a +2 charge. This is theresult of losing two 4s-electrons from elemental cobalt, which results in an electronic configuration for Co2+ ofls22s22p63s23p63d7. There is an odd number ofelectrons, so the species must be paramagnetic. The correctchoice is answer C. Just as was the case with F2> molecular hydrogen (H2) has no unpaired electrons.

22. Choice A is correct. Atomic nitrogen (N) has the electronic configuration ls22s22p3. The last electron is thethird electron to enter the 2p orbital. The principle quantum number is given as 2, and because it is a p-orbital,the angular momentum quantum number (l) is equal to 1. All of the answer choices contain these two values,sonothing is eliminated. To obtain the m/and ms values, the electrons must be filled into their respective p-orbitals. This is drawn below:

@ ©

m^= -1 m^= 0 m^= +1

Last electron is in the third p-orbital, so m = +1;

2Last electron is spin up, so m s = + -

The correct answer is therefore choice A.

23. Choice A is correct. Having six ligands attached results in an octahedral shape. The correct answer is choiceA. Drawn below are generic structures for the other answer choices.

NO

NC

N 3-C

Fe'- CN

CN

Octahedral

.»**L* M"

Generic square planar

M-..N>//

L

..**L

Generic tetrahedral Generic trigonal bipyramidal

24. Choice C is correct. Polarity results from the asymmetric distribution of electron density. Cis compounds arealways polar. The correct answer is choiceC. The structures for the four choices are drawn below.

Cl-

Cl1

CI3-

Fe~ CI

ci

3-FeCfe

CN3-

..,u\\CN'CN

NC Fe

CN

Fe(CN) 3-


Cl-

Cl1

.^\NH3Fe-—nh.

cis-Pt(NH3)2Cl2

ci-

HsN1

,^NH3Fe-—ci

trans-Pt(NH3)2Cl2

Section II Detailed Explanations

25.

26.

27.

Choice B is correct. To determine the number of unpaired electrons (as well as which one has the most), the d-electron count for each transition metal cation first must be worked out. To determine the d-electron count forthe four cations, you must consider the electronic configuration. Neutral chromium is [Ar^s^d5, so when it isCr3+, it has lost the 4s-electron and two 3d-electrons, leaving it with anelectronic configuration of[Ar]3d3. Werefer to Cr3+ as a d3-cation. Neutral manganese is [Ar]4s23d5, so when it is Mn2+, it has lostboth 4s-electron,leaving it with an electronic configuration of [Ar]3d5. We refer to Mn2+ as a d5-cation. Neutral copper is[Ar]4s13d1^1, sowhen it isCu+, it has lost its 4s-electron, leaving it with an electronic configuration of [Ar]3d10.We refer to Cu+ as a d10-cation. Finally, neutral nickel is [Ar]4s23d8, so when it is Ni2+, it has lost both 4s-electron, leaving it with an electronic configuration of [Ar]3d8. We refer to Ni2+ as a d8-cation. From here, itis a matter of placing the d electrons into their respective orbitals. The electron filling is as follows:

3+Cr

2+Ivfri

m]= -2 rri|= -l mi = 0 rri|= +l m]= +23 unpaired electrons

rrij= -2 mj = -1 rrtj= 0 rri|= +1 rnj= +25 unpaired electrons

Cu

Ni2+:

I I I i i

' ' ' '

rri| = -2 mi = -l mi = 0 mi = +l mi = +2

0 unpaired electrons

t i i

mi = -2 rri] = -1 rri| = 0 xr\\ = +1 mj = +22 unpaired electrons

The most unpaired electrons is found with Mn, so choice B is the best answer.

Choice D is correct. The cyano ligand carries a -1 charge, so chromium must have a +6 charge in order forCr(CN)6 to be neutral. The amino ligand is neutral, so chromium must have a +3 charge in order for Cr(NH3)f,to have an overall positive three charge. The charge of chromium is in fact greater in Cr(CN)6 thanCr(NH3)6, so choice A is valid. The water ligand is neutral, so copper must have a +1 charge in order forCu(H20)6 to have an overall +1 charge. When copper is neutral its electronic configuration is [Ar^s^d^0, sowhen copper carries a +1 charge, it has electronic configuration [Ar]3d10. This gives copper ten d-electronswhich makes choice Bvalid. In both Fe(NH3)63+ and FeClf,3", iron has six ligands attached, so the shape ofboth molecules is the same. Choice C is valid. Scandium has theelectronic configuration [Ar^s^d1, so it canlose only three electrons. It is not possible for scandium to have a +4charge (at most it is +3). This means thatchoice D is not true. Pick choice D, and feel a little peppier because of it.

Choice B is correct. To determine the number of electrons on the central metal, the formal charge of the metalfirst must be determined. From the charge, the electronicconfiguration is found, so the d-electron count is found.Fe carries a +3 charge in both Fe(NH3)63+ and FeCl63~, so choice A is eliminated, because they both must havethe same d-count. Co carries a +3 charge inCo(H20)63+, and Mn carries a +2 charge inMnClft4". Neutral Co is[Ar]4s23d7, soCo3+ hasa d-electron count of6 (3d6). Neutral Mn is [Ar]4s23d5, soMn2+ hasa d-electron count of5 (3d5). The d-electron counts are not equal, so choice Bmust be the correct answer. Cr carries a +3 charge inCr(NH3)63+, and Mo carries a +3 charge in MoCl63'. Neutral Cr is [Ar]4s13d5, so Cr3+ has a d-electron count of3 (3d3). Neutral Mo is [Kr^sUd5, soMo3+ has a d-electron count of3 (4d3). The d-electron counts areequal sochoice C is eliminated. Finally, Os carries a +2 charge in Os(NH3)42+ and Rh carries a +3 charge inRhCl3(PR3)3. Neutral Os is [Xe}6s24f145d6, so Os2+ has a d-electron count of 6 (5d6). Neutral Rh is[Kr]5s24d7, so Rh3+ has a d-electron count of 6 (4d6). The d-electron counts are equal so choiceD is eliminated.

PassageV (Questions 28 - 34) Migration through a Membrane

28. Choice A is correct. The relative sizes of ions from smallest to largest radius is: Li+ <Mg2+ <Na+ <Ca2+ <K+.The rate of effusion is: Li+ > Mg2+ > Na+ > Ca2+ > K+. Li+ (the smallest) is the fastest while, K+ (the largest)is the slowest. The best answer is A. Choice D is eliminated, because cations are smaller than neutral atoms.

29. Choice C is correct. Every trend in the periodic table comes back to the nuclear pull on the electrons. Theionization potential is the energy required to remove an electron from the outermost shell, which in the case ofsodium and potassium is the 3s and the 4s, respectively. It requires less energy to remove an electron from thelarger 4s-orbital, because the electron is farther away from the nucleus than the 3s-orbital. This makes choiceC the best choice. Choice D is eliminated, because potassium has a lower electron affinity than sodium.

©Copyright © by The Berkeley Review 151 Section II Detailed Explanations

30. Choice C is correct. The size of an ion or element is a result of the nuclear pull on the electrons orbiting thenucleus. This is best explained by choice C. Because C (carbon) has six protons, F (fluorine) has nine protons,and both have the same principle quantum number for their valence shell, the nuclear pull of fluorine is greaterthan that of carbon. Because the electrons are pulled closer to the nucleus in fluorine, and because size of anatom is determined by the electroncloud, fluorine is smaller than carbon.

31.

32.

33.

34.

C: Is 2s 2p F: ls22s22p5

Choice C is correct. F", Ne, and Na+ all have 10 electrons total, so CI" (with 18 electrons) is eliminated. Of thethree choices left, the largest nuclear charge is found on Na+, making it the smallest (the one with theelectrons held most tightly). Choose C to choose correctness. The following chart of the protons and electronsfor the three choices shows that the greater the proton-to-electron ratio, the smaller the species, assumingthat the outermost electrons are in the same valance shell.

Element Protons Electrons Observation Radius

Na+ 11 10 protons exceed electrons, therefore it contracts 65 pm

Ne 10 10 protons equal electrons 70 pm

F" 9 10 electrons exceed protons, therefore it expands 136 pm

Choice A is correct. The electronic configuration for magnesium is ls22s22p63s2. Magnesium must lose 2electrons (the two 3s electrons) to have a filled outer valance shell (the n = 2 shell). This would makemagnesium a +2 cation. Answer choiceA is a fine selection if your goal is to be correct.

Choice D is correct. According to periodic trends, the size of an atom increases as you descend a column in theperiodic table. I (iodine) is the lowest in the periodic table of the halogen choices, so I is the largest of thehalogen choices. Trust periodic trends and choose D.

Choice D is correct. If the pore were to distinguish by charge, then the greater the charge of the cation, theslower the rate of migration for the cation. For example, if the pore were capable of forming attractiveinteractions with the cations such as polar attraction or hydrogen bonding, then the pore would distinguish bycharge. The cations with the greatest charges are Mg2+ and Ca2+. Because Ca2+ is larger than Mg2*, Ca2+would migrate more slowly than Mg2+ through a pore which distinguishes by both size and charge. Thismeans that overall, Ca2+ would have theslowest migration rate. This makes choice D the bestchoice.

Passage VI (Questions 35 - 42) Ionization Energy

35.

36.

Choice C is correct. The electronic configuration for helium is Is2. The two electrons ofhelium areboth valenceelectrons, so choice A is eliminated. Helium cannot have an effective nuclear charge greater than 2, because itcontains only two protons. Choice B is thus eliminated. The ionization energy does not include anyvaporization energy. Helium is a gas at room temperature, so there is no need to add energy to vaporizehelium. This eliminates choice D. The electron must be removed from the first quantum level. The firstquantum level experiences the greatest nuclear attraction, so the ls-electrons are hardest to remove. It is inyour best interest to choose C.

Choice D is correct. The second ionization energy of the alkali metals (lithium, sodium, and potassium) issubstantially larger than the first ionization energy, because the second electron is being removed from a filledoctet. After the first electron has been lost and the alkali metals are cations, their electronic configuration isns2np6, and they each have a filled valence shell. To remove the second electron would be like removing anelectron from a noble gas, only harder, because the nuclear charge is greater for the alkali cation than theneutral noble gas. Given the answer selections, choice D is the best choice.


37.

38.

39.

40.

41.

42.

Choice D is correct. As stated in the passage, there is half-filled stability for the p-level when each of the p-orbitals contains one electron. Nitrogen as a neutral element has half-filled stability, so when it is ionized, itloses its half-filled stability. Oxygen, on the other hand, has one electron beyond the half-filled state;therefore, when oxygen is ionized, it attains half-filled stability. Losing half-filled stability raises theionization energy of nitrogen (ionization is less favorable), while gaining half-filled stability lowers theionization energy of oxygen (ionization is more favorable). Nitrogen is less electronegative than oxygen, sochoice A is eliminated. Nitrogen has seven protons, while oxygen has eight, so choice B is eliminated.Nitrogen is in fact larger than oxygen, but a larger radius implies that the electrons are farther from thenucleus on average. Being farther from the nucleus, the electrons are not as tightlyheld, so the larger radius ofnitrogen would imply a lower ionization energy for nitrogen than oxygen. The best answer is choiceD.

Choice A is correct. The ionization energies for aluminum, silicon, and phosphorus follow an increasing lineartrend. All three elements are in the same row (period) of the periodic table, so the number of core electrons forall three is the same. This eliminates choices C and D. As you move from left to right across a period of theperiodic table, the atomic number increases, so the number of protons increases, and ultimately the effectivenuclear charge increases. The increasing nuclear charge best explains the trend in first ionization energybetween aluminum, silicon, and phosphorus. Be a hero or heroine by choosing A. Note that there is no half-filled or unfilled p-orbital stability affecting the observed trend.

Choice C is correct. Fluorine and chlorine are in the same column (family) of the periodic table. Fluorine hasits valence electrons in the n = 2 quantum level, while chlorine has its valence electrons in the n = 3 quantumlevel. The larger the quantum number, the easier it is to remove the electrons and thus the lower the ionizationenergy. Fluorine is more electronegative than chlorine, so choice A is wrong and eliminated. The effectivenuclear charge is found from both the number of protons and the number of core electrons. It is true that chlorinehas a greater nuclear charge than fluorine, but chlorine has more core electrons than fluorine. This impliesthat the effective nuclear charge is approximately equal for the two halogens, eliminating choice B. Thelarger radius of chlorine implies that the valence electrons are farther from the nucleus on average than thevalence electrons of fluorine. This makes the ionization energy of fluorine greater than that of chlorine andconsequently makes choice C the best choice. Choice D is nonsensical, so ignore it.

Choice B is correct. The term "oxidation" refers to the loss of an electron. The lower the ionization energy, theeasier it is for an element to lose an electron. The chart lists the first ionization energies, therefore it can beinferred that the element with the lowest first ionization energy is MOST easily oxidized. Of the choices,magnesium has the lowest first ionization energy. Choice B is the choice of winners... be a winner.

Choice B is correct. The relative ionization energies of krypton, bromine, and selenium are predictable, becausethey are in the same row (period) of the periodic table. By comparing the three to chlorine, argon, and sulfur,a trend can be determined. The ionization energies of argon, chlorine, and sulfur follow: I.E.Ar > I-E-Cl > I-E.$,so the first ionization energies of bromine, krypton, and selenium should be LE.Kr > I-E-Br > I-E-Se/ choice B.

Choice A is correct. When sodium loses its first electron, it gains a filled octet and thus stability in its valenceshell. If it were to lose the second electron, the octet would be lost and thus it would become an unstable cation.This explains the drastic difference between the first and second ionization energies for sodium. For magnesiumto have a full octet, it must lose two electrons resulting in relatively low first and second ionization energies.This answer is best explained in answer choice A. The following illustrates the point:

1st ionization: Na(g) -

2nd ionization: Na+(g)

1st ionization: Mg(g)

2nd ionization: Mg+(g)

Na+(g) + e" Na: ls22s22p63s1 and Na+: ls22s22p6 Na+ has a filled octet.Na2+(g) +e- Na+: ls22s22p6 and Na2+: ls22s22p5. Na2+ has lost filled octet.Mg+(g) +e" Mg: ls22s22p63s2 and Mg+: ls22s22p63s1. Mg+ has no filled octet.Mg2+(g) +e" Mg+: ls22s22p63s1 andMg2+: ls22s22p6. Mg2+ has a filled octet.

Passage VII (Questions 43 - 49) Transition Metal Trends

43. Choice B is correct. Table 1 lists room temperature density values of the transition metals. As a metal isheated, it expands. The density of manganese decreases with increasing temperature, because volume increasesasmass remains constant. The density is slightly less than 7.43 grams per cm3, so the answer is choice B.

®Copyright © by The Berkeley Review 153 Section II Detailed Explanations

44. Choice D is correct. The first, second, and third rows of the transitionmetals followsimilar trends as you moveleft to right across any particular row. The metals from the answer choices are all in the second-row of thetransitions metals. No information is given for the second-row transition metals, so their relationship mustbeextrapolated from the information given for the first-row transition metals. The first-row transition metalsfollow the trend Cu >Ni >Cr >Mn for the second ionization energy. According to periodicbehavior trends,Moand Cr should exhibit similar properties, Tc and Mn should exhibit similar properties, Pd and Ni shouldexhibit similar properties, and Cu and Ag should exhibit similar properties. This means that the correctrelationship between the 4d-transition metals is found by substituting the second-row transition metals intothe relationship for the first-row transition metals. The relationship isAg >Pd >Mo >Tc, making choice Dthe best answer.

45. Choice Bis correct. The boiling point and melting point ofanelement increase as the forces holding the atomstogether increase. The greater the forces, the greater the energy required to break the forces. Because zinc hasa lower boiling and melting point than other transition metals, it can be concluded that there are weaker forcesholding zinc atoms together than the other first-row transition metals. Because zinc has a filled d-shell([Ar]4s23d10), there should beno covalent interactions between the zinc atoms. This makes choice Bthe correctchoice.

46. ChoiceC is correct. All of the transition metals listed in Table 1 lose their 4s-electrons prior to losing their 3d-electrons, implying that it is easier to lose the 4s-electrons. Because the 4s-level fills prior to the 3d-level, itcan also be concluded that 4s-electrons are at a lowerenergy level. In theory, the 3d-electrons should requireless energy to remove (since the 3d-level is of higher energy than the 4s-level). However, the 4s-electrons aremore exposed (further from the nucleus), so they are lost more easily than 3d-electrons. This makes choice Cthe best answer.

47. Choice B is correct. Within a row of the periodic table, atomic radius decreases as you move left to right inmain-group elements. Table 2 shows that it is nearly true for the first row of transition metals, with zincshowing the only notable deviation. As you move left to right across a row of the periodic table, the mass ofthe elements increases. The density of an element is measured in terms of mass per volume. As the atomicradius increases, the volume increases, so moving left to right across a row in the periodic table results ingreater mass and reduced volume. This means that the density of the element increases from left to right in theperiodic table. Data inTable 2confirms this inverse relationship, except in the case ofmanganese, where theradius and density both increase. This eliminates choices Aand C. Manganese isanexception, because ithas alarger radius due to half-filled stability, but it follows the same trend with density because itsmass is greaterthan chromium (the previous transition metal). Zinc follows the trend ofgreater radius associated with lowerdensity, so it is not an exception to that rule. Choice Diseliminated, because a general trend is observable,despite the deviation due to manganese. Choice B is the best choice.

48. Choice A is correct. When a transition metal element becomes a cation, it loses electrons from the 4s-level. Aselectrons are lost from the outer shell (valence shell), the radius must become smaller. As a note, the effectivenuclear charge is increasing, because the cation experiences less valence electron repulsion with the absence ofvalence electrons. The best answer is choice A.

49. Choice C is correct. Anegative reduction potential implies that it is less favorable to reduce thedication thanit is to reduce hydrogen ion (a proton). The unfavorable nature ofthe reduction correlates to a low ionizationenergy. The easier it is to ionize (lose two electrons to become the dication), the easier it is to oxidize theelement. The easier it is to oxidize the element, the harder it is to reduce the dication that forms. The sum ofthe first and second ionization energies of Mn is 2226 kj/mole, and the reduction potential is -1.18 volts. Thesum ofthe first andsecond ionization energies ofCo is2405 kj/mole, and the reduction potential is -0.28 volts.The sumof the first andsecond ionization energies ofFe is 2320 kj/mole, which is closer to thevalue ofcobalt(differs by 85 kj/mole) than the value of Mn (differs by 94 kj/mole). The reduction potential should be between-1.18 volts and -0.28 volts, but closer to -0.28 volts. The bestanswer ischoice C. (The actual valueforanyofyouelectromotive force trivia buffs is -0.44 volts.) The data are given below:hfa • Mn2+ +2e" I 1stand 2nd I.E. =2226 kj/mole E°oxidation = 1-18 V .-. E°reduction = "1-18 V

Fe • Fe2+ + 2 e" 11st and 2ndI.E. =2320 kj/mole E'oxidation = ??? V .-. E°reduction = ??? V

Co • Co2+ +2e" 11st and 2nd I.E. =2405 kj/mole E°oxidation = 0.28 V /. E"reduction = -0.28 V

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Passage VIII (Questions 50 - 56) Lasers

50. Choice C is correct. It is grunt work when it comes to figuring out electronic configurations! You can saveyourself some time by eliminating choices with either too many or too few electrons. Aluminum has thirteenelectrons, as do all of the choices. This means that no choices are eliminated. Choice A is out, because an s-orbital can hold no more than two electrons. Choices B and D are eliminated, because the s-orbital fillscompletely before the p-orbital begins to fill. The correct answer is choice C. If it is needed, the Aufbauprinciple can be applied to determine the filling order.

51. Choice D is correct. Monochromatic light is light that is composed of photons with exactly one wavelength (orfrequency). This would occur if every photon emitted came from exactly the same electronic transition betweenenergy levels. However, not all electrons are at exactly the same energy level, due to the close proximity ofrotational and vibrational levels. As a result, not every electron undergoes the exact same energy transition,and thus not every photon that is emitted has the same energy (and thus neither the same frequency norwavelength). This is stated near the end of the passage. Choose D to be a stellar student.

52. Choice B is correct. This question requires determining the photon's energy from its wavelength

(6.6 xlO"34 j-sec x 3.0 x 108 JSL) 198 x 1Q-26 T.mE = he =

X 330 x 10"9mThe more important part of the answer is the power of ten. From that, you must select choice B.

53. Choice Ais correct. The shorthand for the electronic configuration ofvanadium is [Arl4s23d3. For the quantumnumbers of vanadium, we are concerned only with the last electron in vanadium (V). We are thereforeconcerned with the third 3d-electron. Drawn below is the 3d level:

1° I© ©

3.3 x 10"7mE=MxlO"19J

3.3

3d mi = -2 m!= -l rri|=0 mj = +l mj=+2

4sm,= 0

Last electron is in the third d orbital, so m, = 0;Last electron is spin up, so ms = + -

The last electron is in the n = 3 level, because that is given by 3d3. The electron is in the 1= 2 level, becausethatis given by d-orbital in 3d3. The mj andms values must be derived by filling electrons into their respectivelevels. As listed in the box, m\ =0, ms =+*•/% which is choice A.

54. Choice D is correct. Half-filled stability would come into play for atoms that can promote one electron fromthe lower energy s-orbital up to the higher energy d-orbital to yield a d5 species. The term "half-filledstability" is derived from the half-filled d-level (the d-level has a maximum occupancy of ten electrons).Half-filled stability is possible for only chromium (Cr), molybdenum (Mo), and tungsten (W), because they areall in thecolumn that shouldbe s2d4. Tobea quality chemistry student, pick choice D. Copper canexcite one s-electron to fill the d-level completely (to make the species a d-^ atom).

55. Choice B is correct. Whatever goes in,must come out fornormalbehavior of light. Thereare situations wherea high-energy photon is absorbed that results in the excitation of an electron through multiple levels. Fromhere, the electron may relax (fall back to the ground state) by a variety of pathways (either graduallydissipating its energy or releasing the energyall at once). As a general rule, the energy that is absorbed is alsoemitted. Pickchoice B. The exception to this rule is phosphorescence. With a phosphorescing species, lightenergy is absorbed, and the compound undergoes a change in its electronic structure. The excited state istherefore a different complex than the ground state. When the new complex gives off energy to fall to itsground state, the photon released is not of the sameenergy as the original photon absorbed. Phosphorescenceand fluorescence were passage topicson a previousMCAT. Quantum numbers were also a passage topic.


56. Choice Cis correct. Visible light is found approximately in the range 390 x10"9 mto 740 x10'9 m. To be used inthe formula provided, the wavelength must be in terms of 10"6 m. Visible light has arange from 0.39 x10'6mto0.74 x10"6 m. The value for this range in terms of eV is 1-24/o.74 eV to 1-24/o.39 eV. This is a range from somenumber just over 1.5 to some number barely greater than 3.0, a range that fits choice C.

PassageIX (Questions 57- 63) Paint Pigments

57.

58.

59.

60.

61.

62.

63.

Choice A is correct. The reflected color of the four pigments are yellow for alizarin, blue for azurite, red forcinnabar, and blue-green for verdigris. To determine the light absorbed, you must take the complementary colorof the color observed. Theabsorbencies are thus violet for alizarin, orange for azurite, green for cinnabar, andorange-red for verdigris. The shortest wavelength is associated with the highest-energy light. Violet hasthe highest energy in the visible spectrum, so it has the shortest wavelength. The correct answer is choice A.

Choice A is correct. When a sample is heated, its electrons are thermally excited to a higher energy state.When they relax back to their ground state, light is emitted at an exact wavelength. This results in light of aspecific color, best described by choice A.

Choice Discorrect. The absorption spectrum contains black lines where the complementary colors of orange-red(the colors reflected by realgar) should be. The complementary colors are blue-green, so the best answerselection is choice D.

Choice C is correct. The reflected color of verdigris (Cu(C2H302)2-Cu(OH)2) is green-blue, so the absorbedcolor is red-orange. Red and orange lie at the low end of the visible spectrum as far as the energy is concerned,so the wavelength lies at the high end of the visible spectrum. The spectrum goes red-orange-yellow-green-blue-violet, so the wavelength oforange light falls below 700 nm, and the wavelength of red is around 700 ran.This means that the wavelength is less than 700 nm by a small amount. The best answer is 650 nm, whichmakes the best answerchoice C. Therange of visible light in theEM spectrum is givenon page 113.

Choice D is correct. Because white lead appears white in color, no light hasbeen absorbed in the visible range.When no light is absorbed, white light (the incident light) is reflected. The d-d transition associated withwhite lead must lie outside of the visible range. The best answer is thus choice D.

Choice D is correct. A reflected color can be seen only when white light is reflected offof it. This means thatreflected colors cannot be seen in the dark (absence of white light). Candles cannot be seen in the dark, so acolored candle is made from a wax that contains a dye that exhibits reflected color. Clothes cannot be seen inthe dark, so a fabric dye exhibits reflected color. If you wish to argue that there are certain glow-in-the-darkfabric dyes, you're absolutely right. Glow-in-the-dark dyes exhibit emitted color. You are wise in the ways oftrivia. Unfortunately, you get zero credit on this question, because it is not the best answer. Knowing aboutspecial cases like that is a great way to impress your peers, but ithurts you on a standardized exam. Ink frompens cannot be seen in the dark, so ink exhibits reflected color. Gas-filled light tubes (i.e., neon lights) can beseen in the dark, soa glowing gas-filled light bulb exhibits emitted color. The best answer is choice D.

Choice A is correct. The fastest decomposition is observed in the organic pigment. This is stated in thepassage. The inorganic pigments are already oxidized, so they should remain air-stable for some time. Theorganic pigments can oxidize in air, so they do not last as long. The only organic compound among the choices(and in Table 1) is 1,2-dihydroxyanthraquinone, which is the pigment of alizarin. This makes choice Acorrect. Note that organic pigments contain conjugated 7t-networks and are found inoil-based paints.

Passage X (Questions 64 - 69) Fluorescence and Phosphorescence

64. Choice B is correct. As stated in the passage, a spin flip is associated with the process ofphosphorescence. Inthe ground state, two electrons in the same orbital must be spin-paired (have opposite spins). Once an electronhas been excited to a higher electronic level where it occupies the orbital alone, it is free to flip its spin.Equally, the electron that remains in the ground state may also flip its spin. A spin flip changes the totalenergy of the system. The excited electron may not be able to relax back to its original level, since it shares thesame spin as the lower level electron now. When the electron falls back to a lower level, it falls to a differentenergy level, which emits a different frequency of light than it absorbed. Whether or not you know whatphosphorescence is, the answer is given in the passage. The best answer is choice B.


65. Choice A is correct. A diatomic molecule can gain and lose energy in several ways, including changingelectronic levels, vibrating at different frequencies, and rotating at different frequencies. In a diatomicmolecule, when an electron is excited, the valence shell increases, which affects the bond length and bondstrength. The diatomic molecule ismore susceptible to changing its vibrational frequency. This is the physicalcause of fluorescence. When an electron in the diatomic molecule is excited, a photon is absorbed. Becauseenergy is dissipated from the excited state in the form of rotational and vibrational energy, themolecule is in alower energy state. When it finally relaxes back to its ground state, the energy of the photon emitted is lessthan the energy ofthephoton absorbed. The energy ofthe photon absorbed isequal to theenergy ofthephotonemitted plus the dissipated energy (vibrational and rotational transitions). This eliminates choices B and D.As the wavelength of a photon increases, it has less energy. This implies that the photon emitted (being ofless energy)has a longer wavelength than the photon absorbed. Thismakes choiceA correct.

66. Choice C is correct. "Because the electronic energy levels are not singular levels and because the transition israndom, it is impossible for a molecule to absorb or emit light in such a manner that all of the photonssimultaneously have the same frequency (or wavelength). For this reason, monochromatic light is notphysically possible." These last two sentence from the passage state that monochromatic light (light of onewavelength) is not possible, and give the reasoning for that. Because monochromatic light is not possible,choices B and D are eliminated. The reasoning has to do with energy levels, not the quantization of light, sochoiceC is the best answer. This question about a difficultconcept is actually an easy question to answer, if youdon't get intimidated. Learning to manage the intimidation factor associated with seeing new information is apart of your MCAT preparation.For the sake of learning the concept, we shall look at what choice C is stating. The diagram below shows twoscenarios, one where electronic transitions are not coupled with vibrational transitions (on the left) and theother one where electronic transitions are coupled with vibrational transitions (on the right).

67.

Electronic Transition (no vibrational transitions)

~ EExcited

JGround

Electronic transitions occur without vibrationaltransitions, so a single energy transition is possible.Only one photon is emitted, which would result inmonochromatic light.

Electronic Transition (with vibrational transitions

V

EExcitedV2EExcitedVlEExcitedM)

^_^ EGroundV2fipSf EGroundVl

EGroundV0Electronic transitions occur with vibrationaltransitions, so a multiple energy transitionsare possible. Multiple photons are emitted, sothe emission is polychromatic light.

From the diagram, it can be seen that when vibrational energy levels are closer together than electronic energylevels, the transitions can couple. Single transitions between electronic levels are not possible, although singletransitions between vibrational levels appear to be possible. However, vibrational transitions couple withrotational levels, so infrared emissions are not of single wavelength. The best answer is choice C.

Choice B is correct. The arrow in Figure 1 that represents emission due to fluorescence is longer than the arrowthat represents emission due to phosphorescence, so the energy associated with fluorescence is greater than theenergy associated with phosphorescence. This means that the light from fluorescence is of shorter wavelengththan the light from phosphorescence. Choice B is the best answer. Choices A and D are the same answerworded differently (if fluorescence emission is of longer wavelength than phosphorescence, thenphosphorescence must have emission of higher energy than fluorescence), so both choices should have beeneliminated (assuming as we do that there is only one best answer per question).

68. Choice A is correct. The shortest wavelength of light belongs to light of the greatest energy. This occurs withthe transition from the highest excited state relaxing to the lowest ground state. Choices C and D areeliminated immediately, because the transitions they represent are increases in energy, which absorb light,not emit light. The best answer is choice A, because the excited state is the highest of the choices left (A andB),and both of the choices left drop energy to the same ground state.

Copyright© by The Berkeley Review® 157 Section II Detailed Explanations

69. Choice Bis correct. Fluorescence is possible with molecules, because molecules can exhibit vibrational energytransitions. This allows for the dissipation of energy via of heat. Choices A and C true, so they areeliminated. Because atoms do not have vibrational energy transitions (they have no bonds, so they have nobending and stretching modes of their bonds), atoms may not necessarily exhibit fluorescence. It is possible toconvert ultraviolet light to the lesser-energy visible light by fluorescence, but it is not possible to convertvisible light to the higher-energy ultraviolet light by fluorescence. This makes choice Bthe false statement.

Passage XI (Questions 70- 76) Flame Test

70.

71.

72.

73.

74.

75.

Choice D is correct. The definition of paramagnetic is having at least one unpaired electron. The electronicconfiguration for Na+ is ls22s22p6. All electrons are paired, because the octet is complete, eliminating choiceA. The electronic configuration for Sr2+ is ls22s22p63s23p64s23d104p6. All electrons are paired, because theoctet is complete, eliminating choice B. The electronic configuration for Cu+ is ls22s22p63s23p63dlu (copperhas filled d-shell stability, and it loses its 4s-electron before its 3d-electrons). All electrons are pairedbecause each level is filled, eliminating choice C. The electronic configuration for Cr3+ is ls22s22p63s 3p 3d(first-row transition metals lose their 4s-electrons before losing their 3d-electrons). Not all electrons can bepaired, because there is an odd number of electrons. Because Cr3* is paramagnetic, choice Dis the best answer.Choice B is correct. Chromium has half-filled d-shell stability, giving it an electronic configuration ofls22s22p63s23p64s13d5. The correct choice is answer choice B. Without the half-filled stability, the answerwould have been choice A. Molybdenum (Mo) and tungsten (W) also exhibit half-filled stability.

Choice Dis correct. The largest transition is associated with the greatest energy. Vanadium dication (V2+)produces violet light, therefore the highest energy is associated with vanadium dication. Pick choice D.

Choice C is correct. It can be observed from the data in Table 1 that as the first or second columns in theperiodic table are descended, the light emitted from the transition is of progressively higher energy. Thismakes statement I a true statement. Because statement I is not included in choices Band D, choices Band D areeliminated. It can be deduced from the answer choices that remain that statement IIImust be false. To verifythis, copper has the electronic configuration [Ar^s^d10, so Cu+ has the electronic configuration [Ar]3d10.Nickel has the electronic configuration [Ar]4s23d8, so Ni2+ has the electronic configuration [Ar]3d8. StatementIII is in fact a false statement, because Cu+ and Ni2+ do not have the same electronic configuration. Thetransition for potassium cation (K+) emits purple light, while the transition for chromium trication (Cr3+)yields red light. The transition for K+ is of higher energy than the transition for Cr3+, making statement IItrue and choice C correct.

Choice B is correct. Nickel dication (Ni2+) has the electronic configuration ls22s22p63s23p63d8. The lastelectron is the eighth electron in the 3d-orbital. The principle quantum number is given as 3and being in a d-orbital makes 1equal to 2. From the answer choices, this is already known. To obtain the mi and ms values, theelectrons must be filled into their respective d-orbitals. This is drawn below:

<>i © ©, 0 ©i 0 &l ©JNi2+:

mi = -2 rri[ = -l rri|= 0 m! = +l mi = +2

Last electron is in the third d orbital, so rq=0; Last electron isspindown,so ms=- -

The correct answer is therefore choice B.

Choice D is correct. Because violet light is higher inenergy than orange-yellow light, which is in turn higherenergy than red light, the electronic transitions for the group Ications increase as the column is descended. Thetransition for rubidium should therefore be of greater energy than violet light, which makes the transitionemit ultraviolet light. Ultraviolet light is not detected by the human eye, so the flame from heating rubidiumshould appear colorless. The best answer is choice D.

®Copyright © by The Berkeley Review 158 Section II Detailed Explanations

76. Choice C is correct. An emission spectrum shows just the color emitted by the compound after it has beenexcited. The color observed in the flame test is an emitted color, so it is present in the emission spectrum.Crimson was emitted by lithium cation, so the emission spectrum is simply a bright red line. The best answer ischoice C. Otherminoremissions maybeseen, but theywon'tbe as intense as the red emission.


77.

78.

79.

80.

81.

Glyphosate

Choice Cis correct. Two isotopes (of neutral elements) have the same number ofprotons and electrons, but adifferent number ofneutrons. Choice Ahas onemore proton than phosphorus-32, choice Bhas one more electronthan phosphorus-32, and choice Dhasoneelectron less thanphosphorus-32. This eliminates choices A, B, andD. Choice Chas amass that isone less with the same number of protons as phosphorus-32, soit must have oneneutron less. This makes choice C the correct answer.

Choice B is correct. Nitrogen in neutral molecules makes three bonds total (in this case, the three bonds are a 11sigma bonds), and has one lone pair of electrons. The lone pair of electrons repels the electrons in the threesigma bonds to form a trigonal pyramidal orientation about the nitrogen. This can be confirmed when looking atthe hybridization of nitrogen (sp3). The best choice is therefore answer B. Drawn below is a three-dimensional picture of Glyphosate with the nitrogen isolated:

v",,,CH2C02Na+

CH2OP03H+ Na+

Choice Ciscorrect. The concentration of 32P label atconsecutive half-life points along the first-order decay is:188 -» 94 -> 47 -> 23.5. Each arrow represents one half-life, so after three half-lives the concentration is lessthan 25 ppm (it has decayed to 23.5 ppm). To reach a concentration of 25 ppm, it takes a little less than threehalf-lives. The best answer is a little less than 42 (3 x 14)minutes. Choice C, 41 minutes, is the best answer.

Choice C iscorrect. An ionic compound ismade upof ions. The quickest way, without just knowing the answer,is to look for metals suchassodium. Ammonia is held together by three covalent bonds (sigma bonds). Carbondioxide is held together by two covalent bonds (two doubles bonds made up of one sigma bond and one 7t-bondeach). It isonly in compound III, Na2PC>3H, thatwe find ionic bonds. Choice C, III only, is the best answer.

Choice Ais correct. Because 32P remains chemically equivalent after nuclear decay, it is phosphorus. Thismeans that a proton was neither gained nor lost in the process. When a neutron is lost, a phosphorus-32 isotopebecomes a phosphorus-31 isotope. This makes choice A the best choice listed. An alpha particle contains twoprotons and twoneutrons, so the loss ofan alpha particle would form aluminum, eliminating choice B. The lossofa beta particle converts a neutron intoa proton, which would form sulfur, eliminating choice C. The loss of apositronconvertsa proton into a neutron which would form silicon, eliminating choice D. The answer choicesdid not list a gamma ray. Agamma ray is a high-energy photon that, when given off, does not change thechemical behavior, either. This was not listed as a choice, but it is food for thought.

Passage XIII (Questions 82 - 88) Technetium Decay

82. Choice A is correct. Beta decay is the loss of an electron from the nucleus. No mass is lost, therefore the massshould not change. Choices Cand Dare eliminated. Charge must be conserved, so losing a negative chargeshould increase the atomic number by 1. This makes choice A the best answer. The reaction is shown below:

210Po _> 2jgAt + 0e

83. Choice Cis correct. An alpha particle is a helium nucleus (mass =4amu and z =2), so the loss of an alphaparticle decreases themass by 4and theatomic number by2. 250Bk is4mass units less than254Es, so choice A iseliminated. 234Th is 4mass units less than 238U, so choice Bis eliminated. 221Bi is only 2mass units less than223Fr, sochoice C is the best answer. 243Pu is 4mass units less than 247Cm, so choice D is eliminated. Whatelse but berkelium could be right?

Copyright© by The Berkeley Review® 159 Section II Detailed Explanations

84. Choice C is correct. From the passage, the half-life is given as six hours. The twelve-hour duration istherefore a total of two half-lives. The initial concentration should therefore be cut in half two successivetimes to determine the final concentration. The math is as follows: 120 uCi —> 60 uCi —> 30 u\Ci. The finalreadingis 30u.Ci, so thecorrect choice is choice C.

85. Choice Bis correct. Electron capture bya nucleus decreases the positive charge (converting a proton into aneutron), which reduces zby1. This converts element #6(carbon) into element #5(boron).

12C + 5e -> 12BThe question is asking for anuclear process that forms boron. Positron capture increases the nuclear charge byone, so positron capture by boron cannot yield boron. It yields carbon, one atomic number higher, whicheliminates choice A. Positron decay decreases the nuclear charge by one, so positron decay bycarbon yieldsboron, one atomic number lower. This makes choice B thebest answer. Beta capture decreases the nuclearcharge by one, so beta capture by nitrogen yields carbon, one atomic number lower. This eliminates choice C.Alpha decay decreases the nuclear charge by two, so alpha decay by oxygen yields carbon, two atomic numberslower. This eliminates choice D. Thebest answer is choice B, the only choice that didn't form carbon. Theprocesses in choices A,B,C,andDareshown below.Choice A: ijB +Jfi -> aJC; Choice B: igC -> 12,B +?fi; Choice C: 14,N +_?e -> ^C; Choice D: ijO -> ^ +\a

86. Choice D is correct. A gamma ray is high-energy electromagnetic radiation, not a particle. The energyassociated with a gamma ray is greater than the energy associated with an x-ray. Because it is a photon(energy) and not a particle, a gamma ray is massless and without charge. When the nucleus of what emits agamma ray, it drops from anuclear excited state to a lower (and possibly ground) state, as mentioned in thepassage. No mass is lost by gamma emission. Choice D is the best answer. Choice Adescribes an alphaparticle, and choice Bsort of describes abeta particle (the charge is negative one, but it is not necessarily thenuclear charge). Choice Cis not a common nuclear particle and is probably a conglomeration of subatomicparticles.

87. Choice B is correct. The electronic configuration for 99Tc is no different than for any other isotope oftechnetium. Although isotopes have a different number of neutrons, they have an identical number of protonsand an identicalnumber ofelectrons in theirneutralstate. Technetium is elementnumber 43,so it has 43protonsand 43 electrons as a neutral element. The filling ofelectrons follows standard Aufbau principle rules, whichmakes answer Bthe right choice. Itis sometimes ashortcut to look at the periodic table and see where the lastelectron falls. In the case of technetium, it is directly below manganese, so its last electron should be a d -electron (like manganese). The last electron is ina 4d-orbital, sochoice Bisbest.

88. Choice Ais correct. Gamma decay just involves the loss ofnuclear energy, and nota particle. Technetium goesfrom an excited nuclearstate to a lowernuclear stateafterit undergoes gamma decay. No nuclearparticles arelost or gained by technetium, so its mass and atomic number remain the same. This means that the nuclearcomposition of the element remains the same, so the element remains the same. The correct choice is thus A.

Passage XIV (Questions 89 -95) ColdFusion89. Choice Bis correct. Reaction 3andReaction 4show fusion oftwodeuterium atoms. The fusion of twodeuterium

atoms can generate either atritium (3H) and hydrogen (aH) (as shown in Reaction 3), or one neutron and heliumisotope (3He) (as shown in Reaction 4). The only particle in the answer choices that is not shown as aproductis hehum-4, so the correct answer is choice B.

90. Choice Bis correct. We see inthe passage, that the major piece of evidence for believing anuclear reaction hadtranspired was the release of more heat than can be explained by the electrochemical cell reaction. Theexpected nuclear products were observed only in low concentrations however. The best answer is choice B. Theabsorption oremission ofanelectron (beta particle) had noeffect onthe reaction.

91. Choice Dis correct. As described inthe passage, cold fusion is a nuclear fusion reaction that can becarried outat room temperature. The example inthe passage takes place inanelectrode within a test tube. Because moreenergy is released than expected, itcan be assumed that the fusion reaction is exothermic (because of the excessheat that was released). The correct answer is choice D. Pick D, and you'll smile brightly.

Copyright ©by The Berkeley Review® 160 Section II Detailed Explanations

92. Choice A is correct. In the passage, we read that nuclear power plants employ the fission reaction, because todate, fusion reactions require both high temperature and strong magnetic fields. Fusion reactions require moreinputenergy than the fusion reaction releases, making them endothermic and unfavorable as a source ofenergy.The best answer is choice A.

93. Choice C is correct. A fusion reaction results in the combination of the nuclei of the twoparticles undergoingfusion. Answer choice C involves the combining of twoparticles to form one product (with an atomic numbergreater than the two incident particles). This defines fusion. Choices A and B are both fission, and choice Dinvolves the emission of a gamma photon via thedrop from a nuclear excited state to the nuclear ground state.

94. Choice B is correct. An alpha particle has a mass of four and two protons, and a deuterium nucleus has a mass oftwo and oneproton. After losing an alpha particle and deuterium nucleus, themassof an element drops bysixand the number of protons drops by three. Choices Aand C are not sixmass units less, so they are eliminated.Choice D is not three atomicnumbers less,so it is eliminated. Theprocess is shown below.

2gU-»fa + 2H+2g|Ac

95. Choice C is correct. A change in the atomicnumber results from a change in the number of protons in the nucleus.The alpha particle has two protons, so the capture of an alpha particle increases the atomic number by two.This eliminates choice A. The capture of a beta particle converts a proton into a neutron, so the atomic numberdecreases by one. This eliminates choice B. The tritium nucleus carries one proton, so capture of a tritiumnucleus increases the atomic number by one. This eliminates choice D. A neutron capture increases the mass byone, but does not affect the atomic number. Choice C is the best answer.

Questions 96 -100 Not Based on a Descriptive Passage

96. Choice D is correct. "Quantized energy levels" refer to states of finite energy where electrons may exist. Forthis exam, you should know conceptually what behavior is expected. The existence of a neutron or proton at thenucleus may exert Coulombic forces on an orbiting electron, but it does not have any bearing on the quantizationof energy levels. Choice A is therefore eliminated. The scattering of x-rays by thin sheets of material (metalfoil in the Rutherford experiment) shows that matter is mostly empty space, with dense uniformly spacednuclei. Choice B is eliminated. The bending of any particle when moving through a magnetic field simplyindicates that the particle in motion has a net charge of some kind, and that the direction of motion is not inline with the field. Choice C is eliminated. Distinct lines (which can be reproduced in separate trials) showthat the same amount of energy is absorbed when an electron is excited. If the transition between levels is aquantized value (an exact quantity), then it seems logical that the energy levels are also quantized. Thismakes choice D the correct answer.

97. Choice D is correct. As the wavelength of a photon increases, the energy of the photon decreases. This questionis testing your recall of relative energetics of electromagnetic radiation. The lowest energy of the choices givenis associated with infrared light. This makes D the best choice.

98. Choice A is correct. The value of Zeff for H is +1. The value of Zeff for He+, the starting point for the secondionization ofhelium, is +2. Because the ionization energy is proportional to Z2, the second ionization energy ofhelium should be four times as great as the ionization energy of hydrogen. Both electrons in question are beingionized from the ls-orbital, so n = 1 for both ionization energies. You need consider only the effective nuclearcharge. To feel the sensation of correctness, pick choice A. Drawn below are the respective ionizationreactions:

n = l

H • H+ + le"


He

n=l

• He + le"


99. Choice D is correct. Sodium cation has ten electrons, which eliminates choices A and B, both with elevenelectrons. Choice C is the ground state (all electrons fill sequentially) for sodium cation, given that it has theten electrons filled in order. In choice D, there are ten electrons and an electronhas been excited from the 2p-level to the 3s-level. This leaves choice D to be the best answer.

100. Choice Bis correct. The most common shape for a transition metal with five Ugands is trigonal bipyramidal.The best answer is choice B. Square planar has only four ligands attached, so choice A is eliminated.Hexahedral doesnot exist; and if it did, hex is Greek for six, and only five ligands are attached. This wouldalso eliminate choice D. Drawn below is a chart for deterrnining molecular shapes:

Coordination Number = 2

L A L

no lone pairs (sp hybr.)Linear

Bond Angle = 180°


L

IT L

no lone pairs (sp hybr.)Trigonal PlanarBond Angles = 120°


L

I

-N.l*JTv

no lone pairs (sp hybr.)Tetrahedral

Bond Angles = 109.5°

Coordination Number = 2• •

IT L2

one lone pair (sp hybr.)Bent

Bond Angle < 120°


one lone pair (sp hybr.)Trigonal PyramidalBond Angles < 109.5°


L

.....»»»« L

L

no lone pairs (dsp hybr.)Trigonal BipyramidalBond Angles = 90°&120°



* A *

/ XL L

two lone pairs (sp hybr.)Bent

Bond Angle < 109.5°


Lq

two lone pairs (dsp hybr.)Trigonal Planar or T-shapedBondAngles = 90°, 120°, 180°


L

L^^A- L

2 3no lone pairs (d sp hybr.)Octahedral

Bond Angles = 90°


Section in

Equilibriumby Todd Bennett

A(g) + B(g) ^=^ C(g) + D(g)

Keq =

Time

Products = (Pc)(Pp)Reactants (Pa)(Pb)

Time

Fundamentals of Equilibriuma) Definitions and Terminologyb) Equilibrium Constant (Keq)c) Reaction Quotient (Qrx)d) Case Specific K-Valuese) Keq Calculationsf) Using Keq to Calculate Shiftsg) Complex Equilibriumh) Experimental Determination of K

Le Chateliers Principlea) Effect of Stressb) Perturbations and Shifts

i. Direction of Shift

Solubilitya) Definitionsb) Solubility Rulesc) Ionic Structures

i. nomenclature of Salts

ii. Polyatomic Ionsd) Solubility Product and Molar Solubilitye) Relative Solubilityf) Solubility Experimentsg) Common Ion Effecth) Separation by Precipitationi) Ion Exchange Columns

BerkeleyUr'E'V.^E'W8


•>

•>

Equilibrium Section GoalsKnow how to determine an equilibrium constant from experimental data.There are experiments that are designed todetermine achange in total pressure over time for agasequilibrium system. The change inpressure (from the time the reactants are mixed until the reactionreaches asteady equilibriumpressure) canbeused tocalculate anequilibriumconstant. The differencein pressure is theshiftrequired to reach equilibrium.

Understand Le Chatelier's principle and its effects on equilibrium systems.LeChatelier's principle predicts certainbehavior in an equilibrium systemoncea stressis appliedto thesystem. Because changing onevariable canaffect another variable (for instance, changing thepressure may also change thevolume), you must evaluate how thechemical reaction will adjust toalleviatethe applied stress in order to reestablish equilibrium.

Understand the mathematical arrangement of the equilibrium expression.Theequilibrium expression, simply put, is the ratioofproductsover reactants. Themathematicalrules require that you do not put pure liquids (solvents) or solids into the expression. If more thanone mole ofproduct orreactant are involved in the reaction, then the stoichiometric coefficient inthe balanced equation becomes an exponent in the equilibrium expression. The value of theequilibrium constant changes only with temperature.

Know the effects of the system variables P, V, n, and T on reaction equilibrium.It isnotpossible to change justonevariable in anequality. Changing theconditions of thesystemshifts an equilibrium,but it does not necessarily change the equilibriumconstant. The numericalvalue of the equilibrium constant changes onlywith temperature.

Know the relationship between molar solubility and solubility product.The molar solubility of a salt is the concentration of the salt (as measured in molarity) needed tosolvate an aqueous solution completely. The solubilityproduct is an equilibrium constant for thedissociation reaction. What makes solubility products unique is that thereactant is alwaya a solid,so the equilibrium expression has no denominator. Molar solubility is a more useful quantity toknow than solubility product, because it measures the amount of salt in solution.

Understand separation by precipitation and the chelation effect.Chelating is the formation of a Lewisacid-basebond between a lone pair-donor (ligand) and a lone-pairacceptor (central atom). Chelation changes thesolubility ofa saltby changing theconcentrationof free ionsin solution. When a ligand bindsa central metal, thereis a formation constant thatmeasures the strength of the chelation. This allows for specific ions to be removed from solutionbybinding them to form amore soluble complex ion. Ions also can beremoved from solution byadding counter-ions that form an insoluble salt.

Understand the common ion effect.The common ion effect is a twist on LeChatelier's principle as it applies to solubility. The additionof products to an equilibriummixture shifts the reactionin the reverse direction,which in the caseof solubility results m precipitation. This means that addition of an ion to solution or the presenceof an ion in solution reduces the solubility of a salt.

General Chemistry Equilibrium introduction

EquilibriumEquilibrium, as you have learned it, involves the balancing of a chemical reactionbetween reactants and products. Equilibrium is a state that is achieved when thereactants go on to form products at the same rate that the products go back toform reactants. Equilibrium reactions take place in both the gas phase andsolution state, where reactant and product molecules are free to migrate andcollide. The kinetic theory model states that molecules must collide to react.Equilibrium occurs only in a closed system, although steps in a pathway (anopen system) may be an equilibrium reaction.

Equilibrium is an odd yet obvious thing. It is the essence of nature and the foe ofpermanence. We can't beat it...EVER! The concept of equilibrium pervades notonly chemistry, but politics, economics, sociology, health...even ourrelationships. A bank represents a good illustration of this concept. There aredeposits and withdrawals continually going on, but on the whole, the amount ofmoney in the bank remains essentially constant (except following a bankrobbery, after which a new equilibrium will be established). Everything lies in abalance, and all one has to do is read the scale. Fortunately, in chemistry it is easyto read the scale. To understand equilibrium better, consider the following saga:

In a certain house with a large backyard lived an elderly man with a plum tree.Next-door to him lived a young boy who also had a plum tree. Given thatneither of these two were farmers by any means, the fruit generally fell to theground and rotted. One day, while out in his back yard cleaning up the plums,the elderly man was struck by a flash of insight; instead of using all his energy toconsolidate the plums in a trash container, he would instead simply fling themover the fence. The boy next-door saw this and decided that two could play atthis game. The next time he was out in the back yard, he flung a few plums overthe fence. This soon evolved into a daily contest with the advantage going to thelittle boy, for he was younger and quicker and could move around the yardfaster. He was therefore able to fling the unlikely projectiles over the fence atthree times the rate of the old man. At first, there were roughly equal amounts ofplums in each yard, but due to the boy's greater flinging prowess, the old man'syard gradually accumulated more. Finally, when the point was reached wherethe old man's yard had three times as much as the boy's, the overall amount ineachyard becamestable and didn't change. Because the boy had so few plums inhis yard, he spent the majority of his time running around collecting them. Theman, however, could simply stand in one place, scooping and flinging. From theneighbors' perspective, for every one that would sail through the air to the left,there was one that would sail through the air to the right. The plum-flinging hadreached its equilibrium. The ratio of the plums in each yard was equal to theratio of the rates at which the two could fling plums (known as the "plum-flinging rate"). This in essence is as deep as equilibrium gets.This example may not be the most eloquent, but it serves the purpose ofgenerating a memorable analogy to chemical reactions. It is the externalconditions that affect the equilibrium between products and reactants, and askilled chemist knows how to manipulate this relationship. Wewill address howequilibrium plays out in a solvent environment and in the gas phase. We shallview the effect of factors governing the system, such as volume, pressure,temperature, and concentration. Wewill finish by lookingat the equilibrium of asalt dissociating into aqueous solution, and the factors that are involved.Throughout all of the discussion, LeChatelier's principle will play a roll.

Copyright ©byTheBerkeley Review 165 Exclusive MCAT Preparation

General Chemistry Equilibrium Fundamentals of Equilibrium

jjiiiilliijiiM^Definitions and TerminologyEquilibrium occurs ina closed system when the rate of the forward reaction isequal to the rate of the reverse reaction (rate forward =rate reverse). The resultis that from a macroscopic perspective, the system appears not to change.However, a net change of zero does not automatically mean the system is inequilibrium. Reactions may stop when there is no reactant present, which is acase ofno reaction, notequilibrium. As an introduction toequilibrium, considerReaction 3.1, where R= reactant, P = product, kf = forward rate constant, and kr= reverse rate constant.

R ^

Reaction 3.1

The forward rate for this reaction is based on the amount of reactants and theforward rate constant (rate forward = kf[Reactants]). The reverse rate for thisreaction is based on the amount of products and the reverse rate constant (ratereverse = kr[Products]). Equilibrium is achieved when:

kf[Reactants] = kr[Productsl (3.1)

Example 3.1Which of the following graphs represents what is observed over time for areaction starting with all reactants?

Reverse rate

Time Time

SolutionChoices A and Bare eliminated, because the rates must be equal at equilibrium,andneither graphreflects this. Choice D iseliminated, because straight lines arenotvery common for graphs inchemistry. Astraight line would imply that thereaction abruptly stopped, once equilibrium wasreached. The reality is that thereaction gradually slows untilit reaches equilibrium. The best answer is choiceC, which shows equal rates after time and constantly changing rates untilequilibrium is reached.

Copyright ©by The BerkeleyReview 166 The Berkeley Review

Zia Siddiqui

Zia Siddiqui

Zia Siddiqui

Zia Siddiqui

Zia Siddiqui

General ChemiStiy Equilibrium Fundamentals of Equilibrium

Equilibrium Constant (Keq)The equilibrium constant is a mathematical quantity that is calculated for areaction at equilibrium. By definition, the equilibrium constant is theconcentration of products at equilibrium divided by the concentration ofreactants at equilibrium. Equation 3.2 shows this relationship.

Keq =[Pr°duCtel/[Reactants] «•»By applying Equation 3.2 to Reaction 3.1, we find that the equilibrium constant isequal to the forward rate constant divided by the reverse rate constant.

Example 3.2For Reaction 3.1, if the forward rate constant is four times the reverse rateconstant, what is Keq after a catalyst hasbeen addedthatdoubles therateof theforward reaction?

A. Keq =2B. Keq =4C Keq = oD. Keq =16

SolutionAdding a catalyst lowers the activation energy, so the reaction speeds up. In thisexample, the forward rate is doubled, because the forward rate constant isdoubled. However, the activation energy is lowered for the reverse reaction aswell. The reverse reaction rate is also doubled. The ratio of the forward rate tothe reverse rate remains the same. This means that equilibrium is the same, sothe equilibrium constant (Keq) is thesame. Equilibrium is achieved sooner, butthe same equilibrium conditions are reached. The forward rate is four times thereverse rate, so the equilibrium constant is equal to 4, choice B.

Table 3.1 lists some rules about the equilibrium constant you must know.Among them is the rule that the numerical value of Keq changes only withtemperature. Knowing this rule would have made Example 3.2easier to solve.

Stoichiometric values from the balanced equation become exponentsin theKeqexpression.Do not include solids or pure liquids in the Keq expression, onlysolutes (for Kc) andgases (for Kp).The numerical value of Keq varies only with changing temperature,not with catalysts, pressure, volume, or moles.

Table 3.1

When multiple reactants or multiple products are present in the reaction, whichis usually the case, rule # 1 applies. Because the concentration of a solid or pureliquid (solvent) does not change, their values are constant. For this reason, theyare ignored in the Keq determination. As a rule of thumb, only molecules thatare free to move and are rarely in contact with other reacting molecules affect theequilibrium expression. Lastly, because the equilibrium constant is a measure ofenergy distribution, only a change in temperature (a measure of the system'senergy) changes the valueofKeq.

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Consider Reaction 3.2,

kf2A(g) ^ k " B(g) + C(s)Reaction 3.2

Applying the three rules toReaction 3.2 generates thecorrect expression for Keq.TBirciWe startbyplugging inproducts overreactants: Keq =[A]

TBITCIInvalid, because stoichiometric coefficients must be exponents .*. Keq = —

[Al2TBI

Invalid, because Kea does not include solids or liquids/. Keq =—z-4 [A]2

Kea can be written as either Kc =-M-M"1 orKp =—S_ atm."14 [A]2 K (PA)2

Reaction Quotient (Qrx)When a reaction is not at equilibrium, the expression of products over reactantsis said to be the reaction quotient (Qrx)- The relationship betweenKeq and Qrxdictates the direction in which a reaction proceeds to reach equilibrium. WhenK> Q, the denominator of Q (reactants) is too large and the numerator of Q(products) is too small. To establish equilibrium, the reaction must shift to theright. The reaction shifts to the right to increase the products (numerator) anddecrease the reactants (denominator). When K < Q, the denominator of Q(reactants) is too small and the numerator of Q (products) is too large. Toestablish equilibrium, thereaction mustshiftto theleft. The reaction shifts to theleft to decrease the products (numerator) and increase the reactants(denominator). A shortcut to determine the direction the reaction proceeds toreachequilibrium involves drawing the relationship of K and Q alphabetically,and then converting the < or > sign into an arrow. For example: K >Q becomesK—>Q, so the reaction moves right to reach equilibrium, because the arrow ispointing to the right.

Example 3.3When the reaction quotient is greater than theequilibrium constant, which of thefollowing is NOT true?A. The system has toomany products and too few reactants.B. The reaction is displaced from equilibrium.C. The reaction must shift in the forward direction to reach equilibrium.D. The reverse reaction rate is greater than the forward reaction rate.

SolutionWhen the reaction quotient is greater than the equilibrium constant, the systemhas an excess of products and shortage of reactants, relative to equilibrium.ChoiceA is a valid statement. The system is not at equilibrium, so choice Bis avalid statement. To reach equilibrium, the reaction must have a net shift in thereverse direction to reduce the amount of products and increase the amount ofreactants. This means that the reverse reaction rate is greater than the forwardreaction rate,making choice D valid. Thesystem cannothave a net shift in theforward direction, so choice C is an invalid statement.


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Case Specific K-valuesAll equilibrium constants obey the same rules, but depending on the reaction,there may be special features that recur. Different reactions have special K-values. Table 3.2 lists the six typesofKeq values weshalladdress.

Ktype Type of reaction to which the K appliesKp Kgq for the reaction ofgases. Values areinpressure units.Kc Keq for the reaction ofsolutes. Values are in concentration units.Ksp Keq forsaltsdissociating intoions. Measures thesolubility.Ka Kgq for acidsdissociating inwater. Measures theacidity.Kb Keq for baseshydrolyzing in water. Measures thebasicity.Kw Keq forautoionization ofwaterintohydronium andhydroxide.

Table 3.2

Understanding that the rules are the same for all types of K-values will enhanceyour journey through the wonderful world of equilibrium. Knowing that thecommon ion effect is nothing more than Le Chatelier's principle applied tosolubility systems is a perfect example of how the rules apply to all equilibriumsystems. There are just different names to describe the reaction conditions.

Keq CalculationsNow comes the math part, which we will handle through practice! Gettingreacquainted with equilibrium math is a matter of repetition and practice. Onceyou feel sufficiently familiarized, then move on to the next topic.

Example 3.4At 650K, the partial pressures of the component gases were determined for thefollowing reaction:

H2(g) + I2(g) ^=^ 2HI(g)Ph2=0.20 arm., Pi2 =1.50 arm., andPhi =3.00 atm.

What is theKpfor this reaction?A. 10.0 atm.B. 15.0 atm.C. 30.0 atm.D. 45.0 atm.

SolutionPossibleerrors with this problem stem from forgetting to square the numeratoror ignoringl2,becauseyou have seen it as a solidbefore (notethat this reactionisat 650K). In this case, forgetting to square the numerator would yield an answerthat is too small by a factor of 3. As long as you don't forget these things, theproblemmerely involves doing your math quickly. Themath setup is as follows:

Kp = (Phi)2 _ (3)'(PH2)(Pl2) (0.2)(1.5)

Choice C is the correct answer. To make this a more conceptual type of question,they may give the answer choices as ranges rather than exactnumbers.


= _9_=90 = 300.3 3

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Example 3.5At 323°C, there are 0.10 moles H2(g), 0.20 moles HC104(g), 0.10 moles H20(g),and 0.36 moles HCl(g) at equilibrium in a 400-mL flask. What is Keq for thisequilibrium mixture?

lHC104(g) + 4H2(g) ^=^ lHCl(g) + 4H20(g)A. 0.0362B. 0.0724

C 1.8000

D. 44.100

SolutionIn this problem, the system is already at equilibrium, so no determination ofchanges (x-values) is required. This problem is of the plug-'n'-chug (simplesubstitution) type. Moles can be used directly in the equilibrium expression,because although it is technically correct to use concentrations, in this casevolume cancels out of both the denominator and the numerator, leaving just themoles. This is true any time that the number of reactants is equal to the numberof products in either the gas phase or as solutes. The question is really just amath problem, solved in the following way:

Ke _(PHC1)(PH20)4 ,(036)(0.10) 4_0,36 -0.18 . L8^ (PhC104)(Ph2)4 (0.2

This makes choice C the correct answer.

Example 3.6At STP, the partial pressure ofNO is 152torr and the partial pressure ofO2 is 228torr. If the mixture is at equilibrium, what is the Keq at STP for the followingreaction?

NO(g) + Ctyg) ^—*- N03(g)A. 8.333 arm."1B. 4.167 atm.'1C. 1.000 arm."1D. 0.240 atm.'1

SolutionThe key piece of information in this question is STP (standard temperature andpressure). This implies that the total pressure of the systems is 760 torr. The sumof the partial pressures is the total pressure, so 760 = PNO3 + PNO + Po2- Bvsubstitution, 760 =Pno3 + 152 +228, so PNO3 = 38° torr. Because the answer islisted in atmospheres, the values in torr must be converted to atmospheres beforethey are useful. The conversion is 760 torr per atmosphere. The equation forcalculating Keq is:

„ _ Pno3 _ *%,, _ V2 .Vz.so^^"(PNOXPO.) >Y760X228/7J -(1/5X3/10) "3/50 " 6

Of the answer choices given, only choice A is greater than 8. If you pick A, youwill definitely be a star!


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GCIlCral ChemiStry Equilibrium Fundamentals of Equilibrium

Example 3.7For the following reaction, 0.20 moles Bi2S3(s) are mixed with 0.50 moles H2(g).Once equilibrium is established, 0.225moles of H2(g) remain. What is the valueofKeq for thisreaction?

lBi2S3(s) + 3H2(g) ^ * 2Bi(s) + 3H2S(g)

A. 0.57B. 0.82C. 1.22D. 1.81

SolutionThis question is best solved by determining the equilibrium concentrations forboth of the gases. Solids do not affect the equilibrium, so be sure not to includeany solids in the equilibrium expression. The equilibrium expression involvespartial pressures, but the answer is the same whether you use the partialpressures of the gases or the concentrations of the gases. This is because there isthe same number of gas molecules on each side of the reaction. The solids areignored altogether as long as they are not the limiting reagent. Becauseyou needthree times as much H2(g) as Bi2S3(s) and you have only 2.5 times as much,hydrogen is the limiting reagent in this reaction, if it were to go to completion.This means that to determine the value ofKeq, youmustdetermine the ratioofthe two gases. The values are found in the following way:

Reaction: Bi2S3(s) 3H2(g) ^ 2Bi(s) 3H20(g)

Initially: 0.2 0.5Shift: zX zte •Equilibrium: 0.2 - x 0.5 - 3x

0 0

±2x + 3x

2x 3x

In this case, we can solve for the value of 3x from the information given. Atequilibrium, there are 0.225 moles of hydrogen gas remaining, so 0.5 - 3x = 0.225.This means that 3x = 0.275. There is no need to solve for x, because 3x is presentin the gas terms, and solids are not going to be considered. Plugging0.275 in for3x into the equilibrium line of the reaction chart yields:Reaction: Bi2S3(s) 3H2(g) •* 2Bi(s) 3H20(g)

Equilibrium: don't care 0.225 don't care 0.275

These numbers work nicely in determining the equilibrium constant. The mathis shown below:

Ke JPH2S)3 _(molesH2s)3 _(0.275)3 _/p.275\3 =(llf =1.223 =1.22+Gq (Ph2)3 (moleSH2)3 (0.225)3 H).225/ \9/

Only choiceD is a value that is greater than 1.22, so that is the best answer. If.you forget to cube the value, it is easy to choose answer choiceC by mistake.Choices A and B are eliminated, because there are more products than reactants,so the value of Keq must be greater than 1.0. Be sure to use commonsense toeliminate incorrect answer choices. Developing intuition and learning to trustyour common sense is more important in MCAT preparation than honing youralgebra and multiplication skills.

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General Chemistiy Equilibrium Fundamentals of Equilibrium

Using Keq toCalculate ShiftsTheequilibrium constant is used tocalculate thequantityof the productsand thereactants presentat equilibrium. Thisis achieved by following the molesof eachspecies from initial conditions to equilibriumconditions, specifically by settingup a table to keep trackof the components systematically over the courseof thereaction. There are three stages to consider: the initial stage, the shift, andequilibrium. Consider Reaction3.3:

CO(g) + HzO(g) * " C02(g) + H2(g)Reaction 3.3

At 375°C, the equilibrium constant for Reaction 3.3 is 2.51 x101. The equilibriumconstant is unitless when the number of products equals the number of reactants.Consider the reaction to start with 1.0 atm of CO and 1.0 atm. of H2O. Figure 3-1shows the setup for deterrnining the shift and final pressures in Reaction3.3.

Reaction: CO(g) H20(g)

Initially: 1.00 1.00

Shift: -x zJL

Equilibrium: 1.00-x 1.00-x

C02(g) H2(g)

0 0

±£ + x

X X

Figure 3-1

As mentioned, there are three considerations in the setup. The first line is what isinitially given. The second line shows the direction of the shift and thestoichiometric consequences. Youmust be able to determine the direction of theshift by comparing the initial concentrations to the equilibrium distribution. Inthis case it was easy,becausethere are no products, so the reaction must shift inthe forward direction. The third line accounts for what is present onceequilibriumis established. Values from line 3 are plugged into the equilibriumexpression,to solvefor x, the shift in the reaction. The solution is as follows:

K a(PCQ2XPH2) = (x)(x) =^ml ^ 251 =_x?_9 (PcoXPH2o) d-x)(l-x) (1-X)2

25.1 = x2 => 5 = ^L_ =* 5-5x = x => 5=6x .\x=^=0.83(1-x)2 1-x 6

The math was simpler that it first appeared in this case. The MCATdoes notseek to test your algebra skills, as much as it tests your reasoning ability. Beingable to estimate the magnitude of the x (the shift) relative to the initial values isimportant.

Example 3.8At 773K, the Kp for the following reaction is 3.0 x10"5. If the partial pressure ofN2(g) is initially 3.75 atm., and the partial pressure of H2(g) is initially 2.0 atm,;what is the partial pressure of NH3(g) once equilibrium is established, assumingthere is no ammonia in the system initially?

N2(g) + 3H2(g) 1 * 2NH3(g)A. 0.0900 atm.B. 0.0300 atm.C. 0.0100 atm.D. 0.0010 atm.



SolutionThis is a straightforward example. Because the reaction starts with all reactants,it shifts forward to reach equilibrium. The reaction has a very small Keq andstarts with all reactants, so the shift is small. The x-term can be ignored when itis subtracted from or added to numerical values. The setup is as follows:

Reaction: N2(g) 3H2(g) •> * NH3(g)

0

** +2x

2x


Shift: - X -3x

Equilibrium: 3.75 - x 2.00-3x

Ignoring the x and 3x portions of the reactant quantities yields the following:

a (PNH3)2 = (2x)2 =3x 1Q-5 ^ 3x 10-5 =_Jx?_q (Pn^Ph/ (3-75) (2)3 3.75x8

:4xf30

x2 =30 x3x 10"5 = 22.5x 10*5 =>x2 = 2.25x10"4 .\x =Vl25'x 10-2 =1.5x 10"24

The concentration ofNH3 at equilibrium is 2x, which is 3.0 x 10"2 M. The bestanswer is choice B. This is considerably more math than the MCAT requires youto use, so think of this example as merely a step towards getting re-acquainted.

A concern you may recall from your general chemistry class involves whether itis safe to ignore x when it is either added to or subtracted from a numerical term.The x-value is ignored when the initial conditions are like the equilibriumconditions, because the shift is minimal and the value of x is trivial. In Example3.8, Keq isless than 1.0, sothere are fewer products than reactants atequilibrium.This means that hardly anything shifts over to the product side, so x can beignored. Table3.3shows caseswhen x can and cannot be ignored.

InitialConditions

EquilibriumConstant

Shift x-term?

AllReactants

Keq <IO-3 Q and K are SIMILAR .-. SMALL xSMALL shift in the forward direction

Ignore

AllProducts Kgq <10-3 Q and K are DIFFERENT .-. LARGE x

LARGE shift in the reverse directionConsider

AllReactants

Kgq >103 Q and K are DIFFERENT .-. LARGE xLARGE shift in the forward direction

Consider

AllProducts Kgq >103 Q and K are SIMILAR /. SMALL x

SMALL shift in the reverse directionIgnore

Reactants &Products Keq = 1

Q and K are SIMILAR .\ SMALL xSMALL shift in either direction

Ignore

Reactants &Products Keq » or « 1

Q and K are DIFFERENT /. LARGE xLARGE shift in either direction

Consider

Table 3.3

Understanding the math associated with equilibrium is important. Keq canbeused to find concentration values at equilibrium, and equilibrium concentrationvalues can be used to find Keq. This fits nicely into an MCAT experiment,because equilibrium constants result from experiments conducted to determineequilibrium concentrations, in several trials with different starting conditions.



Example 3.9At 92.2'C, the Kp for the following reaction is 0.2000 atm."1. If you were to placeexactly 0.200 atm. of N204(g) into a 1.00 liter vessel, what would the partialpressure ofN02(g)be onceequilibriumwas established?

2N02(g) ^f=^^ N204(g)

A. 0.025 atm. N02(g)B. 0.200 atm. N02(g)C. 0.350 atm. N02(g)D. 0.400 atm. N02(g)SolutionTo solve this precisely, it is necessary to use the quadratic equation. On theMCAT, it is unlikely that you will need to use the quadratic equation. Thisquestion emphasizes the technique of eliminating answer choices throughapproximation. First, you must use the equilibrium expression to estimate themagnitude ofx. In this case, Keq is less than 1.0, and the reaction startswithallproducts. The value of x is going to be significant (more than half shifts over). Ifhalf of the 0.200 atm. of N2O4 shifts over, then the partial pressure of nitrogendioxide is 0.200 atm. Considering that more than half of the N204(g) is going toshift, the value of N02(g) is greater than 0.200 atm. However, not all of theN204(g) can shift over (which would result in 0.400 atm. of NO2), so the answermust be less than 0.400 atm. Only choice C falls within the range of 0.200 to0.400. In a multiple-choice format, this question is rather easy to answer.

In the interest of developing an alternative method to answer the question,substitute one of the four answer choices (the one closest to the x youapproximate) into the equilibrium expression and then compare the answer youget to thegiven value ofKeq. Your answer either will equalKeq or it won'tequalKeq. If it equals Keq, you picked well (go to Las Vegas or Atlantic City withthose skills). If it does not equal K, the error can be used to zero in on the correctanswer, depending on whether your value is too high or too low. For thisreason, always start by substituting a middle value.

Reaction: 2N02(g) fc N204(g)

Initially: 0 0.200Shift: +2x -« oc

Equilibrium: 2x 0.20 - x

These numbers should be plugged into the equilibrium expression.

K _ pN2Q4 _Q.2-x-0.2-xP (PNOzJ2 <2x)2 4x*

Be sure not to substitute the answer choice values directly for x, because thevalues in the answer choices are for Pn02 which is 2x. Choices B and C fall inthe middle of the range, so choose either one to plug in. For numerical ease, let'suse choice B. IfPno2 is0-200 atm-, thenx is 0.100.

K - 0-2- x _ 0.2 - 0.1 - 0.10 _ 10 - 2.5P (2x)2 (0.20)2 0.04 4

This number (2.5) is too large (greater than 0.2000), which means that the x wechose was too small, so we need to choose a larger value for x. Choice C is best.



Complex EquilibriumA complex equilibrium is a balance between two separate reactions that share acommon reagent. In essence, it is two reactions whose equilibrium states dependon one another. In a complex equilibrium, a product in one reaction is a reactantin the other reaction. As a result, if one reaction is displaced from equilibrium,then the other reaction is also affected. When adding the component reactions,the common reagent is absent in the overall reaction and is considered to be anintermediate. The equilibrium constant for the overall reaction is found bymultiplying the individual equilibrium constants of the component reactions.This is because the reagent that disappears when you add the reactions is in thenumerator of one reaction and in the denominator of the other reaction. Toeliminate it, KeqS aremultiplied. Figure 3-2 is a sample ofcomplex equilibrium:Reaction I: 2S03(g) ^ " 2S02(g) + 02(g) KeqiReaction II: CaO(s) + S02(g) * CaS03(s) Keq2

Overall: 2CaO(s) + 2S03(g) * * 2CaS03(s) + 02(g) Keqix(Keq2)2Figure 3-2

Reaction II is multiplied by 2 to balance SO2 when you add the reactions.Because Reaction II ismultiplied by 2, its Kgq value is squared. The addition ofcalcium oxide to an equilibrium mixture of O2, SO2, and SO3 reduces the partialpressure of SO3. This occurs despite the fact that sulfur trioxide and calciumoxide do not react directly. Complex equilibrium questions are essentially justmanipulations of the overall equilibrium and of Le Chatelier's principle.

Example 3.10Consider the following complex equilibrium:

2NO(g) + 102(g) * " 2N02(g) Keqi2N02(g) 1 ' lN204(g) Keq2

2NO(g) + 102(g) •* " lN204(g) KeqixKeq2

When N2O4 gas is removed, how are the partial pressures of NO gas and NO2gas affected?A. Pfsjo and Pno2 both decrease.B. Pno and PN02both increase.C. Pno decreases,while Pno2 remains the same.D. Pno increases, while Pno2 remains the same.

SolutionThe second reaction shifts to the right to compensate for the loss of N2O4, so thepartial pressure of NO2 decreases. A decrease in NO2 causes the first reaction toshift to the right as well, resulting in a decrease in the partial pressure of NO gasand an increase in the partial pressure of NO2. However, the increase in NO2from the forward shift of the first reaction is less significant than the decrease inNO2 caused by the forward shift of the second reaction. This is because the shiftin the first reaction cannot completely replenish the lost NO2 without losing somuch NO that the reaction is beyond equilibrium. A shift never regenerates asmuch as was lost. Both NO and NO2 decrease, making choice A the best answer.


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Experimental Determination of KIn this experiment, a 500-mL closed glass flask is employed, so the system isisochoric (has Vconstant)- The pressure of the system is monitored continuallyover the courseof the experiment. The researcher tests Reaction3.4by observingthe total pressure of the systemduring a fixed interval.

H2(g) + I2(s) ^=^ 2HI(g)Reaction 3.4

Theresearcher mixes together 1.00 atm. of hydrogen gas (H2) with excess iodinesolid (I2). The totalpressureof thesystemismonitoredby a small detectorin thewall of the flask throughout the reaction, which is said to be in equilibrium oncethe total pressure of the system stabilizes. Ample time is given to allow thesystem to reachequilibrium. Thesetup for the reaction is shown in Figure 3-3.

Reaction: H2(g) I2(s)Initially: 1.00 excess

Shift: - x - x

Equilibrium: 1.00 - x who cares?

2HI(g)0

+ 2x

2x

Figure 3-3

The reaction proceeds forward to reach equilibrium. It is unknown how muchthe reaction proceeds, so the shift value is assigned the term x. Hydrogen gasdecreases by x, while hydroiodic acid gas increases by 2x, because of thestoichiometry. Because the number of gas molecules increases, so does the totalpressure of the system. Figure 3-4shows the total pressure of the system as afunction of time. The pressures for H2 and HI are inferred from the totalpressure graph.

Pfinal

Time

Figure 3-4

Ptotalincreases from 1.00 to 1.00 + x

HIincreases from 0 to 2x

H2decreases from 1.00 to 1.00 - x

Keq is calculated from Ptotal> because the shift in the reaction (x) isequal to AP.For instance, if Ptotal is 1-60 atm. and Pjnitial is 1.00atm., then AP is 0.60atm. IfAP is 0.60 arm., then at equilibrium Phi is 1.20 atm. and Ph2 is 0.40 atm. Theequilibriumexpression for the reaction does not include iodine, because it is asolid. Plugging into theequilibrium expression yieldsa numericalvalue forKeq.



Determination ofK

(Ph£=(2x£ =(t2)^=(1.2)(1.2) =1^ 64 PH2 1.0-x 0.4 0.4 0.4

This experiment determines the value of Keq under the ambient reactionconditions. Changes to the system are made to study equilibrium further.

Effect of the Addition ofReactantOnce the pressure stabilizes at 1.60 atm., the researcher alters the system byadding 0.10 atm. of isotopically rich hydrogen gas (predominantly moleculardeuterium, 2H—2H). The total pressure of the system increases as hydrogen gasis added to the system. After the addition is complete, the total pressurecontinues to increase, gradually slowing until it reaches a stable value of 1.76atm. Ptotal f°r me system increases by 0.16 arm., more than the 0.10 atm. added.Analysis using infraredspectroscopy confirms that both 2H—I and 2H—H areformed. From these observations, it is concluded that addition of a reactant gasshifted the reaction in the forward direction to a new distribution of compounds.Substituting the values for Ph2 and Phi in the equilibrium expression revealsthat Keq remains the same, although the partial pressure of each gas haschanged. Thismeans that addition of the reactant gas displacesthe reactionfromequilibrium, following which the reaction shifts in such a way as to reestablishequilibrium. In addition, the equilibrium is dynamic, given that the mixedhydrogen (2H—H) forms by the reverse reaction.

Following the addition of the labeled hydrogen, the researchermakes a secondaddition to the system, after the pressure stabilizes at 1.76 arm.,by adding 10.0gofisotopically rich iodine (predominantly 127I—127I). The system is continuallymonitored, but no change in pressure is detected. Again, the gases in the systemare analyzed by infrared spectroscopy The labels appear to be completelyscrambled, with H—127I, 2H—127I, H—I, and 2H—I all being observed. Fromthis, it is concluded that the addition of a solid reactant does not shift the reactionin either direction. It reconfirms that the equilibrium is dynamic, given that themixture of isotopically labeled compounds.

Amajorpoint of this experiment is to support the concept of dynamic equilibrium,a state where the system is continually reacting in both the forward and reversedirections. On the macroscopic level, there is no net change. This implies thatthe forward and reverse reaction rates are equal and that the concentrationof thereactants and products remains constant. The scrambling of the isotopic labelssupports this idea. If this were a case of static equilibrium, the isotopewould notbe incorporated into the product or back-react to form the mixed isotopereactant. Any set of questions on theMCAT that accompanies the descriptionofan experiment like this,would include a questionon dynamicequilibrium.

Such an experiment is typical for an MCATphysical sciences passage. It is toyour advantage to ponder what types of questions might be asked. Questionscould involve total pressure and its relationship to equilibriumthrough partialpressure. The shift can be rationalized using Le Chatelier's principle, soquestions on Le Chatelier's principle are probable. To round out the questions,there is the possibility of pH questions, solubilityquestions, and the relationshipbetween the reaction quotient (Q) and the equilibrium constant (K). It is morethan worth your time to make up some multiple-choice questions to accompanythis passage. If you know how to write a multiple-choice test, you willunderstand better how to take one.


General Chemistiy Equilibrium Le Chatelier's Principle

Le Chatelier's PrincipleLe Chatelier's principle is justified both mathematically and theoretically. Thebasic rule is that whatever change you make to a system that is in equilibrium,the reaction mixture will react in a way to undo the change and reestablishequilibrium. The formaldefinitionof the principle is as follows:Ifan external stress isapplied to a system at equilibrium, the system will shift itself insuch away that the stress ispartially relieved and equilibrium is reestablished.Effects of StressAn external stress on the system includes changing moles (concentration of acomponent), pressure, volume, and temperature. In all cases, except changingtemperature, the numerical value of K remains constant, and the system willcontinue to react until the K value is obtained once again. When a reactionsystem is not in a state of equilibrium, the same calculation for the equilibriumexpression (K) is used, but it is referred to as the reaction quotient (Q). If K is notequal to Q, then the system is not in equilibrium. It is important to note thatchanging an equilibriumsystemby adding a pure solid or a pure liquid does notdisrupt the equilibrium nor change the equilibrium constant.

This is why in the sample experiment on the preceding page, when iodinecrystals were added (a solid), the system was not moved from equilibrium.However, when molecular hydrogen was added (a gas), the system was movedfrom equilibrium, and it shifted accordingly, to reestablish equilibrium. It isimportant that this conceptual view of equilibrium makes sense to you, becausethe MCAT is a conceptual exam that evaluates your understanding at this level.We shall look at Le Chatelier's principle in terms of mathematics and intuitiveproblem-solving. We shall address the mathematical relationships only to theextent necessary to support our observations. In generic Reaction 3.5, themathematical aspects of LeChatelier's Principle are considered.

A(g) + B(g) ^^ C(g)

Reaction 3.5

Let's assume that Keq for the reaction is 1.00 and that initially, the partialpressure of all threecomponents is 1.00 atm. BY substitution into the equilibriumexpression, we see that the value is 1.0, meaning that the system is inequilibrium. Upon doubling the external pressure (i.e., by applying a stress tothe equilibrium), the system is displaced from equilibrium. According to LeChatelier's principle, it will shift to reduce the pressure of the system, which wepredict will be in the forward direction. Byshifting in the forward direction, thereaction goes from the side with two molecules to the side with one molecule,thereby reducing the number ofmolecules and thus reducing the pressure. Themath to support this prediction is as follows:

Reaction: A(g) + B(g)

Initially: 1.0 1.0

After Stress: 2.0 2.0

Shift: - X -X

Final: 2-x 2-x

C(g) C/A-B State

1.0 1.00 equilibrium2.0 0.5 not equilibrium± _x reacting2 + x 1.00 new equilibrium

The external pressure increased, causing internal pressures to increase. Thisgenerated a systemwhere K>Q, so to reestablishequilibrium, the reaction shiftsforward until once again Q = K.


General Chemistry Equilibrium

The numerical value of K does not change with the change in pressure, althoughthe equilibrium distribution does change.

Kea=—It—=_2+_x_ = x+2 =1=> x2.4x +4 =x +2=>x2-5x +2 =04 (PaXPb) (2-x)2 x2-4x +4

Using the quadratic equation yields a value of roughly 0.44, which means thatthe new partial pressures are Pa = 1-56 atm., Pr = 1-56 atm., and Pc = 2.44 atm.Table 3.4 shows different equilibrium distributions for Reaction 3.5. You shouldnotice that the ratio of thecompounds varies, but thevalue ofKeq isconstant.

Pa Pb PC pc/PA Pc/pAPb0.80 0.80 0.64 0.80 1.000

0.90 0.90 0.81 0.90 1.000

1.00 1.00 1.00 1.00 1.000

1.10 1.10 1.21 1.10 1.000

1.25 1.25 1.56 1.25 1.000

1.50 1.50 2.25 1.50 1.000

1.56 1.56 2.44 1.56 1.003

0.80 1.25 1.00 1.25 1.000

1.20 1.50 1.80 1.50 1.000

1.33 0.50 0.67 0.50 1.000

Table 3.4

Example 3.11What is the observed result of increasing the total pressure under isothermalconditions in the following system initially at equilibrium?

PCl3(g) + Cl2(g) ^=^ PCl5(g)

No change in the PPCI3 to PPCI5 ranoAn increase in the Pq2 to PPCI5 ranoAdecrease in the ratioof PPCI5 to (PPCI3)x (PCI2)An increase in the PPCI5 to PPCI3 ran0

SolutionIncreasing the external pressure shifts the reaction to the right, so it can reducethe pressure to counteract the stress associated with increasing the pressure ofthe system. An alternative way to look at this is to say that when the totalpressure increases, thenumber ofcollisions between molecules increases, forcingthe reaction to proceed in the forward direction (only the forward reactiondepends on collisions). If the system shifts to the right, the partial pressure ofPCI3 decreases, the partial pressure of CI2 decreases, and the partial pressure ofPCI5 increases. This eliminates choice A, because choiceA could be true only ifthe reaction did not shift. Choice B is eliminated, because an increase in the CI2-to-PCls ratio results from a reaction shifting in the reverse direction. ChoiceC iseliminated, because the equilibrium constant does not change unless thetemperature changes. The change is said to be isothermal, so temperature didnot change during the shift. Choice D results from the reaction shifting in theforward direction, so it is the best answer.

Le Chatelier's Principle


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General Chemistry Equilibrium Le Chatelier's Principle

Example 3.12Which of the following reactions at equilibriumwould NOTshift as the result ofan increase in pressure?A. CaO(s) + C02(g) •« *• CaC03(s)B. PCl3(g) + Cl2(g) ^5=^ PCl5(g)C. N2(g) + 3H2(g) ^^ 2NH3(g)D. H2(g) + I2(g) -^=^ 2HI(g)

SolutionThe equilibriumreactiondoes not shift after an increase in pressure when there isan equal number of gasmolecules on each side of the reaction. This is observedin choice D. As a point of interest,CaC03 cannot be stored in an open space forthis reason, because it will fully dissociate into CaO and C02 over time, if thecarbon dioxide partial pressure is not high enough. Also take note that I2is a gasin choice D, which means that the reaction must be at a temperature aboveambient conditions.

Perturbations and ShiftsBefore conducting an experiment, it is important to consider the properties of thecontainer in which a reaction transpires. Two common containers are the closedflask and the closed piston. Flasks are made of a rigid material, so their volumeis fixed (Vconstant)- The pressure varies when using a flask (Pvariable)- Pistonsare flexible (a wall of the piston is free to move), so the volume can vary.(Vvariable)- m piston reactions, the initial pressure equals the final pressure(Pinitial = Pfinal)/ if the volume of the piston exhibits no instantaneous change.This is to say that for a piston that starts with a stationary lid, once the lid isstationary again, the internal pressure equals the external pressure. As a result,reactions are susceptible to environmental changes (perturbations) depending onthe container. Temperature may change in each container, but its effect onpressure and volume depend on the container itself. Le Chatelier's principleshould be applied taking into account the features of the reaction container.

Le Chatelier's principle deals with changes to a system that starts in equilibrium.When you have a system in balance, a change in environment results in thedisturbance of equilibrium. To compensate, the system will shift either left orright (increasing and decreasing concentrations in doing so) to reestablishequilibrium. Thegeneral rule is that the systemwill do whatever it takes to undoor compensate for what you have done to disturb it. Table 3.5 lists the genericshifts a reaction undergoes to alleviate an applied stress.

Applied Stress System's Adjustment Direction of Shift

Add reactant Remove reactant To the RightAdd product Remove product To the Left

Decrease volume Decrease gas volume To the side with fewer molecules

Increase volume Increase gas volume To the side with more molecules

Decrease temperature React to generate heat In the exothermic directionIncrease temperature React to absorb heat In the endothermic direction

Table 3.5


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Direction of ShiftTo observe how these shifts work, each is applied to Reaction 3.6.

N204(g) ^^- 2N02(g) AH=+146 kj/molReaction 3.6

Situation #1 Add N 204(g): This perturbation results in a system with too manyreactants. The value of K is greater than Q. To reestablish equilibrium, thesystem reacts in the forward direction to absorb some of the excess reactants andformmore products. The system shifts to the right. The same result would haveoccurred had N02(g) been removed.

Situation #2 Add N02(g): This perturbation results in a system with too manyproducts. The value of K is less than Q. To reestablish equilibrium, the systemreacts in the reverse direction to form more reactants and absorb some of theexcess products. The system shifts to the left. The same result would haveoccurred had N204(g) been removed.

Situation #3 Increase the external pressure: This perturbation results in a systemwhere the partial pressures are too high. Thismeans that the spacein which thegasescoexist has been reduced, so their molecules aremore crowdedand collidemore often. To reduce this crowding, the system goes from products (twomolecules) to reactants (one molecule) and reduce the total moles of gas in thecontainer. Also, if they collidemore often, this forces the NC^ molecules to formbonds and thus dimerize to N2O4. The system shifts to the left.

Situation #4 Increase the volume: This perturbation results in a system wherethe concentrations are too low. This means that the space in which the gasescoexisthas increased, so they are less crowded and collide less often. Using theinverse of the reasoning from situation #3, the system reacts to make twomolecules of N02(g) from N204(g). The system shifts to the right.

Situation#5 Heat the system: This perturbation results in a system where thereis too much free energy. Byheating the system, we have added energy to thereaction. To consume most of this additional energy, the system reacts in theendothermic direction, which for Reaction 3.6 is the forward direction. Thesystem shifts to the right.

Situation#6 Cool the system: This perturbation results in a system where thereis too little free energy. Bycoolingthe systemwe have takenenergy away fromthe reaction. To regenerate energy to balance this lossof energy to some degree,the reaction moves in the exothermic direction, which for Reaction 3.6 is thereverse direction. The system shifts to the left

Thesescenarios all address a gas-phase equilibrium. The shifts are similar for asolution-phase equilibrium, except that concentrations are considered, ratherthan volume changes of the container. Changes in the concentration can resultfrom changes in the volume (quantity) of solvent. The only effect is thatsituations #3 and #4 are now dilution and evaporation of solvent. The systemstill reacts by asserting the inverse of the stress done upon it. When diluted, itreacts to increase its concentration. When solvent is removed, increasing theconcentration, it reacts to reduce the concentration. The rules of Le Chatelier'sprinciple work very well, if you apply them correctly.


General Chemistry Equilibrium LeChateUefs Principle

Example 3.13Whichof the following is the result of coolingan endothermic reaction that startsat equilibrium?A. The amount of products decreases.B. The amount of reactants decreases.C. The equilibrium constant increases.D. All partial pressures increase.

SolutionCooling the systemcauses it to move in the direction that generates heat. Heat isreleased when a reaction that is endothermic as written goes from products toreactants. Endothermic reactions are viewed as follows:

Reactants + heat ^ *" Products

Heat can be treated as a reactant. As the reaction shifts to the left, the amount ofproducts decrease, the amount of reactants increase, and the equilibriumconstant decreases. This eliminates choices B and C, and makes choice A thecorrect answer. Coolinga system decreases the pressure, so in all likelihood, allpartial pressureswould decrease, not increase. Even if the reactionshifts enoughto offset the pressure decrease due to reduced temperature, that is true only forreactants. Partial pressure of products must decrease. This eliminates choiceD.You should recall that AHis positive for an endothermic reaction.

Example 3.14If the following reaction represents a system at equilibrium, indicate whichstatement is NOT true.

PCl3(g) + Cl2(g) ^^ PCl5(g)A. Increasing the pressure would cause a decrease in PC13.B. Adding PCI5 would cause an increase in PC13.C. Increasing the volume would cause an increase in PC13.D. Removing Cl2would cause a decrease in PC13.

SolutionThe term "NOT true" means false. Do not forget halfway through the problemthat you're looking for the false answer choice. A common mistake on "NOT1problems is to forget that the correct answer is a false statement. To avoid this,write either "T" or "F" next to each answer choice as you run through them. Thenchoose the answer choice with the unique letter.To compensate for increasingexternal pressure, the reaction shifts to the right, soit can reduce the pressure to counteract the increase in pressure felt from thechangeto the system. If thesystemshiftsto the right, PC13 decreases, so choice Ais true. Adding a product to an equilibrium mixture shifts the reaction to thereactant side in order to counteract the increase in PCI5 (the product). If thesystem shifts to the left,PCI3 increases, so choiceBis true. Increasing the volumedecreases the crowding in of the system. To compensate for increasing volume,the reaction shifts to the left, so it fills in the empty space created by the volumeincrease. If the system shifts to the left, PC13 increases, so choice C is true.Removing a reactantshifts the reaction to the reactant side in order to make upfor the lost reactant (Cl2). If the systemshifts to the left, PC13 increases, so choiceD is a false statement, and is thus our answer choice.


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Example 3.15For the following equilibrium reaction, what is the effect of increasing theexternal pressure?

2S02(g) + 102(g) ^^ 2S03(g)A. A decrease in the moles of sulfur dioxideB. An increase in the moles of oxygenC. A decrease in the moles of sulfur trioxideD. An increase the equilibrium constant

SolutionIncreasing the pressure crowds the molecules in a reaction and forces the reactionto shift to the side of the equilibrium with fewer gas molecules (in this case, theproduct side). This means that the moles of products (sulfur trioxide) increaseand the moles of reactants (sulfur dioxide and oxygen) decrease. This eliminateschoices B and C, and makes choice A the best answer. The equilibrium constantdoes not change with pressure changes, so choice D is eliminated.

Example 3.16N02 is a brown gas, while N204 (the product of dimerization) is a clear gas. Thetwo are in equilibrium in a 1.00 liter flask. Upon heating, the contents of the flaskbecame darker brown. What can be said about the following reaction as written?

2N02(g) ^5=^ N204(g)

A. The reaction is endothermic as written.B. The reaction is exothermic as written.C. The reaction is isothermal as written.D. The reaction is adiabatic as written.

SolutionBecause the reaction produced a darker mixture with the addition of heat, thereaction must have shifted to the left, since it formed more of the brown gas(N02(g)). This means that heat is acting as a product in this reaction. If heat is aproduct, then the reaction is exothermic as written. This makes choice B correct.The term isothermal means that there was no change in temperature during thereaction, and the term adiabatic means that there was no change in heat duringthe reaction. Both of these statements (choices C and D) are incorrect.

Example 3.17What is the effect of adding CaO(s) to the following equilibrium mixture?

CaC03(s) ^—!=• COz(g) + CaO(s)A. The products will decrease.B. The reactants will increase.C. The equilibrium constant will increase.D. There is no change in the equilibrium.

SolutionAdding a solid to a reaction mixture that is already at equilibrium has no effecton the equilibrium. This means that the correct answer is choice D. A solidaffects a reaction mixture only if the reaction is not yet at equilibrium. This is tosay that the solid is involved only when it is the limiting reagent for the reaction.


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General Chemistry Buffers and Titration Indicators

SolubilitySolubility is a popular topic on the MCAT. However, for many people, it was atopic covered only briefly during their general chemistrycourse. For this reason,we shall look thoroughly at solubility. Conceptually, solubility is the breakingapart of a lattice to allow particles to move freely in solvent. In this section, wefocus on the dissociation of a salt into water. When a salt dissociates into water,the ionic forces holding the charged species together break, and interactions withthe dipoles of the water form. The fundamental rationale behind solubilityinvolves lattice energy (the interionic forces in the crystal form), the solvationenergy (the strength of the attraction between solvent and dissociated ions), andentropy (the solute form is more disordered than the crystalline form). Whethera salt remains in its crystalline form or dissociates into solution dependscompletely on the relative energetics of the three key features. Figure 3-5showsthe processof dissociation into solvent (water) for sodium chloride.

Figure 3-5

The base of the container holds the undissociated salt, while above the salt areions in solution. Salts dissociate from the surface, where they have the least ionicinteractions with neighboring ions in the lattice, and thus are held least tightly.Surface ions have the greatest contact with solvent, so they are more susceptibleto solvation. Corners dissolve away fastest, followed by edges and then faces.

DefinitionsFor every solute, there is a maximum amount that dissolves into a given volumeof solvent at a set temperature. The solubility of most salts has been measured at25°Cand are compiled in tables (like Table 3.8). You should know how to extractinformation from a table. Information listed in solubility tables measures thedegree to which a salt dissociates into water. As a rule, some salts have a verylow solubility, leading to a low numerical value for the equilibrium constantassociated with dissociation (KSp). The equilibrium constant associated withdissociation is known as the solubility product. However, the relative KSp valuesfor a group of salts is not always a good indicator of their relative solubility. Thisisbecause the units ofsolubility product (KSp) vary withthenumber ofions.


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General Chemistry Equilibrium solubility

The subject of solubility requires being familiar with several terms. Definedbelow are eleven that are regularly encountered in this subject.Dissolving: The breakdown of intermolecular forces between molecules as a solidbecomesa solute within a solvent. The molecule remains intact when dissolvinginto solution. An example is the dissolving of sucrose into water, where atoms inthe sucrose molecule remain covalently bonded, but the forces between sucrosemolecules are eliminated.

Dissociation: The breakdown of ionic bonds between atoms within a latticestructure as a salt turns into a solute within the solvent. The crystal lattice of thesalt breaks apart when it dissociates into solution. An example is sodiumchloride dissociating into water. The ionic bonds between sodium cations andchloride anions break, and the ions are stabilized by the partial charges of water.Solvent: The species in greatest concentration into which the solute dissolves, orsalt dissociates. A solvent must be a fluid (have the ability to flow).Solute: The species not in highest concentration that dissolves into the solvent, orin the case of a salt, dissociates.Solubility: A measurement of the degree of dissolving that a solute undergoeswithin a particular solvent. The driving force for solubility is a preference forsolvation of molecules (or ions) over the lattice strength of the solid. In addition,entropy favors the dissolving process. As the solubility of a compound increases,it is deduced that either the lattice energy of the solid is decreasing, the solvationenergy of the solute form is increasing, or both effects are taking place.Saturated: Describes the state of a solution at the point where no more solid(solute) can dissolve into solution. When an aqueous salt solution is saturated,the rate of dissociation of the salt equals the rate of precipitation.Supersaturated: Describes the state of a solution where the amount of solid(solute) that is dissolved into solution is beyond the maximum amount at a giventemperature. The solution is actually a suspension that when disturbed can forma precipitate rapidly. This state can be achieved by first heating a solvent, thenadding solute to the solution until the solution is saturated at that temperature.Slowly cooling this solution causes the amount of solute in it to exceed whatshould dissolve at the reduced temperature.Solubility product (Ks„): The equilibrium constant for a dissociation reaction,determined from the molar solubility according to standard rules for calculatingequilibrium constants.Molar solubility: The quantitative measurement of the maximum number ofmoles of solid (solute) that can dissolve into enough solvent to make one liter ofsolution under standard conditions. For all practical purposes, the solvent isalways water in inorganic chemistry and thus the calculations are similar innearly every example. Molar solubility can be thought of as the x-value in thecalculation of thesolubility product(KSp).Gram solubility: The quantitative measurement of the maximum number ofgrams of solid (solute) that can dissolve into enough solvent to make onehundred milliliters of solution under standard conditions.

Common ion effect: This results in a reduction in the amount of solid (solute) thatcan dissolve into solution due to the presence in the solution of an ion that is alsopresent in the solid. This concept is similar to Le Chatelier's principle, except thatwith Le Chatelier's principle, the addition of one of the products (ions) causesprecipitation (reduced solubility). With the common ion effect, the ion causingthe reduced solubility is present in solution at the beginning of the reaction,rather than being added once the solution has reached a solubility equilibrium.


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Solubility RulesYou may recall from general chemistry a list of general rules for predictingsolubility. We will emphasize using data from tables todetermine solubility, buthere is a paraphrased version of the solubility rules. The rules as they aretypically written in general chemistry textbooks are listed on page 26 of thisbook. These rules may beused togive a good estimate ofsolubility and shouldbe used when no solubilitydata are available.

1. Most salts composed ofa +1 cation (excluding transition metals)and -1 anion are soluble in water at room temperature.

2. Nitrate (NO3") is a large anion that forms weak latticeinteractions and forms strong hydrogen bonds with water, somost nitrate salts are water-soluble.

3. Most salts containing sulfate anions (SO42") with +1 cations(excluding transition metals) arewater-soluble.

4. Most salts with -2 or -3 anions are insoluble in water, excludingthe sulfate salts.

5. Most oxide (O2-) and hydroxide anion (OH-) salts are onlyslightly water-soluble. KOH andNaOH are notable exceptionsthat are substantially soluble.

Ionic StructuresComposed ofions (charged species) held together by ionic bonds (electrostaticforces). Inionic structures, electrons arenotshared; they are transferred betweenatoms, so that atoms with a deficiency of electrons (cation) and atomswith anexcess ofelectrons (anions) are formed. A typical example of an ionic compoundis sodium chloride (NaCl).Cation: An atom in which the number of protons exceeds the number ofelectrons, thereby resulting in an excess of positive charge. A cation is apositively charged atom. Potassium cation (K+) carries a +1 charge, implyingthat there is one more proton than the number of electrons in the atom.Potassium hasnineteen protons, sopotassium cationhas eighteen electrons.Anion: An atom in which the number of electrons exceeds the number ofprotons, thereby resulting in an excess of negative charge. An anion is anegatively charged atom. Fluoride anion (F") carries a -1 charge, implying thatthere is onemoreelectron than the number ofprotons in the atom. Fluorine hasnine protons, so fluoride anion has ten electrons. Note that the name changesfrom "fluorine" (used when the atom is neutral) to "fluoride" (used when theatom carries a negative charge). Negatively charged species aregiven the "-ide"suffix.

Nomenclature of SaltsScientific convention says that when you name a salt, the name of the cationprecedes the name of the anion. That is why we refer to NaCl as sodiumchloride, rather thanchloride sodium. Therulesfornaminga simplebinarysaltare as follows:

O Name the cation before the anion in the salt.@ The cation name is derived from the element forming the cation (often

ending in "-ium")© The anion name is derived from the element forming the anion with an "-

ide" suffix added.



Example 3.18What is the proper chemical name for the salt BaF2?A. Barium difluorineB. Barium fluorideC. DifluorobaraneD. Fluorinium baride

Solution

The cation of this salt is derived from barium, and the anion is derived fromfluorine. Thus, the name begins with barium and ends with the "-ide" form offluorine (fluoride): barium fluoride, which makes choice B the correct answer.The subscript two in the formula (showing that there are two fluoride anions)need not be named, given that fluoride can be only -1 and barium is +2. The ratiois implied. It was fun to write wrong answers for this question. Granted, I am achemistry-loving geek, so my credibility when it comes to determining what iscooland what isn't is dubious, but nevertheless, choiceC is a coolwrong answer.

Polyatomic IonsOver the course of your academic science career, you may have seen the sameions recurring in a variety of contexts in different subjects. In biology, theinteractions of bicarbonate with carbonate and dihydrogen phosphate withhydrogen phosphate are essential in blood buffering and in the action of thekidney. Kidney stones result from the precipitation of calcium with variouspolyatomic ions. Because of the frequency with which molecular ions appear asa factor in many problems, the MCAT test writers expect you to be familiar withsome common ones. Table 3.6 lists a few that you should know.

-1 Anions -2 Anions

Acetate C2H3O2- Carbonate CO32-Bicarbonate HCO3- Chromate C1-O42-Bisulfite (hydrogen sulfite) HSO3- Dichromate Cr2072-Bisulfate (hydrogen sulfate) HSO4- Hydrogen phosphate HPO42-Dihydrogen phosphate H2P04- Oxalate C2O42-Hypochlorite cio- Oxide 02-Chlorite cio?- Peroxide 022"Chlorate C103- Sulfite so32-Perchlorate C104- Sulfate so42"Cyanide CN"

Hydroxide OH- -3 Anions

Superoxide o2- Phosphate P043-Nitrite N02"

Nitrate NO3- +1 Cations

Permanganate Mn04" Ammonium NH4+

Table 3.6

You should know these twenty-six ions by charge, structure, and nomenclature.Knowing a little about their common applications may help, too.

Solubility


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General Chemistry Equilibrium Solubility

SolubilityProduct andMolarSolubility(Mathematical Applications)With solubility, calculations typically involve converting between molarsolubility and solubility product (Ksp). Molar solubility and solubility productshow up together in calculations, but they are very different terms. If onesalthas a lower Ksp value than another salt, it does not mean that it is less soluble(has a lower molar solubility). Solubility isdetermined by themolar solubility ofa salt, not the solubility product. Ksp values are calculated as products overreactants, but because solids are ignored, there is no denominator. Table 3.7shows different ion combinations for salts, where x represents molar solubility.

Solubility Reaction KSn Expression Ksp Calculation

MX(s) ^=^" M+(aq) + X"(aq) Ksp = M+][xr] _v2KSp =(xXx) =xr2+,MX2(s) ^=^ Mz+(aq) + 2X"(aq) ksd =M2+][X)2 _ av3Ksp=(x)(2x)2=4x;

MX3(s) ^=^ M3+(aq) + 3X"(aq) - 1^3+1Ksp = MJ+][X"] Ksp =(x)(3x)3 =27x4M2X(s) ^=^~ 2M+(aq) + X2"(aq) Ksp =[M+]2[X2'] KSp =(2x)2(x)=4x3M3X(s) ^=^ 3M+(aq) + X3"(aq) Ksp =[M+]3[X3"] KSp =(3x)3(x)=27x

Table 3.7

Example 3.19Whatis themolarsolubility for the hypothetical compoundMX, if it is knowntohave Ksp =4.0 x10"10M2 in water?A. 1.0xlO"5MB. 2.0xlO"5MC. 4.0xlO"5MD. 2.0xlO-10M

SolutionFor an MX salt, Ksp =[M+][X"1 =x2, where xis the molar solubility. This meansthat themolar solubility is thesquare rootoftheKsp value.

Ksp -x ,x=VK^" =V4.0xlO-lu =yi0xVl0-10 =2.0xl0"5This makes choice B the best answer.

Example 3.20Whatis themolarsolubility for thehypothetical compoundMX2, if it is knowntohave Ksp =1.08 x10"7 M3 in water?A. 1.65x10"^B. 2.10xlO"3MC 3.00 x lO-3 MD. 4.80xlO"3M

SolutionFor an MX2 salt, Ksp =[M2+][X']2 =4x3, where xis the molar solubility. Thismeans that themolar solubility is thecube rootofone-fourth of theKsp value.

=^27 xVlCT^ =3.0x 10-3x =3/Ksp ?/1.08 xlO"7 ?l108 xlO

4

This makes choice C the best answer.



It is good to understand the hypothetical examples, because actual salts are donethe sameway, except the numbers are not as clean. Thekey fact to recognize isfound in the units. If Ksp is M2, then square root must be found. Square rootsare easier to solve ifthe power of ten is divisible by two. If Ksp is M3, then cuberoot is necessary. Cube roots are easier to solve if the power often is divisiblebythree. From that point it requires approximatingsquare roots and cube roots byranging the values based on squares and cubes you know.

Example 3.21What is the molar solubility oflead(II) iodide, PM2, ifit isknown to have Ksp =1.42 x 10"8 M3 in water?A. 1.19xlO-4MB. 8.83x10-^C 1.52xlO"3MD. 2.42xlO"3M

SolutionFor Pbl2, Ksp =[Pb2+][I~]2 =4x3, where xis the molar solubility. This means thatthe molar solubility is the cube root ofone-fourth ofthe Ksp value.

Ksp =4x3 /.x-y^E =/yi-42xl0-» ^1*2x10*=211*2. xVlO-^ =7355 xlO"3 =1.5xl0"3±

The cube root of 3.55 is not a common piece of knowledge. However, the cuberoot of 1 is 1 and the cube root of 8 is 2. This means that the cube root of 3.55 fallsbetween 1 and 2. This makes the best answer choice C.

In the previous three questions, you were asked to derive the molar solubilityfrom the solubility product. An alternative way to pose this question is to askyou to derive the solubility product from the molar solubility. Beforeyou attackthese problems, it is important that you make a conscious note of potentialmistakes. Remember to multiply the ions by the coefficient from the balancedequation! This is the most common error on these problems. The second trick isto remember cuberootsby thinking of common numbers cubed, suchas 23 = 8,33=27,and 43= 64.

Example 3.22What is thesolubility product ofanM2X saltwith molar solubility =5.0 x10"2 M?A. 1.25xlO"4M3B. 2.50xlO"4M3C. 5.00 xKHM3D. 1.25xlO"3M3

SolutionFor an M2X salt, Ksp = [M+]2[X2-] =4X3, where x is the molar solubility. Thismeans that theKsp valueis four times themolarsolubility cubed.

Ksp =4x3= 4x(5.0 x10"2)3 = 4x125xlO-6 =500x10'6 =5.0 xlO"4This makes choice C the best answer.

Solubility

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General Chemistry Equilibrium soiubmty

Relative SolubilityIt is difficult to predict relative solubility. The onlyway to compare is throughexperiments ordata analysis. We'll start bycomparing the solubilities ofcalciumcarbonate (CaC03) and calcium fluoride (CaF2). When a question asks for thehighest solubility, it refers to the compound that produces the greatest amount ofdissociated salt,which refers togreatest molarsolubility, not solubility product!

The solubility product of CaC03 is 8.7 x10"9 M2, and the solubility product forCaF2 is4.0 x10"11 M3. Which ofthe twosaltsismoresoluble?With questions like these, keep inmind that you never compare numbers withunlike units. Solubility products are used to determine the molar solubility.CaC03 is an MX salt, so Ksp =x2, while CaF2 is an MX2 salt, so Ksp =4x3. Thedeterminations ofmolar solubility for both calcium salts is shown below.

CaC03:x =/K^" =V8.7xlO"9 =V87xVlO'10 =9.?xlO"5MCaF2:x=^^-^The molar solubility for CaC03 is 9.3 x10"5 Mand for CaF^ it is 2.2 x10'4, butexact values are not necessary, because 9.? x10"5Mis less than 2.? x10"4 M. CaF2is more soluble than CaC03, even though it has a smaller solubility product.Relative solubility questions are asked in many ways such as: "Which saltexhibits greater solvation?" and "Which salt precipitates first?". Use molarsolubility toanswer questions thataddress relative solubility. Solubility productis employed only for calculation-based questions.

Example 3.23Which of the following saltsyieldstheLEAST amount of silver ion in water?A. AgBr(s) Ksp =7.9 x10"13M2B. AgCl(s) Ksp =1.6 x10"10M2C. Ag2Cr04(s) Ksp =9.0 x10"12M3D. Ag3P04(s) Ksp =1.8 x10"18M4SolutionYou must recognize that the choices are not all the same type of salt. Choices Aand BareMX salts, so their solubility productsmay be compared. AgBr has alower solubility product thanAgCl, soAgBr also has a lower molar solubility.This eliminates choiceB. Choice C is an M2X salt, and choice D is an M3Xsalt.To solve thequestion, youneed molar solubility values. The calculation of themolar solubility for the three remainingsalts is shown below.

Choice A: x=fK^ =Y7.9X10"13 =V79 xVlO'14 =8.9 x10"7MChoiceC: x=^^=^^^ =V^"xV^2"=1.3x 10-4MChoiceD: X=^/^=^M^2^ =V^xYl^l.6x lO^M

AgBr, choice A, has the lowest molar solubility. However, the question asked forlowest [Ag+], not lowest molar solubility. For choice C,Ag2Cr04, there are twosilver cations, and choice D,Ag3P04, has three silver cations, so the [Ag+] isactually double and triple their respective molar solubilities, making them muchtoo high to be the actual answer.

Copyright©by The Berkeley Review 190 The BerkeleyReview


Solubility ExperimentsThemolar solubilityof a salt is an empiricalvalue,determinedfromexperimentsinvolving saturated solutions. While forming a saturated solution is easy to do,determining the ion concentration is not necessarily an easy task. To form asaturated solution, enough salt is added to water so that precipitate remains onthe bottom of the container. Determining the amount of dissolved salt may beaccomplished in many ways, of which we shall discuss three. The first methodinvolves spectroscopy,where the amount of absorbed light at a fixed wavelengthdepends on the concentration of ions, according to Beer's law (Equation 1.3).Thismethod works only if one of the ions absorbs electromagnetic radiation. Thesecondmethod involvesmeasuring the amount of salt added and then collecting,drying, weighing, and then subtracting the mass of precipitate from this value.Thismethod works only if the salt is highly soluble. The third method involvesusing an ion exchange column to exchange the cation in solution for hydronium.Theconcentration of hydronium is determined via pH. Thismethod works onlyif the salt does not exhibit acid-base properties and has a cation concentration ofat least 10"5 M. You may encounter passages about all three methods.

Example 3.24What is the molar solubility of calcium carbonate in water given that the additionof 4.00 mg of CaC03(s) to enough water to form exactly 500.0 mL of an aqueousCaC03 solution yields 0.55mg of anhydrous precipitate?A. 6.90xl0-3MCaCO3B. 3.45xl0-3MCaCO3C. 6.90xl0-5MCaCO3D. 3.45xl0-5MCaCO3

SolutionYou must keep in mind the units of the question and then solve for each unit. Inother words, the answer is in terms of moles/liter, which implies that you mustfind both moles and liters to solve the question. The liters of solution are simple,in that 500.0 mL is equal to 0.500 liters. The amount of calcium carbonate thatdissociates into solution is 3.45 mg. The amount of calcium carbonate inmilligrams is converted to grams bymultiplying by 10"3, and then tomoles bydividing by the molecular mass of calcium carbonate (100 grams per mole). Themathematical set-up is shown below

Moles: 3.45 x10-3 gramsCaCOa =̂ ^ ^

Volume: 500.0 mLx 1-00L = 0.500L CaC03(aq)1000 mL

Concentration: 3.45 x10^ molesCaCQ3 =69Qx 1Q-5MCaCo3(aq)0.500 L solution

The correct answer is choice C.

Solubility looks daunting, but is actually quite simple conceptually. Don't beintimidated by difficult-looking topics. They're usually pretty easy. Someapplications of solubility include ion exchange (ion exchange columns), selectivesolubility (precipitation of a selected cation or anion), and the common ion effect.Of ion exchange, selective solubility, and the common ion effect, only commonion effect involves heavy mathematics.


Solubility

General Chemistry Equilibrium soiubmty

Common Ion EffectThe common ion effect is the reduction in solubility of an ionic solid (salt) insolution, because one of the ions in the salt is already present in solution. In away, the concept is similar to Le Chatelier's principle. According to LeChatelier's principle, when you add one of the ions toa saturated salt solution,the reaction shifts to the left (reactant side). This results in the formation of aprecipitate (more solid), so the salt is less soluble. Addition of anion to solutionreduced the solubility. According to the common ion effect, if one of theions isalready present in solution, then less salt is capable of dissociating, so thesolubility isreduced. The results are the same when anion isadded to solution;the difference between Le Chatelier's principle and the common ion effect is thetiming ofwhenthey areadded. Consider Reaction 3.7.

AgCl(s) + H20(1) ^ * Ag+(aq) + Cf(aq)Reaction 3.7

With the addition of either Ag+ or CI" to solution in Reaction 3.7, there is areduction in theamount ofAgCl thatcandissociate intoit, due to thepresence inthe solution ofoneof the ionsconstituting the salt (Ag+ or CI"). Thisis similartoLe Chatelier's principle, except that either Ag+ orCI" ispresent insolution in thebeginning rather than being added once the solution has reached a solubilityequilibrium. Questions on the MCAT may be asked from either a common ionperspective ora Le Chatelier's principle's perspective.Le Chatelier's style question: Which way will the reaction shift when Ag+(aq) isadded to a saturated aqueous silver chloride solution? According to LeChatelier's principle, adding aproduct shifts the reaction tothe reactant side, so aprecipitate forms. This means that the salt becomes less soluble.Common ion effect style question: Issilver chloride more soluble in purewater ora0.10 MNaCl(aq) solution? According to the common ion effect, because thesodium chloride solutionhas chloride ions (also found in silver chloride) alreadypresent in solution, the solubility of the salt is reduced. We shall look at thisquestion from amathematical perspective, later inthe discussion ofthis topic.The common ion effect, simply put, says that the presence of a mutual ionalready dissolved into solution reduces the solubility ofasecond salt sharing thatmutual ion. For instance, stannous fluoride is less soluble in a solution of sodiumfluoride than distilled water because of the presence of the fluoride anion.Common ioneffect questions canincorporate pHandhydroxide concentration.

Example 3.25Magnesium hydroxide isMOST soluble inwhich of the following solutions?A. Vinegar (pH = 2.5)B. Citric acid (pH = 4.5)C. LosAngeles tap water (pH = 5.5)D. Distilled water (pH = 7.0)

SolutionThis question involves the common ion effect, where the common ion ishydroxide anion. Magnesium hydroxide is most soluble inthe solution with theleast hydroxide anion, which corresponds to the most acidic solution. The mostacidic solution has the lowest pH, which makes choice Athebestanswer. This isthe reasonthat acidrain is sucha seriousproblem: it dissolves basicsalts.



Example 3.26At what pH is Ca(OH)2(s) most soluble in water?A. 3B. 5C. 9D. 11

SolutionThe pH at which Ca(OH)2(s) is most soluble is the pH at which there is the leastcommon ion (OH") present in solution. There is the least OH" present at lowerpH values, where the solution is acidic. The best answer is choice A. Note thatbasic salts are most soluble in acidic solutions.

Example 3.27(NH4)3(CoCl6)(s) is MOSTsoluble at which pH?A. 3B. 5C. 9D. 11

SolutionThe pH at which (NH4)3(CoCl6)(s) is most soluble is the pH at which there is themost OH" in solution. This is because of a complex equilibrium. The OH" insolution can deprotonate the ammonium cation (NH4+) to form ammonia, whichreduces the amount of ammonium present in solution. As the amount of NH4+in solution decreases, the (NH4)3(CoCl6)(s) equilibrium shifts to products togenerate more NH4+. The most OH" is present at higher pH values, where thesolution is basic. The best answer is choice D. Note here that acidic salts aremost soluble in basic solutions. The two equilibrium equations are as follows:

(NH4)3(CoCl6)(s) ^ * 3NH4+(aq) + CoCl63"(aq)

NH4+(aq) + OH'(aq) •* * NH3(aq) + H20(aq)As the second reaction shifts to the right, the first reaction is displaced fromequilibrium, so it too shifts to the right. This is a complex equilibrium. The tworeactions are dependent on one another.

Be able to make qualitative predictions about solubility based on both thecommon ion effect and complex equilibrium. Example 3.25and Example 3.26 areexamples of the common ion effect. Example 3.27 is an example of increasedsolubility due to the presence of a complexing ion in solution. Both of theseconcepts play a role in determining solubility. When doing calculations with thecommon ion effect, be sure not to plug variables into the solubility productformula blindly. In the case of an MX salt, for instance, KSp still is equal to[M+1[X"], but the values of [M+] and [X"] are not just x with the common ioneffect. The concentration of the ion that is already present in the solution isfound by summing the initial concentration and the additional ion formed fromthe dissociation of the salt (x). This is [X"]initial + x. In most cases, the x will beinsignificant relative to the [X"]initiai, so it can be ignored. Most common ioneffectquestions should be simple and fun after you have done enough of them.



Example 3.28Whatis themolarsolubility ofCaCl2(s) in 0.01 MNaCl(aq) solution?

CaCl2(s) ^=^ Ca2+(aq) + 2Cl"(aq) Ksp =2.5x 10"10 M3A. 2.5xlO"4MB. 2.5xlO-6MC. 2.5xlO"8MD. 2.5xlO"10M

SolutionThis question ispurely mathematical. The key fact is recognizing that there is0.01 MCI" in solutioninitially. Thesetup and solutionare as follows:

Reaction: CaCl2(s) ^ * Ca2+(aq) + 2C1"Initially: excess 0 0.01Shift: z£ • ±S ±2&Equilibrium: who cares? x 0.01 +2x

Ksp =P*2+][Cr]2 /.Ksp =(x<0.01 +2xfIf we ignore 2x: Ksp =2.5x 10"10 =(x)(0.0l)2 =0.0001(x) .\x=2.5x 10"6M

The correct answer is choice B.

Example 3.29What is themolar solubility ofCa(OH)2 in an aqueous solution at pH =13, giventhat the Ksp for Ca(OH)2 is 6.5 x10"6 M3?A. 6.5xlO"6MB. 6.5xlO"4MC. 1.2xlO"2MD. 6.5xl020M

SolutionThis question is purely mathematical aswell. You must recognize that there is0.10 MOH- present in a pH = 13 solution, because pH + pOH = 14 for anaqueous solution and [OH-] =10"POH. The equation, set-up, and solution are asfollows:

Reaction: Ca(OH)2(s) ^ * Ca2+(aq) + 20H"Initially: excess 0 0.10Shift: ix • +x ±2xEquilibrium: who cares? x 0.10+ 2x

Ksp =eCa2+][OH12 .-.Ksp =(x<0.10+2xfUpon ignoring 2x: Ksp =6.5x 10'6 =(x)(0.10)2 =0.01(x) .\x =6.5x 10"4M


Calculation questions involving the common ion effect are actually rather simple,but inmost cases the average student is not familiar with the process. Simplyplug in the value for the preexisting concentration, andsolve for x from the Kspexpression.



Separation by PrecipitationRemoval of certain ions (cations or anions) from solution can be done in aselective fashion, by precipitation of cations with various anions, according totheir relative solubility values. The least soluble salt precipitates from solutionfirst. Questionsabout selective precipitationinvolvecomparingmolar solubilityvalues. Be careful not to fall into thetrap ofthinking that the smallest Ksp is theleast soluble. Youmust consider what type of ionic system it is. The compoundwith the smallest value for x (molar solubility) precipitates from solution first.

Table 3.8 lists the solubility products for some common salts at standardtemperature. The values are listed without units, as is conventional in generalchemistry textbooks, but you should always consider the units when they areprovided.

Salt Ksp Salt Ksp Salt KspBaF2 2.4 x 10"5 CaF2 4.0 xlO"11 MgF2 6.4 x 10"9PbF2 4.1 x 10"8 SrF2 7.9 x lO'10 ZnF2 2.5 x 10"8BaS04 1.5 x 10"9 CaS04 6.1 x 10"5 MgS04 7.2 x 10"5PbS04 4.1 x 10-8 SrS04 7.9 x 10"10 ZnS04 8.3 xlO"9BaC03 1.6 x 10"9 CaC03 6.1 x 10"9 MgC03 1.1 x 10"15PbC03 1.4 x 10-15 SrC03 8.4 x 10"10 ZnC03 2.2 xlO"12Ba(OH)2 5.3 x 10"3 Ca(OH)2 1.4 xlO"6 Mg(OH)2 1.2 xlO"11Pb(OH)2 1.3 x 10"15 Sr(OH)2 4.1 x lO"4 Zn(OH)2 2.1 x 10"16

Table 3.8

Figure 3-6 shows a precipitation flow chart from a qualitative analysisexperiment designed to identify cations in solution. The cations are selectedfrom Table 3.8. All of the cations are +2 cations, so they all form the same type ofsalts. As such, solubility products can be directly compared. Each step showsthe salt that precipitates.

Ba2+,Ca2+,Pb2+,Sr2+CO,

PbC03(s) Ba2+,Ca2+,Sr2+SO.

SrS04(s)a small amount of BaS04(s)also forms a precipitate

TCaF2(s)

Figure 3-6

Ba2+,Ca2+

2+Ba

Solubility


General Chemistry Equilibrium soiubiuty

Example 3.30According to data in Table 3.8, to remove strontiumby precipitation from asolution with Ca2+, Mg2+, and Sr2+, it isBEST toadd:A. F"B. S042"C. co32-D. OH-

SolutionThe best anion to add to solution to remove a cation is the one that forms a saltless soluble than the salts of the other cations in solution. Of the three cations(Ca2+, Mg2+, and Sr2"*-), the least soluble fluoride salt isCaF2, eliminating choiceA. Of the three cations, the least soluble sulfate salt is SrS04/ so choice B is thebest answer. Of the three cations, the least soluble carbonate salt is MgC03, andthe leastsoluble hydroxide salt isMg(OH)2, eliminating choices C and D.

Example 3.31Mercury IIcations areMOST soluble inwhichof the following solutions?A. 0.10 M NaClB. 0.10MNa2SC. 0.10 MNalD. 0.10MNaNO3

SolutionThis is a question ofrelative solubilities. This question is askingwhat the relativesolubilities of HgCl2, HgS, Hgl2, and Hg(N03)2 are. The key fact to keep inmind is that nitrates are infinitely soluble in water (one of the solubility rules).Of the anions, only nitrate can form hydrogen bonds with water, so it has thegreatest solvation energy of the fouranions. It is alsothe largestof the anions, soit has the weakest electrostatic forces in its lattice structure. The conclusion isthat nitrate salts are the most soluble, so the best answer is choice D. This ispurely a qualitative argument for the solubility. There ismathematical evidenceto back this up aswell,but the numbers were not given in the problem.

Example 3.32Given a solution that combines Ag+(aq), Pb2+(aq), Sr^aq), and Zn2+(aq), what isthe sequenceof precipitation when NaCl(aq) is added to solution?KSp AgCl =2x10"10, Ksp pbCl2 =2x10"5, Ksp SrCl2 =9x10"3, Ksp ZnCl2 =1x10"4A. 1st:AgCl, 2nd:SrCl2,3rd:ZnCl2,4th: PbCl2B. 1st:AgCl,2nd: PbCl2/ 3rd: ZnCl2,4th: SrCl2C. 1st: SrCl2,2nd: ZnCl^ 3rd: PbCl2,4th: AgClD. 1st: PbCl2,2nd: ZnCl2,3rd: SrCl2,4th: AgCl

SolutionThe salt with the lowestmolar solubility forms a precipitate first. AgClhas thelowest solubility product, and themolarsolubility is the square rootofKSp. Forthe other three salts, use the cube root ofKsp to obtain molar solubility. Thismakes AgCl the least soluble, and the first to precipitate, eliminating choices Cand D. Theremaining saltshave+2cations, so the order of precipitation is fromthe lowest Ksp value to the highest Ksp value. This makes choice Bcorrect.



Ion Exchange ColumnAn ion exchange column exchanges one ion in solution for another (which isinitially bound to the column), by precipitating out the ion that forms the lesssoluble salt and releasing a more soluble ion into solution to replace it. A watersoftener is an ionexchange column. Hardwater (rich in Ca2+(aq)) travels downthe column where the anion in the ion exchange resin binds calcium cations toform an insoluble salt while releasing sodium cations (Na+(aq)) into solution.Aqueous sodium cation is referred to as soft water. As calcium precipitatesthrough the ion exchange column, it is filtered from the water, preventing it fromforming precipitates in the plumbing lines of appliances (such as a washer orwater heater). To evaluate the viability of an ion exchange column, the molarsolubility values are compared. Figure 3-7 shows an ion exchange column.

Ca2+(aq) enters at the topofthe ion exchange column.

The column is filled with an ionexchange resin. The cation to beprecipitated must have a lowermolar solubility with the anion ofthe resin than the salt in the resin.

1Na+(aq) exitsfrom thebaseof the ion exchange column.

Figure 3-7

In the water softening example, the removal of calcium ions from tap water isaccomplished by passing the water across rock salt (large chunks of NaCl) andthen filtering out the calcium chloride. Reaction 3.8 is the exchange reaction in awater softener.

2Na+(aq) + CaCl2(s) ^ * Ca2+(aq) + 2NaCl(s)Reaction 3.8

Rock salt is employed to minimize surface area, and to prevent sodium chloridefrom dissociating into solution too rapidly. Sodium chloride is used because itsions are non-toxic, and it does not change the pH of the water.

Selective Precipitation and Ion Exchange: The addition of an ion to solution with theintention of precipitating an existing ion out from the solution. This is achievedwhen the salt being precipitated is less soluble (has a lower molar solubility) thanany other salt combination in solution.

2AgCl(s) + Hg2+(aq) ^ » 2Ag+(aq) + HgCl2(s)Reaction 3.9

Because mercury chloride is less soluble than silver chloride, the addition ofsilver chloride to the mercury solution precipitates mercury out of water. This isthe same principle applied in ion exchange columns and can also be applied inqualitative analysis when we look for ions in solution.

Solubility



Complex Ion Formation: Complex equilibrium occurs when two reactions aresummed to form one overall reaction. This is possible when the product of onereaction is a reactant in a second reaction. A great example involves the additionof ammoniumto salt buildup in sinks in order to remove the deposit. To checkthis, take a small whiff of a bathroom cleaner, and you will detect ammonia.

CaC03(s) + H20(1) ^ ^ Ca2+(aq) + C032"(aq)Reaction 3.10

Ca2+(aq) + 4NH3(aq) ^ * Ca(NH3)42+(aq)Reaction 3.11

Because ammonia forms hydrogen bonds with water, the calcium-ammoniacomplex ismoresoluble than the calcium cation. By the addition of ammonia tosolution, free calcium cation forms a complex ion (the result of Reaction 3.11shifting in theforward direction), which reduces theamountof free calcium andforces theReaction 3.10 to proceedin the forward direction. Asmentionedprior,when reactions are added together, the equilibrium constant for the overallreactionis the product of the equilibriumconstantsof the component reactions.

Complex Ions (Coordination complexes): Complex ions form when a ligand (lonepair donor or Lewis base) donates a pair of electrons to a central atom (Lewisacid), which is typically a metal, to form a coordinate covalent bond. Thisprocess is referred to as chelation, where the Lewis base is a chelating agent. Aprime example is hemoglobin whose porphyrin ring serves as a polydentateligand by chelating the central iron.

Example 3.33Silver chloride isMOSTsoluble in which of the following solutions at 250C?A. 0.10MHgNO3(aq)B. Pure waterC. AgN03(aq)D. NaCl(aq)

SolutionSilver chloride (AgCl(s)) ismostsoluble in a solution whereHg+is present. Thisis because of a complex equilibrium. TheHg+ in solution canbind the chlorideanion (CI") to form a precipitate. As the amount of CI" in solution decreases,more AgCl dissolves into solution to replace it. The best answer is choice A.Both choices C and D are eliminated, because of the common ion effect.

"Chemistry every day keeps yon balanced!"


EquilibriumPassages

14 Passages

100 Questions

Suggested Equilibrium Passage Schedule:I: After reading this section and attending lecture: Passages I,II & VI - IX


Following Task I: Passages III - V, X & XI (34 questions in 44 minutes)Time yourself accurately, grade your answers, and review mistakes.

Review: Passages XII - XIV & Questions 97 - 100Focus on reviewing the concepts. Do not worry about timing.

II

III

R-E-V-I.E-W

Ipeializing ii MGAT Preparation

Eqmlibnum Study Passages

I. Carbon Dioxide and Carbon Monoxide Equilibrium (1-7)

II. Hydrogen and Bromine Equilibrium (8 - 15)

III. Equilibrium Constant Magnitude (17-22)

IV. Gas-Phase Equilibrium Constant Experiment (23 - 29)

V. Equilibrium Reactions (30 - 36)

VI. Le Chatelier's Principle (37 - 42)

VII. Solubility and Qualitative Analysis (43 - 49)

VIII. Solubility Experiment (50 - 56)

IX. Qualitative Analysis Experiment (57 - 62)

X. Solubility Chart (63 - 69)

XI. Calcium Salts Solubility (70 - 75)

XII. Complex Equilibrium (76 - 82)

XIII. Hemoglobin and Acclimation (83 - 89)

XIV. Equilibrium Reaction of N02 and N2O4 (90 - 96)

Questions Not Based on a Descriptive Passage (97 - 100)

Equilibrium Scoring Scale


84 - 100 13- 15

66-83 10- 12

47 -65 7 -9

34-46 4-6

1 -33 1 -3


A researcher studies the equilibrium of carbon monoxidewith carbon dioxide in the presence of oxygen gas at varioustemperatures. The relationship is shown in Reaction 1.

2 CO(g) + Q2(g) •> ^ 2 C02(g)Reaction 1

This is accomplished by monitoring the total pressureofthe systemas it depends on temperature. There are differentstoichiometric amounts of gases on each side of Reaction 1,so the reaction can shift with changes in pressure or volume.To quantify the contents in the flask at equilibrium, thepressure for the system is analyzed. It is assumed thatpressure changesare due to shifts in equilibrium and changesin the system that obey the ideal gas equation (PV = nRT).Changes in the total pressureare evaluatedfor the systemandthe equilibrium constant is derived from the values forpressure. The reaction is monitored for three trials at eachtemperature over a wide range of temperatures. The resultsare summarized in Table 1.

Temperature (K) Kp (atm"1)375 09.0

425 20.8

500 32.2

600 60.5

Table 1

Data in Table 1 demonstrate that the value of theequilibrium increases with increasing temperature. Althoughthe equilibrium systemshifts upon changes in the moles ofreactant, moles of product,external pressure, and volume, theequilibriumconstant remains constant.

1

2.

At 102°C, Pco = 2.0atm. and Po2= 1-0 atm. What isthe partialpressureof CO2at this temperature?A. 2.2 atm.B. 6.0 atm.C. 9.0 atm.D. 18.0 atm.

Addition of C02(g) to the equilibrium mixture results inwhich of the following?A. A decrease in the moles of 02(g).B. A decrease in the moles of CO(g).C. An increase in thevalue of Kp.D. Nochange in thevalue of Kp.


3. According to Table 1, the reaction is which of thefollowing?

A. Exothermic as writtenB. Endothermic as writtenC. Adiabatic as carried outD. Isothermal upon reaction

4. The equilibrium constant (K) varies only withwhich ofthe following?A. A change in the pressure of the systemB. A change in the volume of the systemC. A change in the concentration of the systemD. A change in the temperature of the system

If a flask were filled with pure C02(g) to a total pressureof 1.00atm., then once equilibrium is reached, the totalpressure of the system is which of the following?A. Less than 1.00 atm.B. Exactly 1.00 atm.C. Between 1.00 atm. and 1.50 atm.D. Exactly 1.50 atm.

6. According to Reaction 1, compressing the reactionvessel leads to which of the following?A. An increase in the partial pressure of C02(g)B. An increase in the moles of 02(g)C. An increase in the mole fraction of CO(g)D. An increase in the equilibrium constant

7. Based on the following reaction, the addition ofCa(OH)2(aq) to Reaction 1 at equilibrium results inwhich of the following?

Ca(OH)2(aq) + C02(g)^=^=- CaC03(s) + H20(1)A. No change in the moles of C02(g).B. A decrease in the concentration of 02(g).C. An increase in partial pressure of CO(g).D. An increase in the total pressure.


Saad Azam

Saad Azam

Saad Azam

Saad Azam

Saad Azam


Hydrogen bromide gas is one of the strongest acidsavailable in gas form. In aqueous solution, HBr has a pKavalue of approximately -12. A pKa value of -12 is indicativeof a very strong acid. By comparison, HCl has a pKa ofapproximately -9. Hydrobromic acid (HBr) can be producedby Reaction 1.

H2(g) + Br2(l) ^^ 2 HBr(g)Reaction 1

A student decides to study the equilibrium distributionfor Reaction 1. Under ambient conditions, Reaction 1reaches equilibrium at some temperature T and pressure P.Once at equilibrium, measurements are taken for the partialpressure of each gas and the mass of bromine liquid. Thevalues are listed in Table 1.

Br2(l) 10.0 gramsH2(g) 1.36 atm.

HBr(g) 2.72 atm.

Table 1

The student takes notice that Reaction 1 shifts withchanges in the conditions of the system, such as the additionof bromine liquid, the removal of hydrobromic acid, andincreases in volume. The student irreversibly increases thevolume of the system by opening a valve on the reactionflask that is connected to an evacuated column. Not allchanges shift the reaction. Changes in the equilibriumconcentrations are not accurately recorded.

8. What is the value of K at ambient temperature, T?A. 0.20 atm.B. 0.544 atm.C. 2.00 atm.D. 5.44 atm.

9. Addition of HBr(g) to an equilibrium mixture ofReaction 1 results in which of the following?A. An increase in the partial pressure of H2(g)B. A decrease in the partial pressure of H2(g)C. An increase in the concentration of Br2(l)D. A decrease in the concentration of Br2(l)

10. Which of the following starting conditions results inthe GREATEST amount of H2(g) at equilibrium?

A. 0.80 atm. H2(g) and 20 g Br2(l)B. 1.00 atm. H2(g) and 20 g Br2(l)C. 0.80 atm. H2(g) and 30 g Br2(l)D. 1.00 atm. HBr(g) and 30 g Br2(l)


11. Subsequent experiments show that as the temperatureincreases, the mole percent of the hydrogen bromide gasincreases. Which of the following is TRUE about theequilibrium reaction as written?

A. It is endothermic.B. It is exothermic.C. It is isobaric.D. It is not an equilibrium mixture.

12. How can it be explained that a reaction mixture of 1.0atm. HBr(g) and 1.0 atm. H2(g) results in no change inhydrogen gas partial pressure?

A. The reaction is in equilibrium.B. The reaction cannot move in the forward direction

due to a limiting reagent in zero concentration.C. The reaction cannot move in the reverse direction

due to a limiting reagent in zero concentration.D. The reaction is moving forward and backward at the

same rate.

13. Mixing 3.0 atm. HBr(g) with 1.5 atm. H2(g) overBr2(l) shows which of the following in time?

A. A large increase in the partial pressure of HBr(g)B. A small increase in the partial pressure of HBr(g)C. A large increase in the partial pressure of H2(g)D. A small increase in the partial pressure of H2(g)

14. A flask is initially filled with pure HBr(g). Which ofthe following graphs represents the partial pressure ofH2(g) over time?

time time

15. Addition of which of the following to the equilibriummixture will NOT affect the partial pressure of H2(g)?

A. NaOH(aq)B. Br2(l)C. H2(g)D. HBr(g)


Saad Azam


Chlorine gas has a high reduction potential (largepositive voltage for its reduction half-reaction). Chemicallyspeaking, chlorinegas is veryeasily reduced. Chlorinegas isa strong oxidizing agent that can dissolve away most metalsin short periods of time. The dissolving is the result of anoxidation-reduction reaction. The relative ease with whichmetals are oxidized accounts for this strong tendency to loseelectrons to chlorine. Chlorine gas is not as successful atoxidizing non-metals as it is at oxidizing metals. This isattributed to the decreased desire for non-metals to be oxidizedrelative to metals. Non-metals have a higher ionizationenergy and electronegativity than metals. With non-metals,chlorine may only partially oxidize the correspondingreducingagent. Reactions 1 and 2 are prime examples of theoxidationcapability of chlorine on non-metal compounds.

PCb(g) +Cl2(g) -i * PCl5(g)Reaction 1

2 NO(g) + Cl2(g) ^ 2 NOCKg)

Reaction 2

The equilibrium constant for Reaction 1 is 1.95 atm."1at 100'C, while the equilibrium constant for Reaction 2 is2.1 x 104 atm."1 at 100"C. Both reactions favor products aswritten, but not to the same degree.

In these examples, the non-metals (phosphorus inReaction 1 and nitrogen in Reaction 2) are already in apositive oxidation state in the reactants. The greater thevalue of the equilibrium constant, the more favorable thereaction, and thus the more product that is formed.

16. According to the equilibriumconstants associated withReaction 1 and Reaction 2, which compound is mostreadily oxidized by chlorine?A. PC13B. PC15C. NOD. NOC1

17. If a flask were filled with PCI3 and CI2 to a totalpressure of 1.50 atm. such that the mole fraction ofPCI3 is twice that of CI2, then what is the totalpressureof the system once at equilibrium?A. Less than 1.00 atm.B. Between 1.00 atm. and 1.25 atm.C. Between 1.25 atm. and 1.50 atm.D. Greater than 1.50 atm.


18. What is the equilibrium constant for the followingreaction?

2 NOCl(g)

A. 4.9 x 10"17 atm.B. 4.9 x 10'5 atm.C. 1.4 x 102 atm.D. 2.1 x K^atm.

2 NO(g) + Cl2(g)

19. If there is initially 1.00 atm. of NOC1 in a flask, whatis the partial of NO gas, once at equilibrium?A. 0.0230 atm.

B. 0.0460 atm.

C. 0.0025 atm.D. 0.0050 atm.

2 0. The GREATEST concentration of Cl2(g) is in which ofthe following systems,once equilibrium is reached?A. A system initially with 1.0 atm. PCI3B. A system initially with 1.0 atm. PCI5C. A system initially with 1.0 atm. NOD. A system initially with 1.0 atm. NOC1

21. Which of the following statements is true about theeffectof adding chlorine gas to each equilibrium?A. It will cause the greatest change in Reaction 1.B. It will cause the greatest change in Reaction 2.C. It will cause equal changes in both reactions.D. It will change the equilibrium constant.

22. A flask at 100°C is charged with 0.40 atm. each ofPCI3, PCI5, and CI2. In which direction will it reactto reach equilibrium, according to Reaction 1?A. It is already at equilibrium.B. To the left (reactant side)C. To the right (product side)D. It varies with a catalyst.



The equilibrium constant for a reaction involving gasescan be determined by experimentation, using a glass columnas a closed system. The apparatus used in equilibriumexperiments, shown in Figure 1, is connected to a vacuumpump at one end and is sealed at the other. On the columnare three stopcocks with air tight fittings, to which flaskscontaining gasesmay be attached. At the sealed end of thetube is a pressure gauge. The volume of the glass column isexactly 1.00 liters, when all stopcocks areclosed. Each flaskand attached stopcock contain volumes of exactly 250 mL.This means that when all three stopcocks are open in amannerwhere gas is free to flow between the glass columnandany of the attachedflasks, the total volume for theclosedsystemis exactly 1.75liters. Figure 1 shows the apparatus.

^ ConnectedPressure gauge to vacuum

Flask 1 Flask 2 Flask 3

Figure 1During an experiment, the closed glass cylinder is filled

with 1.00 atm. of hydrogen (H2) according to the gauge inthe cylinder. Flask 1 is filled with 1.0 atm. carbon disulfide(CS2). The reaction starts when the stopcock is opened,allowing the two gases to mix. The temperature of the glasscolumn is maintained at 25°C using an external heat sink.The internal pressure is monitored until it stays constant.Figure 2 shows the internal pressure over time, where t = 0represents the time at which the two gasses were mixed.

1.0-

Time (minutes) •

Figure 2The pressure of each gas in the reaction mixture can be

calculated from the change in internal pressure. The initialpartial pressure of hydrogen gas is 0.8 atmospheres in the1.25 liter closed system. The decrease in partial pressure ofhydrogen gas is double the decrease in the internal pressure,based on the stoichiometry of the Reaction 1, which showsthe reactivity of the compounds.


1 CS2(g) + 4 H2(g) « * 1 CH4(g) + 2 H2S(g)Reaction 1

The final internal pressure is the sum of the partialpressures. Thetotal pressure canbe found using Equation 1.

Ptotal = PCS2 + PH2 + PCH4 + PH2SEquation 1

23. Which of the following is the BEST explanation as towhy thereaction temperature mustbe heldat 25*C?A. A temperature change cancausethecarbon disulfide

to decomposeto carbondioxide.B. The reaction will not proceed unless the

temperature is exactly 25°C.C. The glass will begin to melt at temperatures in

excess of 25°C.D. The equilibrium constant can change with

temperature.

24. The temperature of the closed reaction vessel wasincreased. This resulted in the pressure of the closedsystem decreasing, rather than increasing as expected.Which of the following explanations BEST explainsthis observation?

A. The pressure of a closed system decreases as thetemperature increases according toCharles'slaw.

B. The reaction is endothermic as written.C. The reaction is exothermic as written.D. Heavier molecules are being formed causing the

number of collisions to increase.

25. How can the partial pressures of CH4 and H2S atequilibrium be determinedfrom the experiment?A. At equilibrium, the partial pressure of CH4equals

the value for the change in the pressure of thesystem (APt0tal)» while the partial pressure of H2Sequals twice the value for the change in thepressure of the system (2APtotai).

B. At equilibrium, the partial pressure of H2S equalsthe value for the change in the pressure of thesystem (APt0tal)> while the partial pressure of CH4equals twice the value for the change in thepressure of the system (2APtotai).

C. At equilibrium, the partial pressure of CH4equalsthe value for the change in the pressure of thesystem (APtotal). while the partial pressureof H2Sequals half the value for the change in the pressureof the system (-APtotal)-

D. At equilibrium, the partial pressure of H2S equalsthe value for the change in the pressure of thesystem (APtotal)* while the partial pressureof CH4equals half the value for the change in the pressureof the system (-APtotal)-

2


26. Which of the following relationships between partialpressure for the gases in the mixture at equilibrium willALWAYS be true for the reaction as it is set up?

A. Pcs2>PCH4B. PCH4 >Ph2C. PH2> Pcs2D. Ph2s >PH2

27. Which of the following equations represents theequilibrium expression for the reaction?

A v _(Pch4)(Ph2s)2A. K.p —(Pcs2)(Ph2)4

B K_ (Pcs2)(Ph2)4P (Pch4)(Ph2s)2

c K _(Pch4)(Ph2s)P (Pcs2)(Ph2)

D K - (PCS2)(PH2)P (Pch4)(Ph2s)

28. The final total pressure of the system must be:A. greater than 1.6 atmospheres.B. greater than 1.0, but less than 1.6 atmospheres.C. greater than 0.6, but less than 1.0 atmospheres.D. less than 0.6 atmospheres.

29. Which of the following stresses on the system willchange the value of the equilibrium constant?A. Opening the stopcock to Flask 2, which was filled

with hydrogen gas.B. Turning on the vacuum pump to remove some of

the gas in the column.C. Opening the stopcock to Flask 3, whichwas filled

with hydrogen sulfide (H2S) gas.D. Heating the gas in the glass column by heating the

glass with a hot air blower.



Chemical equilibrium obeys the rules of Le Chatelier'sprinciple. Le Chatelier's principle states that when a systemat equilibrium is stressed, the reaction will shift in a mannerto partially relieve the stress and reestablish equilibrium. Ina gas phase equilibrium, stresses that can be applied includechanges in pressure, volume, temperature, and moles ofeither reactant or product. It is not possible to change justone of the variables, so a change in one variable will lead toin a change in another variable. For instance, if the volumeis increased, the pressure must decrease. Written below arethree reactions and their respective equilibrium constants at500'C:

Reaction 1:

2 S02(g) + 1 02(g) -S"^ 2 S03(g)Kp =5.82 x102 atm.'1 at 500eC

Reaction 2:

1 H20(g) + 1 Cl20(g) ^z^r 2 HOCl(g)Kp =8.61 x 10"3 at 500°C

Reaction 3:

1 HCl(g) + 1 CO(g) -g-*- 1 HCOCl(g)Kp =3.26 x lO'6 atm."1 at 500°C

The equilibrium constant is derived by dividing thepartial pressures of the productgases by the partialpressuresof the reactant gases. In Reaction 1 and Reaction 3, theequilibrium constants weredetermined usinga pistonsystemwhere the total pressure and volume weremonitored, andanychange in pressure or volume can be attributed to a shift inthe reaction equilibrium. Because the number of productsequals thenumber of reactants, the total pressure and volumecannotchange in reaction 2, so equilibrium concentrationswere determined using infrared spectroscopy.

30. Given the initial partial pressure for Ph20 ls 0.5 atmand for Pci20ls 0.5 atm, which graph accurately showsthe rate as a function of time for the reaction?

Time Time


31. Which reactions will shift when the pressure changes?A. Reaction 1 onlyB. Reactions 1 and 2 onlyC. Reactions 1 and 3 onlyD. Reactions 1, 2, and 3

32. Upon increasing the volume at constant temperature ofa vessel holding Reaction 3, how are the partialpressures affected?

A# "HCOC1 increases; HCQC1 increasesPhci-Pco Phci

B# PhCOCI decreases; "HCOC1 decreasesPhci-Pco phci

q 9 PHCOCI remains constant; HCOC1 increasePhci-Pco phci

D. PhCOCI remains constant; HCQC1 decreasesPhci-Pco Phci

33. Which graph depicts partial pressures as a function oftime after 0.75 atm SO2 is mixed with 0.50 atm O2?

S02

Time

Time

•02

B.

D.

3

'S03

Time

Time

34. Which Keq value does NOT match the reaction?A. 2HOCl(g) -^= 1 H20(g) + 1 Cl20(g)

: 1 HOCl(g)Kp =8.61 x10'3 at 500eC

B. lH20(g)+ICl20(g) -^Kp =9.28 x lO'2 at 500"C

C. iHCOCl(g) -^^r iHCl(g) + ICO(g)Kp =5.54 x102 at 500°C

D. 2 HCl(g) + 2 CO(g) -^=Kp=1.06x lO'11 at500°C

2 HCOCl(g)


35. To convert aKp value in terms of ton-1 to aKp valuein terms of atm"1, you should:A. multiply the Kp value by 760.B. divide the Kp value by760.C. multiply theKp value by0.76.D. divide the Kp value by 0.76.

36. Coolingan endothermic equilibriumreaction results inwhich of the following?

A. The mole fraction of products will decrease.B. The mole fraction of reactants will decrease.C. The mole fraction of products will increase.D. The ratio of products to reactants will increase.



Le Chatelier's principle is invoked to explain reactionshifts following the disruption of equilibrium. A system indynamic equilibrium is a closed system where the forwardreaction rate and reverse reaction rate are equal. Componentsin the mixture maintain fixed concentration. Equilibrium isdifferent from steady state. In steady state systems, anintermediate has a static concentration, because the formationreaction and the consumption reaction have equal rates.

A system can be disturbed from equilibrium by applyinga stress, such as changing the temperature, volume, pressure,or moles of a reagent. According to Le Chatelier's principle,the system reacts in a manner to partially alleviate the stressand reestablish equilibrium. Reaction 1 starts with threedifferent starting concentrations on its way to equilibrium.

CO(g) + H20(g) ^=^ C02(g) + H2(g)Reaction 1

Table 1 lists the starting and equilibrium concentrationsfor each of the three trials. The reactions are observed at1000°C and the equilibrium constant is found to be 0.569.

Trial I Trial II Trial III

Molarity Initial Final Initial Final Initial Final

CO 1.00 0.57 0.10 0.014 0.50 0.373

H20 1.00 0.57 1.00 0.914 0.50 0.373

C02 0 0.43 0 0.086 0.50 0.627

H2 0 0.43 0 0.086 0 0.127

Table 1

Once at equilibrium, a separate stresses were applied toeach system. To reaction System I, an additional 0.10molesof CO was added to the 1.00 liter vessel in which the reactionwas housed. To reaction System II, all of the H2O wasremoved from the 1.00 liter vessel in which the reaction washoused. The 1.00 liter vessel containing reaction System IIIwas cooled to 500"C. Once equilibrium was reestablished inthe three reaction vessels, the partial pressure of CO2 hadincreased in reactionssystems I and HIand decreasedin II.

37. All of the following affect the reactant/product ratiosofthe equilibrium mixture EXCEPT:A. increasingthe volumeof the reactionvessel.B. heating the reaction vessel.C. cooling the reaction vessel.D. adding reactant to the reaction vessel.

38. If 1.00 moles of both CO2 and H2 were mixed in a1.00 liter container at 1000°C, what will theequilibrium concentrationsbe for bothCO2and H2?A. [C021 = 0.57 M; [H2] = 0.57 MB. [CO2] = 0.57 M; [H2] = 0.43 MC. [CO2I = 0.43 M; [H21 = 0.57 MD. [CO2I = 0.43 M; [H2] = 0.43 M


39. As heat is added to Reaction 1, what occurs?

A. Both theequilibrium constant (Kgq) and thepartialpressure of CO2 decrease.

B. Both theequilibrium constant (Keq) andthepartialpressure of CO2 increase.

C. The equilibrium constant (Keq) remains constantwhile the partial pressure of CO2 increases.

D. The equilibrium constant (Keq) remains constantwhile the partial pressure of CO2 decreases.

4 0. What effect would an increase in pressure have on theequilibrium system ofReaction 1?A. The reaction would shift to the right and the

equilibrium constant would increase.B. The reaction would shift to the left and the

equilibriumconstantwould decrease.C. The reaction would shift to the right and the

equilibrium constant would remain constant.D. The reaction would not shift in either direction and

the equilibriumconstant would remain constant.

41. Which of the following graphs represents what wouldbe observed over time if equal parts of CO(g),C02(g),andH2(g) weremixedin a closed system?A. B.

Time

p$Cfl ^ *</)

P£

t

Time

223GOCA

a£

Time

= Total Pressure

= H20 Pressure

= CO Pressure

42. For an exothermic reaction at 100°C, the equilibriumconcentration of the products equals the equilibriumconcentration of the reactants. What is true at 25"C?A. Keq> 1; [Products] > [Reactants]B. Keq> 1; [Products] < [Reactants]C. Keq< 1; [Products] > [Reactants]D. Keq< 1; [Products] < [Reactants]



Thesolubility of a salt in water is measured by itsmolarsolubility. The molar solubility is defined as the moles ofsalt that dissociate into one liter of aqueous solution. Thesolubility product (Ksp) is the equilibrium constant for thedissociation reaction. Table 1 lists solubility data for varioussalts of silver, lead, strontium, and zinc.

Compound ^sp Molar solubility

AgCl 1.7 x 10-10M2 1.3 x lO'5 MPbCl2 1.6 x 10'5 M3 1.2 x 10-2MSrCl2 8.8 x lO'3 M3 1.3 x 10"1 MZnCl2 1.1 x 10"4M3 3.1 x 10"2MAg2C03 8.1 x 10"12 M3 1.2 x 10"4MPbC03 8.3 x lO'11 M2 9.1 x 10'6MSrC03 7.1 x 10'10M2 2.6 x 10"5 MZnC03 2.0 x 10"10 M2 1.4 x 10"5 MAgOH 2.0 x 10"8M2 1.4 x 10'4MPb(OH)2 1.2 x 10"15M3 6.5 x 10"6MSr(OH)2 3.2 x 10"4M3 4.3 x 10"2 MZn(OH)2 2.1 x 10"16M3 3.8 x 10"6 MAg2S 1.6 x 10"49M3 3.4 x 10"17 MPbS 2.6 x 10"27 M2 5.1 x 10"14MSrS 4.0 x 10"6M2 2.0 x 10"3 MZnS 2.5 x 10"22M2 1.6 x lO'11 MAg2S04 1.4 x 10"5M3 1.3 x 10-2MPbS04 1.7 x 10"8M2 1.3 x 10"4MSrS04 3.2 x 10"7 M2 5.6 x lO'4 MZnS04 7.6 x lO'8 M3 2.7 x 10"4M

Table 1

The relative molar solubility values can be used todetermine which salts precipitate out from solution whenvarious cations or anions are added. For instance, PbS04precipitates first when sodium sulfate is added to a solutioncontaining lead, strontium, and zinc cations, because leadsulfate is the least soluble (has the lowest molar solubility)of the sulfate salts.

4 3. Which of the following statements are valid?I. Sulfates are more soluble than carbonates.

n. Insoluble cations can be dissolved into solution byadding a complexing agent such as EDTA.

HI. The solubility of chloride salts show pHdependence, while the solubility of sulfide saltsshow no pH dependence.

A. I and II onlyB. I and m onlyC. H and m onlyD. I,n,andm


44. The Ksp value for which of the following salts is theMOST accurate in terms of smallest percent error?A. Ag2S04B. Pb(OH)2C. S1CO3D. ZnS

45. Which anion should be added to selectively removeZn2+ from an aqueous solution that contains Ag+,Sr2+, and Zn2+?A. Cl-B. CO32-C. OH-

D. S042_

46. If a precipitate forms whenNaCl is added to an aqueoussalt solution, what conclusion can be drawn?

A. There is no Zn2+(aq) present insolution.B. There is Ag+(aq) present in solution.C. There is both Sr2+(aq) or Pb2+(aq) present in

solution.D. There is no Sr2+(aq) present in solution.

47. Ifan aqueous solution containing Ag+, Pb2+, Sr2+, andZn2+ is treated first with chloride anion, secondly withsulfide dianion, and lastly with hydroxide anion, thenwhat cation is most likely still present in solution?A. Ag+B. Pb2+C. Sr2+D. Zn2+

48. Adding acid to an aqueous solution containing a basicsalt has what effect?

A. It will increase the solubility.B. It will decrease the solubility.C. It will have no effect on the solubility.D. It will lower the temperature of the solution.

49. Silver hydroxide is MOST water-soluble at which ofthe following pH values?A. 3B. 5

C. 9D. 11



The solubility of salts is often measured in terms ofmoles solute per liter solution in a saturated solution. I tmay also be measured in terms of grams solute per 100mLsolution. The term saturated refers to a solution in which themaximum amount of salt that can dissociated into solution isdissociated. The more salt that dissociates into solution, themore soluble the salt. Solubility measurements cannot beused interchangeably when comparing the solubility of twosalts without first considering the molecular mass of eachsalt. This can be observed in the following experiment.Experiment IWater at 25°C is added to a calibrated burette, to the 25-mL mark. A salt is added in 1.0 grams increments, untilthe salt no longer dissociates into solution. At thispoint, a precipitate should settle to the bottom of theburette. The temperatureof the solution in the burette isslowly increased until the salt completely dissociates.The solution is slowly cooled until the first signs ofprecipitate are observed. At this point, the temperatureand volume of the solution are recorded.

Experiment2The solution continues cooling until it reaches 25°C. Atthispoint, purewaterat 25°C is added quantitatively untilthe last detectable sign of the precipitate dissociate intosolution. The volume of the solution is recorded asprecisely as the burette reads.

Table 1 lists data for this experiment using hypotheticalsalts MX, MY, and MZ. Column 3 lists data fromExperiment 1andcolumn 4 lists data from Experiment 2.

Salt Mass Volume when heated Volume w/ H2O added

MX 7.0 g 26.2mL(@31.6°C) 28.9 mL (@ 25.0°C)MY 9.0 g 25.3 mL (@ 27.4°C) 26.1 mL(@25.0°C)MZ 4.0 g 25.7 mL (@ 30.3°C) 28.7 mL (@ 25.0°C)

Table 1

The molecular mass for MX is 100 grams per mole, forMY is 120 grams per mole, and for MZ is 150 grams permole. The concentration of each solution is measurable inany standard units, including molarity (moles solute perlitersolution), grams per mL, mass percent, and density.Concentration units can be inter-converted. For instance, thepercent solution by mass can be multiplied by the density todetermine the mass solute per volume of solution, which canbe converted to molarity using molecular mass.

Adding solvent to a solution dilutes the solution andthus reduces the concentration of the solution. All of themeasurements of concentration decrease upon the addition ofsolvent, with the exception of the density. The densitychange depends on the relative density of the solvent andsolution. In these experiments, water is the solvent for eachsolution, and the density of water at 25°C is found to be0.9971 grams per mL.


50. How do the solubility products for the three salts in thepassage compare to one another?

A. KspMX > KspMY > KspMZB. KspMY > KspMX > KspMZC KspMZ > KspMX > KspMYD- KspMZ > KspMY > KspMX

51. Which of the following solutions has the GREATESTmolarity?A. Saturated MX at 31.6 °CB. Saturated MX at 25.0 *CC. Saturated MY at 27.4 "CD. Saturated MY at 25.0 °C

52. Roughly how many grams of MX will dissociate into25mLofwaterat25.0°C?

A. Less than 4.0 grams.B. Greater than 4.0 grams, but less than 7.0 grams.C. Greater than 7.0 grams, but less than 9.0 grams.D. Greater than 9.0 grams.

53. What is the molarity of a saturated solution of MY at25°C?

A.

B.

9 grams

120il^ix 0.0261 litersmole

9 grams120grams_x002531iters

molec 9grams x0.0261 liters

120grams_mole

D 9grams x0.0253 liters120grams_

mole

54. As temperature increases, what happens to the variousconcentration measurements of a solution, assuming noevaporation ofsolvent and noaddition of solute?A . The density and molarity both increase.B. Thedensity increases, whilethemolarity decreases.C. Thedensity decreases, whilethemolarity increases.D. The density and molarity both decrease.


5 5. Which of the following relationships at 25°C accuratelyshows the relative density values of saturated solutionsof MX(aq),MY(aq), and MZ(aq)?A. 1.00 > MX(aq) > MY(aq) > MZ(aq)B. 1.00 > MY(aq) > MX(aq) > MZ(aq)C. MX(aq) > MY(aq) > MZ(aq) > 1.00D. MY(aq) > MX(aq) > MZ(aq) > 1.00

56. Which solution has the HIGHEST boiling point?

A. 1.0 grams MX with 10mL water at 25°CB. 1.0 grams MY with 10mL water at 25°CC. 1.0 grams MX with 10 mL water at 50°CD. 1.0 grams MY with 10 mL water at 50°C



The qualitative analysis of an aqueous salt solutioninvolves the systematic addition of reagents designed toidentify certain component ions. For instance, mercury canbe distinguished from alkaline earth metals by adding sulfide(S2") to solution by the formation of an insoluble precipitatewith mercury dication. Silver can be distinguished fromalkali metals by the addition of either chloride, bromide, oriodide. The precipitate that forms between chloride and silvercan be re-dissolved by adding ammonia to solution.

Once a precipitate is formed, it can be removed fromsolution by centrifuging the mixture and decanting away thesupernatant or by filtering away the filtrate. The solutioncan then be further analyzed for other ions. Table 1 showsthe results of a matrix involving the mixture of four cationnitrate solutions mixed with five potassium anion solutionsmixed one at a time. All solutions are 0.05 M, with theexception of KOH(aq), which has a concentration of 0.10 M.Any compound that forms a precipitate is assumed to have amolar solubility less than 0.001.

Ba(N03)2 Sr(N03)2 Ca(N03)2 Mg(N03)2

K2co3 Pcpt Pcpt Pcpt No pcpt

KOH Pcpt No pcpt No pcpt Pcpt

K2Cr04 Pcpt No pcpt No pcpt No pcpt

K2S04 Pcpt Pcpt No pcpt No pcpt

K2C204 Pcpt Pcpt Pcpt No pcpt

Table 1

Table 2 shows the results of a similar experiment whereanions were exposed to a sequence of multiple reactions. Theanions selected for testing are chloride, iodide, carbonate, andsulfate. The test solutions are silver nitrate, barium nitrate,ammonia, and nitric acid. The matrix of Table 2 lists theobservations of sixteen different test tubes. Nitric acid reactswith carbonate anion to form carbon dioxide gas, which thenbubbles out of solution.

HN03 1. Ba2+2. HNO3

l.Ag+2. HNO3

l.Ag+2. NH33. HNO3

Cl" NoRx l.NoRx2. No Rx

1. Whtppt2. No Rx

1. Whtppt2. Dissolves

3. NoRx

I- NoRx l.NoRx2. No Rx

1. Yell ppt2. No Rx

1. Yell ppt2. No Rx

3. No Rx

co32- Gas forms

(Bubbles)1. Pcpt2. Bubbles

1. Whtppt2. Bubbles

1. Whtppt2. No Rx

3. Bubbles

so42- NoRx 1. Pcpt2. No Rx

1. No Rx2. No Rx

l.NoRx2. No Rx

3. No Rx

Table 2


57. What anion should be added to a solution in order toidentify any barium cation present in the solution?

A. Cr042"B. S042"c c2o42-D. CO32-

58. What reaction took place when nitric acid was added tothe precipitate formed from mixing Ba2+ and CO32"?A. Ba2C03(s) + 2 HN03(aq) ->

2 BaN03(aq) + C02(g) + H20(1)B. BaC03(s) + 2 HN03(aq) ->

Ba(N03)2(aq) + C02(g) + H20(1)C. Ba2C03(s) + HN03(aq) ->

BaHC03(aq) + BaN03(aq)D. BaC03(s) + 2HN03(aq) ->

Ba(N03)2(aq) + C02(g) + H2(g)+ 02(g)

59. Why is it NOT possible to use hydrochloric acid insteadof nitric acid in experiment II?A. Chloride anion will interfere with the reactions.B. Nitrate anion will interfere with the reactions.C. Hydrochloric acid is too weak to react.D. Hydrochloric acid is too strong to use.

60. If an unknown mixture was treated with K2SO4 and aprecipitate formed, whatcan beconcluded?A. The unknown contained no calcium or magnesium

cations.B. The unknown contained a barium cation.C. The unknown contained a strontium cation.D. The unknown contained either a strontium cation, a

barium cation, or both.

61. Which of the following anions does form a whiteprecipitatewith silver cation but does NOT react withnitric acid?

A. SulfateB. CarbonateC. ChlorideD. Iodide


62. The information from Table 1 was used to derive thefollowing flow chart:

_, 2+ 2+ „ 2+ 2+Ba , Sr , Ca , Mg

BaV(s) Sr2+, Ca2+, Mg2+W

SrW(s) Ca2+, Mg2+

CaX(s)or

CaY(s)

XorY

2+Mg

Z\MgZ2(s)

Which of the following choices represents anINCORRECT anion-to-letter correlation?

A. V=C032"B. W= S042_c x =c2o42-D. Z = OH-



Water can be analyzed for various inorganiccomponentsusing a seriesof qualitative tests for metals in their variousoxidation states. This process is referred to as qualitativeanalysis. Many of the tests involve the formation of aninsoluble salt. The formation of a precipitate results fromthe addition of an anion to an aqueous cation solution. Someanions may complex with more than one metal, so multipletests are necessary. Molar solubility is different for each salt,so occasionally an anion may precipitate with one cationover another and the second metal can be "masked". For thisreason a schematic of the series of tests is evaluated.Insoluble to the eye is defined as less than one mgdissociating into one mL of solution. By looking at thecombination of precipitates formed, the cations can benarrowed down until one is chosen with reasonable accuracy.

A researcher analyzed an aqueous sample believed tocontain roughly equal concentrations of Ag+, Ca2+, Zn2+,and Na+. Table 1 lists solubility data useful in developing aschematic for the tests. An ideal schematic sequentiallyidentifies one cation at a time, leaving the remaining cationsin solution.

Compound KSp Molar solubility

Ag2S 1.6 x lO'49 M3 3.4 x 10-,7MZnS 2.5 x 10'22 M2 1.6 x 10"11 MCaS 1.1 x lO"11 M2 3.5 x 10'6 MCaC03 8.7 x lO"9 M2 9.3 x lO'5 MAg2C03 8.1 x 10'12M3 1.2 x 10"4MZnC03 2.0 x 10-,0M2 1.4 x 10'5MAgCl 1.6 x 10-10M2 1.2 x lO"5 MZnCl2 1.1 x 10"4M3 3.1 x lO'2 M

Table 1

The values listed in Table 1 are all determined in distilledwater at 25°C. It is occasionally necessary to conduct tests atan elevated temperature, in order to carry out qualitativeanalysis and the separation of cations. Sodium salts are notlisted in the table, because sodium salts are infinitely solublewith nearly all anions, compared to the other metals listed inthe data. To identify sodium cation, a flame test is oftenapplied. By applying a flame to a small sample of thesolution, electrons in sodium are excited. When they relaxback to their ground state, photons are emitted. In the caseof sodium, the light emitted is orange in color.

63. When 0.1 gram ZnC03 is added to 100 mL of water at25°C, which of the following statements is true?A. It dissolves completely.B. It dissolves almost completely, with only a small

portion not dissolving.C. It dissolves slightly, with most of the ZnC03

remaining not dissolving.D. None of the sample dissolves.


64. It is best to visually distinguish a sample tubecontaining Zn2+ from a separate sample tubecontainingAg+ by the addition of:A. 0.10MNa2S.B. 0.10 M NaCl.C. 0.10MNaNO3.D. 0.10MNa2CO3.

65. In which solution is Ag2S MOST soluble?

A. 0.10MAg+(aq)B. 0.01MAg+(aq)C. 0.10MS2"(aq)D. 0.01MS2"(aq)

66. What is the molar solubility of ZnC03 in 0.010 MZnCl2?

A. 2.0 x 10-12MB. 2.0 x 10-10MC. 2.0 x 10"8MD. 1.4 x 10'8M

67. Which of the following salts is MOST soluble inwater?

A. CaC03B. Ag2C03C. ZnC03D. AgCl

68. Which of the following values accurately depicts thesolubility product of Ca3(P04)2 if the molar solubilityof Ca3(P04)2 is represented by x?

A. 18 x3B. 27 x4C. 54 x5D. 108 x5

69. To remove a cation from solution, the solution can beflowed through an ion exchange column where it willprecipitate the cation while releasing a more solublecation. Which of the following filters will NOT work?A. Ag+(aq) through a column containing NaCl(s)B. Zn2+(aq) through a column containing CaS(s)C. Ca2+(aq) through a column containing ZnC03(s)D. Ca2+(aq) through a column containing Na2S(s)



Calcium salts are generally insoluble in water at standardtemperature. Biologically, this is beneficial in that the Ca2+ion is soluble enough to be transported through aqueousmedium, yet insoluble enough to be a major component ofstructural features in the human body, such as bones. Listedin Table 1are some KSp values at 27'C for various calciumsalts:

Calcium Salt Ksp (@ 27°C)CaF2 4.8 x 10-n M3CaS04 6.1 x 10"6M2CaC03 9.1 x 10"9M2Ca(OH)2 4.3 x 10"6M3Ca3(P04)2 1.3 x 10-32M5

Table 1

The solubility products of the given salts can becompared only when the anions bonded to calcium carry thesame negative charge. For instance, the solubility productsof CaF2 and Ca(OH)2 can be directly compared to determinetheir relative solubilities. To enhance the solubility of acalcium salt, an anion that complexes calcium can be addedto solution. The anion competes for calcium in a complexequilibrium. A complex equilibrium is defined as thecoupling of at least two equilibrium reactions where theproduct of one reaction is the reactant in anotherequilibriumreaction. Reaction 1 and Reaction 2 combine to form acomplexequilibrium involving calcium cation.

Ca(OH)2(s) +H20(1) i ** Ca2+(aq) + 2OH"(aq)Reaction 1

Ca2+(aq) +Na2C03(s) ^ *• CaC03(s) + 2Na+(aq)Reaction 2

A change in calcium cation concentration affect bothreactions. If the Reaction 2 shifts to the right, then Reaction1 is forced to shift to the right in order to maintain itsequilibrium.

70. What is observed when sodium carbonate is added to asaturated calcium hydroxide solution with undissolvedcalcium hydroxide on the bottom of the flask?A. The pH of the solution will increase more than it

would increase had it been added to an unsaturatedsolution of calcium hydroxide.

B. The pH of the solution will increase less than itwould increase had it been added to an unsaturatedsolution of calcium hydroxide.

C. The pH of the solution will decrease more than itwould decrease had it been added to an unsaturatedsolution of calcium hydroxide.

D. The pH of the solution will decrease less than itwould decrease had it been added to an unsaturatedsolution of calcium hydroxide.


71. The maximum [Ca2+1 in a pH= 14 solution would bewhich of the following values?

A. 4.3 x 108MCa2+B. 4.3 x 10"2MCa2+C. 4.3 x 10"6MCa2+D. 4.3 x 10"10MCa2+

72. Enough CaCl2 is added to water so that not all of itdissolves, and thus some CaCl2 remains as a solid onthe bottom of the flask. The [CI"] is then measured.Addition of which of the following to the solution willincrease the chloride ion concentration ([CI"])?

A. Water

B. Silver nitrateC. Calcium fluorideD. Sodium phosphate

73. Which of the following graphs depicts the log [Ca2+]versus the buffered pH of an aqueous buffer solution ascalcium hydroxide is dissolved into different solutions?A. B..

Buffer pH Buffer pH

7 4. If addition of salt to water makes the water warmer oncethe salt has dissolved, which of the following are truefor solvation of that salt?

A. AH is positive, AS is negative.B. AH is negative, AS is positive.C. AH and AS are both positive.D. AH and AS are both negative.

7 5. Which of the following, when added to water at 300 K,will yield thehighest [Ca2+]?A. CaF2B. CaS04C. CaC03D. Ca(OH)2



Smog is a collection of several gas pollutants. Mostprevalent in the conglomeration of impurities are the sulfurand nitrogen oxides. The brown colorso familiar in smoggyskies is due to the presence of nitrogen dioxide (NO2). This,along with nitric oxide (NO), make up the majority of theairborne nitrogen oxides. The nitrogen oxides can be inter-convertedby a series of equilibriumreactions.

2 NO(g) + 02(g) -^±=r 2 N02(g)Reaction 1

2 N02(g) ^=^r N204(g)Reaction 2

N02(g) + 02(g) ^^r NO(g) + 03(g)Reaction 3

Reactions 1 and 2 are both exothermic while theReaction 3 is endothermic. Reaction 3 requires the additionof light to transpire. As the day progresses, sunlight carriesout Reaction 3, resulting in the highest ozone concentrationin the early afternoon. Figure 1 depicts the concentration ofpollutants during a hypothetical day:_0.3EQ.Q.ro.2

£0.10uco

JryL _V, _ /> ,•„ ' ,.1,^,'

„—.•"». ^"^ 1111%O

oo

oo

oo

NO

oo

Time

NO,

Figure 1

oo

oo

oo

oo

The amount of the nitrogen oxides in the air can bereduced by bubbling air through an aqueous transition metalhalide solution such as MnCl2(aq). Nitrogen dioxide (NO2)reacts with the water to form nitric acid (HNO3) and nitrousacid (HNO2), both of undergo deprotonation and bind themanganese, forming a complex ion. Reactions 4 and 5 makeup a complex equilibrium, as do Reactions 4 and 6.

2 N02(g) + H20(1) -^^r HN02(aq) + HN03(aq)

Reaction 4

MnCl2 + 2 N02"(aq) ^ *• Mn(N02)2 + 2 Cl"(aq)Reaction 5

MnCl2 + 2 N03"(aq) ^ ^ Mn(N03)2 + 2 Cl"(aq)Reaction 6

When reactions are added to produce an overall reaction,the equilibrium constants for the reactions are multiplied toobtain the equilibrium constant for the overall reaction. Thisis true for any type of equilibrium reaction.


76. Whichof the reactions has a negative entropy change?A. Reaction 2B. Reaction 3C. Reaction 5D. Reaction 6

77. Whichof the followingwillNOT increase the [N2O4]?

A. Decreasing the temperature.B. Increasing the pressure.C. Addition of water vapor to the air.D. Addition of NO2 to the air.

78. If the following reaction is carried out in a closed steelreaction vessel, what will be observed for the pressureof the system if the temperature is doubled?

A(g) + B(g) -^^r C(g) + D(g) + E(g)

AH = -112kJ/mole

A. The pressure will remain the same.B. The pressurewill increaseby less than 100%.C. The pressure will exactly double.D. The pressure will increase by more than 100%.

7 9. What can be concluded about the change in enthalpy forReaction 2?

A. Because a bond is broken, the reaction isendothermic in the forward direction.

B. Because a bond is broken, the reaction isexothermic in the forward direction.

C. Because a bond is formed, the reaction isendothermic in the forward direction.

D. Because a bond is formed, the reaction isexothermic in the forward direction.

8 0. If Reaction 1 is carried out in a closed piston systemwith an external pressure of 1 atm, what occurs when0.1 atm of NO gas is added to the system?A. The volume decreases by more than ten percent.B. The volume decreases by less than ten percent.C. The volume increases by less than ten percent.D. The volume increases by more than ten percent.


81. Which will shift reaction 4 to the left?

A. Addition of sodium hydroxide to the solutionB. Addition of manganese(II) chloride to the solutionC. Removal of nitrate from the solutionD. Removal of water from the solution

82. Which of the following graphs accurately depicts theNO2 concentration in a flask containing reaction 2before and after it has been bubbled through MnCl2(aq)?A. B.

time time

C. D.

time time



Physiological response is correlated with climate. Thiscan be proven by observing the shortness of breath and/orheadaches experienced by mountain climbers and skiers whenthey rapidly change altitudes. These symptoms are associatedwith a disorder referred to as hypoxia, a deficiency in theamount of oxygen that reaches body tissue. Hypoxia is atemporary disorder that in some cases can be fatal in the shortterm. After time, the disorder will disappear because thebody becomes acclimated to the new environment.

The cause of hypoxia is a result of the decrease inoxygen in the environment caused by the increase in altitude.The partial pressure of oxygen gas at sea level in most partsof the world is just over 0.20 atm. At an altitude of onemile above sea level, the partial pressure of oxygen gas isroughly 0.16 atm. Over time, the body adjusts to the loweroxygen content by increasing the production of hemoglobin(Hb), the molecule that binds oxygen. Reaction 1 expressesthe binding of oxygen by hemoglobin.

Hb(aq) + 4 02(aq) •> Hb(02)4(aq)Reaction 1

The equilibriumexpression for Reaction 1 is:K^_[Hb(02)4]

[Hb][02]4Figure 1

Hemoglobin is composed of four separate polypeptidestrands, each capable of binding a ferrous (Fe2+) cation, heldtogether as one molecule. By producing more hemoglobin,the body can shift the equilibrium to the right (forwarddirection). This results in more oxygen diffusing across thecell membrane. Once in the cell, myoglobin transports oneoxygen molecule from the interior of the cell membrane tothe mitochondria.

In acclimation, the amount of myoglobin does notchange. Acclimation takes a period of timethatranges froma week or two to months. For the 1968 Mexico CityOlympics, several athletes trained in high altitude toapproximate the environment of Mexico City (elevation7500 feet above sea level). For those athletes that trainedcloser to sea level, many did not perform well in the"thinner" air. Conversely, many current runners train in themountains for sea-level races so that their blood can provideextra oxygen to starving muscle cells.

83. Once hemoglobin transfers oxygen to the cellmembrane where it is absorbed, how many myoglobinmolecules should be present per hemoglobin?

A. 1B. 2C. 4D. 8


84. Which of the following changes will NOT increase theamount (in moles) of free oxygen in the body?A. Increased ventilation, resulting in an increased

amount of air transferred through the lungs.B. An increase in the production of HbC. Increasing thepartialpressure of COconsumed in a

normal breath.D. Increasing the total amount of blood.

85. What will be observed when a long-time mountainresident travels back to sea level?

A. They will experience hypoxia due to the lowerpartial pressure of oxygen at sea level.

B. They will experience hypoxia due to the higherpartial pressure of oxygen at sea level.

C. They will experience an increase in vitality due tothe lower partial pressure of oxygen at sea level.

D. They will experience an increase in vitality due tothe higher partial pressure of oxygen at sea level.

86. In the presence of carbon monoxide, hemoglobinpreferentially binds CO over O2 in a ratio close to 200: 1. Over time, what occurs in the blood of someonemoving to a carbon monoxide rich environment?

A. The increased CO reduces the amount of Hb thatbinds oxygen, so the Hb(02)4 decreases. Tocompensate, the body produces more Hb andincreases its rate of respiration.

B. The increased CO reduces the amount of Hb thatbinds oxygen, so/ the Hb(02)4 increases. Tocompensate, the body produces more Hb andincreases its rate of respiration.

C. The increased GO reduces the amount of Hb thatbinds oxygen, so the Hb(02)4 decreases. Tocompensate, the body produces less Hb anddecreases its rate of respiration.

D. The increased CO reduces the amount of Hb thatbinds oxygen, so the Hb(02)4 increases. Tocompensate, the body produces less Hb anddecreases its rate of respiration.

8 7. How does hypoxia affect the amount ofmyoglobin?

A. Myoglobin increases initially and then decreases.B. Myoglobin increases initially and then remain at

higher concentration.C. Myoglobin decreases initially and then remain at

lower concentration.

D. Myoglobin remains constant.


88. What activity will most likely result in hypoxia?A. Repelling down a mountainB. SCUBA divingC. Water skiing on the oceanD. Snow skiing

8 9. Why should an athlete choose to train at high altitudes?A. At higher altitudes, the partial pressure of oxygen

is greater, thus the athlete's body will increase itshemoglobin count.

B. At higher altitudes, the partial pressure of oxygenis less, thus the athlete's body will increase itshemoglobin count.

C. At higher altitudes, the partial pressure of oxygenis greater, thus the athlete's body will decrease itshemoglobin count.

D. At higher altitudes, the partial pressure of oxygenis less, thus the athlete's body will decrease itshemoglobin count.



The equilibrium of a gas-phase reaction can vary withthe dimensions of the container. This is observed when areaction has a different number of gas molecules on the twosides of the reaction equilibrium. Although the equilibriumconstant remains constant when pressure, volume, and molesare altered, the ratio of product to reactant can vary. Thefollowing reaction is a typical reaction which shows adependence on the dimensions of the container:

N204(g) ^ * 2 N02(g)

Reaction 1

The reaction may be carried out in a glass flask, wherethe volume of the container remains constant while theinternal pressure may vary. The reaction may also be carriedout in an enclosed piston, where the internal pressure remainsconstant while the volume of the container may vary. LeChatelier's principle predicts that when the internal pressurerises, Reaction I will shift to the left to alleviate the increasein pressure. Le Chatelier's principle also predicts that whenthe volume of the container is increased, Reaction 1 willshift to the right to fill the volume. The equilibriumconstant for Reaction 1 at 25°C is 4.72 x 10"3 atm.

Independent of the starting conditions and presence orabsence of an inert gas, the ratio of the square of the partialpressure of nitrogen dioxide to the partial pressure ofdinitrogen tetraoxide is constant, given that the temperaturedoes not change. The ratio of nitrogen dioxide to dinitrogentetraoxide increases as the temperature is increased.

9 0. The reaction as written is:

A. endothermic with AS > 0.B. endothermic with AS < 0.C. exothermic with AS > 0.D. exothermic with AS < 0.

91. The reaction will shift to the right with all of thefollowing changes EXCEPT:A. addition of N204(g).B. an increase in volume at constant pressure.C. a decrease in pressure at constant volume.D. addition of He gas to the system at constant

volume.

92. Which of the following accurately shows Keq in termsofAG?

A. Keq =

B. Keq =^ InRT

AG/C. Keq =e 'RT-AG/D. Keq = e 'RT

Copyright © by The Berkeley Review®

AG

InRT

AG

217

93. What is NOT true for a reaction with an equilibriumconstant of 1.0 x 105?A . There is mostly product present at equilibrium.B. When starting with all reactants, the shift to reach

equilibrium is insignificant.C. The Keq for the reverse reaction is 1.0 x10"5D. The same reaction when carried out with a catalyst

will have the same equilibrium constant.

94. If the value of Keq at a certain temperature is greaterthan one, what will be observed after radiolabeled15N204 is added toanequilibrium mixture ofNO2 andN2O4 in a rigid container?

A . The amount of radio-labeled NO2 increases, whilethe amount of radio-labeled N2O4 decreases.

B. The amount of radio-labeled NO2 decreases, whilethe amount of radio-labeled N2O4 increases.

C. The amount of radio-labeled NO2 increases, whilethe amount of radio-labeled N2O4 remainsconstant.

D. The amount of radio-labeled NO2 decreases, whilethe amount of radio-labeled N2O4 remainsconstant.

95. As the handle of a piston container filled with anequilibrium mixture of NO2 and N2O4 is lifted, whatoccurs?

I. The PNO2 ratio increases.

pN204II. The piston cools down.HI. The mole percent of N2O4 increases.

A. I onlyB. II onlyC. I and II onlyD. II and HI only

96. The Keq for the reaction aswritten is:pN02A. Keq =

B. Keq

PN204

(Pno2)2pN204

C K -PN2°4PN02

D. Keq =J^t.(pNo2r


Questions97-100 areNOTbasedon a descriptive passage.

97. If the molar solubility of anMX typeof salt is definedas y, then the solubility product is:A. yB. 2yC. y2D. 4y2

98. MgF2(s) would be most soluble in which of thefollowing solutions?A. 0.10MCaF2(aq)B. O.lOMNaF(aq)C. O.lOMNaCl(aq)D. 0.10MMgCl2(aq)

99. If the forward rate constant for a one-step reaction isfour times the reverse rate constant, then which of thefollowing is true?A. Keq= 0.0625B. Keq= 0.25C. Keq= 4.00D. Keq =16.00

100. The amount of a salt that dissociates into water isALWAYS increased by which of the followingchanges?A. Increasing the temperature.B. Decreasing the temperature.C. Increasing the amount of water.D. Decreasing the amount of water.


"It's all about balance you sec, you must keep balance!"

1. B 2. D 3. B 4. D 5. C

6. A 7. B 8. D 9. A 10. B11. A 12. B 13. D 14. C 15. B

16. C 17. B 18. B 19. B 20. B

21. B 22. B 23. D 24. B 25. D

26. C 27. A 28. C 29. D 30. A

31. C 32. D 33. C 34. A 35. A

36. A 37. A 38. D 39. A 40. D41. B 42. A 43. A 44. A 45. C46. B 47. C .48. A 49. A 50. B

51. C 52. B 53. A 54. D 55. D

56. C 57. A 58. B 59. A 60. D

61. C 62. A 63. C 64. B 65. D

66. C 67. B 68. D 69. C 70. A

71. C 72. D 73. C 74. B 75. D

76. A 77. C 78. B 79. D 80. C

81. D 82. A 83. C 84. B 85. D

86. A 87. D 88. D 89. B 90. A

91. D 92. D 93. B 94. A 95. C

96. B 97. C 98. C 99. C 100. c

THAT'S IT!

Equilibrium Passage AnswersPassage I (Questions 1-7) CarbonDioxide andCarbonMonoxide Equilibrium

Choice B is correct. This question involves first determining the value of the equilibrium constant (Keq) ateq;102°C, and then solving for the partial pressure of CO2 at 102°C. Converting 102°C into Kelvin yields 375 K.According to Table 1, at 375 K, Keq =9atm"1. The solution is as follows:

Keq -9 - (Pco2)2 (Pco,)2 (Pco,)2(Pcor(Poo) (2)2(i;

=9 .-. (PCO2)2 =36 .-. Pco2 =6atm., choice B

2. Choice D is correct. Adding CO2 (a product) to an equilibrium mixture forces the reaction to the left. When thereaction shifts to the left, reactants (02(g) and CO(g)) increase. This eliminates choices Aand B. Because Keqchanges only with temperature, and the temperature did not change in this case, choice C is invalid. Bydefault, (the elimination of three wrong choices), choiceD is the correct answer.

3. Choice Bis correct. Because the value ofKeq increases as the temperature increases, products increase upon theaddition of heat to the reaction. This means that heat lies on the reactant side of the reaction and thus thereaction is endothermic as written. The AH is positive for an endothermic reaction. Pick B and feel warm.

4. Choice D is correct. The equilibrium constant is just that, a constant. It is specific for a given temperature, so itvaries only with a change in temperature. The equilibrium may shift with changes in either the pressure,volume or concentration, but the value of K remains constant. This makes choice D the best answer.

5. Choice C is correct. The following chart summarizes the partial pressures during the course of the reaction:

6.

7.

Reaction:

Initially:Shift:

Equilibrium:

2CO(g)

0

±2x

2x

02(g)0

+ x

2C02(g)

1.00

-2x

l-2x

The total pressure of the system is the sum of the partial pressures of each component at any given time duringthe reaction. The question here asks for the total pressure at equilibrium, so the equilibrium partial pressuresmust be added. To solve this question, use the equation:

Ptotal = PC02 + pCO +Po2= (! "2x) +2x+ x= 1+ x-The value of x can be no larger than 0.5 atm., because you can lose no more C02(g) than you start out with (1.00arm.). The value of x must be somewhere between 0 and 0.5, because some of the carbon dioxide is lost, but notall of it. This makes the following relationship 1.5 > Ptotal > 1 true> which is choice C, your best choice.

Choice A is correct. Compressing the reaction vessel results in reduced volume and increased pressure. Upondecreasing the container volume, prior to any shifting of the reaction, the partial pressure and concentration ofeach gas increases, but the mole ratio is the same until the reaction shifts. This is why no answer choicescontained the term concentration. Compressing the container decreases the free area in which the gases canexist. The more crowded environment favors the formation of CO2, because carbon dioxide lies on the lesscrowded side of the reaction. This means that moles of CO2 increase, moles of CO decrease, and moles of O2decrease. This eliminates choice B. When moles of CO decrease, the mole fraction of CO decreases, so choice Cis eliminated. The temperature did not change, so the equilibrium constant remains the same, so choice D iseliminated. The best answer is choice A. As a note of interest, all of the partial pressures increase, but carbondioxide increases by the greatest amount. The shift to re-establish equilibrium never completely offsets thestress, so the result is that CO and O2 have slight increases in partial pressure overall, while CO2 shows asignificant increase in partial pressure.

Choice B is correct. Addition of Ca(OH)2(aq) to an equilibrium mixture of Reaction 1 serves to remove C02(g)from the reaction mixture. To compensate for lost 002(g), Reaction 1 proceeds in a way so as to regenerate002(g), but it is not completely regenerated. This eliminates choice A. To produce C02(g), the reaction mustshift right, which consumes both 02(g) and CO(g). This results in a decrease of both 02(g) and CO(g). The bestanswer is choice B. Total moles of gas decrease, so the total pressure of the system decreases.

Copyright © by The Berkeley Review® 219 Section III Detailed Explanations

Passage II (Questions 8 -15) Hydrogen and BromineEquilibrium

9.

10.

11.

12.

13.

Choice Dis correct. Keq for the reaction is determined by plugging the values from Table 2into the equilibriumexpression. Upon doing so, the value is found to be 5.44 atm., making choice Dthe best answer.

Ke =(PHBr)2 =(^l =(2.72)(2.72) =171x172 =272x2= 5M atm.PH 1.36 1.36 1.36

Choice A is correct. Addition of the gaseous HBr (a product in Reaction 1) to the equilibrium mixture pushesthe reaction to the reactant side (right), and thus increases both the partial pressure and mole percent of H2. Ifyou are of sound mind and body, then you should pick choice A. As long as the temperature remains constant,the concentration ofBr2(l) (or any other pure liquid) remains constant. The concentration ofa liquid ismeasuredas density, which changes with temperature. This eliminates choices Cand D, so don't pick them.

Choice B is correct. The greatest amount of Pb(g) forms from the reaction that generates the greatest backreaction or smallest forward reaction. Choices A and C are eliminated, because less H2(g) is present initiallythan in choice B, therefore less H2(g) exists at equilibrium. In addition, the difference between choices AandCis the amount of Br2(l), and liquids do not affect the equilibrium distribution, so both yield the same amount ofH2(g). There is only one correct answer per question, so identical answer choices should both be eliminated. Ifchoice Dwent one hundred percent in the reverse direction, then 0.50 atm. ofH2(g) would be generated. Thisvalue is halfof the amount ofH2(g) in choice B, sochoice D iseliminated. To generate an equivalent amount ofH2(g) at equilibrium asstarting with 1.00 atm. H2(g) and excess Br2(l), requires starting with 2.00 atm. HBr(g).

Choice A is correct. An increase in temperature results in the addition of heat to the equilibrium mixture.Because themole percent of product (HBr) increases, heatmustbe viewed as a reactant in the reaction.

Heat + H2(g) + Br2(l) ^=^ 2 HBrfe)When heat is viewed as a reactant, the reaction is defined as being endothermic, so choice A is the correctanswer. An endothermic reaction has AH value that is a positive number. Choice C is eliminated, because thepressure changed with both the influx ofheat and the shift of the reaction, so it cannot be an isobaric (constantpressure) process. Choice D iseliminated, because thepassage states thatReaction 1 is an equilibrium mixture.

Choice B is correct. The definition of equilibrium is a state where the forward and back reactions have equalrate. Choices A and D are equivalent, so they are both eliminated. The reaction lacks Br2(l), and despiteliquids not affecting equilibrium, keep in mind that thesystem is not in equilibrium. Liquids can be a limitingreagent. Because there isno Br2(l) present, the reaction cannot move in the forward direction, even if it wantedto. This makes choice B the best answer. At this point, we are uncertain it is correct, but it is the best answerchoice. To determine the direction in which a reaction proceeds, it is necessary to compare the value of thereaction quotient (Qrx) to the equilibrium constant (Keq). The two possible scenarios are: Qrx >Keq, in whichcase the numerator is too large in the equilibrium quotient (products are in excess of reactants), making thereaction shift to the left (reactant side) to reach equilibrium. The second scenario is where Qrx <KeC[, in whichcase the denominator is too large in the equilibrium quotient (reactants are in excess of products), making thereaction shift to the right (product side) to reach equilibrium. We solve for Q as follows:

s2

Qrx =_(PHBr)2 _12 _1 1.36 W'lb

Because Qrx <Keq, Reaction 1should shift right (to the product side). The problem here is that there is no Br2present, so there can be no reaction. This confirms that Br2 is the limiting reagent. To quickly determine thedirection ofa reaction, putKeq and Qrx alphabetically and turn the greater than or less than sign into an arrow(i.e., the Keq >Qrx becomes Keq —> Qrx, meaning the reaction proceeds forward to reach equilibrium).

Choice D is correct. All reagents are present,so a reaction is possible. To answer this question, solve forQrx

.(PHBr)2 _32 (2.72)2Qrx =

Pr-b 1.5 1.36.*. Keq <Qrx bya small amount

Because Keq <Qrx, the reaction proceeds to the left (reactant side). This decreases HBr(g) and increases H2(g),so choices A and B are eliminated. The shift is small, because Q and K are close, so choice D is the best answer.

®Copyright © by The Berkeley Review 220 Section III Detailed Explanations

14.

15.

Choice C is correct. If the flask is initially charged with only HBr(g), then the graph of H2(g) over time startsat 0, eliminating choice A. Choice Dis a titration curve, so it is an incorrect answer choice. The graphs depictthe progression of a reaction. Reactions proceed with a gradual tapering off (choice C), not an instantaneousshut-off (choice B). Pick C, because H2(g) increases most rapidly initially, then levels off until equilibrium.

Choice B is correct. "...NOT affect the partial pressure of H2(g)?" Addition of NaOH(aq) removes HBr andindirectly shifts the Reaction 1 equilibrium to the right, through a complex equilibrium. Addition ofH2(g) andHBr(g) directly affect the equilibrium, shifting the equilibrium to the opposite sideof the compound added. Ifthe reaction is already at equilibrium, then neither a solid nor a liquid (such as Br2(l) in this case) affects theequilibrium. For this reason, choice B is a particularly swell answer choice. A solid or liquid can only affectthe net results (equilibrium conditions) if the system is not at equilibrium and the compound being added (thesolid or liquid) is the limiting reagent. When added in, thesystem can then achieve equilibrium.

Passage III (Questions 16 - 22) Equilibnum Constant Magnitude

16.

17.

18.

Choice C is correct. According to the two reactions, only PCI3 and NO react with CI2, so PCI5 and NOCl are notoxidized by CI2. This eliminates answer choices Band D. By looking at the oxidation states of the reactantsand products, PCI3 (P = +3), PCI5 (P = +5), NO (N = +2), and NOCl (N = +3), it shows that the forwarddirection in bothReaction 1 and Reaction 2 as drawn are oxidation half-reactions for the non-metal compounds.The reactants get oxidized by chlorine in the forward reaction as written. Comparing the two values of Keqshows that reaction 2 lies further to the right (product side) than does Reaction 1. Thus, Reaction 2 favors theformation of products more so than Reaction 1. This implies that NO reacts with CI2 more readily than PCI3reacts with CI2. The greater the reactionwith CI2 means more oxidation by CI2. Choose C to feel good.

Choice B is correct. Initially, there is twice as much PCI3 as CI2, and their sum pressure is 1.50 atm. When thereaction shifts in the forward direction, total pressure decreases, so choice D is eliminated. According to themole ratio, the initial partial pressure of phosphorus trichloride is 1.0 atm. and the initial partial pressure ofchlorine gas is 0.5 atm. The chart below summarizes the partial pressures during the course of the reaction.

Reaction: PCl3(g) Cl2(g)Initially: 1.0 0.5

Shift: ^x -x

Equilibrium: 1-x 0.5-x

PCl5(g)

0

• ±x

x

The total pressure of the system is a sum of the partial pressures of each component at any given time duringthe reaction. The question here asks for the total pressure at equilibrium, so the equilibrium partial pressuresmust be added. To solve this question, use the formula X Pj = Ptotal- ln this case the following holds true:

Ptotal = PPCI3 + PC12 + pPCl5 = (1 " x) + (0-5 - x)+ x= 1.5 - x.If everything reacted to the limiting reagent capacity, then x is 0.5. But the reaction does not go to completion,so the value of x falls between 0 and 0.5. This is because some, but not all, of the reactant is lost. Choice A iseliminated, because 1.5 > Ptotal > 1- To solve this question, you must estimate the magnitude of x. The value ofx is greater than 0.25, because over half shifts to products to reach an equilibrium constant of 1.95. This makeschoice B the best answer. To solve it more precisely, assume Ptotal at equilibrium is 1.25 (the borderlinebetween answer choices B and C). If Ptotal at equilibrium is 1.25 atm., then x = 0.25 atm., and at equilibrium,PpCl3 =0.75 atm., Pq2 =0-25 arm., and PPCI5 =0.25 atm. The following Qwould result:

PPC15 _Qrx = 0.25 _ 1 _=1.33 .'. Keq >Qrx, somore mustshift over.(PPC13XPC12) (0.75)(0.25) 0.75

This means that x is between 0.25 and 0.50 atm., and thus Ptotal at equilibrium is between 1.00 and 1.25.

Choice B is correct. The reaction listed in the question is the reverse reaction of Reaction 2. When a reaction isreversed, the equilibrium expression is inverted, because reactants and products are inter-converted. Theresulting equilibrium constant is the inverse of the original value. The inverse of something times 104 issomething times 10"5, making choice B the best answer. Choice A is an incorrect answer that could have beenreached if you inadvertently decided the values must multiply to 1.0 x 10"14 (a common value in acid-basechemistry). Choice C was the result of taking the square root, which is wrong in this case. Choice D iseliminated because the forward and back reactions are different.


19.

20.

21.

22.

Choice Bis correct. Welcome to the wonderful world of math. The following chart summarizes the changes inpartial pressures as the system goes to equilibrium. You must solve for the x-value. Keep in mind that theanswer you seek is actually 2x, but you must first solve for xbefore you determine 2x.

^ S 2 NOCl(g)Reaction:

Initially:Shift:

Equilibrium:

2 NO(g)

0

±2x

2x

Cl2(g)0

+ x

1.0 atm.

-2x

1 -2x

From here, it is a matter ofplugging theequilibrium values into theequilibrium expression.(PNOCl)2 _(l-0-2x)2_ t2

K =

>q (PNO^Pciz) (2x)Z(>0 (2x^i i=J- =2.1 xlO4

3

The easiest thing to do here is to take the reciprocal of each side and then solve for x, keeping in mind that theanswer choices are 2x.

4X3 = 1 =4.9 xlO"5 =49 xlO"6 .\x3 =12.25 X10"6 .\x= T12.25 x 10"22.1 x 104

The cube root of 12.25 lies between 2and 3(because 23 =8and 33 =27), so let's call it2.? PNo =2(2.? x 10"2) =4.?x 10"2, which is choice B. The math is not that bad, if you approximate.Choice B is correct. Immediately, choices A and C are eliminated, because there is no reaction possible, whenthe reaction has only one reactant (and lacks CI2). From here, it comes down to interpreting the magnitude ofKeq values. The reaction that forms the greatest amount of CI2 is the reaction with the Keq that most favorsformation of reactants (because CI2 is a reactant in Reaction 1 and Reaction 2). Reaction 1, with the smallerKeq, most readily favors the formation of reactants out of the two reactions. Because of the squared term in keqfor Reaction 2, we need to look more closely in most cases, but the Keq values are so drastically different for thetwo reactions, that you are safe ignoring the squared terms in the Keq of Reaction 2. The larger Keq isassociated with Reaction 2, so choice D is out and the best answer is choice B.

Choice B is correct. This is a rehash of an ongoing theme in this passage. Addition of CI2 shifts both Reaction1 and Reaction 2 in the forward direction. The question here entails the magnitude of the shift. Reaction 1,with a Keq value of approximately 1, shifts the equilibrium to the product side to use roughly half of the CI2gas that was added. Adding CI2 to reaction 2, with a KCq that is extremely large, will shift the equilibrium tothe productside to use nearly all of the CI2 gas that wasadded. Choice B is the best answer.Choice B is correct. To determine the direction of a shift requires first solving for the equilibrium quotient (Q)and then comparing that value to the Keq.

Qrx =Ppci5 0.40 _ 1 _=2.5 .". KL.q < Qrx, so the reaction shifts left.

(PPC13XPC12) (0.40)(0.40) 0.40When K<Q, the Q is too large, so the products must decrease and the reactants must decrease (K <- Q). Thisforces the reaction to move to the left (reactant side). You really should pick choice C. Choice D is eliminated,because a catalyst affects only the kinetics (rate), and not the thermodynamics of a reaction. A catalyst simplylowers the activation energy, but has no effect on theheat of reaction (AH), the free energychange (AG) for thereaction, or the equilibrium constant (K) for the reaction.

Gas-Phase Equilibrium Constant ExperimentPassage IV (Questions 23 - 29)

This is an experiment that is used to determine theequilibrium constant for the gas phase reaction listed in thepassage. In interpreting the apparatus in Figure 1, you must note the two reactants are being reacted once thevalve is opened between the gas line and the flask. The reactants are mixed in a ratio of4H2 to 1CS2, andthey just-so-happen to react in a ratio of 4 : 1. This means that the reactants always exist in a 4 : 1 ratio andthe products always exist in a 2 : 1 ratio. The reactants are never completely depleted, thus the total pressure(Pjnit - 2x) of thesystem can never reach a value as low as0.60 atm. Overall, thenumber of molecules of gas aredecreasing as the reaction proceeds, which accounts for the decrease in pressure experienced by the system as i tapproaches equilibrium. The final total pressure is not given on the graph, but a AP is a negative value. Thispassages seems hard, but it is really testing your organization skills.


23. Choice D is correct. The purpose of the experiment is todetermine the equilibrium constant (K) for Reaction 1.The value of theequilibrium constant varies with the temperature, so the temperature mustbeheldconstant toaccurately evaluate the value of the equilibrium constant. This means that the best answer is choice D.Hopefully the other answer choices did not seem that tempting.

24. Choice B is correct. Upon heating a closed, isochoric system, the internal pressure increases according to theideal gas law (PV = nRT). This in essence is Charles' law, eliminating choice A, which has Charles' lawlisted incorrectly. Because the pressure decreases rather than increases, another variable must be changing.The system is in a flask, so the volume can't change. This means thatmoles must be decreasing, causing thepressure to decrease. The only way for moles to change in a closed system, is for a chemical reaction to takeplace. Reaction 1 shifted from the reactant side (five gasmolecules) to the product side (three gas molecules).This means that the reaction proceeds in the forward direction to consume the heat added to the system. Thismeans that the reaction as written must be endothermic,making choiceB correct.

25. Choice D is correct. This question requires setting up the equilibrium relationship that transpires during thecourse of the reaction. The reaction goes as follows:

Reaction: CS2(g) 4H2(g)


Shift: -x -4x

Equilibrium: 0.2-x 0.8 - 4x

CH4(g) 2H2S(g)

0 0

+ x ±2x

X 2x

At any point, including equilibrium, the total pressure of the system is found according to Equation 1.Ptotal = PCS2 +PH2 + PCH4 +Ph2S = (0-2 - x)+ (0.8 - 4x) + x +2x= 1.0- 2x

The initial pressure of the system was 1.0 atmospheres when the valve was first opened to start the reaction.The pressure of the system at equilibrium is 1 - 2x atm., therefore the change in internal pressure from the startof the reaction until equilibrium is 2x. This means that the partial pressure at equilibrium due to H2S (whichis also 2x) equals the change in internal pressure during the reaction. The partial pressure due to CH4 atequilibrium is one-half of the change in internal pressure. This makes choice D the best answer.

26. Choice C is correct. This question requires that you deduce that the value of x can never exceed 0.2. This upperlimit of 0.2 is due to the initial pressure of CS2being only 0.2 atm. You can lose no more CS2 than you haveavailable as a reactant. Because of this, no matter what value for x is used (any value between 0 and 0.2), 0.8 -4xwillalwaysbe greater than 0.2 -x. This means that Ph2 is always greater than Pcs2 because 4xmust alwaysbe less than 0.8. The best answer is choice C. If the reaction goes completely in the forward direction, thenproducts will exceed reactants and Pcs2<PCH4/ eliminating choice A. Because there is no CH4 or H2S in theinitial reaction mixture, the only CH4 or H2S present during the reaction or at equilibrium, is the result of thereactionproceeding in the forward direction. Forevery oneCH4that forms, two moleculesof H2Sform, so Ph2Sis always greater (not less) than PCH4/ eliminating choice B. If the reaction barely proceeds in the forwarddirection, then the reactants will exceed the productsand Ph2> Ph2S- Thiseliminates choice D.

27. Choice A is correct. The equilibrium constant is the pressure of the products divided by the pressure of thereactants. Answer choices B and D (which both have reactants divided by products) are eliminated. Thecoefficients from the balanced equation must be included as exponents in the equilibrium expression, so choice Ais the correct choice. All of the reactants and products are gases, thus they all belong in the Keq expression.Keep in mind that solids and pure liquids are not included in the equilibrium expression. The equilibriumconstant for this reaction is referred to as Kp, because it is a gas system and the values are partialpressures.

28. Choice C is correct. The total pressure of the system at equilibrium is a sum of the partial pressure of eachcomponent gas at equilibrium. The total pressure of the system is found as follows:

Ptotal = PCS2 +PH2 + PCH4 +PH2S = (0.2 - x) + (0.8 - 4x) + x +2x=1.0- 2xChoices A and D are eliminated, because the total pressure is less than 1.0 atm. The final pressure of thesystem at equilibrium is 1 - 2x. The value of x cannot be greater than 0.20 atm because the CS2 gas is thelimiting reagent and only 0.20 atm. of CS2 is initially present. Because the value of 2x is subtracted, the finalpressure drops, but it cannot drop to a value less than 0.6atmospheres. The best answer is choice C.

Copyright© by The Berkeley Review® 223 Section in Detailed Explanations

29. Choice D is correct. The equilibrium constant (a numerical value), changes with the temperature, but it doesnot change with volume, pressure, moles of reactants, or moles of products. Choice Ainvolves adding hydrogengas (a reactant). The reaction shifts to the right to reestablish equilibrium, but the equilibrium constantremains constant. This is to say that the equilibrium position may change, but the equilibrium constant doesnot. Choice Binvolves removing all four gases. To determine which escapes most readily involves molecularmass and effusion rates, but that is irrelevant in this question. Losing both products and reactants causes thereaction to shift to reestablish equilibrium, but the equilibrium constant remains constant. Choice B iseliminated. Choice C involves adding hydrogen sulfide gas (a product). The reaction shifts to the left toreestablish equilibrium, but the equilibrium constant will remain constant. Choice D involves increasing thetemperature. The equilibrium position and the equilibrium constant both change with a change in temperature.Thevalue of the equilibriumconstant can only be changedby temperature.

Passage V (Questions 30 - 36) Equilibrium Reactions

30. Choice A is correct. Given that the initial partial pressures of H2O and CI2O are both 0.5 atmospheres, andboth are reactants, it is possible initially for the reaction to proceed in the forward direction. As such, theinitial forward rate cannot start at zero. This eliminates choices C and D. There is no HOC1 present initially,therefore the reverse reaction rate must start at zero. This reaffirms that choices C and D are invalid. As thereaction proceeds to equilibrium, the reactant pressures decrease, thus the forward reaction slows graduallyuntil equilibrium is reached. Equilibrium occurs when the rate forward is equal to the rate reverse. Becausethe forward rate diminishes with time, the best answer is choice A.

31. Choice C is correct. Reactions that can shiftwhen the pressure of the system changes are reactions that havean unequal number ofgas molecules on the reactant and product sides of the reaction. InReaction 1, there arethree reactant gas molecules and two product gas molecules, soReaction 1 shifts in the forward direction if thetotal pressure increases on the equilibrium system. In Reaction 2, there are two reactant gas molecules and twoproduct gas molecules, soReaction 2cannot shift in either the forward or reverse direction if the total pressureincreases. In Reaction 3, thereare two reactant gasmolecules and oneproduct gasmolecule, so Reaction 3 shiftsin the forward direction if the total pressure increases on the equilibrium system. This means that bothReaction1 and Reaction 3 shift when the total pressure of the system changes. The best answer is choice C.

32. Choice D is correct. The equilibrium constant of a reaction remains constant unless the temperature of thesystem ischanged. Changing the volume ofthe system does not change the equilibrium constant (although thereaction may shift and the equilibrium ratios change). This eliminates choices A and B, which show the Keqin its equation form. As the volume of the system increases, the reaction shifts to the side of the reaction withmore molecules (in this case the reactant side), thus the ratio of HCOC1 (a product) to HCl (a reactant)decreases as the reaction reestablishes equilibrium following a change in volume. The best answer is choice D.

33. Choice C is correct. Sulfurdioxide and oxygen are both reactants, according to Reaction 1. When mixing 0.75atm. SO2 with 0.50 atm. O2, Reaction 1 proceeds in the forward direction to establish equilibrium. The valueof Keq for this reaction (582), as listed in the passage, is well above one. At equilibrium, there is significantlymore SO3 than both SO2 and O2. This eliminates choices Band D. TheSO2 should start higher than the O2,but it diminishes twice as fast as O2, thus the SO2 line should drop below the O2 line at equilibrium. If SO2 iscompletely depleted (all 0.75 atm. areconsumed), 0.375 atm. ofO2 would be consumed, leaving 0.125 atm. O2.This is because SO2 is the limiting reagent, if the reaction goes to completion. This makes choice C the bestanswer.

34. Choice A is correct. The equilibrium constant for reaction 2 is 8.61 x 10"3 at 500°C Choice A is the reversereaction from Reaction 2, so the equilibrium constant is the reciprocal of8.61 x 10"3 at 500°C. This leads to avalue with 102, not 10"3 in it. This makes choice A the mismatch of Keq and reaction. Choice Bis the same asReaction 2 except the values have been cut inhalf. The result is that the Keq as written should be the squareroot of 8.61 x 10"3. This leads to a value with 10"2 in it, therefore choice B seems reasonable and thus valid.The equilibrium constant for Reaction 3 is3.26 x10"6 at 500°C. Choice C is the reverse reaction from Reaction 3,whereall the valueshave beencut in half. The equilibrium constant as writtenshould be the square root of thereciprocal of3.26 x 10"6 at 500°C. This leads to a value with 102 in it, so choice C seems reasonable and valid.Choice D is Reaction 3, where all the values have been doubled. The equilibrium constant as written should bethe square of 3.26 x10'6 at 500°C. This leads to a value with 10"11 or 10"12 in it, so choice Dseems reasonableand valid. The only mismatch is found with choice A.


35. Choice Ais correct. The conversion from the pressure unit of torr to the pressure unit of atmospheres involvesdividing the value in torr by 760 (the conversion factor from torr to atm.). Because the Kp value is given interms oftorr1, the value must be multiplied by 760 inorder to convert it into a value in terms ofatm-1. The bestanswer is choice A.

36. Choice A is correct. An endothermic reaction is set up as follows:heat + reactants -^ *- products.

The cooling of the reaction can be treated as the removal of heat. When heat is removed from an endothermicreaction, because the heat is acting like a reactant, the reaction shifts to the left (reactant side) to regeneratethe lost heat. This shift causes an increase in reactants and a decrease in products. The product to reactantratio therefore decreases. Choices B, C, and D all account for the same shift, so the three choices can beeliminated. The best choice is thus choice A.

Passage VI (Questions 37 - 42) Le Chatelier's Principle

37.

38.

39.

40.

41.

42.

The interesting feature of this reaction is that the ratio of the reagents is 1 : 1 : 1 : 1, so the total pressure of thesystemnever changes. The reactants always decrease by the same amount, and the products always increase bythesameamount as the reaction proceeds from the initial conditions to theequilibrium conditions. Thepurposeof this experiment is to prove that equilibrium is pathway independent, implying that whether the reactionstarts with all reactants or all products, it will reach the exactly the same equilibrium.

Choice A is correct. Increasing the volume of the reaction vessel has no effect on the equilibrium, because thereare equal moles of gas molecules on both sides of the reaction. Adding moles of either reactant or product willshift the equilibrium to the opposite side. Because the reaction is exothermic, it will shift forward with adecrease in temperature and reverse with an increase in temperature. Thus, the best answer is choice A.

Choice D is correct. The equilibrium constant is less than 1.00, so the reactants are more abundant than theproducts. This means at equilibrium, both CO2 and H2 have concentrations less than half of their initialconcentration (1.00 M). The only answer that shows this is choice D. Note that the final concentrations are thesame as in trial I. Regardless of the starting concentrations, the equilibrium concentrations will be the same ifthe total pressure and the temperature are the same.

Choice A is correct. When reaction vessel III was cooled from 1000°C to 500°C, the partial pressure of carbondioxide gas (CO2) increased. This increase in partial pressure for carbon dioxide indicates that the reactionshifted to the right (product side). The reaction must therefore be exothermic, because as heat is removed fromthe system, the reaction shifts to the product side to produce heat. Any reaction that releases heat in theforward direction is exothermic. This means that as heat is added to the reaction, the reaction shifts to theleft (to the reactant side). This results in a decrease in the partial pressure of CO2 and a decrease in theequilibrium constant. The correct answer is choice A.

Choice D is correct. A change in pressure will never change the equilibrium constant. This eliminates choices Aand B. Because there is the same number of gas molecules on both sides of the equation, a change in pressure hasno effect on the equilibrium. This means that the equilibrium does not shift so the best choice is D.

Choice B is correct. Because there is an equal number of gas molecules on both sides of the reaction (twomolecules on each side), the total pressure of the system will not change with shifts in the reaction. Thiseliminates choices A and C. Because there is no water present initially, the reaction must shift left to reachequilibrium. This results in an increase in both water and carbon monoxide, therefore both slopes shouldincrease equally. This is shown only in choice B.

Choice A is correct. When the equilibrium constant (Keq) is greater than 1, the concentration of products isgreater than the concentration of reactants. As such, choices B and C are incorrect statements, and can beeliminated. For an exothermic reaction, the equilibrium of the system shifts to the right (product side) as thetemperature of the system decreases. This means that the products increase and the reactants decrease as thetemperature of the system decreases. Both choices B and D are eliminated because of this, leaving only choiceA still standing. If the products out number the reactants, then the equilibrium constant is greater than 1.0, sothe best answer is choice A.


Passage VII (Questions 43 - 49) Solubility and Qualitative Analysis

43. Choice A is correct. The molar solubility values of the sulfate salts are greater than the molar solubilityvalues of the carbonate salts, as shown by the values in Table 1. This makes statement I valid. The addition ofcomplexing agents helps to dissolve cations into aqueous solution by forming ligand bonds to the metal. Themost common example is hemoglobin, which serves to make iron dication and trication more water soluble.This makes statement II valid. Given that both statements I and II are true, choices B and C are botheliminated. Hydrochloric acid is a strong acid, therefore chloride anion is not affected by the presence ofhydronium ion. As aresult, chloride salts show no pH dependence. Sulfide is aweak base, therefore sulfideshows apH dependence. This makes statement III invalid, so the best answer is choice A.

44. Choice Ais correct. The most accurate value for Kgp is found when the most accurate measurement of the molarsolubility is used in the calculation. The most accurate measurement of the molar solubility is found with thesalt having the greatest molar solubility. As a general rule, the larger the value, the less significant the errorin its measurement. The highest molar solubility is found with Ag2S04/ at 1.3 x10"2 M. The best answer ischoice A. This question is just adifferent way of asking which compound has the greatest molar solubility.

45. Choice C is correct. The anion chosen to remove Zn2+ selectively is the anion that forms the least soluble saltwith zinc dication, rather than with silver cation or strontium dication. According tomolar solubility values,the least soluble chloride salt is AgCl, so chloride anion is not a good choice. According to molar solubilityvalues, the least soluble carbonate salt is ZnC03, so carbonate anion is a possible choice. According to molarsolubility values, the least soluble hydroxide salt is Zn(OH)2/ by a substantial margin, thus hydroxide anionis the best choice so far. According to molar solubility values, the least soluble sulfate salt isZnSO^ so sulfateanion is a possible choice. The greatest difference between zinc and the next most soluble cation comes withhydroxide anion. Thebestansweris thuschoice C.

46. Choice B is correct. Silver chloride is the only chloride salt that is highly insoluble. A precipitate formingwhen chloride is added to solution indicates that silver cation must be present. This makes choice B the bestanswer. Choices AandDcannot beconcluded, because noprecipitate is expected between chloride andeitherzinc or strontium. This means that zinc or strontium may or may not be present.

47. ChoiceC is correct. When the cation mixture is treated with chloride anion, the least soluble chloride salt issilver chloride, therefore chloride will remove the Ag+ from solution. When the solution with the threeremaining cations is treated with sulfide dianion, the least soluble sulfide salt is lead sulfide, thereforesulfide will remove the Pb2+ from solution. When the solution with the two remaining cations is treatedwithhydroxide anion, the least soluble hydroxide salt is zinc hydroxide, therefore hydroxide will remove the Zn2+fromsolution. Thecationthat will remain in solution is strontium (Sr2-1-), thus the best answer is choice C.

48. Choice Ais correct. An acid can protonate abasic salt insolution (which insome cases serves toform the moresoluble ionic form of the product). The acid serves to reduce the hydroxide concentration which allows more ofthe basic salt to dissociate. This is a complex equilibrium that develops as a result. A complex equilibriumusually enhances the solubility in aqueous solution. Pick A to feel perky.

49. Choice A is correct. Because the solution has OH" present already, the solubility ofAgOH will behindered bythe common ion effect where hydroxide is the common ion. AgOH will be most soluble inthe solution with theleast hydroxide anion present. The least hydroxide is found in the solution with the most H+ present. Thissolution consequently has the lowest pH. Pick A to feel spunky.

Passage VIII (Questions 50 - 56) Solubility Experiment

50. ChoiceB is correct. Because all three salts dissociate into one cationand one anion, the compoundwith thegreatest molar solubility also has the greatest value for its solubility product (Ksp). The MZ salt is theheaviestand has the leastmass dissociate intosolution, soMZhas the lowestmolesof salt in solution and thushas the lowest solubility product. This eliminates choices Cand D. Deciding between MX and MY requiresmorework. MoremassofMY dissociates thatMX, but MX has a lowermolecular mass. ThemolesofMX equal7over 100 while the moles ofMY equal 9over 120. This means that there are 0.07 moles MX and 0.075 moles MY.MY is present in a smaller volume of solution than MX, so MY has a greater molarity and solubility productthan MX at 25°C The correct answer is choice B.

Copyright ©by The Berkeley Review® 226 Section III Detailed Explanations

51.

52.

53.

54.

55.

Choice C is correct. Because the solubility ofa salt increases with temperature, answer choices Band Dareeliminated. It can alsobe read from the third column in Table 1 thatmoreMX and MY dissociate intowaterata temperature greater than 25°C than what dissociates at 25.0°. This also eliminates choices B and D. Themolarity of MXat 31.6°C and the molarity of MY at 27.4qC are found as follows:

7g 9g

MMX03L6-C =1°0g'mOle"1 =O^ZQmol =_7_0.0262 L 0.0262 L 2.62

M Mmy@274'C =120g'm°le"1 =0-075mol =^5_M2.530.0253 L 0.0253 L

There are more moles of MY than MXin a smaller volume of solution, therefore saturated MY at 27.4°C is moreconcentrated than saturated MX at 31.6°C. The best answer is choice C.

Choice Bis correct. Exactly 7.0 grams ofMX dissociates into 28.9 mL ofwater at 25.0°C according to Table 1.The mass thatwill dissolve into 25.0 mL isproportionally less. The volume ratio is justunderninety percent, soroughly ninety percent of 7.0 grams MXwill dissociate. The best answer is choice B.

Choice A is correct. As with any question like this, first consider units, liters must be in the denominator, sochoices C and D are eliminated. At 25°C, 9.0 grams ofMY requires that enoughwater be added to reach26.1 mLof solution when fully dissociated. This makes choice A the best answer. To determine the molarity of thesolution, the 9.0gramsmust first be converted intomoles, by dividing 9.0 grams by the molecularmass ofMY(120 grams per mole). This value is divided by the volume of the solution as measured in liters (0.0261 L). Thevalue of 0.0253 preys upon the possibilityyou read thewrong column. Thebest answer is choiceA.

Choice D is correct. As the temperature rises, the volume of the solution increases according to Table 1.Liquids, as a general rule, expand with increasing temperature. As the volume increases, both the molarityand the density of a solution decrease. The best answer is therefore choice D.

Choice D is correct. Density is defined as mass per volume. The densities of all of the solutions are greaterthan pure water (1.00), so choices A and Bare eliminated. It is easy to forget to consider the mass of the waterwhen determining the density. At 25°C there are 7.0grams ofMXand 25.0 grams of H2O in 28.9mL of solution,9.0grams ofMYand 25.0grams of H2Oin 26.1 mLof solution, and 4.0grams ofMZ and 25.0grams of H2O in 28.1mL of solution. The densest solution (greatest mass in the least volume) is found with the saturated MYsolution. This makes choice D the correct choice.

56. Choice C is correct. The solution with the highest boiling point is the solution with the highest molality ofimpurities. Molality is defined as moles solute per kg solvent. MX has a lower molecular mass than MY, thus1.0 grams of MX generates a greater moles of solute than 1.0 grams of MY. Because the density of waterdecreases as the temperature of water increases, the mass of 10 mL water is greater at the lower temperature.This means that the largest molality is found with 10 mL water at 50°C, because the mass solvent is least. Thebest answer is therefore choice C.

Passage IX (Questions 57 - 62) Qualitative Analysis Experiment

57. Choice A is correct. An anion can be used to identify a cation when it reacts differently with that cation thanthe other cations. Of the five anions used in table I, all five form a precipitate with barium cation. Thismeans that to identify barium (distinguish barium from the other cations) the anion chosen must precipitateexclusively with barium. Only chromate Cr042" forms a precipitate with barium and no other cation. Todistinguish barium cation from the others, chromate should be added to solution. If a precipitate forms, thenbarium was present in the solution. If no precipitate forms, then barium was not present in the solution. Thebest answer is choice A.

58. Choice B is correct. Nitric acid protonated carbonate anion to form carbonic acid (H2CO3), which subsequentlydecomposed to form water and carbon dioxide. Becausethe carbonate anion carries a -2 charge, two equivalentsof nitric acid are required. The two nitrate anions and barium cation are spectator ions. The best answer istherefore choice B. This can also be inferred from the passage, where it is stated that nitric acid is added toreact with carbonate anion to form carbon dioxide gas. The barium cation carries a +2 charge and carbonateanion carries a -2 charge, therefore the two must be in a one-to-one ratio in the salt. This eliminates choices AandC.

Copyright© by The Berkeley Review® 227 Section III Detailed Explanations

59. Choice Ais correct. Hydrochloric acid is not used, because it would introduce the chloride anion, whichaccording to table II, forms a precipitate with silver cation. This means that chloride interferes, makingchoice A the correct choice.

60. Choice D is correct. The salt K2S04 delivers the anion S042" (sulfate) to solution. From table I, sulfateprecipitates with both barium and strontium cations. The conclusion reached upon observing the formation of aprecipitate should be that either strontium or barium is present in solution. It cannot be determined whether ornot calcium or magnesium cations are present because they do not form any detectable precipitate with sulfate.The best answer is choice D.

61. Choice C is correct. Looking at the third reactivity column of table II shows that a white precipitate formsupon the addition of silver cation with both carbonate and chloride anions. Addition of nitric acid dissolvesthe carbonate away, but it does not react with the chloride. The best choice is the chloride anion, choice C.

62. Choice A is correct. The first anion added precipitates with only barium. This according to the chart ischromate (Cr042'). Answer choice Ashows this to be carbonate (CO32-), making choice Aanincorrect choice.Just tobecertain, anion Wprecipitates with strontium butnotcalcium andmagnesium. According to the chart,this is sulfate (SO42-) sochoice Bis valid. Anion Xprecipitates with calcium but notmagnesium. According toTable 1, it is either carbonate (CO32-) or oxalate (C2042"), so choice C (oxalate) is valid. Anion Z precipitateswith magnesium. According to the chart, this is hydroxide (OH-), sochoice D (hydroxide) is valid.

Passage X (Questions 63 -69) Solubility ChartThe focus of this passage is on the solubility of different cations in the presence ofanions. Solubility isone ofthe more favored topics on the MCAT so far, so be certain to make sense of this passage. The key fact toremember is that when comparing solubility, lookat themolarsolubility, not theKspvalue.

63. Choice C is correct. This question boils down to whether Qsp orKsp is larger. When Q>K, there are too manyproducts, inwhich case the salt cannot completely dissociate. When Q<K, there is room for more products, inwhich case the salt completely dissociates. When Q = K, it is exactly saturated. In this question, it is fareasier to decide whether the molarity of ZnCC>3 in the event it fully dissociates exceeds the molar solubilityconveniently listed in Table 1. The molarity if Z1-1CO3 fully dissociates is calculated as follows:

ai &ams =OJLmole =-QA.m0le =0.0008 mole => Q-0008 mole =0.008M1?[-. grams/ 125 1000 0.10LiZD-* /mole

0.008 M > 0.000014 by a largeamount. This means that very little of the ZnCC>3 dissociates. On this question,you must goeven further. It will barely dissociate according to the numbers which implies choice C. Do whatis expected of you: be kind to strangers, brush after every meal, and pick choice C.

64. Choice Bis correct. To distinguish one thing from another visually, you mustbe able to seea difference in theirbehaviors. The behavior here is the formation of a precipitate. To visually distinguish a tube with Zn2+ froma tube with Ag+, an anion thatprecipitates with only one of them must be used. The passage defines the cut-offfor seeing a precipitate as one mg permL solution. Assuming the average salt tohave a molecular mass around100 grams per mole, means that the cut-off for visual delegability is roughly 0.010 M. You must locate theanion that shows a greater molar solubility than 0.01 Mfor one cation, but less for the other. Sulfides show thebiggest ratio of numerical values, but in a practical sense, neither dissolves to a detectable extent to visuallydistinguish. The numbers are different in a multiplicative manner, but they are close in an additive sense.Choice A is eliminated. With the chloride salts ([Ag+] =1.2 x10"5, while [Zn2+] =3.1 x10"2), there is a casewhere the silver chloride precipitates while the zinc chloride dissociates. The zinc concentration isdetectable to the eye. In other words, we can see the quantity ofsalt thatdissociates. Choose B, and feel nice.

65. Choice D is correct. This question addresses the common ion effect from aconceptual perspective. Silver sulfideis most soluble in a solution with the least amount of common ion. This eliminates choices A and C. Thequestion is whether or not silver cation or sulfur dianion more affects the calculation. In this case, we aredeciding which leads to a greater xvalue, Ksp =(.01)2(x) or Ksp =(2x)2(0.01). In a solution with 0.01 MAg+,the solubility is 104 xKsp. In asolution with 0.01 MS2", the solubility is 5xsquare root of Ksp. Because theKsp for Ag2S is so small, the square root of Ksp is alarger number than Ksp. The best answer is choice D.

Copyright ©by The Berkeley Review® 228 Section III Detailed Explanations

66.

67.

68.

69.

Choice C is correct. This question is purely mathematical. The key fact is recognizing that there is 0.01 MCI"present in solution initially. The set-up and solution are as follows:

^ Zn2+(aq)0.01

Reaction: ZnC03(s) ^

Initially: excessShift: zX • +x

Equilibrium: who cares? 0.01 + x

Ksp =[Zn2+][C0321 •• Ksp =(0.01 +x)(x)

co32-0

± x

x

Upon ignoring x: Ksp =2.0 x10"10 =(0.01 +x)(x) =0.01(x) .-. x=2.0 x10"8MThe correct answer is choice C.

Choice B is correct. The most soluble salt is the salt with the highest molar solubility (not necessarily thehighest solubility product). This is a "read-the-chart" type of question. According to data in Table 1, thelargest molar solubility of the choices is found with Ag2CC>3 (1.2 x 10~4 M). The trick here is picking thecorrect column in the table to reference (molar solubility rather than Ksp). To bemore thanyoucanbe,pick B.

Choice D is correct. To calculate the solubility product for the calcium phosphate salt, the solubility reactionis needed:

Ca3(P04)2 r 3Ca2+(aq) +2P043"(aq)Ksp = [Ca2+l3[P043-l2 = (3x)3(2x)2 = (27 x3)(4 x2) =108 x5^sp

The best answer is choice D.

Choice C is correct. An ion exchange column involves competition by two cations for an anion bound to thecolumn. The cation of the less soluble salt will precipitate with the anion, which is bound to the column. Thetwo cations to consider are the one in solution and the one originally bound to the column. The columns works ifthe cation in solution is less soluble with the anion in the column than the cation originally coupled with theanion in the column. Silver chloride (AgCl) has a lower molar solubility than sodium chloride (NaCl),therefore silver cation precipitates with chlorine anion preferentially over sodium cation. Choice A is a validstatement. Zinc sulfide (ZnS) has a lower molar solubility than calcium sulfide (CaS), therefore zinc cationprecipitates with sulfide anion preferentially over calcium cation. Choice B is a valid statement. Calciumcarbonate (CaCC>3) has a higher molar solubility than zinc carbonate (ZnCC^), therefore calcium cation doesnot precipitate with carbonate anion preferentially over zinc cation. The column of choice C will not work,thus choice C is the best answer. Calcium sulfide (CaS) has a lower molar solubility than sodium sulfide(Na2S), therefore calcium cation precipitates with sulfide anion preferentially over sodium cation. Eventhough the two salts have different stoichiometry, they can be compared directly through their molarsolubility values. Choice D is a valid statement.

PassageXI (Questions 70 - 75) Calcium Salts Solubility

70. Choice A is correct. Because there is solid calcium hydroxide at the base of the flask, it can safely be assumedthat the solution is saturated. As sodium carbonate is added to the flask, calcium carbonate will begin toprecipitate. This reduces the calcium ion concentration, causing more of the calcium hydroxide solid can beginto dissociate in order to regenerate calciumion in solution. As calciumhydroxide dissociates into solution, thehydroxide concentration increases. As the hydroxide ion increases, the solution's pH increases, eliminatingchoices C and D. Had the solution not been saturated, there would not be any solid calcium hydroxide presentto dissociate into solution. This means that the pH increases more in the solution with solid calciumhydroxidepresent than the solution with no calcium hydroxide present. The best choice is therefore answer A.

71. Choice C is correct. The maximum calcium ion concentration at pH = 14 can be determined using the solubilityproduct of calcium hydroxide (Ksp =[Ca2+][OH-]2). Because the pH is 14, the pOH is 0, so the hydroxide anionconcentration is 1.00M. This means that when plugging into the solubility product to determine the calcium ionconcentration, it can be determined that the calcium ion concentration equals the numerical value of thesolubility product (Ksp =[Ca2+1). The solubility product is 4.3 x10"6 M3. The best choice is C, 4.3 x10"6 M.


72. Choice D is correct. In order to increase the chloride concentration in solution, an anionmust be added thatprecipitates calcium cation out from solution. This allows the calcium chloride solid present on the bottom ofthe flask to dissociate into solution to regenerate the lostcalcium cation. As the calcium chloride dissociates,the chloride anion concentration increases. ChoiceC should be eliminated immediately, because the solubilityofcalcium chloride is reduced by the common ion effect. Choice Bis eliminated, because thesilver cation willprecipitate chloride anion from solution. This is one of the solubility rules you may want to know. Silverhalides are insoluble in water. The answer is either choiceA or D. The sodium phosphate adds phosphateanion into solution, which then precipitates out of solution as calcium phosphate and thus decreases thecalcium cation concentration. The solid calcium chloride dissociates to counteract the decrease in calciumconcentration and indoing soincreases the chloride ion concentration. The bestanswer ischoice D.

73. Choice C is correct. It is important to realize that the solution is buffered, so the calcium hydroxide thatdissociates does not drastically affect the pH. The buffer absorbs the hydroxide that is released as calciumhydroxide dissociates. The amount of calcium hydroxide that dissociates is controlled by the pH of thesolution. At a high pH, there is a large amount ofhydroxide present in solution, so only a small amount ofcalcium hydroxide dissociates. As the pHdecreases, the hydroxide anion concentration decreases, resulting inan increasing calcium ion concentration. The overall effect is that as the pH of the buffer increases, calciumhydroxide is less soluble due to the common ion effect. This results inan inverse relationship, sochoices AandB are eliminated.

The question may also beaddressed from a complex equilibrium perspective. As the buffer pH decreases, theamountofhydronium insolution increases. Increasing thehydronium concentration forces the second reaction inthe forward direction. This reduces the hydroxide concentration. To compensate, the first reaction shiftsforward to regenerate hydroxide. In doing so,calcium cation is released into solution. This confirms that thecalcium ion concentration increases as the pH decreases.

Ca(OH)2(s) + H20(1) -^-^ Ca2+(aq) + 2OH"(aq)H3Q+(aq) + OH-(aq) ^^r 2H2Q(1)Ca(OH)2(s) + 2H30+(aq) •» ^ Ca2+(aq) + 4H20(1)

The net result is that as the [H3<D+1 increases, the [Ca2+] increases. As the calcium cation concentrationincreases, the log of the calcium cation concentration increases, therefore the log of the calcium ionconcentration increases as the pH decreases. Because both the x-axis and y-axis are based on a log scale, therelationship is linear, but not necessarilywith a slope of one. The slope is negative (due to the negative sign indetermining pH). The best answer is choice C.

74. Choice B is correct. If the dissociating of a salt into water increases the temperature of the water, thendissociation (the solvation process) releases heat. The release of heat makes the solvation process exothermicwhich makes the enthalpy change (AH) a negative number. This eliminates choices A and C. The entropyincreases when a salt dissociates into solution, because the ions have more freedom to randomize within thesolution than within the lattice. The change in entropy (AS) for dissociation is therefore positive. The correctanswer is choice B.

75. Choice D is correct. This question requires viewingTable 1. The solubility of calcium fluoride and calciumhydroxide are directly comparable using the solubility product, becausefluoride and hydroxide carry the sameanionic charge. The solubility of calcium sulfate and calcium carbonate are directly comparable using thesolubility product, because sulfate and carbonate carry the same anionic charge. Choices A and C areeliminated due to their lower solubility product values. Because calcium hydroxide forms three ions whendissociating into solution, the solubility product is 4x3, where x is the calcium ion concentration ([Ca2+]). Thevalue ofx ([Ca2+]) is some number times 10-2 for Ca(OH)2- For calcium sulfate, the solubility product is x2,where x is thecalcium ion concentration ([Ca2+]). The value ofx ([Ca2+]) is some value times 10"3 for calciumsulfate. The value is greater for calciumhydroxide,so the best answer is calciumhydroxide, choiceD.


Passage XII (Questions 76 - 82) Complex Equilibrium

This is a difficult passage because as one reaction shifts, it affects the other reactions. It is easiest to view thereactions as independent, and observe how specific reagents are changing during each step. For instance, theshift of reaction 1 will affect the partial pressure of NO2 which will in turn affect reactions 2, 3, and 4. Theresult is a complex equilibrium.

76. Choice A is correct. A negative change in entropy results when the system becomes more ordered. The systembecomesmore ordered when it decreases in volume or changes phase to a more ordered phase. Reaction 2 goesfrom two gas molecules to one gas molecule, therefore in has lost in entropy. Reaction 3 goes from two gasmolecules to two gas molecules, therefore entropy has not changed. Reaction 5 goes from three ions in solution tothree ions in solution, therefore entropy has not changed. Reaction 6 also goes from three ions in solution tothree ions in solution, therefore entropy has not changed. The best answer is therefore choice A.

77. Choice C is correct. Decreasing the temperature shifts reaction 2 in the forward direction, because it isexothermic. This results in an increase in [N2O4]. Increasing the pressure shifts reaction 2 in the forwarddirection. This results in an increase in [N2O4]. Water vapor in the air reacts with the NO2 in the air(according to reaction 4), and thus reduces the amount of N02- Reaction 2 shifts in the reverse direction to reestablish equilibrium. This decreases (NOT increases) the [N2O4]. Choice C is the correct choice. Addition ofNO2 shifts reaction 2 in the forward direction which increases the [N2O4].

78. Choice B is correct. According to the ideal gas law, when the temperature of a gas system is doubled, thepressure doubles, as long as the volume of the system and moles of gas remain the same. The reaction vessel is aclosed steel container, so the volume of the system cannot change. The problem here is that the moles of gaschange with the increase in temperature. Because the reaction is exothermic, an increase in temperaturepushes the reaction in the reverse direction. This increases the reactants and decreases the products. There aremore products than reactants in the balanced equation, so a shift in the reverse direction results in fewer molesof gas in the system. The increase in pressure due to the ideal gas law is not be as great as expected, due to thedecrease due to the shift in equilibrium. The ideal gas law predicts that change in temperature doubles thepressure, while the shift in reaction (reduction in moles of gas) predicts that the pressure decreases. Thechange in moles is a smaller factor than the increase temperature, because at the least, the moles of gas wouldbe two-thirds of their original value. The moles cannot be cut in half according to the balanced equation.Therefore, the overall change in pressure is a little less than double. This is best described as choice B.

79. Choice D is correct. A bond must be formed in reaction 2, because the two NO2 molecules are combining to formone N2O4 molecule. This eliminates choices A and B. Bond formation is an exothermic process, thereforechoice D is the correct answer.

80. Choice C is correct. The total pressure of reaction 1 at equilibrium is 1.00 atmospheres, because the internalpressure can equilibrate with the external pressure in a piston system. The pressures equilibrate by having thepiston plunger either rise up (increasing the volume) or drop down (decreasing the volume). The addition of0.10 atmospheres of NO gas initially increases the total pressureof the system to 1.10 atmospheres, before thesystem can equilibrate. The piston rises to accommodate the change in pressure (and equilibrate the internaland external pressures), thus increasing the volume of the piston. The addition of NO gas displaces thereaction from equilibrium. The reaction is no longer at equilibrium, so it reacts to re-establish equilibrium.Excess reactant is present, so the reaction shifts in the forward direction (to use up the excess reactant). Thereare three moles of reactants and only two moles of products, so the number of moles decreases, causing thevolume to decrease slightly. The overall result is a slight increase in volume, making choice C correct. If therewas no shift in the reaction, the volume would increase by exactly ten percent, therefore the increase in volumemust be less than ten percent.

81. Choice D is correct. Addition of sodium hydroxide to solution deprotonates HNO2 (HNO3 is a strong acid andhas already fully dissociated), and thus shifts reaction 4 to the product side to re-establish equilibrium.Addition of manganese(II) chloride (M1-1CI2) to solution removes both NO2" and NO3" from solution throughcomplexing of the ligands. To re-establish equilibrium in reaction 4, the reaction must shift right to make moreNO2" and NO3". Removal of nitrate (NO3") from solution results in a shift in the product direction to reestablish equilibrium. Removal of water (a reactant) results in a shift in the reverse direction (left) to reestablish equilibrium. Choice D is the best answer available to you.

©Copyright © by The Berkeley Review 231 Section III Detailed Explanations

82. Choice A is correct. At equilibrium, theNO2 concentration is constant. When bubbled through water, it reactsto form nitrous acid (HNO2) and nitricacid (HNO3), which can both deprotonate and form ligand bonds to themanganese cation. This means that the N02(g) is depleted when it is bubbled through an aqueous solution ofmanganese chloride. Graphs C and D are eliminated, because they show that the concentration of NO2increases with the addition of aqueous manganese chloride solution. The N02(g) is gradually regenerated byboth the reverse reaction of reaction 2 and the forward reaction of reaction 1. This means NO2 is regenerated(and thus increases), but not to the level it was initially at. This is best shown in graph A.

PassageXIII (Questions 83 - 89) Hemoglobin and Acclimation

83. Choice C is correct. The passage states that hemoglobin binds four oxygen molecules while myoglobin bindsonlyoneoxygen molecule. Totransfer allof theoxygen, there mustbe four myoglobin molecules per hemoglobin.Thisshould be common knowledgefrom biology. Pick choice C, and start the passageoff on the right foot.

84. Choice B is correct. Increasing the rate of respiration increases the uptake of oxygen. This results in an increasein free oxygen, which lessens slightly by the shift in equilibrium to the oxygenated hemoglobin. Overall, theamount of free oxygen increases, so choice A is eliminated. An increase in the amount of hemoglobin results in ashift in the equilibrium in the forward direction. This reduces the amount of free oxygen. Choice B is thereforethe best answer. An increase in CO consumed results in more bound sites on the iron of hemoglobin, so less oxygencan bind. Less oxygen bound results in more free oxygen. Choice C is eliminated. More blood results in moremoles of all components in the equilibrium including oxygen. Choice D is thus eliminated. Select choice B.

85. Choice D is correct. A long-time mountain resident has more hemoglobin in their blood than a long-time sea-level resident. When the long-time mountain resident descends to sea level where there is a higher abundanceof oxygen, they experience increased vitality due to the increased partial pressure of oxygen gas. Pick D.

86. Choice A is correct. Because CO binds the iron of hemoglobin preferentially over O2, any CO present in the airbind iron cation, and thus reduce the amount of O2 that can bind. To compensate, respiration increases, in orderto increase the amount of air consumed. Over time, the body produces more Hb (this is a result of acclimation).The best answer is choice A.

87. Choice D is correct. Because hypoxia does not directly affect the cellular uptake rate of oxygen, once it isabsorbed from the lungs, myoglobin (present in cells) is not affected directly by hypoxia. Without consideringany other factors, the myoglobin should remain constant. The best answer is choice D.

88. Choice D is correct. Hypoxia results from a drastic decrease in the amount of oxygen present in the air. Thisdecrease is associated with an increase in elevation. Repelling down a mountain and scuba diving both involveincreases in the oxygen present. In scuba tanks, the gas may be mixed with helium (an inert gas) to compensatefor the greater amount of air consumed per breath (due to the increased pressure under water). If the tanks werenot partially filled with helium, too much nitrogen and oxygen would enter the body. Of the last two choices,only snow skiing involves a high elevation, so choice D is the best answer.

89. Choice B is correct. According to the passage, there is less oxygen present at higher elevations. By training athigher elevations, where the air has less moles of oxygen, the body acclimates by producing more Hb. Once anathlete returns to a lower elevation, the increased amount of Hb remains for a short time, before the body canre-acclimate. For a short period of time, an athlete can increase their oxygen carrying capacity (and thusincrease their cellular metabolism). The best answer is choice B.

r®Copyright © by The Berkeley Review 232 Section III Detailed Explanations

Passage XIV (Questions 90 - 96) Equilibrium Reaction of NO2 and N2O4

This passage mimics a passage that appeared on theMCAT years ago. The reaction is a typical example of gasphase equilibrium, and is found in most every general chemistry text book. The reaction is often used todemonstrate Le Chatelier's principle. In this passage, the reaction is chosen because it has an unequal numberof reactant molecules as product molecules. Thismeans that changes in the condition of the system (volume andpressure) can shift the equilibrium reaction, but the equilibrium constant remains the same, as long as atemperature change does not accompany the volume or pressure change. The passage also points out that aninert gas does not disturb an equilibrium, as long as the container is rigid (which a glass container is assumed tobe). You may wish to note that if inert gas is added to an expandable container, it can disrupt the equilibrium,because the partial pressures of the component gases are changed.

90. Choice A is correct. Because the reaction involves the breaking of a bond, it must be an endothermic reaction.This is confirmed in the last sentence of the passage, which states that the ratio product to reactant increasesas the temperature increases. Because there are two molecules formed from just one, the AS for the reaction aswritten is greater than zero (positive). The best answer is therefore choice A.

91. Choice D is correct. Addition of reactant (N2O4) shifts the reaction to the right (product side), thus choice Ais valid. An increase in volume disrupts equilibrium, and results in the reaction shifting from the side with onemolecule to the side with two molecules. This results in a shift to the right, so choice B is valid. A decrease inpressure disrupts equilibrium, and results in the reaction shifting from the side with fewer molecules (one) tothe side with more molecules (two). This results in a shift to the right, thus choice C is valid. Choices B and Cshould both have been eliminated, because they are the same answer and cannot both be correct. When addinghelium gas (an inert gas) to the reaction at constant volume, the partial pressures of N2O4 and NO2 do notchange. Because the partial pressures do not change, the reaction is not displaced from equilibrium, thus itdoes not shift in either direction. This makes choice D the correct answer.

92. Choice D is correct. The equation relating AG and Keq is AG =-RT In Keq. To isolate Keq, both sides of theequation are first divided by -RT. To eliminate the natural log function, the two sides of the equationmust beexponentsof e. Thismakeschoice D the correct answer. Thederivation is shown below.

AG =-RT InKeq .'. lnKgq =-—.-. K^=e"AG/Rr93. Choice B is correct. When the equilibrium constant has 105 associated with it, it is said to be large. As such,

the reaction distribution at equilibrium is almost exclusively products, making choiceA valid. If the reactionstarts as mostly reactants, it shifts nearly one hundred percent to form products, making the shift significant.Choice B is NOT true. The equilibrium constant for the reverse reaction is the reciprocal of the equilibriumconstant for the forward reaction. This makes choice C valid. A catalyst increases the reaction rate (in boththe forward and the reverse directions), but it does not affect the equilibrium constant. This makes choice Dvalid. The best answer is choice B.

94. Choice A is correct. Because the equilibrium is dynamic, the forward and reverse reactions are continuallytranspiring. When the equilibrium is disturbed by the addition of N2O4 (labeled or not), the equilibrium islost and the reaction must undergo a net shift forward to compensate for the excess reactant. The reversereaction continues as well, but not to the degree of the forward reaction. The result is that the amount of N2O4decreases and the amount of NO2 increases. The radiolabeled nitrogen will eventually be evenly distributedbetween the products and reactants, once equilibriumhas been re-established. This makes choice A the bestanswer. This is referred to as scrambling of the label.

95. Choice C is correct. Because there are more molecules on the product side than the reactant side, the reactionshifts to products as the volume of the piston increases. This makes statement I a valid statement. Thereaction is shifting in the forward direction, which is endothermic, therefore heat is absorbed by the reaction.The result is that the temperature decreases. This makes statement II a valid statement. Do not mistakenlythink of PV = nRT,because the volume changes and the pressure changes, the temperature was not changed tocause the volume or pressure change. As the reaction shifts to the right, N2O4 decreases and the amount ofNO2 increases, causing the mole fraction of N2O4 to decrease. This makes statement III an invalid statement.The best answer is choice C, both statements I and II.


96. Choice B is correct. In this case, the equilibrium constant is the product squared (because of the stoichiometriccoefficient) divided by the reactant. This makes choice B the best answer.

Questions 97-100 Not Based on a Descriptive Passage

97. Choice C is correct. When an MX salt dissociates into water, it forms M+ cation and X' anion. The solubilityproduct (Ksp) is equal to [M+][X"] =y2. You really have no alternative but to pick C.

98. Choice C is correct. This is a question involving the common ion effect. Because F" is present in solution inchoices A and B, and Mg2+ is present in choice D, allof thechoices except C are eliminated due to thecommonion effect. Itis only in choice Cthat the compound does not have acommon ion (either Mg2+ orF") present.

99. Choice C is correct. Equilibrium is the state in which the forward reaction rate equals the reverse reactionrate. For a one-step reaction at equilibrium, kf[R] =kr[P]. Keq is defined asproducts over reactants, which ismanipulated as follows:

Given kf[R] =kr[P] and Keq =—, Keq =M[RJ kr

Because the forward rate is four times the reverse rate, 4kr canbe substituted for kfand thus Keq is4kr dividedby kr, which is 4. Choose with dignity; choose C.

100. Choice C is correct. Choice D is eliminated, because less solvent reduces the amount of a salt that candissociate into solution. Because a solvation reaction can be either exothermic or endothermic, the effect on thesystem by a change in temperature varies, and is thus unpredictable. This eliminates choices A and B. In mostcases, an increase in temperature results in an increase in the amount of salt that dissociates into solution, butthere are some exceptions. Only choice C, increasing the solvent, always increases the amount of a salt thatdissociates into solution. As more solvent is added, more compound can dissociate. Be sure that you realizethat this question is asking about the amount (in mass or moles) of salt dissociated, not the concentration (molarsolubility).

®Copyright © by The Berkeley Review 234 Section III Detailed Explanations

Section IVAcids and Bases

by Todd Bennett

Strong Acid:pH =-log[HX]

Strong Base:pOH =-log[MOH]

Weak Acid:

PH=ipKa-ilog[HA]Weak Base:

pOH =IpKb-llog[A"]Buffer:

pH = pKa + log

or

[A"][HA]

pH =pKa +log m0leS Con)ugate Basemoles Conjugate Acid

Terminologya) Fundamental Definitions

b) Water-based Acid-Base Chemistryi. Acid Dissociationii Base Hydrolysis

c) Determination of Reagent Strengthi. Strong Acidsii. Weak Acidsiii. Very Weak Acidsiv. Strength and the pK Scalev. Strong Basesvi. Weak Bases

vii Very Weak Bases

Types of Acids and Basesa) Haloacidsb) Oxyacidsc) Metal Oxides and Metal Hydroxidesd) Organic Acidse) Polyprotic Acids

Calculating pHa) Determining pHb) Log Reviewc) pH for Strong Reagentsd) pH for Weak Reagents

Conjugate Pairsa) Typical Conjugate Pairsb) Relationship of pKa and pKbc) Henderson-Hasselbalch Equation

BERKELEYUr-e-v-i^e>w®


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Acids 8t Bases Section Goals

Know the definitions for an acid and some common examples of acids.Anacid canbedefined aseither a proton donor, anelectron-pair acceptor, ora compound thatyieldsH30+ when added to water. Typical examples includeacetic acid (H3CCO2H), hydrochloric acid(HCl), and sulfuric acid (H2SO4). You are expected to recognize the common acids.

Know the definitions for a base and some common examples of bases.Abase can be defined as either a proton acceptor, an electron-pair donor, or a compound thatyieldsOH"when added to water. Typical examples include sodium nydroxide (NaOH), ammonia (NH3),and potassium tert-butoxide ((CH3)3COK). You are expected to recognize the common bases.

Be able to calculate the pH of aqueous solutions of base or acid.The pH of a solution is defined as the negative log of the hydronium ion concentration in the solution.You must be able to determine the pH of the solution, knowing the concentration and strengthofthespecies. ThepHofaweakacidisgreater thanthepHofa strongacidwhenthe twoarein equalmolar concentrations.

Understand what is meant by the strength of a reagent.The strengthofa reagent is the measure ofitsdegree ofdissociation inwater. Acompound thatfullydissociates inwater issaid tobestrong, while a reagent thatonly partially dissociates inwater issaid to be weak. You should be able to aetermine the relativestrengths of acid fromchemicalfeaturesand its pH in the aqueous solution.

0<§j^ Know how conjugate pairsand buffers work.Abuffer is formedwhen a weak acid and its conjugate base are combined in an aqueous solution.An equilibrium exists between the two species, so as long as both are present m solution, thehydronium ion concentration will remain fairly constant, andthe pHwill also remain constant. Thiseffect isknown as"buffering." You must understand buffers ana how pHisdetermined using theHenderson-Hasselbalch equation.

Recognize and specific types of compounds.Youshould understand why metal oxides and metal hydroxides are basic. Youshould understandwhynon-metal oxides and non-metal hydroxides areacidic. Befamiliar with typical examples suchas acid rain and soil pH. Recognize that the conversion from a non-metal oxide (Lewis acid form)to a non-metal hydroxide (Br0nsted-Lowry form) involveshydration of the acid. The strength ofthe reagent is not affected by this conversion.

Understand the terms associated with polyprotic acids.Polyprotic acids have multiple pKa and pKbvalues. You must understand the conceptual andmathematical relationships of the variables to one another. Youmust be familiar with terms suchas "normaility" and "equivalents." Knowthe typicalexamplesofpolyproticacidsand their formulae.

General Chemistry Acids and Bases

Acids and BasesAcid-and-base chemistry is a foundation for understanding organic chemistry,biochemistry, and physiology. As such, we shall address it from severalperspectives, observing what effect the varying of a solvent has on the nature of areaction. Water and other protic solvents solvate charged species, so acid-and-base chemistry can be monitored by the gain and loss of charge, associatedspecifically with the gain and loss of an ionized hydrogen atom (H+). The terms"protonated" and "deprotonated" are derived from the ionized hydrogen atom,which is just a proton. In a lipid environment, charges cannot be stabilized, soacid-and-base chemistry is typically considered from an electron-transferperspective, where bonds are broken and formed. The different definitions ofacids and bases are the result of differences in solvent, not chemical reactivity.

Once the definitions are established, reactivity will be considered. There are twocommon misconceptions to clear up, if we are to understand acid and basechemistry better. The first common misconception is that the pH scale has fixedlimits. The second common misconception is that weak acids automatically havestrong conjugatebases. Let us address both of these before anything else.Acidity is typicallymeasured in terms of [H+] on the pH scale. The pH value of asolution is determined by taking the negative log of the hydroniumconcentration. Because neutral water has a hydronium concentration of 10~7 M,due to autoionization of water, neutral water has a pH = 7.0. Acidic solutionshave pH values less than 7.0, while basic solutions have pH values greater than7.0. The only limits to the value of pH are associated with concentration andsolubility, so the pH scale is limitless in theory. In practice, there are limits.The pHscale doesnot range from 0 to 14. The pHofacompound can be negativefor highly concentrated strong acids and greater than 14 for highly concentratedstrong bases (e.g., 10MHCl has pH= -1,and 10MNaOH has pH= 15).

The relative strength of an acid can be determined from the relative strength ofits conjugate base. This relation is reversible: The relative strength of a base canalso be determined from the relative strength of its conjugate acid. Using therelationship between pKa and pK^, for conjugate pairs in water at 25°C, either pKvalue can be determined from knowing the other. For instance, a weak acid witha pKa of 7 has a conjugate base with a pKbof 7. Note that the conjugatebase ofthe weak acid is a weak base. This is not a typo! It is the relative reactivities thatare compared, not the absolute reactivities. If acid HA is stronger than acid HB,then base A" is weaker than base B". This can be demonstrated by comparingReaction 4.1 with Reaction 4.2:

HA(aq) +pKa = 7

HB(aq) +pKa =9

H20(1)

H20(l)

Reaction 4.1

Reaction 4.2

H30+(aq)

H30+(aq) +

A-(aq)pKb =7

B-(aq)pKb = 5

The lower pKa value indicates that HA is a stronger acid than HB while thelarger pKb value indicates that A" is a weaker base than B".

The stronger the acid, the weaker its conjugate base. It is not necessarily the casethat weak acids have strong conjugate bases. A prime example is a weak acid withpKa = 7. The pK[, of the conjugate base is 7, which is not strong.

Introduction


General Chemistry Acids and Bases Terminology

TerminologyFundamental DefinitionsA perfect place to start a discussion of acids and bases is with the definitions ofan acid, a base, and strength. There are three definitions proposed for both acidsand bases, although each considers an acid as the opposite of a base. The firstdefinition is the Arrhenius definition, which states that an acid yields H30+when added to water, while a base yields OH" when added to water. An acidicaqueous solution therefore has a higher hydronium concentration thanhydroxide. This definition is useful when doing calculations of pH and pKa. Thesecond definition is the Bronsted-Lowry definition, which states that an acid is aproton (H+) donor and that a base is a proton (H+) acceptor. This definition isbased on what occurs in a protic solvent, where reactions are viewed as proton-transfer reactions. The third definition is the Lewis definition, which states thatan acid is an electron-pair acceptor while a base is an electron-pair donor. Thisdefinition is based on what occurs in an aprotic solvent, and is typically used inorganic chemistry. For this section, only the Arrhenius definition will beemphasized. Table 4.1 lists the different definitions of acids and bases.

Term Definition ExampleArrhenius acid yields H30+ when added to H2O aq: [H30+] > [OH"]Arrhenius base yields OH"when added to H2O aq: [OH"] > [H30+]

BrOTisted-Lowry acid Proton donor HX in protic solventBr0nsted-Lowry base Proton acceptor KOH in protic solvent

Lewis acid Electron pair acceptor BF3 in aprotic solventLewis base Electron pair donor NH3 in aprotic solvent

Table 4.1

Water-based Acid-Base ChemistryFor our purposes, acid and base chemistry is to be considered as reactions thatonly occur in aqueous solution. Therefore, to understand water-based acidityand basicity, it is vital to understand the properties of water. Water isamphoteric, meaning that it may act as either an acid or a base. Neutral waternaturally dissociates into hydronium and hydroxide, according to Reaction 4.3.

2 H20(1) -^-^ H30+(aq) + OH"(aq)Reaction 4.3

Only a small fraction of water dissociates into solution. Water at 25"C, in theabsence ofan acid ora base, dissociates enough togenerate a solution with 10"7MH30+(aq) and 10"7 MOH"(aq). This means thatneutral waterhasboth a littlehydronium and a little hydroxide. A neutral aqueous solution is defined as onein which [H30+] = [OH"], and therefore pH = pOH. If the solution is at 25°C,then the pH and pOH are both equal to 7.0. For an acidic aqueous solution,[H30+] > [OH"], pH < pOH, and the pH of the solution is less than 7.0. Equally,for a basic aqueous solution, [OH"] > [H30+], pH > pOH, and the pH of thesolution is greater than 7.0. There are several perspectives according to which asolution may be deemed acidic or basic, and regardless of the reasoning behindeach one, you must know how all of these different perspectives relate to oneanother. Any acidic solution has a pH of less than 7.0, hydronium in it beingmore abundant than hydroxide, and it can turn blue litmus paper red.


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Acid DissociationWhen an acid is mixed with water, it is said to dissociate. This is the reason weuse the term acid dissociation, expressed quantitatively with an acid dissociationconstant (Ka). The acid dissociation constant is nothing more than theequilibrium constant for the dissociation reaction of an acid in water. A moleculeof an acid when added to water, dissociates to form hydronium ion (H30+) andits conjugate base (expressed generically as A") upon reaction with one watermolecule. Reaction 4.1 is an acid dissociation reaction in water.

HA(aq) + H2O0) ^ H30+(aq) + A'(aq)

Reaction 4.1

Equation 4.1 is the equilibrium expression (used to solve for the equilibriumconstant) for the acid dissociation reaction shown in Reaction 4.1.

K [H3Q+][A-] (41)[HA]

Equations 4.2and 4.3are the equations for convertingbetween pKa and Ka.

pKa = - log Ka (4.2)

Ka = 10"PKa (4.3)

As the relative strength of an acid increases, its Ka increases and its pKadecreases. This means that stronger acids have higher Ka values and lower pKavalues. The Ka and pKa of an acid depend on the strength of the acid, but not itsconcentration.

Base HydrolysisWhen a base is mixed with water, it is said to undergo hydrolysis. This is thereason we use the term base hydrolysis, expressed quantitatively with a basehydrolysis constant (Kb). The base hydrolysis constant is nothingmore than theequilibrium constant for the hydrolysis reaction of a base in water. Eachmolecule of a base when added to water, hydrolyzes one water molecule to formhydroxide ion (OH") and its conjugate acid (expressed generically as HA).Reaction 4.4 is a base hydrolysis reaction in water.

A'(aq) + H2O0) ^ HA(aq) + OH'(aq)

Reaction 4.4

Equation 4.4 is the equilibrium expression (used to solve for the equilibriumconstant) for the base hydrolysis reaction shown in Reaction 4.4.

[HA11QH-] (44)[Al

Equations 4.5 and 4.6are the equations for convertingbetweenpKb and Kb-

pKb = - log Kb (4.5)

Kb=10"PKb (4.6)

As the relative strength of a base increases, its Kb increases and its pKbdecreases. This means that stronger bases have higher Kbvalues and lower pKbvalues. As seen with the acids, the Kb and pKb of a base depend on the strengthof the base, but not its concentration.



Tying everything together for a conjugate set, we observe that as the acid getsstronger, its conjugate base gets weaker. The overall correlation is shown inFigure 4-1.

Overall RelationshipAs acid strength T, Ka T, pKa i, conjugate base strength i, Kb I, pKb T

Figure 4-1

Determination of Reagent StrengthThe strength of a reagent is determined strictly from the Ka (or pKa) and the Kb(or pKb). Strength is a measure of the completeness of a reaction in water. Termsthat are sometimes used instead of "dissociation" are ionization and electrolyticnature. The stronger the acid, the more electrolytic it is, because it conductselectricity better due to the greater number of ions in solution. For water toconduct electricity, there must be ions in solution to transfer the electron charge.Because the equilibrium expressions are comparable, the values can be correlatedfor a conjugate pair. At 25°C, Equation 4.7 describes the relationship betweenpKa and pKb-

pKa (HA) +pKb (A-) =14 (4.7)

The strength of a reagent is measured by its ability to carry out a reaction inwater. The stronger an acid, the more readily it dissociates into water. Thestronger the base, the more readily it undergoes hydrolysis when mixed withwater. Be careful not to confuse the concentration of a reactant with its strength.A highly concentrated weak acid may have a lower pH than a strong acid in lowconcentration.

Example 4.1What can be said of the pKb associated with the conjugate base of the moreelectrolytic acid of a pair of acids?A. The pKb associated with the conjugate base of the more electrolytic acid is

greater than the pKb associated with the conjugate base of the lesselectrolytic acid.

B. The pKb associated with the conjugate base of the more electrolytic acid issmaller than the pKb associated with the conjugate base of the lesselectrolytic acid.

C. The pKb associated with the conjugate base of the more electrolytic acid isequal to the pKb associated with the conjugate base of the less electrolyticacid.

D. The pKb values associated with the conjugate bases of two electrolytic acidscannot be compared.

SolutionThe acid that is more electrolytic is the acid that forms more ions, and thus isbetter able to conduct electricity. The stronger acid dissociates more, and indoing so, produces a greater ion concentration. Themore electrolyticacid is thestronger acid, and according to Figure 4-10, the stronger acid has a conjugatebase with a higher pKb(associated with the weaker conjugate base.) Thismakeschoice A the best answer.


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Strong AcidsStrong acids are acids that dissociate fully when mixed with water. That is to saythat strong acids ionize completely into hydronium and a conjugate base whenadded to water. Reactants convert completely into products, so the Ka value of astrong acid is very large. It is safe to assume that an acid that does not have ameasurable Ka (i.e., it has a Ka that is too large to measure, making it greater than1.0)is a strong acid. Reaction 4.5 shows the dissociation of a strong acid, HX.

HX(aq) + H2OP) 1 H30+(aq) + X"(aq)

Reaction 4.5

Theequilibriumconstant for the acid dissociationreactionshown in Reaction 4.5has a largenumerator and a minusculedenominator. Assuch,Ka is significantlygreater than 1.0, and pKais negativeforHX. Thisis summarizedin Figure4-2.

Ka=[H3Ol[>C]>>1[HX]

pKa<0

Figure 4-2

It happens that all of the strong acids can be classified as either haloacids oroxyacids. Sometypicalstrong acids are listed in Table 4.2.

Acid Name pKaHCl Hydrochloric acid -7

HBr Hydrobromic acid -7

HI Hydroiodic acid -9

H2SO4 Sulfuric acid -9(pKai)HNO3 Nitric acid -2

HCIO4 Perchloric acid -10

Table 4.2

Weak AcidsWeakacids are acids that only dissociate partially when dissolved into water.They do not fully ionize into conjugate base and hydronium ion in water.Reactants areconverted partially intoproducts, sodividing products by reactantsindicates a smallvalueforKa (avalue that is lessthan1.0). Anyacid withKa lessthan 1.0 is a weak acid. Reaction 4.1 shows the dissociation of a weak acid.

HA(aq) + H20(1) ^ ^ H30+(aq) + A"(aq)

Reaction 4.1

The equilibrium constant for the acid dissociation reaction shown inReaction 4.1is less than 1.0. However, because our definitions must apply to biochemistry aswell as general chemistry, weak acids are classified as weak orvery weak. Forinstance, aspartic acid issaid tohave anacidic side chain, while the side chain ofleucine is considered to be neutral. We will define any acid with a Ka less than10"14 (and thus a pKa greater than 14) asa very weak acid, meaning that the pKarange foraweakacid is from 0 to14. This is summarized inFigure 4-3.

=[H3Q+][A-];wherel >K >10-i4 . 0<pKa <14[HA]

Figure 4-3

Terminology


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Weak acids are members of a number of different classifications, includingoxyacids of low oxidation state, one haloacid, carboxylic acids, alkylammoniums, and phenols. Table 4.3 lists some typical weak acids and theircorresponding pKa values.

Acid Name PKaCI3CCO2H Trichloroacetic acid 0.64

CI2HCCO2H Dichloroacetic acid 1.27

H2SO3 Sulfurous acid 1.82 (pKai)

HCIO2 Chlorous acid 1.90

C1H2CC02H Chloroacetic acid 2.82

HF Hydrofluoric acid 3.15

HN02 Nitrous acid 3.41

HC02H Formic acid 3.74

H3CCO2H Acetic acid 4.74

2,4-(H3C)2C6H3NH3+ 2,4-dimemylanilinium 5.08

4-H2NC6H4NH3+ 4-aminoanilinium 6.18

H2CO3 Carbonic acid 6.36 (pKai)4-02NC6H4OH 4-nitrophenol 7.15

HCIO Hypochlorous acid 7.46

HBrO Hypobromous acid 8.72

NH4+ Ammonium 9.26

HCN Hydrogen cyanide 9.32

HIO Hypoiodous acid 10.66

Table 4.3

Very Weak AcidsVery weak acids dissociate less than water. The Ka value of a very weak acid isless than 10~14, because products divided by reactants is less than 1.0 x 10"14(Kw). Given thatwater is thought tobeneutral, an acid witha Ka less than10"14(and pKa greater than 14) is a very weak acid. This is summarized in Figure 4-4.

[H30+][A-] r <10_14 > ^[HA]

Figure 4-4

Strength and the pK ScaleThe same rules that apply to acids also apply to bases, except that hydrolysis,rather than dissociation, is considered, and Kb replaces Ka. As the pK value for acompound decreases, its strength increases (this is true for both acids and bases).As a rule, the stronger the acid, the weaker its conjugate base. Strong acids havevery weak conjugate bases, and strong bases have very weak conjugate acids.The odd sounding relationship is that weak acids have weak conjugate bases. Itmay seem peculiar, but the conjugate base of a weak acid is most often a weakbase. You have seen this relationship before with buffers, although it is unlikelythat it was emphasized in your general chemistry courses.


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Table 4.4 defines the strength of an acid and a base in relative terms.

Compound NumericalData

Conjugate NumericalData

Strong acid(Fully dissociates in H20)e.g.,HCl(PKa=-7)

Ka»l

pKa<0

Very weak base(Hydrolyzes less than H20)e.g./Cl-(pKb = 21)

Kb <10'14pKb > 14

Weak acid

(Partly dissociates in H2O)e.g.,RC02H(pKa = 3-5)

10'14 < Ka < 10 < pKa < 14

Weak base

(Partly hydrolyzes in H20)e.g., RC02-(pKb = 9-11)

10"14 < Kb < 10 < pKb < 14

Very weak acid(Dissociates less than H20)e.g.,CH4(PKa=49)

Ka < 10"14pKa > 14

Strong base(Fully hydrolyzes in H20)e.g.,CH3-(pKa = -35)

Kb»l

pKb<0

Table 4.4

The strength of a reagent is measured by its ability to react in water. The morereadily an acid dissociates into water, the stronger it is. Themore readily a baseundergoes hydrolysis, the stronger it is. Do not confuse the concentration of areactant with its strength. Both affect pH. Within conjugate pairs, to calculatethestrengthof one reagent from the strength of itsconjugate, use Equation 4.7.On the exam, one of the required skills will be determining what the question isasking. In termsof acid and base chemistry, manyof the questions will be morecomplicated versions of thebasicquestion of"which acid isstronger?"

Example 4.2Whichof the following acids would yield the highest pH in water?A. OAOMHCKaq)B. 0.10 M HC104f^)C. 0.10 M HBrfaq)D. 0.10 M HC02H(aq)

SolutionThe highest pH inwater results from thelowest hydronium concentration. Thisis associated with the weakest acid, which in this case is carboxylic acid, choiceD. Choices A, B, and C are all strong acids that fully dissociate in water.

Example 4.3Which of the following is the BEST choice to titrate 0.10 MH3CNH2M?A. 0.50 M KOHfoq)B. 0.05 M NH3H;C. 0.10 MHBr(aq)D. 0.10MHCO2Hfa^

SolutionTo ensure complete reaction, the titrantmustalways bea strong reagent. Thiseliminates choice B(a weak base) and choice D (a weak acid). In this case, thesolution being titrated is a weakbase solution (H3CNH2 is a weakbase), sostrong acid is added. Choice A is a strongbaseand choice C isa strongacid.

Terminology


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Example 4.4Whichof the following acids dissociatesto the GREATEST extent when added towater?

A. HNO3B. HNO2C. H3PO4D. H2CO3

SolutionThe most complete dissociation is associated with the strongest acid. Thestrongest acid in this case is nitric acid (HNO3), choiceA. These questions showa few of the many ways relative acidity can be compared. In addition todissociation, acidities can also be compared by pKa values (lower is more acidic),Ka values (larger is more acidic), pH values (lower is more acidic for equal acidconcentrations), electrolytic strength (more electrolytic is more acidic), andreactivity with bases (stronger acids react with weaker bases).

Strong BasesStrong bases, like strong acids, react completely when they are added to water.Strong bases fully hydrolyze water, so they completely ionize when dissolvedinto water. Reactants are completely converted into products; thus, the Kb valueof a strong base is much greater than 1.0, because products divided by reactantsis very large. Any base that does not have a measurable Kb (or has a Kb that ismuch greater than 1.0) is a strong base. Reaction 4.6 shows the hydrolysis of thestrong base, KOH:

MOH(s) + H20(1) 1 M+(aq) + OH'(aq)

Reaction 4.6

The equilibrium constant for the base hydrolysis reaction shown in Reaction 4.6has a large numerator and a minuscule denominator. Its value, therefore, issignificantly greater than 1.0. This results in a negative pKb value for a strongbase. This is summarized in Figure 4-5.

[M+][OH][MOH]

»1 .-. pKb<0

Figure 4-5

It happens that all of the strong bases are either hydrides, hydroxides, alkoxides,amides, or carbides. Some typical strong bases are listed in Table 4.5. No pKbvalues are listed, because all of these bases when added to water will formhydroxide.


Base Name

KH Potassium HydrideNaOH Sodium HydroxideKOCH3 Potassium Methoxide

NaNH2 Sodium Amide

Li(CH2)3CH3 Butyl Lithium

Table 4.5


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Weak BasesWeak bases, like weak acids, partially react when added to water. Weak basespartially hydrolyze water, so they only partially ionize when dissolved intowater. Reactants are only partially converted into products, so dividing productsby reactants is placing a small number over a large number, leading to a smallvalue for Kb (less than one). As with the weak acids, because water is consideredto beneutral, we will define any base witha Kb less than 10"14 (and thus pKbgreater than 14) to be a base that is too weak to consider. This means that thepKb range for a weak base is from 0 to 14. It should be safe to assume that anybase on the test for which you are given a Kb value (or pKb value), is likely to bea weak base. A base with less than 100% hydrolysis is a weak base. Reaction 4.7shows the hydrolysis of a weak base, A".

A-(aq) + H2OO) 1 HA(aq) + OH"(aq)

Reaction 4.7

The equilibrium constant for the base hydrolysis reaction shown in Reaction 4.7has a small numerator and a larger denominator, making it less than 1.0. Thisresults in a small positive pKb value for a weak base. This is summarized inFigure 4-6.

Kb =[HA][OH]; where 10"14 <Kb <1.\ 14 >pKb >0[A"]

Figure 4-6

There are many different types of weak bases. Some typicalweak bases are theconjugate bases of the acids listed in Table4.3. In addition,other commonweakbases include the carboxylates (RCO2"), the alkyl amines (RNH2), bicarbonate(HCO3-), carbonate (C03*"), phosphate (PO43-), and phenoxides (C6H5O-).Very Weak BasesVery weak bases do not undergo any significant (or detectable) hydrolysis inwater. The Kb value for a very weak base isless than 10"14, because the productsdivided byreactants is less than 1.0 x10"14 (Kw). Given that water isconsideredtobeneutral, a basewitha Kb less than10"14 (and pKb greater than14) is a veryweak base. This is summarized in Figure 4-7.

Kb =[HA][OH-];whereKb <10_i4 .pKb >14[A"]

Figure 4-7

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Example 4.5Which of the following compounds when added to water does NOT yield asolution where hydroxide ion is in greater concentration than hydronium ion?A. Na2C03B. HC02NaC. HNO3D. C6H5ONa

SolutionA base is a compound that hydrolyzes water to generate hydroxide, and thusform a solution where hydroxide is more concentrated than hydronium. Thisquestion is in essenceasking, "Whichcompound is NOT a base?" Choice C, nitricacid, is NOT a base, making choice C the answer.

Example 4.6Which of the following bases has the STRONGEST conjugate acid?A. 0.50 M KOHB. 0.50MLiN(CH3)2C. 0.50 MNaHD. 0.50MNaHCO3

SolutionThe concentration does not affect the strength. No matter what concentration isadded to solution, the value of Kb is constant. This means that this questionsimply addresses the strength of the compound. The strongest conjugate acid isassociated with the weakest base. In this question, choice A is a metal hydroxide,choice B is a metal amide, and choice C is a metal hydride. Choices A, B, and Care strong bases. The only weak base is the bicarbonate base, choice D.

Example 4.7Which of the following is the BEST choice to titrate 25 mL of 0.10M HC02H(aq)?A. 0.10 M KOH

B. 0.05MH3CNH2C. 0.10 M HBrD. 0.05MH3CCO2H

SolutionTo ensure complete reaction, the titrant must be a strong reagent. This eliminateschoice B (a weak base) and choice D (a weak acid). In this case, the solutionbeing titrated is a weak acid solution (HCO2H is a weak acid), so strong basemust be added. Choices C is a strong acid, so the only strong base is choice A,KOH.

Copyright © by The BerkeleyReview 246 The Berkeley Review

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General CliemiStry Acids and Bases Types ofAcids and Bases

Types of Acids and BasesHaloacidsThe haloacids are the series of HX acids, whose X represents a halogen. Theacidity of haloacids is predictable based on the size of the halide ion. The largerthe halide, the more acidic the haloacid. This means that HI > HBr > HCl > HF.This trend in acidity is attributed to the increased stability of the conjugate baseas it increases in size. The negative charge is more diffuse on the larger anion.The more diffuse, the less basic the anion, and thus the more acidic its conjugateacid. A second, and perhaps easier, way to view the acidity of haloacids involvesestimating their bond strengths. The longer the bond, the weaker the bond, as ageneral rule in chemistry. As you descend the halogen column in the periodictable, halogen size increases. As halogen size increases, the bond length of an H-X bond must also increase. As the bond length increases, the bond strengthdecreases, and the H+ can be removed more readily. This approach works eventhough bond dissociation energies are determined from homolytic bondbreaking, as opposed to the heterolytic cleavage associated with acid-basechemistry. Ions (H+ and A") are the result of heterolytic bond breaking, but thecorrelation between bond strength and acidity still holds for haloacids. Figure 4-8 summarizes the effect of halide size on acidity.

H—F

H—CI

H Br

H 1

Descending a column:Halogen size increases.*. Bond length increases.'. Bond strength decresases/. Dissociation increases/. Acidity increases

Figure 4-8

Example 4.8Which of the following conclusions can be made concerning the relative aciditiesof haloacids?

A. Acid strength increases with increasing electron affinity of the halide.B. Acid strength increases with increasing electronegativity of the halide.C. Acid strength increaseswith increasing ionic radius of the halide.D. Acidstrength increaseswith decreasing isotopicabundanceof the halide.

SolutionAs you descend a column in the periodic table (such as the halogen column),acidity increases due to the increase in atomic size. The anion formed upondeprotonation is more stable as it increases in size, so acidity increases as thecolumn is descended. Electron affinity and electronegativity decrease as thecolumn is descended, and they have no bearing on acidity. Isotopic abundanceand atomic mass do not affect acidity, either, so choice C is the best answer.

An important fact to recall about haloacids is that they are all strong acids, exceptfor hydrofluoric acid (HF). HF has a pKa of roughly 3.3, so it does not fullydissociate when added to water. As we have observed in studying other topicsrelated to the properties of this chemical family, this weaker acidity is attributedto the smaller atomic radius of fluorine relative to the other halogens.


General Chemistry Acids and Bases Types of Acids and Bases

Example 4.9Whichof the following acidshas the largest Ka value?A. HFB. HClC. HBrD. H2S

SolutionThe largest Ka value is associated with the strongest acid. Because Br is thelargest anion of thehalogens listed, it dissociates from the protonmost readily.This makes HBr the most acidic, so you should pick choice C. Choices D (H2S)can be eliminated, because S is of roughly comparable size with, but lesselectronegative than,CI. This meansthat HCl is a stronger acid than H2S.

Within a period (row) of theperiodic table, it is electronegativity that dictates thestrength of an acid, not atomic radius. A prime example of this idea is therelationship between ammonia (H—NH2), water (H—OH), and hydrofluoricacid (H—F). Thestrongest acid of the three compounds is the hydrofluoric acid,because fluorine is more electronegative than both nitrogen and oxygen. Theatomicsizedoesnot changethat noticeably betweenN,O, and F,because they allhave the same valence level (n = 2). The periodic trend that most changes is theelectronegativity.

Oxyacids (Non-metal Hydroxides and Non-metal Oxides)Oxyacidsare also an acid type to be familiarwith. They differ from haloacids, inthat the hydrogendissociates froman oxygen rather than a halide. In an oxyacid,the acidic hydrogenis bonded to an oxygen, which in turn is bonded to a centralatom (which in some cases can be a halogen). The simplest rule is that the moreoxygen atoms there are bonded to the central atom, the more the oxygen atomswithdraw electron density from the central atom, and thus the more acidic theoxyacid. This can be summarized as the resonance effect (which is typicallyclassified as an organic chemistry concept). Unlike the haloacids, where size isimportant, in oxyacids, the electronegativity of the central atom is mostimportant. As a general rule, for every additional oxygen on the central atom,the pKa of the acid will drop by approximately 5 pKa units. Table 4.6 lists sometypical oxyacids.

Oxyacid Name Oxyacid Name

HNO2 Nitrous acid HNO3 Nitric acid

H2SO3 Sulfurous acid H2SO4 Sulfuric acid

H3PO3 Phosphorous acid H3PO4 Phosphoric acidH2CO3 Carbonic acid H02CC02H Oxalic acid

HCIO Hypochlorous acid HCIO2 Chlorous acid

HCIO3 Chloric acid HCIO4 Perchloric acid

HBrO Hypobromous acid HBr02 Bromous acid

HBr03 Bromic acid HBr04 Perbromic acid

HIO Hypoiodous acid HI02 Iodous acid

HIO3 Iodic acid HIO4 Periodic acid

Table 4.6



When comparing the relative strength of oxyacids, the number of excess oxygenatoms and the electronegativity of the central atom must both be considered. Forcarbonic acid, there is an excess oxygen count of one. What is meant by excessoxygencount is the number of oxygen atoms exceedingthe number of hydrogenatoms. The excess oxygen count often turns out to be the number of oxygenatoms double-bonded to central atom. The more rc-bonds to oxygen from thecentral atom, the more resonance withdrawal from the central atom, and thus themore acidic the compound. This explains why H2SO4 is a stronger acid thanH2SO3. When two compounds have the same excess oxygen count, the nextfactor to consider is the electronegativity of the central atom. The moreelectronegativethe central atom, the more it withdraws electrondensity from theacidic proton, increasing the acidity. This is in essence the inductive effect.Sulfur is more electronegative than carbon, so H2SO3 is a stronger acid thanH2CO3. As you learned in organic chemistry, the resonance effect is greater thanthe inductive effect.

Example 4.10Which of the following acids has the LARGEST pKa value?A. HC102B. HI02C. HCIO3D. HIO3

SolutionThe largest pKa value is associated with the weakest oxyacid, which is associatedwith the acid having the smallest number of excess oxygens and the leastelectronegative central atom. Choices C and D are eliminated, because theycontain two excess oxygens each. Choice B is better than choice A, because theyboth have one excess oxygen, but iodine is less electronegative than chlorine.This question could have easily asked for the strongest acid. It is important torealize that most questions will be asking for either the strongest acid or theweakest acid among the answer choices. The wording may be in reference toconjugate bases, electrolytic nature, ionizability, Ka values, pKa values, pHvalues, or reactivity. The secret is to do enough practice questions so that youencounter all the different possibilities at least a few times before your test.

As a point of interest, oxyacids result from the hydration of non-metal oxides.This is the cause of acid rain, where most often nitrogen oxides and sulfur oxides(Lewis acids) react with moisture in the air to form Bransted-Lowry acids. Alowering in pH for rainfall may also be observed in environments rich in carbondioxide (which hydrates to become carbonic acid). Reaction 4.8 shows thehydration of carbon dioxide, while Reaction 4.9 shows the hydration of sulfurtrioxide.

C02(g) + H20(1) ^

non-metal oxide

Reaction 4.8

S02(g) + H20(1) ^=

non-metal oxide

Reaction 4.9

H2C03(aq)

oxyacid

H2S03(aq)

oxyacid

Types of Acids and Bases



Non-metal oxides act as Lewis acids (electron-pair acceptors), while theirhydrated counterparts (non-metal hydroxides, commonly known asoxyacids) actas Bronsted-Lowry acids. Both SO3 and H2SO4 reactwith hydroxide to formHS04", so theyare equivalent in terms of strength. Thedifference is that SO3 ishydrated to become H2SO4. By a similar reaction, sulfur dioxide (SO2) formsH2SO3 (sulfurous acid) and nitrogen dioxide (NO2) forms both HNO2 (nitrousacid) and HNO3 (nitric acid). These are major components of acid rain. Acidrain is a combination of rain and air-borne pollutants, such as sulfur oxides andnitrogen oxides. It is often treated with steam to convert from the Lewis acidform (non-metal oxide) intotheBransted-Lowry form (non-metal hydroxide) andthen neutralized with calcium oxide.

Metal Hydroxides and Metal OxidesJust as non-metal oxides are Lewis acidsand non-metalhydroxides are Bronsted-Lowry acids, metal oxides are Lewis bases and metal hydroxides are Bronsted-Lowry bases. You should be familiar with these general classifications. Metaloxides are basic and will form metal hydroxides when treated with water. Aprime example is calcium oxide (CaO), which forms calcium hydroxide(Ca(OH)2) whenhydrated. Reaction 4.10 shows theLewis acid-base reaction ofametal oxide and a non-metal oxide, while Reaction 4.11 shows the Bransted-Lowry acid-base reaction ofa metal hydroxide and a non-metal hydroxide. Thereactants in Reaction 4.11 are the hydrated form of the reactants in Reaction4.10.

S03(g) + CaO(s) ^ " CaS04(s)non-metal oxide metal oxide neutral saltLewis acid Lewis base

Reaction 4.10

H2S04(aq) + Ca(OH)2(aq) ^ fc Ca2+(aq) +S042"(aq) + 2H20(1)non-metal hydroxide metal hydroxide cation anionBren-Lowry acid Brem-Lowry base

Reaction 4.11

Organic AcidsLet us considerthree typesof organic acids: carboxylic acids,phenols, and alkylammonium salts. It is importantthat you recognize these functional groups andknow their pKa ranges. Forcarboxylic acidsand alkylammoniums, you shouldknow their organic pKa rangeand their range in aminoacids. In carboxylic acidsand phenols, the proton comesoffof an oxygen that is involved in resonance. Inan alkyl ammonium cation, the proton comes offof a nitrogen. Figure4-9 showsa generic carboxylic acid and its pKa range, a genericphenol and its pKa range,and a generic alkyl ammonium cation and its pKa range. TheAAin parenthesisrefers to amino acid terminals, while the Rdesignation refers to an alkyl group.

:o: ^^ H H

\N©

R OH R H

Carboxylic acid *QH Alkyl ammonium cationpKa(R) =3-5 Phenol pKa(R) =9-UpKa(AA) =2-3 PKa =9-5"10-5 PKa(AA) =9-10

Figure 4-9


General ChemiStry Acids and Bases Types ofAcids and Bases

Polyprotic AcidsPolyprotic acids are acids which yield multiple equivalents of hydronium(H30+) when treated with a base. The three most common examples fromgeneral chemistry are carbonic acid (H2CO3), sulfuric acid (H2SO4), andphosphoric acid (H3PO4). In addition to these three acids, there are also commonamino acids that qualify as either diprotic (no active proton on the side chain) ortriprotic (an active proton on the side chain). Concentrationsof polyprotic acidsare often given in terms of normality (N). Normality is defined as moles ofequivalents per liter solution. This is to say that a 1.0molar diprotic acid solutionwould be listed as "2.0 normal," because there are two equivalents of acid.

Example 4.11Which of the following acid solutions is 3.0 N?A. 1.00 M alanineB. 1.50 M carbonic acidC. 1.50M phosphoric acidD. 1.25 M sulfuric acid

SolutionNormality (N) is found by multiplying the molarity of an acid by the number ofprotons per molecule. In choice A, alanine (H3NCH(CH3)C02H) is diprotic, sothe normality is 1.00 x 2, which equals 2.0 N. Choice A is eliminated. In choice B,carbonic acid (H2CO3) is diprotic, so the normality is 1.50x 2, which equals 3.0N. Choice B is the correct answer. In choice C, phosphoric acid (H3PO4) istriprotic, so the normality is 1.50 x 3, which equals 4.5N. Choice C is eliminated.In choice D, sulfuric acid (H2SO4) is diprotic, so the normality is 1.25 x 2, whichequals 2.5 N. Choice D is eliminated.

Polyprotic acids have multiple pKa values, one for each dissociable proton. Bydefinition, the first proton removed is more acidic than the second one removed,so pKai is always lower than pKa2- Some difficulty may arise when you considerthe pKa values of a polyprotic acid and its conjugate base. For a diprotic acidsuch as H2CO3, the first proton removed corresponds to the second protongained by the conjugate base. H2CO3 and HCO3" are a conjugate pair with pKaiand pKb2 summing to 14. The full dissociation of carbonic acid is shown inReaction 4.12 and Reaction 4.13.

PKaiH2C03(aq) + H20(1) * „ * H30+(aq) + HC03"(aq)

PKb2

Reaction 4.12

HC03"(aq) + H20(1) •* *2 " H30+(aq) + C032"(aq)PKbl

Reaction 4.13

Determining the amount of base needed to neutralize a polyprotic acid is atypical question from general chemistry. While the MCAT does not emphasizesolving mathematical problems, understanding the setup is still important,because understanding equivalents can help to determine the pH of mixtures.



Example 4.12How many mL of 0.40 MNaOH are required to neutralize 100 mL 0.25 MH2S04?A. 62.5 mLB. 80.0 mLC 100.0 mLD. 125.0 mL

SolutionIn order to neutralize an acid, an equal mole quantity of hydroxide must beadded to thehydronium source. The general relationship is shown in Equation4.8.

molesOH" =molesH+ .-. (Moh-)(Voh-) = (Mh+)(Vh+) (4.8)

Substituting intoEquation 4.8 yields the following results:

(MrjH-XVoH-) = (MH+)(VH+)(0.40 M)(V0H-)= 2 x (0.25 M)(100 mL)

VoH- =2x(°-25M)xl00mL =(°-50) x100 mL = 125 mLV0.40M/ V0.40/

Choice D is the best answer. Choice A would have been the result of yourcalculation, if you had forgotten to multiply the molarity by the number ofequivalents, a commonmistake with these typesofquestions.

Example 4.13How manymilliliters of0.60MHCl arerequired toneutralize 3.0 grams CaC03?A. 50 mLB. 100 mLC. 200 mLD. 300 mL

SolutionAccording to thebalanced equation, twomolecules ofHClare required for everyone molecule of CaC03- The balanced equation is shown below:

CaC03(s) + 2HCl(aq) ^ * CaCl2(aq) + C02(g) + H20(1)The mathematical setup is:

moles OH" =2 xmoles CO32- = (Mh+)(Vh+)

2x 3-0g =(0.60 M)(VH+)1007mole

0.06moles = (0.60M)(Vh+) •'. Vh+ = 0.10 L = 100 mL

Choice B is the best answer. Choice A would have been the result of yourcalculation, if you had forgotten to multiply the molarity by the number ofequivalents, a common mistake with these types ofquestions.

Copyright©by The Berkeley Review 252 The BerkeleyReview

General ChemiStry Acids and Bases Types of Acids and Bases

Example 4.14Citricacid (C6Hg07) has three dissociable protons with pKa values of 3.14,4.79,and 5.20. Which of the following solutionswould have the lowestpH values?A. 50mL0.10 M citric acid(aq) + 75mL 0.20MNaOH(aq)B. 25mL 0.10M citric acid(aq) + 50mL 0.10M NaOH(aq)C. 25mL 0.20M citric acid(aq) + 75 mL 0.20M NaOH(aq)D. 50mL 0.20M citric acid(aq) + 50 mL 0.10M NaOH(aq)

SolutionForquestions involving mixtures, it is important to think of the reagents in termsof equivalents. The lowest pH belongs to the solution that is most acidic. Themost acidic solution is the solution where the fewest equivalents of base relativeto citric acid have been added. In choice A, the NaOH(aq) solution is 1.5 timesthe volume and twice the concentration of the citric acid solution, so there arethree equivalents of NaOH(aq). In choice B, the NaOH(aq) solution is double thevolume of the citric acid solution and of equal concentration, so there are twoequivalents of NaOH(aq). In choice C, the NaOH(aq) solution is three times thevolume of the citric acid solution and of equal concentration, so there are threeequivalents of NaOH(aq). In choice D, the NaOH(aq)solution is of equal volumeand half the concentration of the citric acid solution, so there is only one-half ofan equivalent of NaOH(aq). The fewest equivalents are found in choice D, sochoice D is the best answer.


General Chemistry Acids and Bases Calculating pH

Calculating pHDetermining pHAsolution's acidity ismeasured in terms ofhydronium concentration ([H30+l)using the pH scale. The pH value ofa solution isdetermined using Equation 4.9.

pH=-log [H30+] (4.9)By manipulating Equation 4.9 soas toisolate hydronium concentration, Equation4.10 can be derived.

[H30+1 =10"PH (4.10)Because neutral water has an H30+ concentration of 10"7 M, due to theautoionization of water, neutral water has a pH of 7.0. Acidic solutions have pHvalues less than 7.0, whilebasicsolutionshave pH values greater than 7.0. Thereare no limits to the pH scale other than those imposed by the strength andconcentration of the acid or base in solution.

On the test, youcan almost assume that they will have questions about bothpHandpOH for anaqueous solution. Just asEquation 4.9 defines pH,Equation 4.11defines pOH.

pOH=-log[OH"] (4.11)

This means that in order to calculate pH or pOH, it is necessary to determine theH30+ concentration or the OH" concentration. Once this is accomplished, it issimply log math (negative logs, actually). There areno calculators allowed onthis test. The volume of a solution does not matter in determining pH; only theconcentration is important. Equation 4.12 may be used to interconvert betweenpH and pOH in an aqueous solution at 25°C

pH + pOH=14 (4.12)

Example 4.15What is the pH of a solution where the hydroxide concentration is 106 timesgreater than the hydronium concentration?A. 4

B. 6C. 10D. 13

SolutionBecause hydroxide is in greater concentration than hydronium, the solution isbasic, so thepHmust be greater than 7. This eliminates choices A and B. Thetrick here is to use the logscale correctly. Because the concentrations differ by afactor of106, the pHandpOH differ bylog 106, which is 6. If thepH is 10, thenthepOHis4,which aredifferent by 6. If thepH is 13, then the pOHis 1,and thedifference is 12. This means that the best answer is choice C. This is a trickyquestion where understanding the conversion between the log scale andconcentration scale is pertinent. Most studentschoose D,because they add 6 tothe neutral pH of 7. It is important to recognize why choice D is a typicalmistake.



Log ReviewBecause pH isbased on logs, it is helpful todo a quick review of logs. When youmultiply numbers, you add their logs. When you divide numbers, you subtracttheir logs. You can solve for any log given that log 2 =0.3 and log 3 =0.48.Drawn inTable4.7are examplesof how to solve for a logvalue.

Number Mathematical Calculation Log Value2 Given value 0.301

3 Given value 0.477

4 Log 4 = Log (2x2) = log 2 + log 2 = .301 + .301 = .602 0.602

5 Log 5 = Log (10 * 2) = log 10 - log 2 = 1.00 - .301 = .699 0.699

6 Log 6 = Log (3x2) = log 3 + log 2 = .477+ .301 = .778 0.778

7 Log7 is approximated as being closer to log 8 than log 6 0.845 ± .01

8 Log 8 = Log (2 x 2 x 2) = 3 (log 2) = 3(.301) = .903 0.903

9 Log 9 = Log (3 x 3) = log 3 + log 3 = .477 + .477 = .954 0.954

0.33 Log 0.33 = Log (1 * 3) = log 1 - log 3 = 0 - .477 -0.477

0.50 Log 0.50 = Log (1 * 2) = log 1 - log 2 = 0 - .301 -0.301

1.20 Log 1.2 = Log (6*5) = log 6 - log 5 = .778 - .699 = .079 0.079

1.25 Log 1.25 = Log (5*4) = log 5 - log 4 = .699 - .602 = .097 0.097

1.33 Log 1.33 = Log (4*3) = log 4 - log 3 = .602 - .477 = .125 0.125

1.40 Log 1.40 = Log (7*5) = log 7 - log 5 « .845 - .699 = .146 0.146

1.50 Log 1.50 = Log (3*2) = log 3 - log 2 = .477 - .301 = .176 0.176

1.60 Log 1.60 = Log (8*5) = log 8 - log 5 = .903 - .699 = .204 0.204

1.67 Log 1.67 = Log (5 * 3) = log 5 - log 3 = ..699 - .477 = .222 0.222

1.75 Log 1.75 = Log (7*4) = log 7 - log 4 « .845 - .602= .243 0.243

1.80 Log 1.80 = Log (9*5) = log 9 - log 5 = .954 - .699 = .255 0.255

Table 4.7

Logcalculations should be carried out only to the level of approximation. Do notforget that log of 10* = x, so knowing the power of 10 that some quantity is canbe a useful hint when selecting the correct log value. Often, only one answerchoice relates to the correct power of 10, which saves a great deal of time spent incalculating. All sorts of shortcuts and tricks will be presented during thefollowing sample problems. Before moving on in this section, however, make aconcerted effort to work through Table 4.7 and understand thoroughly how toestimate log values.

Determining pH for Strong ReagentsCalculating the pH for strong acid and strong base solutions is based on theconcept of full dissociation. The calculation of the pH for a strong reagentfollows an easy pattern. For a strong acid, Equation 4.9 is employed, where theconcentration of the strong acid ([HX]) is substituted for the concentration ofhydronium (IH30+J). For a strong base, Equation 4.11 is employed, where theconcentration of the strong base ([MOH]) is substituted for the concentration ofhydroxide ([OH")). Calculating pH and pOH from concentrations is most easilydone when the concentration is written in scientific notation. This is becausetaking the log of a number in scientific notation is more convenient in terms ofbookkeeping than taking the log of a number in standard notation.

Calculating pH



Example 4.16What is the pH of0.0020MHCl(aq)?A. 2.00B. 2.70C 3.00D. 7.00

SolutionDetermining an exact numerical value involves calculations. The pH for anysolution isdefined aspH=- log [H30+], sofor a strong acid, thepHis- log [HX].It is best to use scientific notation for the concentration.

pH=- log (2xl0-3)pH =-(log 2+log lO'3) =- log 2- log 10"3

pH = -log2-(-3) = 3-log2pH = 3-0.3 = 2.7

The correct answer is choice B. Choice D should have been eliminated, becausethe solution is acidic, so pH is less than 7.00. Choices A and C could also havebeen eliminated, ifyou noted that thelogvalues endedin ".00". For a logtobeawhole number, the concentration must be a power of ten. The fact that theconcentration was .002 tells us that the log could not be a whole number. Fromthisexample, youshould derive a shortcut for use in the future. You mighttakenotice that the - log of2x10"3 isequal to3- log 2,sowhy notremember this andskipa few steps in thefuture. Use therelationship: - log(zx 10_y) = y - logz.

This shortcut applies to all negative log calculations, including the conversionfrom Ka to pKa. For instance, the pKa for aweak acid with Ka equal to 4.1 x10"6is6- log4.1. This value canbeestimated tobegreater than5.0 (which is equal to6 - log 10), but less than 5.5 (which is equal to 6 - log 3). A range of 5.0 to 5.5shouldbe good enough to choose the correct answer from four choices. Withexact numerical questions on the MCAT, your goal should be to narrow theanswer choice range enough so that threewronganswers maybe eliminated.

Example 4.17What is the pH of 100mL of 0.030 MHBr(aq)?A. 1.30B. 1.52C. 2.30D. 7.00

SolutionThe volume of the solution does not affect the pH, unless another solution isadded. Using the shortcut, the pH is found as follows:

pH =- log (3 x10"2) =2- log 3pH = 2 - 0.48 = 1.52

The correct answer is choice B. Choice D should have been eliminated, becausethe solution is acidic, so pH is less than 7.00. Choices A and C could alsohavebeeneliminated, if you noted that the log values ended in ".30". This wouldcome from logarithmic insights.


Zia Siddiqui


Example 4.18What is the pOH of 0.050M KOHfa^?A. 1.30B. 1.70C. 12.30D. 12.70

SolutionThe pOH of a basic solution is found in a manner similar to getting the pH of anacidic solution. Using the shortcut, the pOH is found as follows:

pOH =- log(5 x lO'2) =2- log 5pOH = 2-0.7= 1.30

The correct answer is choice A. Choices C and D should have been eliminated,because the solution is basic, so pH is greater than 7.00, and therefore pOH is lessthan 7.00. The pH of the solution can be found using pH = 14 - pOH, wherepOH = 14-1.30 = 12.70.

Example 4.19What is the pH of 200 mL of 0.00391M KOH(aq)?A. 2.41

B. 2.61C. 11.39D. 11.59

SolutionThis question would seem to be quite difficult at first glance; but if you followthe rules, it is easy. Because KOH is a strong base, it will fully dissociate whenadded to water. Plugging values into our shortcut method yields the following:

pOH =- log (3.91 x lO'3) =3- log 3.913 - log 10 < 3 - log 3.91 < 3 - log 3 .-. 2 < pOH < 2.5

If 2 < pOH < 2.5, then 12 > pH > 11.5The correct answer is choice D. Choices A and B should have been eliminated,because the solution is basic, so pH is greater than 7.00. Choice C is eliminated,because it does not fit into the range for the correct number. Some of you mayhave chosen to approximate 3.91 as 4, and solved accordingly. This method isfine, too.

You should be able to determine pH or pOH for strong compounds in less thanfifteen seconds. While the MCAT does not offer up many calculation questions,if you are fortunate enough to get a pH calculation question, you should finish itquickly, and carry the time you save over to more difficult questions.


General Chemistry Acids and Bases Conjugate Pairs

Determining pH for Weak ReagentsBecause weak acids and weak bases do not fully dissociate, it is necessary to usethe Ka and Kb values to determine their H30+(aq) and OH"(aq) concentrations.Ka and Kb are equilibrium constants for the respective compounds in water.Equation 4.7 shows the relationship between the pKa and pKb for anacid and itsconjugate base. Equation 4.13 is based on the relationship of equilibriumconstants to Kw (the dissociation constant for water).

KaxKb =10-14 (4.13)

According to Reaction 4.1, weak acids dissociate intohydronium and conjugatebase when added to water. Equal parts of conjugate base and hydronium ionform. To determine the [H30+], the dissociation reaction and the dissociationconstant (Ka) must be employed. This too, like the strong acids, is a systematicprocess tomaster, in the course ofwhich weshall discover another shortcut. Tounderstand the process, consider someconcentration of a weakacid (HA) with apKa between 2and 12. The setup for the reaction isshownin Figure 4-10.Reaction: HA(aq) H20(1) „ H30+(aq) A"(aq)Initially: [HA]jnit excess negligible 0Shift: ix ix • +x ±xEquilibrium: [HA]init-x irrelevant x x

Figure 4-10

Substituting values into the acid dissociation expression leads to Equation 4.14,whichcan be applied if the concentration is greater than the Ka, and if pKa fallsbetween 2 and 12. Equation 4.14 does not generate precise answers, but it doesgive a very close approximation.

Given: [Al =H30+], then Ka =1™^ JH30+][H30+] JH^lJ [HA] IHA] [HA]

JH30+] . [h30+]2 =Ka x[HA] .-. [H30+] =VKax[HA][HA]

Plugging thisvalueintoEquation 4.9 for [H30+],yieldsEquation 4.14

pH=-log VKax[HA] (4.14)Equation 4.14 canbe further manipulated to generate Equation 4.15 (theshortcutequation), which will save you time, once it is understood and mastered.

pH =-log VKax[HA] =- log/Ka" + (- log V[HA])pH =-log(Ka)V2 +(-log [HA]V2) =-1 log Ka -1log [HA]

pH =-i log Ka -1 log [HA] =1pKa -1 log [HA]

pH =1pKa -1 log [HA] (4.15)2 2

To ensure that Equation 4.15 makes conceptual sense, let's consider the pH of aweak acid solution. Ifyou add more acid, the pH should decrease. According toEquation 4.15, increasing [HA] lowers the pH. Stronger acids have moredissociation,so they should form solutions of a lower pH. Equation 4.15 showsthisby including thepKa term. Getting thepHofa weakacid shouldbe easy.

Copyright ©byThe Berkeley Review 258 The Berkeley Review

vieneral ChemiStry Acids and Bases Calculating pH

Example 4.20What is the pH of 1.00M HF(aq) with pKa = 3.32?A. 0.50B. 1.66C. 2.00D. 6.64

SolutionThis question is made easy by applying Equation 4.15. Two requirements forEquation 4.15 to work are that: 1) the weak acid concentration must be greaterthan the Ka and 2) the pKa must lie between 2 and 12. Both of these criteria aremet, so Equation 4.15 may be applied.

pH =1 pKa -1 log [HA] =1 (3.32) -1 log 1.0 =1.66 -1(0) =1.662 2 2 2 2

The correct answer is choice B. Because [HA] is 1.0M, the pH is half of the pKa.If the acid concentration is 0.10M, then the pH is half of the pKa + 0.5. Thismeans that the pH of a weak acid can be estimated quickly.

Example 4.21What is the pH of 0.07562 M HC02H with a pKa of 3.642?A. 2.259B. 2.383C. 2.759D. 2.883

SolutionThe pH can be estimated from one-half of the pKa. If the acid concentration were1.00M, then the pH would be 1.821 (half of the pKa). If the acid concentrationwere 0.10 M, then the pH would be 2.321 (half of the pKa + 0.5). If the acidconcentration were 0.010M, then the pH would be 2.821 (half of the pKa + 1.0).The concentration falls between 0.10M and 0.010M, so the pH must fall between2.321 and 2.821. This eliminates choices A and D. Because the concentration isjust less than 0.10M, the pH should be slightly higher than 2.321,making choiceB the best answer.

You should not expect numbers to be this difficult. But given that these numbersnow can be handled without difficulty, any weak acid pH calculation can bemade easy. The answer to a question like this one in a multiple-choice exam canbe approximated quickly and simply using this technique. The answer choicesmay be considered as other expressions of.A. 1.821+0.438B. 2.321+0.062C. 2.321 + 0.438D. 2.821 + 0.062

Equation 4.16 is the equivalent equation for the pOH of a weak base solution. Itis derived in the same fashion as Equation 4.15, but Kb replaces Ka, [A"] replaces[HA], and base hydrolysis is considered instead of acid dissociation.

pOH =1 pKb -1 log [A-] (4.16)



Example 4.22What is the pH of 0.20Msodium propionate, if ithas Kb =7.2 x10"10?A. Less than 3B. Between 3 and 7C. Between 7 and 11D. Greater than 11

SolutionThesolution is basic, so the pH is greater than 7.0, eliminating choices A and B.Theweakbaseconcentration is greaterthanKb, and pKbfalls between 2 and 12,soEquation 4.16 can beapplied. The pKb for thebase is 10 - log7.2, which isslightlymore than 9. A good estimate is 9.2.The pOHcanbe estimated from one-half of the pKb- If the baseconcentrationwere 1.00 M, then the pOH would be 4.6 (half of the pKb). If the baseconcentration were 0.10 M, then the pOH would be 5.1 (half of the pKb + 0.5).The concentration is 0.20 M, so the pOH lies between 4.6 and 5.1. This meansthat pH liesbetween 8.9 and 9.4, makingchoice C the bestanswer.


General ChemiStry Acids and Bases Conjugate Pairs

Conjugate PairsA conjugate pair consists of an acid and a base that exchange one proton; onegains a proton and the other member of the pair loses that proton. An acid whendeprotonated forms its conjugate base, and a base when protonated forms itsconjugate acid. Reaction 4.1 and Reaction 4.4 show that HA and A"are a genericconjugate pair. The charge difference between the acid and the base in aconjugate pair is +1.

Example 4.23The conjugate base of HC03" is which of the following?A. C032"B. H2CO3C. CO2D. HC02"

SolutionA conjugate base is formed when an acid loses one proton. This eliminateschoice B,which happens to be the conjugate acid of HCO3". The conjugate baseof HCO3" is CO32", so choice A is the best answer. Thedissociation reaction ofbicarbonate into hydronium and carbonate is shown below:

HC03"(aq) + H2CKI) ^ H30+(aq) + C032"(aq)acid conjugate base

Example 4.24Which of the following pairs of compounds is NOT a conjugate pair?A. NH3/NH4+B. H2C03/HC03"C. H2SO3/HSO3-D. P0437H2P04-

SolutionIn choiceA, NH4+ is formed when a proton is added to NH3, so NH4"1" and NH3differ by one proton. This makes them a conjugate pair, and it eliminates choiceA. In choice B,H2C03 is formed when a proton is added to HC03", so H2CO3and HCO3" differ by one proton. This makes them a conjugate pair, and iteliminates choice B. In choice C, H2SO3 is formed when a proton is added toHS03", so H2S03 and HS03"differ by one proton. Thismakes them a conjugatepair, and it eliminates choice C. Choice Dis the best answer because, PO43" andH2PO4" differ by two protons, so they do not constitute a conjugate pair.

Recognizing conjugate pairs should be effortless. Knowing how to apply theconcept of a conjugate pair to solving problems is the more useful ability.


Zia Siddiqui


Typical Conjugate PairsTypical conjugate pairs either areobserved inbiological examples orarecommonlaboratory buffer systems. There aremany weak acids and conjugate bases, so itis a good idea for you to recall classes of compounds, rather than specificcompounds. Common conjugate pairs where both components areweak includethecarboxylic acids/carboxylates (RCO2H/RCO2"), thealkyl ammoniums/alkylamines (RNH3+/RNH2), and phenols/phenoxides (C6H5OH/C6H5O-) Specificpairs that are likely to appear on theMCAT with great frequency are carbonicacid/bicarbonate (H2C03/HC03") and phosphoric acid/dihydrogen phosphate(H3P04/H2PC>4"), due to their presence in physiological systems. It is a goodidea to know the acid-base properties of compounds that are common inphysiology.The distribution within a conjugate pair is dictated by the pH of the solution.Theconjugate pair favors the conjugate acidform in the presence ofhydronium.Theconjugate pair favors the conjugate baseform in the presence of hydroxide.The exact distribution is determined by the relationship between pH of thesolution and thepKa of theconjugate acid. Figure 4-11 shows this relationship.IfpH>pKa, thesolution isbasic relative to thecompound, so it is deprotonated.IfpH <pKa/ thesolution isacidic relative to thecompound, so it is protonated.

Figure 4-11

The pH refers to the surrounding solution (environment) in which thecompounds exist. The pKa refers to the conjugate acid that exists in solution.Thecompound responds to thepH of the solution. Because of the importance ofrelatingpH to pKa, common pKa ranges should be known. SomecommonpKarangesare shown in Figure 4-9. Carboxylic acidshave pKa values of 2 to 5. Inamino acids, the carboxyl terminal has a pKa between 2 and 3. Because of thelowpKa valuerelative to physiological pH, the physiological form of carboxylicacids is the deprotonated form. Ammonium and alkyl ammoniums have pKavalues of 9 to 11. In amino acids, the amino terminal has a pKa between 9 and 10.Because of the high pKa value relative to physiological pH, the physiologicalform of amines is the protonated form. Figure 4-12 summarizes some commonphysiological compounds and their natural states:

Acid Conjugate BaseRC02H (pKa = 3-5) RC02- (pKb= 9-11)

physiological form, because 7.4> 5

RNH3+ (pKa =9-10) RNH2(pKb = 4-5)physiological form, because 7.4< 9

H2P04- (pKa2 =7.2) HPO42- (pKb2 =6.8)physiological form,because 7.4> 7.2

H2C03 (pKai = 6.4) HC03" (pKb2 = 7.6)physiological form, because 7.4> 6.4

Figure 4-12



Relationship of pKa and pKbBecause the members of a conjugate pair exchange one proton, their respectiveequilibrium constants are related. An acid when deprotonated forms itsconjugate base, and a base when protonated forms its conjugate acid. Equation4.7 shows the relationship of the pKa for the acid and the pKb for its conjugatebase. This relationship is useful for interconverting between pK values.However, when polyprotic acids are involved, it gets a little more complicated.The relationship is emphasized in Reaction 4.12 and Reaction 4.13. Also keep inmind that Equation 4.7 applies only at 25°C in water.

Example 4.25What is the pKa for ammonia, given that the pKb for ammonia is 4.7?A. 4.7

B. 7.0C. 9.3D. 33

SolutionIt is not 9.3! The compound with a pKa of 9.3 is ammonium (NH44"), theconjugate acid of ammonia. Ammonia is a weaker acid than ammonium, so ithas a pKa greater than 9.3 The best answer is choice D, 33, indicating thatammonia is such a weak acid that when added to water, there is no detectabledissociation. The mistake of choosing 9.3 is easy to make, one that most studentsmake routinely. But the equation pKa (HA) + PKb (A") = I4 is f°r conjugate pairs,not for the same compound.

The test-writers are more likely to apply this concept to carbonic acid (H2C03),which has two dissociable protons, and thus has both pKai and pKa2- Thecorrect relationships between pKa and pKb for the two respective conjugate pairsare pKai + pKb2= 14 and pKa2 + pKbi = 14. Also note that pKai is always lessthan pKa2, because the first proton is more acidic than the second one, bydefinition.



Henderson-Hasselbalch EquationThe pHofa solution comprised ofbothcomponents in a weak conjugate paircanbe determined using the Henderson-Hasselbalch equation, which is shownbelow as Equation 4.17. For the Henderson-Hasselbalch equation to hold true,both the acid and its conjugatebase must be present in appreciable concentrationin solution.

pH =PKa +logIC°"jugatebase] ^IConjugate acid]

Theequation shows that as the conjugate base concentration ([conjugate base])increases, the pH of the buffer increases. It also shows that as the conjugate acidconcentration ([conjugate acid]) increases, the pH of the buffer decreases.Equation 4.17is derived from the Ka equation.

JA-HH^l + |IHA]|[HA] \ [Al |

taking -log ofeverything yields: -log [ H30+] = -logKa - log ——\ IA ] /

-log[H30+] =-logKa-log(!j±^pH =pKa-log(^) =pKa +logjiJljEquation 4.17 can be used with either concentration units or mole quantities forHA and A". This means that Equation 4.17 can be rewritten as Equation 4.18.

.. . moles Conjugate Base ,,10,pH = pKa + log *-2 (4.18)moles Conjugate Acid

According to Equation 4.17, the addition of water to a conjugate pair mixture hasno effect on the pH of the solution. The compounds are diluted; and thus lessconcentrated, but the pH remains the same because the [A"] : [HAJ ratio remainsthe same.

Example 4.26Which of the following solutions has the GREATESTpH?A. 10mL 0.10M NH3(/u/) with 15mL 0.10M NH4+(aq)B. 15mL 0.10M NH3(aq) with 10mL 0.10M NH4+(aq)C. 10 mL 0.10 M HCQ2Nafaj) with 15 mL 0.10 M HC02H(aq)D. 15 mL 0.10 M HCC^Nafog) with 10 mL 0.10 M HCC^Hfaq)

SolutionThe pH of a conjugatemixture can be determined using Equation 4.18. To makethe pH high, the pKa must be high, and the mixture must be rich in conjugatebase. The pKa of ammonium is greater than the pKa of a carboxylic acid (in thiscase, formic acid), so choices C and D are eliminated. Choice A has more acidthan conjugate base, so the pH is less than the pKa for the acid. Choice B hasmore base than conjugate acid, so the pH is greater than the pKa for the acid.This means that choice B has the greatest pH.


Acidsand

Bases

Passages13 Passages

100 Questions

Suggested Acids and Bases Passage Schedule:I: After reading this section and attending lecture: Passages I, III, V, VII, & IX

Grade passages immediately after completion and log your mistakes.Following Task I: Passages II, IV, VI, & VIII (27 questions in 35 minutes)Time yourself accurately, grade your answers, and review mistakes.Review: Passages X - XIII & Questions 92 - 100Focus on reviewing the concepts. Do not worry about timing

II

III

mmm^mSpecializing in MCAT Preparation

Acids and Bases Study Passages

I. Acid and Base Definitions

II. Dissociation and Colligative Properties

III. Oxyacids

IV. Acidity of Thiols and Alcohols

V. Organic Acids

VI. Electron-Withdrawing Effect and Acidity

VII. Weak Acid pH Equation

VIII. Aspirin and Antacids

IX. Household Acids and Bases

X. Stomach Acid and pH

XI. Tooth Decay and pH

XII. Acid Rain and Scrubbers

XIII. Amino Acids pKa Values

Questions not Based on a Descriptive Passage

Acids and Bases Scoring Scale


84 - 100 13- 15

66-83 10- 12

47 -65 7 -9

34 -46 4-6

1 -33 1 -3

(1 -7)

(8- 13)

(14-21)

(22 - 28)

(29 - 34)

(42 - 48)

(35-41)

(49 - 55)

(56 - 62)

(63 - 70)

(71 -76)

(77 - 84)

(85-91)

(92 - 100)


A base can be defined in three different ways, dependingon its solvent. The definitions are:

1. The Arrhenius definition states that a base yieldsOH"(aq)when added to water.

2. The Br0nsted-Lowry definition states that a base is aproton acceptor.

3. The Lewis definition states that a base is an electronpair donor.

The same base can fit the description of all threedefinitions. The perfect example of a base that fits the threedefinitions is ammonia (NH3), which can donate its lone pairof electrons to accept a proton from water, to yield anaqueoushydroxide anion (OH"(aq)).

Although a base may simultaneously fit the threedefinitions, each definition has its own unique application.For the calculation of the pH of water-based solutions, theArrhenius definition is the most applicable. The Br0nsted-Lowry definition is applicable in a more general sense,because it accounts for acid-base chemistry that takes place ina protic solvent other than water. The Lewis definition ismost commonly applied to organic chemistry where basesdonate their lone pairs to empty p-orbitals. A Lewis base canreact in any solvent, including aprotic ones.

A Lewis base also may be referred to as a nucleophile.The electrons of the nucleophile are donated to the partiallypositive site on an electrophile (which also may be a protonin addition to a partially positive carbon). The partial or fullcationic charge is often the result of an excess of protons onthe electrophile. The term "nucleophile" is derived from thestrong affinity of the substances for positive charge (thecharge of protonsin an atomic nucleus). All threedefinitionscan account for the nucleophilic nature of a base.

1. If the [OH"l of a pH = 8.0 solution is tripled, the newpH will be:

A. 5.0.B. 7.5.C. 8.5.D. 11.0.

2. Which of the following bases would have theLARGEST pKb value?

A. A base that undergoes 20% hydrolysis in waterB. A base that undergoes 15% hydrolysis in waterC. A base that undergoes 10% hydrolysis in waterD. A base that undergoes 5% hydrolysis in water


3.

4.

6.

Which of the following is an Arrhenius base?

A. HCIOB. HB1O2C. Li2C03D. HN02

Which of the following would be the BEST choice toneutralize 25 mL 0.10 M HCIO3?

A. 25mL0.10MNaOH(aq)B. 250 mL 0.010 M NH3(aq)C. 25mL0.10MHCl(aq)D. 250 mL 0.010 M HCQ2H(aq)

As the conjugate acid for a base gets stronger, the base:A. exhibits a decreasing pKb value.B. can react with weaker acids.C. acquires a higher pOH value in water.D. requiresmoremolesof acid to be neutralized.

Which of the following definitions does NOT describean acid?

A. A electron-pair acceptorB. A proton donorC. A compound that produceshydronium ion in waterD. A nucleophilic molecule

As the strength of an acid increases, which of thefollowing does NOT happen?A. The acid becomes more electrolytic.B. The acid dissociates more.C. The acid has a lower pKa.D. The acid has a lower Ka.



The strength of an acid in water is defined by its abilityto dissociate into hydronium and conjugate base. Strongacids generate moreions in solution. As the number of ionicimpurities in the water increases, so does the solution'selectrical conductivity. Current does not readilypass throughdistilled water, so the degree of an acid's dissociation can beestimated by a solution'sability to conductelectricity.

To determine the correlation between acid strength andelectricalconductivity, a researcherplaces the two ends of anopen circuit into a container of water so that the solutionbecomes part of a closed circuit. Two volumetric tubes arepoised above the solution, allowing for an exactquantity ofsolution to be added to the circuit solution. Figure 1 showsthe apparatus used in the experiment.

Tube A TubeB

Jl -i-o

Aqueous solution

12V

Figure 1

In different trials, 25 mL aliquots of acid are added, andthe current is measured at different points in the wire. Anaverage current was recorded. Any deviation in current atdifferent sites can be attributed to errors in measurement, asthe circuit is a single loop, so current should be uniformthroughout. Table 1 lists the average current in the wire andthe contents of each tube in six separate trials.

Trial Tube A Tube B Current

1 0.10 M HCl Nothing 5.94 amps2 0.10MHC1O Nothing 0.42 amps3 0.10 M HCl 0.10 M KOH 6.03 amps4 0.10MHC1O 0.10 M KOH 5.98 amps5 0.10 M KOH Nothing 5.89 amps6 O.lOMKOAc Nothing 5.92 amps

Table 1

The voltage of the battery is constant for the duration ofthe experiment.

8. What current would be expected, if 25.0 mL of 0.10 MHF were added to the aqueous solution?

A. 0.02 ampsB. 0.48 ampsC. 5.42 ampsD. 6.17 amps


9. How can it be explained that there is no differencebetween the current readings in Trial 5 and Trial 6?A. Both KOH and KOAc are strong bases.B. Both KOH and KOAc are weak bases.C. Neither compound dissociates into ions in water.D. The number of ions in solution does not depend on

the base strength.

10. In Trial 2, the water is acting as:A. aBr0nsted-Lowrybase.B. an Arrhenius acid.C. a Lewis acid.D. an amphoteric species.

11. Which of the following is NOT true?A. The solution in Trial 1 has a higher boiling point

than the solution in Trial 2.B. The solution in Trial 1 has a higher freezing point

than the solution in Trial 2.C. The solution in Trial 1 has a higher osmotic

pressure than the solution in Trial 2.D. The solution in Trial 1 is more electrolytic than the

solution in Trial 2.

12. All of the following would show a current of roughly 6amps when added to the solution EXCEPT:A. O.IOMHNO3.B. O.IOMH2SO4.C. 0.10 M NaOH.D. 0.10 MNH3.

13. What might be the reason for the similarities betweenthe results in Trial 3 and Trial 4?

A. A complete reaction transpiresonly in Trial 3.B. A complete reaction transpires only in Trial 4.C. After reaction, whether it is a strong acid with a

strong base or a weak acid with a strong base, thesame concentration of spectator ions remains.

D. Only anions conduct electricity, and both solutionshave the same amount of anions after reaction.


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A B C are strongD is a weak base

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KOAc is a weak base.they’re both salts so they both dissociate. So D is the correct answer


The term oxyacid is coined from the oxygen, the keycomponent of this type of acid. Oxyacids contain a non-metal atom (highly electronegative) bonded to oxygens, andthe acidic hydrogen is bonded to an oxygen. A typicalexample is nitric acid (HNO3), where nitrogen is theelectronegative atom that is bonded to the oxygens, one ofwhich is also bonded to a hydrogen. Some oxyacids ofinterest are those containing halides, sulfur, phosphorus, andnitrogen.

The primary rule for predicting the strength of an oxyacidis that the greater the number of oxygens attached to thecentral atom, the more acidic the compound. For example,HCIO4 is more acidic than HCIO3. The secondary rule forpredicting the strength of oxyacids is that it increases as theelectronegativity of the central atom increases when thenumber of oxygens is equal between two oxyacids. Table 1shows the trend for the halide acids:

Acid Formula Ka value pKaPerchloric HCIO4 1.4 x 109 -8.8

Perbromic HB1O4 2.6 x 105 -4.6

Periodic HIO4 1.5 x 102 -1.8

Chloric HCIO3 8.9 x lO'1 0.1

Chlorous HCIO2 1.3 x lO'2 1.9

Hypochlorous HOC1 5.4 x 10"8 7.3

Hypobromous HOBr 2.3 x lO"9 8.6

Hypoiodous HOI 1.7 x lO"11 10.8

Table 1

Table 1 shows that the more dominant of the two causesof strengthened acidity is the increase in oxygens attached tothe central atom of the acid. This exceeds the effect ofchanging the halide central atom. A rough approximation isthat each additional oxygen will lower the pKa of the acid byapproximately 5. Fluorine cannot expand its octet toaccommodate multiple oxygens, so it is not among theelements that form halogen-based oxyacids.

14. The acidity of halide-containing oxyacids increasesdirectly with which of the following?A. The increasingelectronegativityof the halideB. The increasingbond length of the H-O bondC. The increasing size of the halideD. The increasing bond angle of H-O-X

15. Which sequenceaccuratelydescribes the relativestrengthof oxyacids?A. HIO4 > HCIO4 > HCIO3 > HBr03B. HCIO4 > HBr03 > HIO4 > HIO3C. HIO4 > HCIO4 > HBr03 > HCIO3D. HCIO4 > HCIO3 > HBrQ3 > HI02


16. If HCIO has a pKa = 7.26, then the pH of a 0.10 MHCIO solution is which of the following?

A. 3.63B. 4.13C. 7.26D. 8.26

17. If pKai for a diprotic acid is 7.8, then the BEST choicefor pKa2 for its conjugate base, formed when the acidloses a proton, is which of the following?A. 3.9B. 6.2C. 7.8D. 12.9

18. If the dissociation of an acid is exothermic, then(assuming that entropy is negligible):A. Kashould increase as the temperature increases.B. Kashouldremainconstantas temperature increases.C. Kashoulddecrease as the temperature increases.D. Ka never changes with varying temperature.

19. Which of the following values MOST accuratelydescribes the pKa for HB1O2, based on Table 1?

A. 0.63B. 1.44C. 1.92D. 2.85

20. Which of the following acids would show theGREATEST dissociation in water?

A. H3PO4B. HN02C. HIO2D. HCIO4

21. What is the pH for a 125-mL sample of 0.10 MHCIO4, given thatHCIO4 is a strong acid?A. -1

B. 0.1

C. 1

D. 7


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The acidityof thiols (RSH) is observedto be greaterthanthe acidity of alcohols (ROH). This is determined bycomparing the relative pKa values for corresponding alkylgroups attached to thiols and alcohols. Table 1 lists pKavalues for various alcohols and thiols with comparable alkylsubstituents. From these values, it is possible to determinethe relative acidities of the two classes of compounds.

Alcohol pKa Thiol pKaH3COH 16.1 H3CSH 10.3

H3CCH2OH 16.3 H3CCH2SH 10.6

(H3Q2CHOH 17.0 (H3Q2CHSH 11.0

(H3Q3COH 17.8 (H3O3CSH 11.4

Table 1

The difference in acidity between alcohols and thiols isattributed to the polarizability of their respective conjugatebases. The larger the anion (more correctly, the atomcarrying the negative charge in the conjugate base), the morediffuse the electrons will be, and thus the more polarizablethe electron cloud of the anion. The result is that the electroncloud is spread over more area, increasing the stability of theanion (conjugate base).

As the stability of the conjugate base increases, thebasicityof the conjugate base decreases. The final correlationis that as the basicity of the conjugate base decreases, theacidity of the conjugate acid increases. This leads ultimatelyto the conclusion that as the size of the atom to which theacidic hydrogen is attached increases, the acidity of thecompoundincreases. This can also be correlated to the bondlength of the bond between the acidic hydrogen and the atom.The longer the bond, the weaker that bond will be.

2 2. Which of the following compounds would be MOSTacidic?

A. H3CSCH3B. H3COCH3C. H3CNHCH3D. H3CCH2CH3

23. From the table of pKa values, what conclusion can bedrawn about the role of alkyl groups?A. Alkyl groups are electron-withdrawing and decrease

acidity.B. Alkyl groups are electron-donating and decrease

acidity.C. Alkyl groups are electron-withdrawing and increase

acidity.D. Alkyl groups are electron-donating and increase

acidity.


24. Which of the following halogen-containing acids is theSTRONGEST acid?

A. HFB. HClC. HBrD. HI

25. HCl is considerably more acidic than H2S, because therelative acidities of the compounds formed by:A. atoms in the same row of the periodic table depend

on the electronegativity of the non-hydrogen atomin the compound.

B. atoms in the same column of the periodic tabledependon the electronegativityof the non-hydrogenatom in the compound.

C. atoms in the same row of the periodic table dependon the size of the non-hydrogen atom in thecompound.

D. atoms in the same column of the periodic tabledependon the size of the non-hydrogen atom in thecompound.

26. Howdoes ethanol comparewith its corresponding ethylthiol?

A. Ethanol has a greater value of Ka.B. Ethanol dissociates more completely in water.C. H3CCH2O- is a stronger base than H3CCH2S-.D. Ethanol yields a greater [H30+l.

27. When an atom in question is not directly attached to theacidic hydrogen, then the acidity of that compoundcorrelates to the electronegativity of that atom, not itssize. This is known as the inductive effect. Accordingto the inductive effect, which of these acids is theSTRONGEST?

A. H3CCH2C02HB. H3CCCI2CO2HC. H3CCSCO2HD. H3CCI2C02H

28. Whichsequenceaccuratelydescribes the relativestrengthof these acids?

A. H3CCH2SH > (H3Q3CSH > H3CCO2HB. (H3Q3CSH > H3CCH2SH > H3CCO2HC. H3CCO2H> H3CCH2SH > (H3O3CSHD. H3CCO2H > (H3O3CSH > H3CCH2SH



Common organic acids include carboxylic acids(RCO2H) and phenols (C6H5OH). Electron-withdrawinggroups along the backbone of an acid increase its acidity, andthus lower its pKa. Chloro, fluoro, and nitro groups areamong the common electron-withdrawing groups. Table 1lists some organic acids with their respective pKa values.They are all monoprotic organic acids.

Structure Formula pKa

0

II

CI3C OH

CI3CCO2H 0.64

/ V °°HJK>H p-02NC6H4C02H 3.40

Q^OH C6H5C02H 4.21

O

II

H3C OH

H3CCO2H 4.74

02N—4^-OH p-02NC6H4OH 7.18

O~0H C6H5OH 10.01

Table 1

When a weak acid dissolves into water, it partiallydissociatesaccording to its Ka. Equation I, can be applied tocalculatethe pH of an aqueous solution of a weak acid.

PH =ipKa-llog[HAl2 2

Equation 1

Equation 1is derived from the equilibrium expression forthe dissociation of a weak acid in water. It applies only if theacid concentration exceeds Ka by 100 fold and if the pKa ofthe weak acid is between 2 and 12.

log 2 =0.3 log 3 = 0.48

29. What is the pH for an acid solution with a concentrationof0.50MandaKa = 8x 10"8?A . 0 < pH < 3.0B. 3.0<pH<3.5C. 3.5 < pH < 7.0D. pH>7.0


30. What is the mass percent of carbon in phenol(C6H5OH)?

A. 66.7%B. 72.0%C. 76.6%D. 82.4%

31. Which of the following acids would yield the lowest pHvalue once completely neutralized by strong base?A. Acetic acid (H3CCO2H)B. Trichloroacetic acid (CI3CCO2H)C. p-Nitro benzoic acid (O2NC6H4CO2H)D. Benzoic acid (C6H5CO2H)

3 2. When comparing equal molar aqueous solutions of acidsfrom Table 1, what is TRUE?

A. 1.0 M acetic acid has a higher boiling point than1.0 M p-nitro benzoic acid.

B. 1.0 M trichloroacetic acid has a higher boilingpoint than 1.0 M p-nitrophenol (P-O2NC6H4OH).

C. 1.0 M trichloroacetic acid has a higher freezingpoint than 1.0 M benzoic acid.

D. 1.0 M acetic acid has a higher freezing point than1.0 M phenol.

33. Which of the following acids has the STRONGESTconjugate base?A. Phenol (C6H5OH)D. p-NitrophenoI (P-O2NC6H4OH)C. Acetic acid (H3CCO2H)D. Trichloroacetic acid (CI3CCO2H)

34. After 0.10 moles of an unknown weak acid HA aredissolved into 100 mL of H2O, the solution's pH is3.7. What is the concentration of the conjugate base(A")?

A. [A-]>2x 10"4MB. [A'l = 2x 10"4MC. [A1<2x 10'4MD. The Ka of the weak acid must be known to

determine the value of [A"].


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Passage VI (Questions 35-41)

The strength of an organic acid varies directly with thestrength of any electron-withdrawing groups attached to itsorganic backbone. Forinstance, theacidity of acetic acid canbe increased by adding an electron-withdrawing group to themethyl carbon. Theacidity of acetic acid canbedecreased byadding an electron donating group to themethyl carbon. Theeffect on the strengthof the acid varieswith the location andelectronegativity of the electron-withdrawing group and ismost substantial when the functional group withdraws byway of resonance. The following set of phenols demonstratesthe effect of these substituents on acidity:

Organic Acid £&

02N—4V-OH 7.2

3>-Oh~ohO

8.4

O~0H 10.0

H3C—4^—OH 10.4

H3CO—£ ^—°H 11.2

Figure 1

Increased acidity can be attributed to the withdrawal ofelectron density (through either the inductive effect orresonance) from the O-H bond of phenol. The loss ofelectron density weakens the bond, causing it to break in aheterolytic fashion more easily. This makes the compoundmore acidic. Resonance withdrawal is most pronounced whenthe substituent is in the ortho or para position, so thestrongest acids are those with electron-withdrawing groupsattached to the phenyl ring at the ortho or para positions.Resonance is substantially weaker from the meta position.The inductive effect is weaker than resonance from anyposition.

3 5. The LOWEST pH value would be associated with:A. 0.10 M O2NC6H4OH (p-nitrophenol).B. 0.50 M O2NC6H4OH (p-nitrophenol).C. 0.10 M C6H5OH (phenol).D. 0.50 M C6H5OH (phenol).


36. A 100mL 1.0M phenol solution has a pH of:A. 10.0.B. 7.0.C. 5.0.D. 1.0.

37. According to the pKadata in Figure 1,whichcompoundis the STRONGEST electron-withdrawing group?A. 02N—B. H3CO—C. H3C—D. H3CCO—

38. The conjugate baseof p-nitrophenol has a pKbof:A. 10.4.B. 7.2.C. 6.8.D. 3.6.

39. Howmany grams of methoxy phenol must be added to100mLof pure water to reach a pH of 6.1?A. 12.4 gramsB. 2.48 gramsC. 1.24 gramsD. 0.62 grams

4 0. What is the Ka associated with p-methyl phenol?

A. 4.0 x 10"10B. 2.5 x 10'10C. 4.0 x lO"1 ]D. 2.5 x 10"11

41. What is the pH of a solution made by mixing 20 mL0.15 M phenol with 10 mL water?A. 10.0B. 7.0C. 5.8D. 5.5


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A strong acid fully dissociates when dissolved in water,and the pH of the resulting solution can be found by takingthe negative log of the strong acid concentration:

pH = -log [strong acid]

Equation 1

For a weak acid, the determination of the pH is not assimple as it is with a strong acid, because weak acids do notdissociate completely when added to water. To find the pH ofa weak acid in water, the [H30+] must first be determinedfrom the weak acid equilibrium:

HA(aq) - ^ H+(aq) + A-(aq)

Initial: [HAJinitiai lO'7 0Dissociation: - x + X + X

Equilibrium: [HAlinitiai - x x + 10"7 x

10~7 is assumed to be negligible compared to x, andx is assumed to be negligible compared to [HA]jnjtia|

The [H30+] can be determined from the Ka value. Thefollowing is a derivation of Equation 2, used to determine thepH foran aqueous solution of a weakacid.

K=[H30+][A-1 =[H3Q+12 ... Ka x[HAJ =[H30+]2[HA] [HA]

[H30+] = VKax[HA]pH = -log [H30+] = -log VKax[HA]-log VKax[HA] =-logVKa" - logV[HA]=- i-logKa - i-log [HA] =ipKa - llog [HA]

2 2 2 2

.-. pH =ipKa - i-log [HA]2 2

Equation 2

Equation 2 works only when the value of Ka is less than[HA], and when pKa is between 2 and 12.

42. For which of the following acid solutions does Equation2 NOT work?

A. HCIOB. H2C03C. HID. H5C6C02H

43. What assumption is inherent in the derivation ofEquation 2?

A. [HA] < [H30+]B. [H30+] = [A1C. [HA] = [A"ID. [HA]<[A"]


44,

45

46.

47.

48.

If the pH of a 0.10 M aqueous weak acid solution is2.70, what is the concentration of the dissociatedconjugate base in solution?

A. 0.I0M

B. 2.0 x 10'2MC. 2.0 x 10"3MD. 2.0 x 10"4M

Which of the following equations can be used todetermine the pOH of an aqueous weak base solution?

A. pOH = -log [OH']B. POH =IpKa-llog[A-]

2C. pOH =ipKb

2-Llog IA-]2D. pOH =14 -(i-pKa -i-log [HA])

If a weak acid is titrated with enough strong base so that[A"] > [HA], then for the resulting compound:

A. [H30+]<Ka.B. pH < pKa.C. [H30+]>[A-].D. [HA]<[H30+].

Which acid yields the GREATEST conjugate baseconcentration when 0.010 moles HA are added toenough water to make the final volume 100mL?A. HFB. HNO3C. HCND. H3CC02H

A series of 0.10 M weak acid solutions, each containinga different acid are compared. What is true for the acidin the series that has the GREATEST Ka value?

A. Its aqueous solution has the lowest conjugate baseconcentration.

B. Its aqueous solution has the lowest pOH.C . Its aqueous solution has the highest [H30+].D. Its aqueous solution has the highest ratio of HA to

A".



When the pH of the stomach drops to a value less than2.0, the excess acidity may be reduced by consuming anantacid. The role of the antacid is to neutralize the HCl inthe gastric fluid. Somecommon bases found in antacids areCaC03, MgC03, Mg(OH)2, and NaHC03- Calciumcarbonate, magnesium carbonate, and magnesium hydroxideneutralize acid in a two-to-one ratio. Sodium bicarbonateneutralizes only in a one-to-one ratio.

The gas buildup experiencedafter consumingan antacidis due to carbon dioxide gas, which forms in the stomachwhen carbonic acid (formed upon the full protonation ofcarbonate) decomposes into water and carbon dioxide. Theprincipal antacid ingredients of somecommon, commerciallyavailable products are listed in Table 1.Alka-Seltzer Aspirin, NaHC03, citric acidBufferin Aspirin, MgC03, aluminum glycinateMilk ofMagnesia Mg(OH)2Rolaids Dihydroxyaluminum sodium carbonateTurns Calcium carbonate

Table 1

Aspirin (acetylsalicylic acid) can irritate the stomachlining. In an acidic environment, aspirin does notdeprotonate, so the molecule remains neutral and thus lipidsoluble. As a result, aspirin is able to work its way into themembrane of the stomach lining and then into hydrophilicpores within the membrane. It deprotonates inside thehydrophilic pore, causing ion levels to build up in thepocket, until osmotic pressure forces water into the interiorregion of the membrane, where parietal cells swell andrupture. The net effect is to eat away the stomach lining.Figure 1 shows the structure of aspirin and its ionizationequilibrium.

O^CH3 O^CH3

O O

Figure 1

49. At what pH is aspirin MOST soluble in water?

A. 1.5B. 3.4C. 7.4D. 9.2

+ H +

5 0. How many acidic sites are there in acetylsalicylic acid?

A. 1

B. 2C. 3D. 4


51. Which antacid gives the MOST neutralizing strengthper gram?A. Magnesiumcarbonate (84.3 grams per mole)B. Calcium carbonate (100 grams per mole)C. Magnesium hydroxide (58.3 grams per mole)D. Sodium bicarbonate (83.9 grams per mole)

5 2. Which of the following compounds would cause theMOST damage to stomach lining?

OCH«

53. How many acidic protons can be neutralized permolecule of dihyroxyaluminum sodium carbonate(Al(OH)2NaC03)?

A. 1

B. 2C. 3D. 4

54. The Keq for the ionization reaction of acetylsalicylicacid in water is:

A. much greater than 1.00.B. barely greater than 1.00.C. barely less than 1.00.D. much less than 1.00.

55. Which of the following mixtures results in a buffersolution?

A. 1 equiv. acetylsalicylic acid + 1 equiv. HClB. 2 equiv. acetylsalicylic acid + 1 equiv. HClC. 1equiv. acetylsalicylate + 1 equiv. HClD. 2 equiv. acetylsalicylate + f equiv. HCl



Many common products used every day have acidic andbasic properties. An example is the antacid taken by manyheartburn sufferers to neutralize excess stomach acid (HCl).Antacids often contain hydroxide anion, carbonate anion, orboth in their salt forms. The contents of a normallyfunctioning stomach can reach a pH as low as 1.7 (highlyacidic) and in extreme cases can reach a pH as low as 1.0.Because the pH scale is a log scale, a decrease of 0.7 pH unitsrepresents a hydronium ion (H30+) concentration that is500% greater. Equation 1 is used to determine the pH of asolution.

pH = -log[H30+]

Equation 1

Table 1 lists some common household acids.

Acids Formula Product

Acetic acid H3CC02H VinegarAscorbic acid C6H806 Vitamin C

Hypochlorousacid HCIO Bleach

Acetyl salicylate H3C2OC6H4CO2H AspirinSulfuric acid H2SO4 Battery acid

Table 1

Table 2 lists some common household bases.

Bases Formula Product

Ammonia NH3 Windex

Sodium Bicarbonate NaHC03 Baking SodaSodium Hydroxide NaOH Drano

Sodium Lauryl sulfate H3C(CH2)i6S04Na Shampoo

Bases when added to neutral water raise the pH of thesolution to a value greater than 7.0. Acids when added toneutral water lower the pH of the solution to a value lessthan 7.0. A pH of 7.0 is considered to be neutral, becausepure (distilled) water has an [H30+] = 1.0 x 10"7.

Acid in solution may be represented as either H30+ orH+, depending on the solvent. Both representations areequivalentways of describing an acid.

56. A hydroxide anion is formed when water loses a proton(H+) to another water (known as autoionization). If theconcentration of dissociated hydroxide anion (OH") indistilled water is 1.0 x 10-6 M, the pH of the solutionmust be:

A. 7.0, because [H30+] = [OH'].B. 7.0, because water always has a pH of 7.0.C. 6.0, because [H30+] = [OH"].D. 8.0, because the solution is rich in base (OH").


57. What is the hydronium ion concentration ([H30+]) ofan aqueous solution with a pH value of 1.7?A. 1.7MH30+B. 2.0 x 10-2MH3O+C. 1.0x 10-7MH3O+D. 5.0 x 10-,3MH3O+

58. If 10 mL of an aqueous solution of a strong acid withpH = 2.0 were mixed with 100 mL of pure water, thenthe final pH value would be:A. less than 2.0.B. between 2.0 and 3.0.C. exactly 3.0.D. greater than 3.0.

59. Howmany grams of CaC03 are needed to neutralize 50mL of stomach acid at pH = 2.0 completely, if thefollowing equation represents the neutralizationreaction?

CaC03(aq) +2H+(aq) —> Ca2+(aq) +C02(g) +H20(1)A. 25 mgB. 50 mgC. 100 mgD. 1.0 grams

6 0. Which of the following household products wouldNOTundergo an acid-base reaction withWindex?A. VinegarB. AspirinC. DranoD. Bleach

61. Which of the following when added to an aqueoussolution at pH = 6.0 would NOT raise the pH of thesolution?

A. Distilled waterB. ShampooC. AntacidD. Vitamin C

62. How many milliliters of 0.20M H30+ are required toneutralize 1.68 gramsof baking sodacompletely?NaHC03(aq) + H+(aq) —>Na+(aq) + C02(g) + H20(1)A. 10 mLB. 25 mL

C. 50 mLD. 100 mL



The average human being produces between two andthree liters of gastric fluid in thenormal day. Gastric fluid ishighly acidic (due to the presence ofhydrochloric acid). It issecreted by the mucous membrane of the stomach lining toaid in the digestion of food. The average pHof this fluid isaround 1.5. The acid concentration necessary for this pH is0.030M, if the acid is a strongacid. The acid in gastric fluidis strong enough and concentrated enough to dissolve(oxidize) metalswith a positiveoxidation potential.

The liningof the stomach includes parietalcells that aretightly fused to form junctions in the stomach wall. Thesecells have a cell membrane that is permeable to neutralmolecules (such as water), but not to ions (such as Na+ andCI"). Hydronium ion, responsible for the acidity of thestomach, is a byproduct of metabolism. Carbon dioxide, abyproduct of metabolism, is a non-metal oxide that convertsto an oxyacid when combined with water. The hydration ofcarbon dioxide (CO2) to carbonic acid (H2CO3) is shown inReaction 1. Carbonic acid is a weak acid that partiallydissociates into bicarbonate anion and a proton, as shown inReaction 2.

C02(g) + H20(1) ^ H2C03(aq)

Reaction 1

H2C03(aq) , H+(aq) + HC03"(aq)

Reaction 2

The proton, along with a chloride anion, is carried acrossthe membrane into the stomach via active transport.Enzymes assist the migration of both the proton and chlorideanion from the blood plasma into the interior of the stomach.These ions remain in the stomach until removed, because ofthe impermeability of the cell membrane to ions.

Eating stimulates the production of the enzymeresponsible for active transport and thus the release of acidinto the stomach to hydrolyze food molecules. Some protonsare absorbed by the mucous lining of the stomach wall,resulting in small localized hemorrhages. About 30 millioncells are destroyed per hour in normal stomach activity. As aresult, the stomach's entire lining is regenerated roughly onceevery 72 hours. Excess acid production increases thishemorrhaging and, in the worst case scenario, an ulcerdevelops.

6 3. What can be concluded about the following reaction?H2C03(aq) + NaCl(aq) -^-^ NaHC03(aq) + HCI(aq)

A. AG > 0; requires active transport in vivo.B. AG < 0; requires active transport in vivo.C. AG > 0; does not require active transport in vivo.D. AG < 0; does not require active transport in vivo.


6 4. Which solution has the LOWEST pH?

A. Carbonated waterB. Distilled waterC. SaltwaterD. Lime water (CaO(aq))

65. Which of the following metals can be oxidized bygastric fluid?A. CopperB. GoldC. SilverD. Zinc

66. Which of the following reactions is NOT catalyzed byacid?

A. Hydrolysis of disaccharidesB. Hydrolysis of polypeptidesC. Hydrolysis of estersD. Hydrolysis of alkanes

67. Which will NOT reduce the hydronium ion (H30+)concentration in the stomach?

A. The consumption of waterB. The consumptionof baking soda (NaHC03)C. The consumption of aluminum metalD. The consumption of solid food

68. What is the pH of 0.030M HCl(aq) solution?A. 0.7

B. 1.5C. 3.0D. 7.0

69. Which of the following beverages when consumeddoesNOT promote the decay of the stomach lining?A. Orangejuice (citric acid)B. Lemonade (citric acid)C. Milk (lactose)D. Coca-Cola (phosphoric acid)

7 0. Which of the following has the LOWEST pH?A. 0.03MHCl(aq)B. 0.01 MHCl(aq)C. 0.03 M H2C03(aq)D. 0.01 M H2C03(aq)



As a general rule, the solubility for basic salts (conjugatebases of weak acids) increases as the pH of the solutiondecreases. This is to say that basic salts are more soluble inacidic solution than in neutral solution. The calcium saltssuch as calcium hydroxide (found in cement), calciumcarbonate (marble), and hydroxyapatite (tooth enamel) areperfect examples. Hydroxyapatite is CasfPO^OH but canalso be written as 3Ca3(P04)2 • Ca(OH)2- Cavities resultwhen the hydroxyapatite dissolves away. When smallcrevices are formed in teeth, acid-producing bacteria build upand further dissolve away the worn region in the enamel. Theacid that dissolves the enamel is formed when aldoses(aldehyde sugars, such as glucose) are oxidized to theirrespective aldonicacid compounds. To reduce the dissolvingof the enamel, toothpaste and mouthwash containing fluorideare recommended. The fluoride will substitute for hydroxidein the salt to form the less soluble fluorapatite Ca5(P04)3F(also known as 3Ca3(P04h • CaF2). This lengthens thelifetime of tooth enamel.

The solubility product (Ksp) for calcium fluoride (CaF2)is 3.4 x 10" M3, for pure calcium phosphate (Ca3(P04)2)is 1.4x 10"26 M5, and for calciumhydroxide (Ca(OH)2) freeof calciumoxide is 2.3 x 10"8 M3. The lower solubility ofcalcium fluoride compared to calcium hydroxide is mimickedin the reduced solubility of fluorapatite compared tohydroxyapatite.

The following chart lists the molar solubility of calciumhydroxide at varying pH along with the gram solubility per100mL solution at varying pH. The temperature was heldconstant at 25°C for all values:

pH Molar solubility Gram solubility

1 2.3 x 1018 1.7 x 10192 2.3 x 1016 1.7 x 10173 2.3 x 1014 1.7 x 10157 2.3 x 106 1.7 x 10711 2.3 x 10"2 1.7 x lO"112 2.3 x 10"4 1.7 x 10"3

Table 1

Table 1 shows that the effect of pH on the solubility ofbasic salts is drastic. This is because pH is measured on alogscale, while solubilities aremeasured on a linear scale.

71. As sodium fluoride (NaF) is added to an aqueoussolution with undissolved hydroxyapatite in it, whathappens to the solution?A. The pH of the solution increases.B. The pH of the solution decreases.C. The pH of the solution remains constant at a value

greater than 7.0.D. The pH of the solution remains constant at a value

less than 7.0.


72. Calcium hydroxide is BEST described as which of thefollowing?

A. A strong acidB. A strong baseC. A weak acidD. A weak base

73. If the solubility of enamel in a 1.00 M solution of acidHA is greater than the solubility of enamel in a 1.00 Msolution of acid HB, which of the followingconclusions can be drawn concerning to the two acids?

A. HA when dissolved into 100 mL pure water has ahigher pH than HB.

B. The conjugate base of HA is a stronger base thanthe conjugate base of HB.

C. HA has a higher Ka value than HB.D. HA has a higher pKa value than HB.

74. Which of the following changes will ALWAYSincrease the amount of hydroxyapatite that dissolvesinto solution?

A. Adding a strong acid to the solutionB. Adding a strong base to the solutionC. Increasing the pH of the solutionD. Lowering the amount of water in the solution by

evaporation

75. An example of an aldonic acid is drawn below. Withwhich of the answer selections listed should the aldonicacid share a similar pKa value?

O OH

H—C—OH

IHO—C—H

IH—C—OH

IH—C—OH

ICH2OH

Acetic acid (H3CCO2H)Hydrochloric acid (HCl)Nitric acid (HNO3)Sodium carbonate (Na2C03)

A.

B.C.D.

76. In which of the following solutions does tooth enameldissolve the FASTEST?

A. 100 mL 0.10 M HClB. 100 mL 0.0010 M HClC. 100 mL 0.10MHF(Ka = 6.8x 10"4)D. 100 mL 0.0010 M HF(Ka = 6.8x 10"4)



Normal rainfall hasa pHbetween 5.6 and 6.0, dependingon the amount of carbon dioxide in the air. Excessiveamounts ofatmospheric carbon dioxide lower the pH of rain.However, even with high carbon dioxide concentration, thepH ofrain does not drop much below 5.4. While rain rich incarbon dioxide is more acidic than normal rain, it is notconsidered to be acid rain.

Acid rain is attributed to nonmetal oxides present in theair, like sulfur trioxide, and the nitrogen oxides. Thesenonmetal oxides combine with atmospheric moisture to formnonmetal hydroxides, known as oxyacids. Reaction 1 showsthe hydration of sulfur trioxide, converting it from anonmetal oxide into a nonmetal hydroxide.

S03(g) + H20(g) H2S04(g)

Reaction 1

Sulfur oxides are common by-products produced duringthe combustion of coal. A scrubber (a chamber at the base ofan exhaust stack filled with high pressure steam and calciumoxide) can help neutralize the non-metal oxides in theindustrial emissions of coal-burning factories. As theexhaust passes through the chamber, steam hydrates thenonmetal oxide and converts it from an airborne Lewis acidintoan aqueous Br0nsted-Lowry acid. Then as this nonmetalhydroxide solution flows across the calcium oxide in thescrubber, it is converted into calcium sulfate and water. Thisis shown in Reaction 2.

H2S04(aq) + CaO(s) Z, CnS04(s) + H20(i)

Reaction 2

If the acidic industrial waste generated by burning fossilfuel is not treated in this manner, it can produce rain with apH as low as 4.0. Acid rain is known to damage bodies ofwater by changing the bacteria population. Acid raindamages plants by changing soil pH. Changes in pH affectthe structure of plant enzymes needed for the uptake andtransport of nutrients. The enzymes become ineffective atlower pH because of their denatured structures.

77. What chemical reaction is indicated in Reaction 1?

A . The neutralization of sulfur trioxideB. The acidification of sulfur trioxideC. The conversion of a Br0nsted-Lowry acid into a

Lewis acid.D. The conversion of a Lewis acid into a Br0nsted-

Lowry acid.

78. A nonmetal oxide can be characterized as:

A. amphoteric.B. an Arrhenius base.C. a Br0nsted-Lowry acid.D . a Lewis acid.


79. Which of the following compounds would create asolution with the LOWEST pH when added to water?A. S02B. C02C. N205D. MgO

80. What is the pH of 0.10 M HOClfa^, which has a pKaof 7.46?

A. 1.00B. 4.23C. 6.46D. 7.46

81. Which of the following could be a component of acidrain?

A. Na20B. N2C. N02D. H2

82. All of the following qualifications affect the acidity ofan oxyacid EXCEPT:

A . the oxidation state of the central atom.B. the electronegativity of the central atom.C . the size of the central atom.D. the number of 7i-bonds to the central atom.

83. All of these statements about H2SO4, H2SO3, andtheir constituents are true EXCEPT:

A. SO32" has a lower pKbi than SO42".B. 1.0 M H2S04(aq) has a lower pH than 1.0 M

H2S03(a^.C. H2SO4 has a higher Ka| than H2S03.D. H2S03 is more electrolytic than H2SO4.

84. Which sequence correctly lists the four acids shownbelow according to their INCREASING pKa value?

A. pKai h2S04 < pKaHN03 < PKal H2S03 < PâHN02

B. pKaHN03 < PKal H2S04 < Pâ HN02 < PKalH2S03

C PKal H2S03 < pKaHN02 < PKal H2S04 < PKaHNO3

D- PKaHN02 < PKal H2S03 < Pâ HN03 < PKalH2S04


Passage XIII (Questions 85-91)

Amino acids can be classified as either diprotic ortriprotic acids, depending on their side chain. A genericamino acid is shown below in Figure 1:

O

H3N+c

4\H R

Figure 1

O"

The chemical structure in Figure 1 represents an aminoacid in a pH = 7 aqueous environment. At a pH less than thepKa of the carboxyl terminal (pKai), the carboxylate group isprotonated and exists in its neutral carboxylic acid form.Similarly, at a pH greater than the pKa of the amino terminal(pKa2 or pKa3), the ammonium group is deprotonated andexists in its neutral amine form.

If the R group (side chain) of an amino acid exhibitsacid-base properties, then the amino acid is triprotic. Theside chain may be deprotonated at pH = 7, depending on thenature of the functional group. The carboxyl terminal pKalies between 1.8 and 2.6, and the amino terminal pKa liesbetween 8.8 and 10.6 for all twenty common amino acids.The pKa for the protonated form and the pKb for thedeprotonated form add up to a value of 14at 25°C.

pKa (protonated form) + P^b (deprotonated form) = 14Equation 1

The pKa values for common side chains range from 3.9to 13.2. Table 1 lists the R group and pKa values for sevencommon triprotic amino acids.

Amino acid Side chain group pKaAspartic acid -CH2C02H 3.88

Glutamic acid -CH2CH2CO2H 4.32

Histidine -CH2C=CH-N+H=CH-NH- 6.05

Cysteine -CH2SH 8.36

Tyrosine -CH2C6H4OH 10.07

Lysine -CH2CH2CH2CH2NH3+ 10.80

Arginine -(CH2)3NHC(NH2)=NH2+ 13.21

Table 1

Physiological pH is considered to be 7.4, althoughgastric fluids and the fluid contained in lysosomes haveconsiderably lower pH. At physiological pH, a diproticamino acid exists predominantly in its zwitterion form.

8 5. What is the relationship between the pKa and pKb forlysine?A. pKai +pKbi = 14B. PKai+pKb2=14C. PKa2 + pKb2=14D. pKa2 + pKb3 = 14

Copyright © by The Berkeley Review®

86. At physiological pH, which of the following aminoacids exists predominantly as a cation?

A. ArginineB. Glutamic acid

C. GlycineD. Histidine

87. What explains the lower pKa for the side chainassociated with cysteine than the one associated withserine (R = -CH2OH)?

A. Oxygen is less electronegative than sulfur.B. Sulfur is less electronegative than oxygen.C . Oxygen has a greater atomic radius than sulfur.D. Sulfur has a greater atomic radius than oxygen.

8 8. The carbon chain associated with aspartic acid is shorterthan the one associated with glutamic acid, whichresults in a:

A. stronger inductive effect, making aspartic acid's sidechain more acidic.

B. weaker inductive effect, making aspartic acid's sidechain more acidic.

C. stronger inductive effect, making aspartic acid's sidechain less acidic.

D. weaker inductive effect, making aspartic acid's sidechain less acidic.

89. At pH = 7, what are the applicable values for histidine?A. pKai = 1.81; pKa2 = 6.05; pKb3 = 4.85B. pKa3 = 9.15; pKb2 = 7.95; pKbi = 4.85C. pKb2 = 7.95; pKa2 = 6.05; pKa3 = 9.15D. pKa3 = 9.15; pKb2 = 7.95; pKb3 = 12.19

90. What is the normality of 0.50 M glutamic acid inwater?

A . 0.50 N H3N+CH(CH2CH2C02H)C02H(aq)B. 1.00 N H3N+CH(CH2CH2C02H)C02H(aq)C . 1.50N H3N+CH(CH2CH2C02H)C02H(aq)D. 2.00 N H3N+CH(CH2CH2C02H)C02H(aq)

91. All of the following amino acids have a neutral sidechain at pH = 7 EXCEPT:A. cysteine.B. histidine.

C. lysine.D. tyrosine.

279 GO ON TO THE NEXT PAGE

Questions 92-100 are NOT basedon a descriptive passage.

92. Which of the following solutions has the LOWESTpH?A. 0.01 M HCIO (pKa = 7.5)B. 0.05 M HC02H (pKa = 3.6)C. 0.10 MHF(pKa = 3.3)D. 1.00 MHCN(pKa = 9.1)

93. Acid rain can possibly contain all of the followingcompounds EXCEPT:A. oxidized non-metals.B. hydrated non-metal oxides.C. electron pair acceptors.D. metal hydroxides.

94. Which of the following conjugate pair-mixtures has theLOWEST pH?

A. 10 mL 0.10 M HF(aq) with 15 mL 0.10 M KF(aq)B. 15 mL 0.10 M HF(aq) with 10 mL 0.10 M KF(aq)C. 10 mL 0.10 M NH3(aq) with 15 mL 0.10 M

NH4+(aq)D. 15 mL 0.10 M NH3(aq) with 10 mL 0.10 M

NH4+(aq)

95. Which of the following relationships applies tophosphoric acid?A. pKai +pKbi = 14B. PKa2 + pKb2 = 14C. pKa3 + pKb3=14D. PKal+pKa3 = 14

96. What is the normality of a solution made by mixing50.00 mL of 1.5 M H3PO4 with 50.00 mL of purewater?

A. 3.OONH3PO4B. 2.25NH3P04C. I.5ONH3PO4D. O.75NH3PO4


97. A 0.10 M solution of unknown material has a pH of4.62. It can BEST be described as a:

A. strong acid.B. weak acid.C. strong base.D. weak base.

98. The HIGHEST pH is observed in which of thefollowing solutions?A. 0.10MHNO2(aq)B. 0.10MHNO3(aq)C. 0.10MNaNO2(aq)D. 0.10MNaNO3(aq)

99. What is the pH of 0.050M H3CCH2C02H? (pKa forH3CCH2CO2H is 4.89)

A. 1.30B. 3.10C. 4.89D. 9.78

100. Which pH is INCORRECT for the correspondingsolution?

A. 0.10 M HBr(aq) has pH = 1.00B. 0.10 M HC02H(aq) has pH = 2.32C. 0.10 M NaOAc(aq) has pH = 5.13D. 0.10 M KOH(aq) has pH = 13.00

1. C 2. D 3. C 4. A 5. C

6. D 7. D 8. B 9. D 10. A

11. B 12. D 13. C 14. A 15. D

16. B 17. D 18. C 19. D 20. D

21. C 22. C 23. B 24. D 25. A

26. C 27. B 28. C 29. C 30. C

31. B 32. B 33. A 34. B 35. B

36. C 37. A 38. C 39. C 40. C

41. D 42. C 43. B 44. C 45. C

46. A 47. B 48. C 49. D 50. A51. C 52. C 53. D 54. D 55. D

56. C 57. B 58. D 59. A 60. C

61. D 62. D 63. A 64. A 65. D

66. D 67. D 68. B 69. C 70. A

71. A 72. B 73. C 74. A 75. A

76. A 77. D 78. D 79. C 80. B

81. C 82. C 83. D 84. A 85. C

86. A 87. D 88. A 89. D 90. C

91. C 92. C 93. D 94. B 95. B

96. B 97. B 98. C 99. B 100. C

THAT'S ENOUGH CHEM FOR NOW.

Acids and Bases Passage AnswersPassage I (Questions 1-7) Acid and Base Definitions

Choice C is correct. A quick look says that the pH should increase when the hydroxide ion concentrationincreases, which eliminates choices A and B. The pH should not increase by a full 3 units. An increase by 3units would mean 1000 times the base concentration. This implies choice C is the most intuitive choice. Thisproblem can be solved longhand as well. The question deals with [OH"] and pH, so some form of conversion mustoccur. First,pH must be converted to pOH, using the relationship pOH = 14- pH (14 - 8 = 6.0). Then, pOHmustbe converted to [OH"], using the relationship [OH'] =10-POH ([OH'] = 10"6). Tripling the amount ofOH" yieldsa concentration of 3 x 10"6 M. The new pOH is - log (3 x 10"6), which equals 6 - log3, which is approximately5.5. The new pH is equal to 14 - 5.5 = 8.5, so choice C is in fact correct.

Choice D is correct. The larger the value of pKb, the weaker the base. The weaker the base, the less itundergoes hydrolysis to form hydroxide anion when added to water. This makes choice D the correct choice.

Choice C is correct. An Arrhenius base is defined as a base that yields OH'(aq) upon addition to water. Of theanswer selections, only choice C is basic. Choice C must be correct.

Choice A is correct. HCIO3 is a weak acid that can be neutralized using a base of some sort. This eliminatesboth choice C and choice D. Choice B is a weak base, so it cannot fully react, eliminating choice B. Pick choiceA. For titration, 25 mL of strong base is ideal, because there are 25 mL of equimolar acid present. Although wehaven't reviewed it yet, titration is the quantitative addition of an equal mole quantity of a reactant.

Choice C is correct. As the conjugate acid gets stronger, the conjugate base gets weaker. The weaker theconjugate base, the higher its pKb value, so choice A is eliminated. The weaker the conjugate base, the strongerthe acid that is required to react with it, so choice B is eliminated. The weaker the conjugate base, the lowerthe hydroxide anion concentration, so the higher the pOH and the lower the pH. This makes choice C correct.The weaker base requires the same moles of acid as a stronger base to be neutralized, as long as they are in equalconcentration. The strength of the acid required for reaction is different for the two bases, but the moles ofhydroxide are the same from equimolar bases, regardless of their strengths.

Choice D is correct. The passage provided the three definitions for bases. The definitions for acids is theopposite of that for bases. This means that the Lewis definition of an acid is an electron-pair acceptor. ChoiceA is thus valid. The Bronsted-Lowry definition of an acid is a proton donor, making choice B valid. TheArrhenius definition defines an acid as a compound that produces hydronium ion upon addition to water. Thismakes choice C valid. Choice D must therefore be the best answer. A Lewis base, not a Lewis acid, is definedas a nucleophilic molecule.

Choice D is correct. By definition, as the strength of an acid increases, it dissociates to a greater extent whenmixed with water. This eliminates choice B. By dissociatingmore, the acid produces more ions, making thesolution more electrolytic. ChoiceA is eliminated. A stronger acid produces more hydronium ion, so Ka shouldincrease and pKa should decrease. This both eliminates choice C and makes choice D correct.

Passage II (Questions 8 -13) Dissociation and Colligative Properties

8. Choice B is correct. Hydrofluoric acid, HF, is a weak acid, so it partially dissociates in water. The currentwould be minimal. HCIO is also a weak acid, so an HF solution has conductivity similar to an HCIO solution.The best answer is choice B. Pick choice B, and be a happy camper.

9. Choice D is correct. Both KOH and KOAc are salts. As such, they will both completely dissociate into water,so the ion concentration is the same in both solutions. This eliminates choice C. KOH is a strong base, whileKOAc is a weak base, so choices A and B are also eliminated. The best answer is choice D.

10. Choice A is correct. HCIO is a weak acid, so water must be acting as a base by removing a proton fromhypochlorous acid. This eliminates choices B and C. Water can be an amphoteric species, meaning it can act aseither an acid or a base, but in this reaction it acts only as a base. Choice D is eliminated. By gaining a protonfrom HCIO, water is acting as a proton acceptor, making it a Bronsted-Lowry base. Choice A is the best answer.

Copyright © by The Berkeley Review® 281 Section IV Detailed Explanations

11. Choice Bis correct. The solution inTrial 1shows a greater current than thesolution inTrial 2,so thesolution inTrial 1 is more electrolytic than the solution in Trial 2. Electrolytic is defined as the ability of a solution toconduct electricity. This eliminates choice D. HCl is a strong acid, while HCIO is a weak acid. As such, moreions are present in solution during Trial 1 than during Trial 2. This explains the greater conductivity in Trial 1than Trial 2. Other colligative properties include boiling point elevation, freezing point depression, andosmotic pressure. Because the solution in Trial 1has more ionic impurities (and thus more total impurities), thesolution in Trial 1 has a higher boiling point, lower freezing point, and greater osmotic pressure. This makeschoice B the correct answer.

12. Choice D is correct. As seen in the experiment, full dissociation leads to a current of approximately 6.0 amps.Nitric acid is a strong acid, so choice Ashould generate a current of roughly 6.0 amps. Sulfuric acid has onestrong proton, so choice Bshould generate acurrent of roughly 6.0 amps. Sodium hydroxide is a highly solublesalt, so choice C should generate a current of roughly 6.0 amps. Ammonia is a weak base with no charge, soaddition of ammonia towater forms only a few ions. The current cannot reach 6.0 amps, so choice D is thebestanswer.

13. Choice C is correct. A strongbase is mixed with a strong acid in Trial 3. A complete reaction occurs, leavingbehind potassium cation (K+) and chloride anion (CI"). Astrong base is mixed with a weak acid in Trial 4. Acomplete reaction occurs, leaving behind potassium cation (K+) and hypochlorite anion (CIO"). Acompletereaction occurs in both Trial 3 and Trial 4, because in each case a strong base has been added. This eliminateschoices Aand B. Because equal amounts ofstrong base (KOH) areadded toequal mole quantities ofacid in bothTrial 3 and Trial 4, both reactions go tocompletion, generating thesame number of ions in solution. This favorschoiceC. Choice D is invalid, because cations are also in equal concentration, and cations conduct electrons.

Passage III (Questions 14 - 21) Oxyacids

14. Choice A is correct. From the relative acid strengths derived from the pKa values, HCIO4 is more acidic thanHBr04,which in turn is more acidic than HIO4. This same trend is seenwith HOC1, HOBr, and HOI. BecauseCI is more electronegative than Br which is more electronegative than I, the trend indicates that theelectronegativity of the halide dictates the relative acidity of an oxyacid. This fact was also given in thepassage. Pick answer A, and feel what it is to becorrect. Choice Capplies tohaloacids, not oxyacids.

15. Choice D is correct. To determine the relative strength of the oxyacids, both the number of oxygens and theelectronegativity of the halide must be considered. HCIO4 is a stronger acid than HIO4, because chlorine ismore electronegative than iodine, so choices A and C are both eliminated. Choice B is eliminated, becauseHIO4 is a stronger acid than HBr03, due to the extra resonance structure associated with the additionaloxygen. The best answer is choice D, since it follows both the decline in number of oxygens trend and thedecreasing electronegativity trend.

16. Choice B is correct. For a weak acid solution with [HA]jnjtial > Ka, use the shortcut equation to determine thepH. The shortcut equation, pH =ipKa -ilog [HA], applies if the pKa lies between 2 and 12. The math is asfollows:

pH =IpKa - llog [HA] =1(7.26) - llog (0.10) =3.63 - I(-l) =3.63 +0.5 =4.132 2 2 2 2


17. Choice D is correct. The pKa2 valuemust be largerthan the pKai value, because the first proton is more acidicthan the second proton (by definition), and the stronger the acid, the lower its pKa value. Given that pKa] is7.8, the value of pKa2 must be a number larger than 7.8. The only answer larger than 7.8 is 12.9, so pick choiceD, and feel satisfied that you did. The pKb (not pKa) for the conjugate is 6.2, but the question asks for pKa2-

18. Choice C is correct. Since the dissociation reaction is exothermic, heat is given off when forming the products.Thus when heat is added to the system, it acts as an inhibitor of product formation according to Le Chatelier'sprinciple. This means that products decrease and reactants increase, as the system is heated. The Ka(equilibrium constant) is a measure of the products over the reactants, so the value of Ka decreases with theaddition of heat to the system. The best answer is choice C. Choice D should be eliminated immediately.

Copyright © by The Berkeley Review0 282 Section IV Detailed Explanations

19.

20.

21.

Choice D is correct. We know HBr02 is less acidic than HC102, since both acids have the same number ofoxygen atoms, and chlorine is more electronegative than bromine. Given that HBr02 is the weaker acid, thepKa for HBr02 must be greater than 1.9 (the pKa of HC102), so the only answer possible is choice D.

Choice Dis correct. Dissociation refers to the breaking of the bond between the acidic proton and the conjugatebase, so that the conjugate base and a hydronium ion are formed. The greatest dissociation is associated withthe strongest acid. This question is asking for the strongest acid. The strongest acid is the acid with thegreatest number of excess oxygens and the most electronegative central atom. This makes choice D the bestanswer. This answer could also have been determined by recalling the six strong acids listed in the text. Onlychoice D is one of these strong acids. A strong acid fully dissociates, while weak acids partially dissociate.The amount of dissociation can be determined from the Ka and vice versa. For instance, a 1.0M weak acid with10% dissociation results in 0.90 MHA undissociated acid, and theformation of0.10 MA" and 0.10 MH3O+ Theequilibrium constant for the dissociation reaction (Ka) is shown below:

K _[H3Q+][A] _ (0.1X0.1) _o.Qi ;[HA]

0.01110.9 0.9

Choice C is correct. The volume of the solution does not affect the pH; only the concentration and strengthaffect the pH. The first step is to identify the type of acid. Since HCIO4 is a strong acid, it fully dissociates,so the major source of protons in the aqueous solution is from the dissociation of HCIO4.

pH = - log [H30+] = - log [HCIO4] = - log(0.10) = - (-1) = 1This best answer is choice C.

Passage IV (Questions 22 - 28) Acidity of Thiols and Alcohols

22. Choice C is correct. Themost acidic compound is thecompound thatmost readily loses H+. In choices A,BandD, the hydrogens are bonded to carbon. In choice C, there is a hydrogen that is bonded to nitrogen. The sizedifference between atoms is significant only when the atoms are in different rows of the periodic table.Nitrogen and carbon are in the same row of the periodic table, so they are comparable in size. When atoms arein the same row of the periodic table, the most important factor to consider when looking at acidity iselectronegativity. Nitrogen is more electronegative than carbon, making a hydrogen on nitrogen more acidicthan a hydrogen on carbon, so choice C is the most acidic. It is a common mistake to not notice that thehydrogens are all on carbons in choices A and B. Be careful not to make mistakes like this.

23. Choice B is correct. It can be observed from the data in Table 1 that as the number of methyl groups increases,the pKa value increases. An increase in pKa is indicative of decreased acidity. Acids are defined as electron-accepting, so electron-withdrawing groups increase acidity, while electron-donating groups decrease acidity.This makes choice B correct.

24. Choice D is correct. In haloacids (binary compounds), the acidic proton is directly bonded to a halogen, sorelative acidity can be discerned from the features of the halide. Halogens are in the same column of theperiodic table, so the important factor when considering bonding is the size of the halide. Because iodine is thelargest of the halogens, the HI bond is the weakest, so HI is the strongest haloacid. Pick choice D, and smile tothe world.

25. Choice A is correct. Sulfur and chlorine are adjacent to one another in the same row of the periodic table, sochoices B and D are eliminated. Because chlorine is smaller than sulfur, the increasing size does not correlatewith acidity. This eliminates choice C. Chlorine is more electronegative than S, so CI draws electrons from Hmore than S does. This makes HCl a stronger acid that H2S. The electronegativity predicts the acidity bestfor atoms in the same row, so pick answer choiceA to get this one correct.

26. Choice C is correct. Table 1 shows that thiols have lower pKa values than their corresponding alcohols, soethyl thiol is more acidic than ethanol. Because ethanol is a weaker acid (has a lower Ka value than ethylthiol), ethanol dissociates less than ethyl thiol when added to water, and ethanol yields a lower [H30+] thanethyl thiol. These three facts eliminate choices A, B, and D. Because ethanol is a weaker acid than ethylthiol, the conjugate base of ethanol (CH3CH2O") is a stronger base than the conjugate base of ethyl thiol(CH3CH2S"), making choice C the correct answer.


27. Choice B is correct. The most electronegative atom that differs from choice to choice is chlorine. This highdegree of electron withdrawal due to chlorine results in an increase in acidity. This implies that becausechlorine is more electronegative than the other atoms, it withdraws electron density the most and thusincreases the acidity to the greatest extent. The answer ofanswer choices is choice B.

28. Choice C is correct. Of the answer choices, the carboxylic acid is the most acidic, because of the resonanceassociated with the carbonyl bond. This eliminates choices Aand B. From this point, data in Table 1must beanalyzed. Table 1shows that as the substitution of the thiol decreases, the acidity increases, so the primarythiol is more acidic than the tertiary thiol. This makes choice C the best answer.

Organic AcidsPassage V (Questions 29- 34)

29.

30.

31.

32.

33.

34.

Choice C is correct. The solution is acidic, so the pH is less than 7.0, eliminating choice D. For a weak acidwith pKa between 2and 12 in an aqueous solution with [HA]initiai >Ka, use the shortcut equation to determinethe pH. If Ka is 8x10"8, then pKa is 8- log 8=7-0.9 =7.1. The [HA] is 0.5 M, and log 0.5 =- log 2=-0.30.

pH =IpKa -llog [HA] =1(7.1) - ±-log (0.50) =3.55 -I(-0.3) =3.55 +0.15 =3.72 2 2 2 2

The value falls in the range of choice C, so thebest answer is choice C.

Choice Cis correct. Mass percent is defined as the mass of one component atom relative to the total mass of thecompound. In this case, we are interested in carbon within phenol. Units cancel, so we can use atomic masses.

72gCmass%carbon= massofcarbon = x 100% => 72% =^- <22- < Z2. = 8o%100 94 90mass of molecule 94 g C6H5OH

The mass percent of carbon is between 72% and 80%. Choices Aand Bare too small, while choice Dis too large.Only choice C,76.6%, falls into the range, sochoice C is thebest answer.

Choice B is correct. When an acid is titrated to its equivalence point, it has been completely converted into itsconjugate base. The lowest pH value after neutralization is found with the weakest conjugate base, assumingthat the concentrations are all equal. The weakest conjugate base corresponds with the strongest acid (that is,the acid with the lowest pKa). This is because the stronger the acid, the weaker its conjugate base. Using thedata from Table 1, the acid with the lowest pKa value is choice B, trichloroacetic acid. Pick choice B please.

Choice Bis correct. Increasing the number of impurities dissolved into an aqueous solution raises the boilingpoint and lowers the freezing point of a solution. The total impurity concentration depends on the soluteconcentration, the number of particles it dissociates into, and the degree to which it dissociates. All of thechoices are monoprotic weak acids of equal concentration (1.0 M), so it depends only on the dissociation of theacid (acid strength). Acetic acid is weaker than p-nitro benzoic acid, so choice A is eliminated. Tricloroaceticacid is stronger than p-nitrophenol, so choice Bis correct. Tricloroacetic acid is stronger than benzoic acid, sochoice C is eliminated. Acetic acid is stronger than phenol, so choice D is eliminated.

Choice A is correct. The acid with thestrongest conjugate base is theweakest acid. The highest pKa value isassociated with the weakest acid. Referring to the data in Table 1, the acid with the highest pKa is phenol.Choice A is the correct answer, since the highest pKa value indicates the weakest acid.

Choice Bis correct. This question appears to be difficult at first glance. However, if you realize that when anacid dissociates, equal parts of hydronium and conjugate based are formed, then you know that [H30+] =[A"].All that is required is converting from pH to [H30+], using [H30+] =10"PH. The pH of the solution is 3.7, so the[H30+] is 10"^-7 =1003 x10"4. Because log 2=0.3, it is true that 1003 =2. This means that 1003 x10"4 =2x10"4.The [H30+] is 2x10"4 M, and [H30+] =[A"], so[A"] =2x10"4. The correct answer ischoice B.

Passage VI (Questions 35 - 41) Electron-Withdrawing Effect and Acidity

35. Choice B is correct. The lowest pH is found in the solution with the greatest amount of hydronium. Thehydronium concentration is affected by the strength and concentration of the acid. The most acidic solutionresults from the strongest acid in highest concentration. In this question, p-nitrophenol has a lower pKa thanphenol, sochoices Cand Dare eliminated. Choice Bhas the higher concentration ofp-nitrophenol, sochoice Bis the best answer. Pick choice B for best results and greatest test satisfaction.


36.

37.

38.

Choice C is correct. The solution is acidic, so the pH is less than 7.0, eliminating choices A and B. For a weakacid with pKa between 2 and 12 in an aqueous solution with [HA]initiai >Ka, use the shortcut equation todetermine the pH. The pKa of phenol is 10, and the [HA] is 1.0 M. The log of 1.0 is zero, so the pH is simplyhalf of the pKa. Half of ten is five, so the pH is 5.0. Choice C is the best answer.

Choice A is correct. A stronger electron-withdrawing group on the aromatic ring of the phenol makes thephenol more acidic, thus decreasing its pKa value. As a result, a substituted phenol with an electron-withdrawing group has a lower pKa than phenol. From the acids listed in Table 1, the acid with the lowestpKa is p-nitrophenol. The electron-withdrawing group on p-nitrophenol is 02N—,so choice A is terrific.

Choice C is correct. For a monoprotic acid and itsconjugate base, pKa +pKb =14. From Table 1,the pKa valuefor p-nitrophenol is 7.2. This means that pKb for the conjugate base (p-nitrophenoxide) is 14 - 7.2 = 6.8. Thecorrect answer is choice C.

39. Choice C is correct. This problem is a twist on the normal approach (i.e., it's harder than typical problems).This problem ismost easily solved using the shortcut equation, pH =ipKa -llog [HA], to solve for [HA].

pH =IpKa -llog [HA] .-. 6.1 =1(11.2) - llog [HA] =5.6 -llog [HA] .-. 0.5 =-llog [HA], so log [HA] =-1log [HA] = -1 .-. [HA] = 10"1 M=0.10 M

The concentration of the acid is 0.10 M, so the next step of the solution is to determine the moles of weak acidneeded to make 100mL of a 0.10M solution. To make 100mL 0.1M H3COC6H4OH, 0.01moles of the acid mustbe added to enoughwater tomake100 mL aqueous solution. The final step is to convert0.010 moles into grams.The molecular mass of H3COC6H4OH is 124grams per mole, so 0.01 moles is 1.24grams H3COC6H4OH. Thecorrect answer is choice C.

40.

41.

Choice C is correct. To convert from pKa toKa, the relationship is Ka = 10"PKa. The pKa of p-methylphenol is10.4 (as given in Table 1). The Ka is therefore 10"10-4 = 100-6 x 10"11, when expanded into scientific notation.Given log2=0.3 and log 4= log2+ log 2=0.3 +0.3 =0.6. According to theanti logrelationship, 100-6 = 4. Thismakes the value of Ka= 4 x10 . Pick choiceC and be a successstory in acids and bases.

Choice D is correct. The addition of water dilutes the aqueous phenol solution, but it does not react with thephenol. Using the relationship, M1V1 = M2V2, the new molarity is found to be 0.10M. As is becoming thenorm, this problem ismost easily solved using the shortcut equation, pH=ipKa -llog [HA].

pH=IpKa -llog [HA] =1(10.0) - llog (0.10) =5.0 - i-(-l) =5.0 +0.5 =5.52 2 2 2 2

The correct answer is choice D. Because the solution is acidic, choices A and B should have been immediatelyruled out.

Passage VII (Questions 42 - 48) Weak Acid pH Equation

42. Choice C is correct. The shortcut equation, as presented in the passage, works only for weak acids. It does NOTwork with a strong acid. The only strong acid in the choices is hydroiodic acid (HI), which you should know isa strong acid, because HI is stronger than HCl, a common strong acid. Choice B (carbonic acid) and choice D (acarboxylic acid) should immediately be recognized as weak acids. Pick choice C for the greatest success here.

43. Choice B is correct. In the first step of the derivation, [H3O4"] is substituted for [A"], as shown in the conversionfrom [H30+][A"] to [H3O"4"]2. Thebasic assumption is thatwhen the acid dissociates, equal amounts of H30+and A" are formed in solution, and the little (10"7 M) H30+ formed from water is insignificant. This makeschoice B the best answer.

44. Choice C is correct. Regardless of the molarity of the weak acid, when the pH of the solution is 2.70, the[H30+] in solution is10"2-7 M. This means that the [H30+] is equal to something x 10"3 (where "something" is avalue less than ten). Because [H30+] = [A"] when a weak acid dissociates, it should be concluded that [A"] =something x 103, which makes choice C the best answer.


45. Choice C is correct. Starting from the Kb relationship for the hydrolysis of a weak base in water, theequilibrium expression can be manipulated to yield the relationship pOH =-log VKbx[A"]. This relationshipcan be manipulated in the same way that the pH relationship was manipulated in the passage to yield thesame type of equation as the shortcut equation, except that it's for a weak base. The logic is that the pOHdepends on the base, hence the formula contains Kb rather than Ka, and [A"] rather than [HA]. This makeschoice C the best answer. Choice A is applied in general to all basic solutions, but it is not the best answer.Choice Dmay sound tempting, but that is how the pOH for a weak acid solution is found, not for a weak basesolution.

46.

47.

48.

Choice A is correct. According to the Henderson-Hasselbalch equation, when [A-] > [HA], the pH is greaterthan the pKa, because the log of the [A-] to [HA] ratio is a positive value. This eliminates choice B. If pH isgreater than the pKa, then by definition -log [H30+] is greater than the -log Ka, hence log Ka is greater thanthe log [H30+] (when both sides of an inequality are multiplied by -1, the inequality sign must be reversed).Because log Ka is greater than the log [H30+], Ka is greater than the [H30+], making choice A the best answer.Choice Ccannot be true, because aspH goes up, [A"] increases as [H30+] decreases, making [A"] > [H30+] at allpoints of a reaction after the initial mixing of the weak acid into water. This eliminates choice C. Choice Dcan be true only if [HA] = 0, which is physically impossible with the equilibrium constant favoring HA. Itshould always be true that [HA] > [H3O4-], which eliminates choice D. The best answer ischoice A.

Choice Bis correct. The greatest conjugate base concentration is found with acid thatdissociates to the greatestextent. The acid that dissociates most completely is the strongest acid, which describes choice B (HNO3), theonly strong acid among the choices. This question is just another way ofasking, "Which acid is stronger?"

Choice C is correct. As the value of Ka increases, the strength of the acid is increasing. As the acid getsstronger, the degree of dissociation when added to water increases. This causes the conjugate base concentrationto increase, which eliminates choice A. The amount of H30+ increases, which causes the pH to go down andthus the pOH to increase. This eliminates choice Band confirms that choice Cis the best answer. Because theacid dissociates more, the amount of HA decreases and the amount of A" increases, causing the ratio of HA to A"to decrease. This eliminates choice D.

PassageVIII (Questions 49 -55) Aspirin and Antacids

49. Choice D is correct. Aspirin (acetylsalicylic acid) when added to water dissociates into the anionic conjugatebase and a proton, just like all other carboxylic acids. The dissociation obeys the laws of equilibrium, resultingin an equilibrium constant known asKa. Because it obeys the laws ofequilibrium, the anion form (more water-soluble form) ismost abundant when the H+ (proton) is removed from solution. The lowest concentration ofprotons ([H30+]) is found at high pH. The highest pH results in the greatest water solubility for the aspirin,because it exists in its anionic form. This, in essence, is the common ion effect. The best answer is thus choice D.

50. Choice A is correct. Acetylsalicylic acid has only one hydrogen on the carboxylic acid functional group that isacidic. All of the other hydrogens are bonded to carbon, which does not make them acidic. The correct answerfor the number of acidic protons is thereforeonly one. ChoiceA is the best answer.

51. Choice C is correct. The first three choices can all neutralize two equivalents of acid, while choice D, sodiumbicarbonate, can neutralize only one equivalent of acid. This means that the most neutralizing strength pergram is found with the compound having the lightest molecular mass of the compounds capable ofneutralizingtwo equivalents ofacid. Magnesium hydroxide has the lowest molecular mass of the first three answer choices.The correct answer is therefore choice C.

52. Choice C is correct. The molecule that can cause damage to stomach lining is the acidic molecule that ionizesonce within the membrane pocket of parietal cells. The only acidic compound of the choices is benzoic acid,answer choice C. It is the most similar in structure to aspirin, so it is the best answer.

53. Choice D is correct. Dihydroxyaluminum sodium carbonate is composed of the dihydroxyaluminum cation(which loses two hydroxide substituents when hydrolyzed), sodium cation (with no basic properties), andcarbonate dianion (which can neutralize twoacidic protons). The net result is that the two hydroxides and onecarbonate can neutralize four equivalents ofprotons, making choice D correct. You might also consider that Alcarries a +3 chargeand Na carries a +1 charge, so thesumcharge of the bases must be -4.


54.

55.

Choice D is correct. Acetylsalicylic acid is a carboxylic acid, and carboxylic acids are weak acids. Weak acidshave Ka values that are less than one. The pKa for a carboxylic acid ranges from 2.0 (as seen with the carboxylterminal ofamino acids) to 5.0 (as seen with regular alkyl chains). This results in a range for Ka of 10"2 to10"5,which yields a Ka that ismuch less than 1.00. This vague question isbest answered by choice D.

Choice Dis correct. Abuffer forms when the conjugate acid and base of a conjugate pair are present in roughlyequal molar ratio. Choices Aand Bshould immediately be eliminated, because the mixtures are made up oftwo acids, and not ofan acid and base pair. Choices CandD involve a strongacid and weak base, so thecorrectchoice involves partial titration of the weak base. It is in choice D that an equalmolar ratio of the weak baseand its conjugate acid are present. The one equivalent of HCl will convert one of the two equivalents ofacetylsalicylate into acetylsalicylic acid, while one equivalent remains as acetylsalicylate. The equalquantity of the twocomponents of theconjugate pair results in a buffer (with a pH equal to the pKa of theweakacid). The best answer is choice D.

Passage IX (Questions 56 - 62) Household Acids and Bases

56. Choice C is correct. The first word in the question is "If", so keep that in mind. The mantra that you chantabout water having a pH of 7 does not apply in this speculative question. If the only hydroxide ion in solutionis formed when onemolecule ofwater loses a proton to another molecule ofwater, then the hydroxide anionconcentration [OH"] must equal the hydronium cation concentration ([H3O"1"]). This is true within distilledwater. This means that if [OH-] = 1.0 x10"6 M, then [H30+] must also equal 1.0 x10"6 M. The negative log of1.0x10"6 is 6.0, so the pH of the water must be 6.0. The best answer is choice C. For this hypothetical solution, pH=6 and pOH= 6, which means that pH + pOH= 12, not 14. This may bother your sense of what is right in theworld of acid and base chemistry, but keep in mind that pH + pOH = 14 applies only at 25°C. At highertemperatures, there is more autoionization, so more hydronium and hydroxide are generated. At 37°C forinstance, pH of distilled water is 6.8, so pH + pOH = 13.6.

57. Choice B is correct. The hydronium ion concentration in solution can be-found from the pH of the solution.Because the pH is defined as pH = -log [H3O"1"], the hydronium ion concentration can be found by manipulatingthe relationship to yield [H^O*] = 10"PH. The hydronium ion concentration in a pH = 1.7 solution is 10"1-7 M,which is therefore roughly equal to 10"2 M. The best answer selection is choice B.

58. Choice D is correct. The volume increases from 10 mL to 110 mL when 100 mL of distilled water is added to thesolution, so the concentration of the acid must decrease by a factor of 11. The log of 10 is equal to 1, so the log 11is slightly greater than 1. This implies that the pH increases by just a little more than 1.0, because the [H30+]goes down by a factor of 11. This makes the final pH greater than 3.0. The best answer is greater than 3.0,which is choice D. This is true only of a strong acid solution. If the solution were an aqueous weak acid, the pHincrease would not be as significant as with the strong acid.

59. Choice A is correct. The hydronium ion concentration in a pH = 2.0 solution is 1.0 x 10"2 M. There are 50 mLpresent, so the moles of H+ are found bymultiplying 0.010 Mby 0.050 L = 5.0 x 10"4 moles H+. Only half themoles of CaC03 are necessary to neutralize the H+, because it is a 2 : 1 ratio of H+ to CaC03 in the balancedequation. Thismeans that 2.5 x 10"4 moles CaC03 are required to neutralize all of the hydronium ion (H3O"4").2.5 x 10"4 moles when multiplied by 100 grams per mole yields 2.5 x 10"2 grams CaC03, which equals 25milligrams. Choice A is the best possible answer you can choose in a situation like this.

60. Choice C is correct. Windex contains ammonia, which according to Table 2 is a base. An acid can react with abase, while a base cannot react with another base. The question is thus asking: "Which of the followingcompounds is not an acid?" Vinegar, aspirin, and bleach are all listed as acids in Table 1, so they all can reactwith ammonia (Windex). The best answer is choice C, Drano, because Drano is a base and not an acid.

61. Choice D is correct. Raising the pH of a solution requires reducing (or diluting) the concentration of hydroniumion (H30+). Adding distilled water reduces the hydronium ion concentration by diluting the solution. Thismakes choice A invalid. Shampoo and antacid are bases, which when added to solution, react with the H3O4"ion and thus reduce the hydronium ion concentration and raise the pH of the solution. Choices B and C aretherefore invalid. Adding an acid lowers ~ not raises —the pH of the solution, so choice D (vitamin C), anacid, is the only answer choice that will NOT increase the pH of the solution when added. Pick choice D witha smile.


62. Choice D is correct. 1.68 grams of baking soda is equal to 0.020 moles NaHC03 as determined by dividing the1.68 grams by 84 grams per mole. The equation for the reaction shows a1:1 ratio between baking soda and thehydronium ion for full neutralization, so 0.020 moles of H30+ are required to reach full neutralization. Themolarity of the acid solution when multiplied by the volume of the acid solution must equal 0.020 moles H3O+The molarity of the acid solution is given as 0.20 M, so the volume of the acid solution must be 0.10 liters, whichis 100 mL. Selecting answer choice D ina situation like this, is the best thing to do.

Stomach Acid and pHPassage X (Questions 63- 70)

63.

64.

65.

66.

67.

68.

69.

70.

Choice A is correct. The formation of a strong acid from a weak acid and neutral salt is unfavorable.Hydrochloric acid is stronger than carbonic acid, so the reaction as written has AG >0. To carry this out znvivo, it must be coupled with some other favorable reaction. As stated in the passage, this process is activetransport. The correct answer is therefore choice A.

Choice A is correct. The lowest pH is associated with the most acidic solution. Both salt water and distilledwater are neutral (have a pH of 7.0), so choices Band Care eliminated. Carbon dioxide (a non-metal oxide)when dissolved into water gets hydrated to form carbonic acid (H2C03, a non-metal hydroxide). Calciumoxide (a metal oxide) when dissolved into water gets hydrated to form calcium hydroxide (Ca(OH)2, ametalhydroxide). Non-metal hydroxides are acidic and metal hydroxides are basic, so the best answer is choice A.Choice Dis correct. Acid is capable of oxidizing any metal with a low ionization energy. In this question, onlyone answer is correct, so it is better to ask yourself which metal is most easily oxidized (or least stable).Relying on either trivial knowledge or experience, you know that copper, gold, and silver are relatively air-stable, as evident by their commercial uses in wiring and currency. All three tarnish over enough time. Zincmetal is most readily oxidized (more so than hydrogen gas even). The best answer is choice D. If you areinterested in smuggling zinc coins across the border, it is recommended that you refrain from ingesting them to doso. If you don't know about the relative reactivity of these metals, you just have to take aguess and move on.Choice D is correct. This is really a misplaced organic chemistry question. But then again, the passage is on aphysiology topic, emphasizing that acidity and basicity are applicable to many areas of study. Alkanescannot be hydrolyzed whether an acid is present in solution or not. Disaccharides, polypeptides, and esters areall broken down in the stomach by acid-catalyzed hydrolysis. The best answer is choice D.

Choice D is correct. Consuming water dilutes all solutes (including H30+ in an acidic solution), thus thehydronium ion concentration is reduced by consuming water. Baking soda can consume some of the hydronium ionby way of an acid-base reaction. Aluminum metal can be oxidized by hydronium ion, resulting in the formationof aluminum trication and hydrogen gas. Therefore, choices A, B, and Care all eliminated. Consumption ofsolid food induces the release ofhydrochloric acid into the stomach. The correct answer is choice D.

Choice B is correct. This answer is taken straight from the first paragraph of the passage. 0.030 Mhydroniumhas a pH of 1.5. However, in the event you skipped reading the passage, you can solve for the pH bydetermining the negative log of the hydronium concentration (pH =- log [H30+]). The pH of a 0.030 MHClsolution is - log (3 x10"2) =2-log 3. The value is less than 2, but greater than 1. The correct answer ischoice B.

Choice C is correct. Stomach lining is deteriorated by reacting with acid. All of the answer choices exceptmilk contain acid, so they all promote the decay of the stomach lining. The best answer is choice C. Thisquestion is asking for you to identify which compound is not an acid, so look for th choice that is different.Lactose is a disaccharide composed of galactose and glucose.

Choice A is correct. The lowest pH results from the presence of the strongest acid in the highest molarconcentration. Hydrochloric acid is the strongest acid (stronger than carbonic acid), and 0.030 Mis a higherconcentration than 0.010 M. The correct answer is therefore choice A.

Passage XI (Questions 71 - 76) Tooth Decay and pH

71. Choice A is correct. The first question is whether the pH of the solution changes. Because fluorapatite is lesssoluble than hydroxyapatite, hydroxide anion is released as fluoride anion exchanges for the hydroxide ofhydroxyapatite. When hydroxide anion is released, the pH of the solution increases. Although the pH isgreater than 7.0, choice C is incorrect, because the pHdoes not remain constant. The best choice isA.

Copyright ©by The Berkeley Review® 288 Section IV Detailed Explanations

72.

73.

Choice Bis correct. Calcium hydroxide when added to water yields hydroxide anion (OH"), making it basic.This eliminates choices A and C. The question is whether it is a strong or weak base. These questions on theMCAT are tough, because who knows what the test-writer wants? Because calcium hydroxide does not fullydissolve into water at high enough concentrations, it can be considered a weak base. However, because it reactswith weak acids, and it does fully dissociate at low concentrations, it must be a strong base. The best answerbeyond any questioning and rational argument you can present is choice B (unless,of course, you wish to chooseD). On somequestions, like this one, there are two reasonable answers, but the writer of the question wants youto pick just one best answer.

Choice C is correct. Enamel is more soluble in the more acidic solution. Because enamel is more soluble in theHA solution than the HB solution, the HA solution must be more acidic than the HB solution. Both the HAand HBsolutions are of equal concentration, so it must be that HA is a stronger acid than HB. This means thatif HA is dissolved into water, it has a lower pH than when HB is dissolved into water. The conjugate base ofHB is a stronger base than the conjugate base of HA. HA has a higher Ka and a lower pKa than HB. Thiseliminates choices A, B, and D. Choice C is the best choice.

74. Choice A is correct. The hydroxyapatite dissolves most readily in acidic medium, so the addition of acidincreases its solubility. Thus, the amount of hydroxyapatite that dissolves into solution increases as acid isadded to solution. Addition of base increases the pH of the solution, meaning that choices Band C are the sameanswer. The MCAT presents questions like this on occasion. When they have two identical answers wordeddifferently, eliminate them both. Decreasing the solvent reduces the amount of salt that has dissolved intosolution. The best answer is choice A.

75. Choice A is correct. Sodium carbonate (choice D)should be eliminated immediately, because it is a base. Ofthe three remaining choices, both hydrochloric acid (choice B) and nitric acid (choice C) are strong acids. Thealdonic acid drawn is not a strong acid, so choices B and C are eliminated. The structural similarity because ofthe CO?H group on both acetic acid and gluconic acid (the aldonic acid drawn) makes answer A the best choice.

76. Choice A is correct. Tooth enamel is a basic substance, so it dissolves fastest in an acidic solution. Table 1 showsthat the solubility increases as pH decreases. This means that the most acidic solution (solution with thelowest pH) dissolves tooth enamel the fastest. The lowest pH results from the strongest acid in the highestconcentration. The highest concentration is found in choices A and C, so choices B and D are eliminated. Thestronger of the two acids is HCl, leading us to conclude that the 0.10 M HCl solution dissolves the tooth enamelthe fastest. The best choice is answer A.

Passage XII (Questions 77 - 84) Acid Rain and Scrubbers

77. Choice D is correct. In Reaction 1, sulfur trioxide is hydrated. The addition of water (hydration) does notchange the acidic or basic nature of a compound. This eliminates choices A and B. As stated in the passage, thehydration of a non-metal oxide (a Lewis acid) converts the compound into a non-metal hydroxide (a Bronsted-Lowry acid). This tells us that choice C is wrong and makes choice D the best answer.

78. Choice D is correct. A non-metal oxide acts exclusivelyas an acidic (and not a base), so it is neither amphotericnor an Arrhenius base. This eliminates choices A and B. A non-metal oxide has no protons, so it cannot be aBronsted-Lowry acid. This eliminates choice C By having 7i-bonds, a non-metal oxide is a good electrophile,which makes it a Lewis acid. The best answer is choice D.

79. Choice C is correct. The most acidic compound when combined with water creates the solution with the lowestpH, although the compound's concentration has an effect on the pH of the solution as well. Metal oxides arebasic, so choice D (MgO) should be eliminated. The three acids formed when choices A, B, and C are added towater are H2S03 (from S02 and H20), H2C03 (from C02 and H20), and HNO3 (from N205 and H20),respectively. If you knew that nitric acid is stronger than either sulfurous acid or carbonic acid, then you couldhave picked choice C. However, if you didn't know that tidbit of trivia, then you could have determined theirrelative strength from their oxidation states and from the electronegativities of their central atoms. Sulfurhas a +4 oxidation state in S02 (and H2SO3), carbon has a +4 oxidation state in C02 (and H2C03), andnitrogen has a +5 oxidation state in N2O5 (and HNO3). Nitrogen is the most electronegative and has thehighest oxidation state. The nitrogen-based oxyacid has the greatest propensity to gain an electron pair,making HNO3 the most acidic. Pick choice C.


80. Choice B is correct. The solution is acidic, so the pH is less than 7.0, eliminating choice D. For a weak acidwith pKa between 2and 12 in an aqueous solution with [HA]injtiai >Ka, use the shortcut equation to determinethe pH. The pKa is 7.46 and the [HA] is 0.10 M.

pH =IpKa -llog [HA] =1(7.46) -llog (0.10) =3.73 -U-l) =3.73 +0.50 =4.23The best answer is choice B. Choice Acan be eliminated, because HCIO is a weak acid. A 0.10 Mstrong acidhas a pH of 1.0. An equimolar weak acid, because it dissociates less than a strong acid, would have a pHgreater than 1.0. Choice Ccan be eliminated, because the pH is less than the pKa.

81. Choice C is correct. This is a freebie question, ifyou read the passage carefully. Non-metal oxides are found inacid rain, particularly the oxides of sulfur and nitrogen. Choice Ais eliminated, because it is a metal oxide,making it is basic, not acidic. Choice Bshould be eliminated, because nitrogen is present in the air all the timeand can be found in all rain, not just acid rain. Choice D is eliminated, because it is neither acidic nor veryabundant in our atmosphere. This means that thecorrect answer ischoice C.

82. Choice C is correct. As theoxidation stateof the central atom in an oxyacid increases, the compound becomesmore acidic. This is because thenumber of rc-bonds between oxygen atoms and thecentral atom increases. Thisincrease in 7t-bonds results in more electron withdrawal (by way of resonance) from thecentral atom, making itelectron-poor. The compound therefore becomes more acidic. This means that both the oxidation state and thenumber of Tt-bonds to the central atom affect the acidity of the oxyacid, so choices A and D are eliminated. Asthe electronegativity of the central atom in an oxyacid increases, the compound becomes more acidic. This isbecause of the inductive effect. The increase in electron withdrawal from the central atom, due to the inductiveeffect, makes the central atom electron-poor. This means that the electronegativity of the central atom affectsthe acidity of the oxyacid, so choice Bis eliminated. The size of the central atom has no bearing on the acidityof the oxyacid, because the acidic proton is not directly bonded to the central atom. Tins makes choice C thebest answer.

83. Choice D is correct. Because H2S04 has more oxygen atoms than H?S03, it is more acidic. The more acidiccompound has a lower pKa, a higher Ka, a weaker conjugate base (with a higher pKb), and generates a lowerpH in water. This makes choices A, B, and Cvalid statements. The fact that H2S04 is more acidic thanH2S03 means that H2S04 dissociates more when added to water, resulting in a higher ion concentration, andthus a higher conductivity. H2S04 is more electrolytic (has higher conductivity) than H9SO3, so choice D isan invalid statement, which makes it the correct answer.

84. Choice A is correct. The stronger the acid, the lower its pKa value, so the lowest pKa is found with thestrongest acid. Having more oxygen atoms on an oxyacid increases its acidity, so HN02 and H2S03 cannot bethe most acidic compounds. This eliminates choices Cand D. Sulfuric acid is stronger than nitric acid, whichyou can determine from their oxidation states (S is +6 is H2S04, while N is +5 in HNO3). This makes choice Athe best answer.

85.

86.

PassageXIII (Questions 85 - 91) Amino Acid pKa Values

Choice C is correct. Lysine is a triprotic acid. The first proton released (from the carboxyl terminal)corresponds to the third proton gained by the conjugate base. This means that from a conjugate pair perspective,the pKa-to-pKb relationship is: pKal +pKb3 =14. This is not one of the answer choices, so we next consider thesecond proton on lysine. The second proton released (from the amino terminal) corresponds to the second protongained. This means that from a conjugate pair perspective, the pKa-to-pKb relationship is: pKa2 +pKb2 =14.Thismakes choice C the correct answer;but to finish our discussion of the correlationbetween pKa and pKb, wemust consider the third proton. The third proton released (from the sidechain) corresponds to the first protongained. This means that from a conjugate pair perspective, the pKa-to-pKb relationship is: pKa3 +pKbi = 14.

Choice A is correct. At physiological pH, the carboxyl terminal isdeprotonated (pH > pKa (carboxyl terminal))'so it carries a negative charge. At physiological pH, the amino terminal is protonated (pH < pKa (aminoterminal))' so ^ carries apositive charge. In order for the amino acid to be a cation, the side chain must carry apositive charge. Arginine has a side chain pKa of 13.21, so pH <pKa. This means that the side chain isprotonated, so itcarries a positive charge. Choice Ais the best answer. Choice Bis eliminated, because it is ananion. Choice C is eliminated, because the side chain cannot carry a charge. Choice D is eliminated, becausehistidine is deprotonated (pH> pKa (sjde chain))/ so it carries no charge.


87. Choice D is correct. The difference between thesidechains of serine and cysteine is that serinehas an alcoholgroup (O-H), while cysteine has a thiol group (S-H). This means that the difference in acidity lies in thedifference between oxygen and sulfur. Choices A and C are eliminated immediately, because we know fromtheir relative positions in the periodic table that oxygen has a smaller atomic radius and is moreelectronegative than sulfur. When comparing the acidity ofprotons bonded to different atoms that occupy thesame column of the periodic table, we find that the most significant factor influencing the acidity is atomicradius. The best answer is choice D.

88. Choice A is correct. Based on the pKa values in Table 1, aspartic acid (with a side chain pKa of 3.88) is moreacidic than glutamic acid (with a side chain pKa of 4.32). This eliminates choice C and D. A shorter chainresults in the electron-withdrawing NH3+ group's being closer to the side chain carboxylic acid group, whichincreases the acidity of the carboxylic acid group. This makes choice A the best answer.

89. Choice D is correct. According toTable 1 in the passage, the pKa values for histidine are: pKai between 1.8and2.6, pKa2 = 6.05, and pKa3 between 8.8 and 10.6. At a pH of 7.0, the carboxyl terminal and the side chain aredeprotonated, while the amino terminal is protonated. This is because pKa3 > pH > pKa2 > pKa]. Since thecarboxyl terminal is deprotonated, we consider pKb3 rather than pKa-j. Since the side chain is deprotonated,we consider pKb2 rather than pKa2. And since the amino terminal is protonated, we consider pKa3 rather thanpKbi- This means that the pK values of interest are pKa3, pKb2, and pKb3, making choice D the best answer.

90. Choice C is correct. Normality is defined as the moles of equivalents per liter. Because glutamic acid istriprotic, it yields three acidic protons per molecule. For each mole of glutamic acid, three moles of acidicprotons can be generated. The normality in this case is the molarity multiplied by a factor of three (N = M x3).The molarity is 0.50, so the normality is 1.50. The best answer is choice C.

91. Choice C is correct. Cysteine is neutral when protonated. The side chain pKa for cysteine is 8.36, so at pH = 7.0,pH < pKa. Under such conditions, the side chain of cysteine is protonated, and thus neutral. Choice A iseliminated. Histidine is neutral when deprotonated. The side chain pKa for histidine is 6.05, so at pH = 7.0,pH > pKa. Under such conditions, the side chain of histidine is deprotonated, and thus neutral. Choice B iseliminated. Tyrosine is neutral when protonated. The side chain pKa for tyrosine is 10.07, so at pH = 7.0, pH <pKa. Under such conditions, the side chain of tyrosine is protonated, and thus neutral. Choice D is eliminated.Lysine is neutral when deprotonated (as is the case with the three basic amino acids). The side chain pKa forlysine is 10.80, so at pH = 7.0, pH < pKa. Under such conditions, the side chain of lysine is protonated, and thuscationic. Choice C is the best answer, because the side chain is charged.

Questions 92 -100 Not Based on a Descriptive Passage

92. Choice C is correct. The lowest pH is associated with the solution with the greatest hydronium concentration.The greatest hydronium concentration depends on the concentration and the strength of the acid. To lower pH,the concentration of an acid may be increased, or a stronger acid, with a lower pKa value, may be employed.Tins means that the correct answer is a combination of lowest pKa and greatest concentration. Choice A getseliminated for having the lowest concentration and a high pKa value. Choice B gets eliminated, because it hasa lower concentration and greater pKa than choice C. The pH in choice C is half of the pKa + 0.5, which is 1.65+ 0.5 = 2.15. The pH in choice D is half of the pKa, which is 4.55. Choice C has the lowest pH.

93. Choice D is correct. Acid rain is contains airborne Lewis acids. Non-metal oxides such as sulfur oxides andnitrogen oxides make up most of the acid rain we study. This means that choice A is an example of acid rain.Once these non-metal oxides react with moisture in the air, they become hydrated, so acid rain does containhydrated non-metal oxides (also known as non-metal hydroxides). A Lewis acid is an electron-pair acceptor,and non-metal oxides qualify in this category. Choice C is a valid description of an acid rain component. Metalhydroxides are basic, so they are not found in acid rain. Choice D is the best answer.

94. Choice B is correct. The pH of a conjugate pair is found using the Henderson-Hasselbalch equation. The lowestpH is attributed to the conjugate pair having the acid with the lowest pKa and the mixture that most favorsconjugate acid. The pKa of ammonium is 9.26 (although you really just need to know it's between 9 and 11),while the pKa of hydrofluoric acid is 3.17. This eliminates choices C and D. Choice B has more acid thanbase, while choice A has more base than acid. The best answer is choice B.


95. Choice Bis correct. For a conjugate pair at 25°C, the pKa of the acid when added to the pKb ofits conjugatebase is equal to 14. This question boils down to which of choices represents aconjugate apair. Choice Dcan beeliminated immediately, because both values are pKa values, and if they happen to sum to 14, it's purelycoincidental. Phosphoric acid is triprotic, so there are three pKa values. The first proton off (pKai)corresponds to the third proton back on (pKb3), so pKai +pKb3 =14. The second proton off (pK^) corresponds tothe second proton back on (pKb2), so pKa2 +pKb2 =14, making choice Bthe correct answer. The third proton off(pKa3) corresponds tothe first proton back on (pKbi), so pKa3 +pKbi =14.

H3P04(aq) +H20(1) * ^ *" H30+(aq) +H2P04-(aq)First proton off/Third proton on .-. pKai +pKb3 =14

pKa2 ,,H2P04-(aq) + H20(1) •* „ * H30+(aq) + HP042-(aq)

PKb2

Second proton off/Second proton on ,\ pK^ +pKb2 =14

pKa3 „HP042-(aq) +H20(1) •* R * H30+(aq) +P043-(aq)Thirdprotonoff/Firstprotonon .*. pKa3 +pKbi=14

96. Choice B is correct. 1.5 MH3PO4 =4.5 NH3PO4, because there are 3 equivalents of H+ perH3PO4. Thenormality is found by multiplying the molarity by the number of equivalents. Doubling the volume by addingpure water cuts the concentration (measured in either normality or molarity) in half. This makes normality onehalf of 4.5 N, so choice B, 2.25 N, is the best answer.

97. Choice B is correct. Because the pHis less than seven, the solution must contain an acid rather than a base.This eliminates choices C andD. If the acid were a.strong acid, it would fully dissociate in water. This wouldlead to ahydronium concentration of 0.10 M, which would make the pH =1. Given that the pH is greater than1.0, the acid does not fully dissociate, so it is a weak acid. This eliminates choice Aand makes choice Bthebest answer.

98. Choice C is correct. The highest pH is found in the solution with the highest concentration ofhydroxide.Hydroxide concentration depends on the compound and its concentration. Choice Ais aweak acid, so the pH islow. This eliminates choice A. Choice Bis a strong acid, sopH is very low. This eliminates choice B. ChoiceCis the conjugate base of aweak acid, so it is aweak base. Choice Ccould be the correct answer. Choice Disthe conjugate base of astrong acid, so it is avery weak base, meaning ithas no significant impact on the pH ofthe aqueous solution. Because the pH of choice Dis 7.0, choice Diseliminated. The best answer ischoice C.

99. Choice Bis correct. The solution is acidic, so the pHis less than 7.0, eliminating choice D. The acid isweak, soit partially dissociates in water. If the acid were strong, it would fully dissociate, leading to a hydroniumconcentration of0.050 Mand thus a pHof 1.30. Carboxylic acids are weak acids, so the pHis higher than 1.3,This eliminates choice A. The pHis less than the pKa, sochoice Cis eliminated. The only choice remaining ischoice B. Ifyou choose to do so, you can solve for the pH exactly. For aweak acid with pKa between 2and 12 inan aqueous solution with [HA]initiai >Ka, use the shortcut equation to determine the pH. The pKa is 4.89. The[HA] is 0.05 M,and log0.05 = - log20= - (log 10+ log3)=- (1 +0.3) = -1.3.

pH=IpKa -llog[HA] =i(4.89) - llog (0.050) =2.45 -1<-13) =2.45 +0.65 =3.10The value confirms that choice B is the best answer.

100. Choice Cis correct. Hydrobromic acid (HBr) isa strong acid, so a0.10 Msolution has a pH of1.00. Choice Aisvalid and thus eliminated. Formic acid (HCO2H) is a weak acid, so it doesnot fully dissociate. It has a pHgreater than 1.0 but less than 7.0. Choice Bcould be valid, soit is unlikely. To eliminate it with certainty, weneed to do a calculation. We shall hold off oncalculating until wehave evaluated theother answers. Sodiumacetate (NaOAc) is a weak base, so it does not fully hydrolyze. It has a pHgreater than 7.0 but less than 13.0.Choice C is invalid, so it is thebestanswer. Potassium hydroxide (KOH) is a strongbase, so a 0.10 Msolutionhas a pH of 13.00. ChoiceD is valid and thus eliminated.

Copyright ©by The Berkeley Review® 292 Section IV Detailed Explanations

Buffers

^* ^^^-t #-•£ rV1>1 ^T a) Buffer Composition3CC110I1 V b) pH Range for Buffers

c) Buffer Reciped) Physiological Buffers

Buffers and „„__»TTi f"l*C|f"1f\K| a) Quantitative Reactions1 1LI CILIV711 b) fundamental Curve Shapes

by Todd Bennett c) Plotting a Titration Curved) Concentration Effects

pH ( e) Strength Effects/ 0 Polyprotic Acids

8 85Eq" Point"06 J Indicators

( a) Composition and Functionb) Detecting the Equivalence Pointc) Estimating Solution pH

^ nc\

pH=pK^^^^-^^

3.00

0.5 1.0

Equivalents 0.10 M KOH added

n BERKELEYSpecializingy in MCAT Preparation

°%

Buffers & Titration Section GoalsKnow how to approximate the pli of a solution using a titration curve.By knowing the relative molequantities ofwhat has beenmixed in solution, a pH value can beestimated fromthe positionon a titrationcurvecorresponding to themixture. This requiresbeingable to identifythe shape ofa titrationcurvebasedon the components in themixture.

gfrifc Understand how polyproticacids affect the pH of a solution.*™ Polyprotic acidshavemultiple equivalent points, ofwhich only the lastwill fluctuate with concentration.

Know that amino acids are a subset of polyprotic acids. Be able to determine the pH at middleequivalents points by applying the equation for pi (averaging the two respective pKa values).

'fr©*

Know the key points along all titration curves.On titration curves associated with stronc reagents, the pH at equivalence is always equal to 7.0.On titration curves associated with a weak acid or weak base, titrated by a strong reagent, the pHat the half-titrated point is equal to the pKa of the weak acid. The pH at equivalence can beapproximated by averaging the pKa of the weak acid and the pH of the titrant.

Know the role of an indicator in titration and solution pH.Anindicator is used tomake theendpoint of the titration visible. Anindicator is a species that hasa different color for the conjugate acid and conjugate base. Most indicators are organic compoundswith a great degree of conjugation, and the color is caused by a transition from the k level to the n*level. Indicators can also oe used to predict the pH of a solution.

Know how conjugate pairs and buffers work.A buffer is formed when a weak acid and its conjugate base are added to the aqueous solution.Because there exists an equilibrium between the two species, as long as both are present in solution,the hydronium ion concentration will remain fairly constant, therefore the pH will also remainconstant. The effect is known as "buffering." Youmust understand buffers and now the pH is foundfrom the Henderson-Hasselbalch equation.

Recognize and be able to generate titration curves for any titration.Youmust be able to identify the titrant and the species being titrated when you look at the titrationcurve. Features to note are the weak acid lip, the strong acicTsigmoidal shape, the ascent (associatedwith the titration of an acid by a base) or descent (associated with the titration of a base by an acid)of the curve, and the number of inflection points (indicative of whether the compound is polyproticor polybasic).

General Chemistry Buffers and Titration

Buffers and TitrationA buffer is a solution where pH remains relatively constant after the addition ofeither strong acid or strong base. The pH may vary slightly, but for all intentsand purposes, it does not change significantly. Buffers play a major role inphysiology and biochemistry, so understanding how they work is critical. Theycan be made in one of two ways. The first method involves combining aconjugate pair in roughly equal mole portions. The second method involvespartially titrating a weak acid with roughly half of an equivalent of strong base,or by partially titrating a weak conjugate base with roughly half of an equivalentof strong acid. The pH of a buffered solution is determined using theHenderson-Hasselbalch equation. The data associated with buffers are generallyeasy to work with in a conceptual sense. Titration curves have buffering regions,so understanding buffers can help you to understand titration better.

Neutralization is the mixing of equal mole portions of an acid with a base,regardless of their concentrations and strengths. To neutralize an acid, an equalmole quantity of base must be added to solution. To neutralize a base, an equalmole quantity of acid must be added to solution. A neutralized solution hasmoles H30+ equal to moles OH". Addition of a base to an acid (or acid to a base)yields water and a salt upon neutralization. This is shown in Reaction5.1:

HX(aq) MOH(l) H20(1) MX(aq)Reaction 5.1

Depending on the strength of the acid and base, the pH at the neutralizationpoint (also referred to as the endpoint and the equivalence point in titration)varies. Neutralization does not mean to make the pH of the solution equal to7. When the base is stronger than the acid, the neutralized solution is slightlybasic, so the pH is greater than 7.0. When the base is weaker than the acid, theneutralized solution is slightly acidic, so the pH is lower than 7.0. When the baseand acid are equally strong, the neutralized solution is neutral, so the pH is equalto 7.0. The three possible combinations are summarized below.

HX(aq) + MOH(l)

Strong acid + Strong baseStrong acid + Weak baseWeak acid + Strong base

H20(1) + MX(aq)

P**at equivalence = ' -0pHatequivalence < ' -0pHatequivalence >7.0

The pHis not always 7 at theequivalence pointof a titration. For aweak acidtitrated by a strong base, the equivalence point is the point atwhich it is completelyconverted into its conjugate base. The conjugate base will yield a pHgreater than 7so the pHisgreater than 7at the equivalence point.

The last section in this chapter involves detecting the equivalence point. Visualindicators that change color upon changing pH are a typical component of anygeneral chemistry curriculum. They are often highly conjugate organicmolecules. The color associated with an indicator is a reflective color, resultingfrom the absorbance of a photon accompanied by the excitation of an electronfrom a rc-bonding molecular orbital to a rc-antibonding molecular orbital. Thetransition energy changes when the compound gets deprotonated, so the energyof the photon absorbed, and ultimately the color of light reflected also changes.Hence, any pH-sensitive chromophore is an indicator.

Introduction


General Chemistiy Buffers and Titration Buffers

Buffer CompositionBuffers are solutions that resist drastic changes in pH. Buffers are made of aroughly equal mole mixture ofa weakacid and its weak conjugate base in anaqueous solution. Both the acid andthe base oftheconjugate pairmustbeweakin order to form a buffer solution. This is so that the equilibrium between thetwo species can be controlled by the environment. With approximately equalmolar quantities ofconjugate acid and conjugate base in solution, the solution isresistant topHchange caused by theaddition ofeitherstrongacid or strongbaseto solution. When bothof the species in the conjugate pair are weak, the buffercan equilibrate in both the forward and reverse directions of the reaction toabsorb anyhydronium orhydroxide thatmay be added tosolution. Addition ofa strong acid to solution converts theweak base into its conjugate acid. This haslittle to no effect on the pH. Likewise, addition of a strong base to the solutionconverts the weak acid into its conjugate base and has little to no effect on thepH.Toemphasize theneedforroughly equalportions, the following experiment maybe studied. In this study, three mixtures of acid and conjugate base aregenerated. In the first system, weak acid is in extreme excess relative to itsconjugate base. In the second system, the two species are in roughly equalconcentrations. In the third system,weak base is in extreme excess relative to itsconjugate acid. The results are shown in Figure 5-1.

System I: Mix 999 partsHAwith1partA". Tothismixture, add 1partOH".

Initially: -^J-=-±-=> after addition of 1part OH":-i—L=_2_J [HA] 999 [HA] 998

Because Ka =[H+] x-^-J-, and Ka is aconstant, when -i—- doubles, [H+][HA] [HA]

must be cut in half,and the pH changes. SystemI is NOTa buffer.

System II: Mix 500 partsHAwith500 part A". Tothismixture, add 1part OH".

Initially: i^J-=52Q.=> after addition of 1part OH_:-^J-=^217 [HA] 500 [HA] 499BecauseKa = [H+] x-—-, and Ka is a constant, when -—- barely changes,

[HA] [HA][H+] is constant, and thepH doesn't change. SystemII is a buffer.

System III: Mix 2partsHAwith998 part A". Tothismixture, add 1part OH".

Initially: i^l=998 rafter addition of 1part OH-:-I^J-=2997 [HA] 2 [HA] 1

Because Ka =[H+] xJ—-, and Ka is a constant, when -—— doubles, [H+][HA] [HA]

must be cut in half,and the pH changes. SystemHI is NOTa buffer.

Figure 5-1

The conclusion is that the pH remains constant only if the weak acid and itsconjugate base areinroughly equal concentration. The addition ofstrongbase orstrong acid shifts the ratio ofweak acid (HA) to its conjugate base (A"), but thepH does not change, if the A": HA ratio is close to 1.0.


General Chemistry Buffers and Titration Buffers

pH Range for BuffersThe experiment in Figure 1 explains why the weak acid and weak conjugate basemust be present in roughly equal parts. If they are not relatively close in molequantity, then the system does not act as a buffer. According to convention, theratio can not exceed 10 : 1. Substituting 10 : 1 and 1 : 10 into the Henderson-Hasselbalch equation shows us that the range of a buffer is the pKa of the weakacid ± 1. The pH of a buffer solution obeys the Henderson-Hasselbalch equation,which is shown in Figure 5-2.

, [A-] TT ,, , Moles conjugate basepH = pKa + log-^- .\pH = pKa + log }—2[HA] Moles conjugate acid

Lowest pH =pKa +log-i- =pKa -1 Highest pH =pKa +log!2. =pKa +110 1

/. pH range = pKa ± 1

Figure 5-2

The derivation of the Henderson-Hasselbalch equation from the Ka equation ison page 264. The Henderson-Hasselbalch equation shows that as [conjugatebase] increases, buffer pH increases. The Henderson-Hasselbalchequation alsoshows that as [conjugate acid] increases, buffer pH decreases. It also offersquantifiable verification of the concept that when pH is greater than the pKa, thesolution is rich in conjugate base.

Example 5.1Ifwater is added to a buffer solution with pH = 3.96, what happens to the pH?A. The pH increases slightly.B. The pH decreases slightly.C. The pH remains the same.D. If the pH is greater than 7, then it decreases. If the pH is less than 7, then it

increases.

SolutionAddition of water to a buffer equally dilutes the concentration of the weak acidand its weak conjugate base. This means that the mole ratio of the weak base tothe weak acid does not change upon the addition of water. According to theHenderson-Hasselbalchequation, the pH of the solution does not changebecausepKais constantand the fraction has not changed. Theresult is that the pH of abuffer does not change when it is diluted. This is why the Henderson-Hasselbalch equationcan bewritten as moles A" overmoles HA, as well as [A"]over [HA].

Knowing the buffer range is important when making a buffer, which isaccomplished in two steps. First, an acidwithin the range must be chosen. ThepKa of the acid should be as close as possible to the desired pH. Second, amixture containing both weak acid (HA) and its weak conjugatebase (A") shouldbe formed so that the two species are in roughly equal concentration. Select anacid for the buffer whose pKa value lies within the ±1 range of the desired pH.For buffering at pH values between 2 and 5, carboxylic acids are typical. Forbuffering at pH values between 8 and 11, amines are typical.



Buffer RecipesBuffers can be made either by mixing the conjugate pair together, or by partiallytitrating either component in a conjugate pair of weak reagents. The partialtitration method can be either to half-titrate the weak acid with strong base, or tohalf-titrate the weak conjugate base with strong acid. First, a weak acid must bechosen with a pKa value close to the desired pH. Buffers can be mixed by any ofthe methods shown in Figure 5-3.

1. Weak acid + the salt of the conjugate base in roughly equal moleproportions (e.g.,HCO2H with HC02Na)

2. Weak base + the salt of the conjugate acid in roughly equal moleproportions (e.g., NH3 with NH4CI)

3. Weak acid and roughly half of an equivalent of strong base (e.g.,HOAc with half equivalent KOH)

4. Weak base and roughly half of an equivalent of strong acid (e.g.,H3CNH2 with half equivalent HCl)

Figure 5-3

When using either of the last two methods, titration takes place until the desiredpH is achieved within the buffering region. The buffering region (where the pHdoes not change appreciably) is found in the middle area of the titration curve,between the starting point and the equivalence point. Buffering occurs only withthe titration of a weak reagent by a strong reagent. A strong acid combined withits conjugate base or a strong base combined with its conjugate acid do notproduce a buffer, so strong acid and strong base titration curves have nobuffering region.

Example 5.2Which of the following solutions results in a buffer with a pH of 5.0, given thatHA has a pKa of 4.7?A. HA with one-half equivalent of A"B. A"with one equivalent of HAC. HA with one-third equivalent of OH"D. A"with one-third equivalent of H30+

SolutionThe pH is greater than the pKa, so the solution must be rich in the deprotonatedspecies. In choice A, [HA] > [A-], so the pH is less than pKa (4.7), meaning thatchoice A can be eliminated. In choice B, [HA] = [A-], so the pH equals pKa (4.7),meaning that choice Bcan be eliminated. In choice C, one-third of an equivalentof A"forms from the reaction, and two-thirds of an equivalent of HA is left over.In choice C, [HA] > [A~], so the pH is less than pKa (4.7),meaning that choice Ccan be eliminated. In choice D, one-third of an equivalent of HA forms from thereaction, and two-thirds of an equivalent of A" is left over. In choice D, [HA] <[A-], so the pH is greater than pKa (4.7). Choice D is the best option for thosewho like to pick the correct answer. To solve for the exact numerical value, theHenderson-Hasselbalch equation must be employed. The values differ by 0.3,and the antilog of 0.3 is 2, so the correct answer must form a 2 : 1 ratio ofconjugate base to acid.



Example 5.3Which of the following combinations, when mixed in the correct ratio, produce abuffer?

A. NaOH/NaClB. NaN03/HN03C. HC1/KOHD. NH3/HCI

SolutionChoice A is a strong base and a neutral salt, which does not make a buffer.Choice B is a neutral salt and a strong acid, which does not make a buffer.Choice C is a strong acid and a strong base, which also does not make a buffer.By elimination, choiceD is correct. To be a buffer, the weak base (NH3) must behalf-titrated by the strong acid (HCl). Ammonia and hydrochloric acid make abuffer when mixed in the correct ratio (2:1).

Example 5.4Which of the following does NOT form a buffer when added to NaHC03(s)?A. NaOH(aq)B. HCl(aq)C. H2C03(aq)D. H20(l)

SolutionAdding half of an equivalent of NaOH to a sodium bicarbonate solutionconvertshalf of theHCO3" to its conjugate base CO32". The mixture of the two forms abuffer, so choice A is correct. Adding half of an equivalent of HCl to a sodiumbicarbonate solution converts half of the HCO3" to its conjugate acid H2CO3. Themixture of the two forms a buffer, so choice Bis correct. Adding an equivalent ofH2CO3 to a sodium bicarbonate solution results in equal portions of the HCO3"and its conjugate acidH2CO3. Themixtureof the two forms a buffer, so choice Cis correct. Adding water does not make a buffer, becausewater is amphoteric, soit does not convert sodium bicarbonate to either its conjugate acid or conjugatebase. To be a buffer, there must be both a weak acid and its weak conjugate basepresent in solution at the same time. Thismeans that choiceD is correct.

Example 5.5If the ratio of base to acid in a conjugate pair is 3:1,and the weak acid has a Ka =1.0 x10"5, what canbe said about the pH of thebuffer solution?A. pH<5B. pH = 5C. 5<pH<7D. pH>7

SolutionIf the ratio of base to acid were 1:1, the pH would equal the pKa. Because thebase is inexcess, the pHisgreater than the pKa. The pKa is -log 1.0 x10"5 =5.This eliminates choices A and B. According to the Henderson-Hasselbalchequation, the pH is log 3 greater than the pKa. Log 10 = 1, and log 10 is greaterthan log 3, so pH (= pKa + log 3) < 6. The best choice is C.



Example 5.6If 1.0 moles of a weak acid in 1.0 liters of water are treated with 0.4 moles ofstrong base, what is the pH of the solution? (Ka for the weak acid is 2.0 x10"4)A. pH<3.7B. pH = 3.7C. 3.7<pH<7D. pH>7

SolutionIf the ratio of base to acid were 1:1, the pH would equal the pKa. The pKa is - log2.0 x10"4 =4- log 2=4- 0.3 =3.7, ashinted at by theanswer choices. To behalf-titrated, it would require0.5 moles of strong base. At the half-titrationpoint, thepH =pKa. With only 0.4 moles ofbase, thehalfway point is not yet reached andthere is excess weak acid relative to conjugate base (0.6 moles to 0.4 moles).According to the Henderson-Hasselbalch equation, the pH is less than the pKa.Thebest answer is choice A. ThepH is less than pKa, which is 3.7. Beaware thatthis questioncouldabout the [H+] as well as the pH of the solution. If a solutionis half-titrated, then the Ka = [H+]. This can be tricky, but the answer is foundfrom the Henderson-Hasselbalch equation. pH =pKa +logO^4-, so pH<pKa.

0.6

Example 5.7Abuffer made by mixing 100 mL of0.5 MHOAc (Ka =1.8 x10"5) with 25 mL of1.0MKOHhas a pH approximately equal to which of the following values?A. 0.2B. 4.7C. 7.0D. 9.3

SolutionThere are 0.05 moles of HOAc present and 0.025 moles of KOH present. Thismeans that exactly halfof an equivalent of strong base has been added to a weakacid, converting halfof theoriginal weakacid to its conjugate base. Half of theoriginal weakacid remains unreacted. This means that pH =pKa. ThevalueforpKa is solved for as follows:

pKa =-log (1.8 x10"5) =-log 1.8 - log 10"5 =-log 1.8 - (-5) =-log 1.8 +5Log 1.8 is lessthan1,so thepH isgreaterthan 4 and less than5. Choice Bisbest.

Example 5.8A buffered solution initially has a pH of 8.31. When five drops of 12MHCl areadded to a 500-mL beaker filled with this buffered solution, what would beexpected for the final pH value?A. 3.31

B. 8.26C. 8.31D. 8.36

SolutionAddinga little acid decreases thepH slightly. Choice Ashowsa lowerpH value,but a drasticallylower pH value. The change is small, so the answer is choiceB.



Physiological BuffersPhysiological pH is considered to be 7.4 (although venous and arterial pH vary,and gastric fluids are highly acidic). The pKa value for physiological acidsshould be considered relative to this value in order to determine the structure ofthe compound in vivo. When the pH of the environment is greater than the pKaof the species, it exists predominantly in its deprotonated state. Equally, whenthe pH of the environment is less than the pKa of the species, it existspredominantly in its protonated state.A prime example of physiological buffering involves carbonic acid, which is usedto buffer the blood of biological systems. Arterial blood is oxygen rich (and thuscarbon dioxide-poor), so it has a slightly higher pH than venous blood which iscarbon dioxide-rich. Carbon dioxide when added to water undergoes thecomplex equilibrium shown as Reaction 5.2.

C02(g) + H20(1) ^-^ H2C03(aq) ^=^ HC03-(aq) + H+(aq)Reaction 5.2

This equilibrium regulates blood pH, so any conditions (or disorders) that affectcarbon dioxide levels in physiological systems also affect the pH and buffering ofblood. For instance, emphysema hinders the uptake of oxygen from the lung andthe release of carbon dioxide into the lung. The blood can compensate for thereduced uptake of oxygen by increasing heart rate and producing more redblood cells. But the release of carbon dioxide is not as easily adjusted for. Theconsequence is that carbondioxide levelsin thebloodincrease, causingthe bloodto be more acidic than normal. This condition is known as respiratory acidosis.This can inhibit the binding of oxygen, as is demonstrated by the Bohr effect.Using the same logic, it canbe concluded that respiratory alkalosis results fromtheexcessive loss of carbon dioxide.

Gastricfluids are highly acidic (rich in HCl), so the loss of gastric fluids results inthe loss of acid from the body, producing a condition known asmetabolic alkalosis.Loss ofHCl can accompany vomiting or the pumping of the stomach. When thestomachis pumped, the acidic solution that is removedmust be replenished toreduce the risk of metabolic alkalosis. As a point of interest, food poisoning anddrug overdose patientsare givena solutionofcharcoal and water to drink,whichabsorbs the organic toxin. The porous carbon matrix binds organicmoleculesbetter than the water, especially compounds with rc-bonds.

Waste products exit the body in a chemically neutralized state (and believe methat if they didn't, you'd notice), so the lower end of the GI trackmust be basic.Losing lower intestinalfluids results in the lossofbasic metabolites, in particular,HCO3" (bicarbonate). Diarrhea (which can be caused by Olestra consumption)results in the loss of basic metabolites, producing a condition known as metabolicacidosis. The intertwining of acid-base chemistryand physiology is a perennialfavorite on the MCAT. Figure 5-4 summarizes the pH-related disorders of thebody.

Retention ofCO2: Blood pH i .-. Respiratory acidosisLoss ofCO2: Blood pH T .*. Respiratory alkalosisLoss ofHCO3": Blood pH I .*. Metabolic acidosisLoss ofH30+:Blood pH t .*. Metabolic alkalosis

Figure 5-4


General Chemistry Buffers and Titration Titration Curves

Titration CurvesQuantitative Reactions (Titration Theory)Acid-base titration is the quantitative addition of a titrant acid to a base insolution, or of a titrant base to an acid in solution. What is meant by quantitativeaddition is that the volumes of acid and base are measured precisely. Thetitration is said to reach its equivalence point when the moles of acid are equal tothe moles of titrant base. The volumes must be known precisely in order todetermine the relative concentrations, using the relationship that moles are equalto the product of volume and molarity. We shall consider two versions oftitration. The first case is the titration of a strong reagent by a strong reagent,such as strong acidby strongbaseor strong base by strong acid. The second caseis the titration of a weak reagent by a strong reagent, such as weak acid by strongbase or weak base by strong acid. There is no titration of a weak reagent with aweak reagent,becausethe twoweak reagents do not react with one another.We shall first consider what is in solution at different points during the titration.The two species canbemixedin three differentways: excess of one, excess of theother, or in equal portions. Letus address eachof these scenarios,alongwith theinitial point of titration. When less than one equivalent of the titrant is added tothe original reagent, this mixture is found in a region on the curve before theequivalence point. When exactly one equivalent of the titrant is added to theoriginal reagent, this is the equivalence point. Whenmore than one equivalent ofthe titrant is added to the original reagent, this mixture is found in a region onthe curve beyond the equivalence point. Table5.1 shows the different conditionsand pH calculations along a strong-by-strong titration curve.

Strong Acid titrated by a Strong BasePoint on Curve Species in Solution pH Range pH Calculation

Initial Pure strong acid pH«7 pH = -log[HX]initiaiBefore equivalence Leftover H30+ pH<7 pH = - log [H30+]excessAt equivalence H2O & neutral salt pH = 7 pH = 7 at equivalencePast equivalence Leftover OH" pH>7 pOH = - log [OH-]excess

Table 5.1

When a weak reagent is titrated by a strong titrant, it's different than the strong-by-strong titration. The difference when dealing with a weak reagent is that aweak conjugate is formed as the product, so there is an equilibrium betweenproducts and reactants. ThepH calculations must consider all species in solutionthat affect the pH. Table 5.2shows the different conditions and pH calculationsalong a weak-by-strong titration curve.

Weak Acid titrated by a Strong BasePoint on Curve Species in Solution pH Range pH Calculation

Initial Pure weak acid pH<7 PH=lpKa-llog[HA]

Before equivalence HA and A" (Buffer) pKa±l pH =pKa+log[A]V v [HA]At equivalence A" (diluted) pH>7 POH =IpKb-Ilog[A-]

Past equivalence Leftover OH" pH>7 pOH = - log [OH"]excess

Table 5.2


Zia Siddiqui

Zia Siddiqui


Fundamental Curve ShapesThe mixtures listed in Table 5.1 can be studied quantitatively with the help ofgraphs. Strong acid and strong base titration curves have distinct features.Figure 5-5 shows the titration of a strong acid by a strong base. We shall call thisCase la.

pH

iEquivalence point (pH = 7)moles HXinit = moles OH"added

mL strong base addedCase la: Strong acid titrated by a strong base

Figure 5-5

Figure5-6shows the titration of a strong base by a strong acid. We shall call thisCase lb.

Note that strong curves sharean equivalence point at pH = 7.

Equivalence point (pH = 7)moles OH-init = molesH\dded

mL strong acid addedCase lb: Strong base titrated by a strong acid

Figure 5-6

In Cases la and lb, the equivalence point is at pH = 7.0, because the neutralizedproduct is a neutral salt formed from the reaction of a strong acid with a strongbase. The graph, although not extended far enough to tell, is symmetric aboutthe equivalence point. The shape of the curve is referred to as sigmoidal. Thereagent when initially added to water fully dissociates (or in the case lb,hydrolyzes with a base), giving the highest concentration at first. Over thecourse of the titration, the concentration is reduced. This means that the pHgradually increases during the entire titration, with rapid pH change near theequivalencepoint. Because pH is measured on a log scale, the graph assumes thecharacteristic sigmoidal shape. The reasoning behind this is that as youapproach pH = 7, each change of 1.0 in the pH requires ten times less titrant. Forinstance, in going from pH = 1 to pH = 2, the hydronium concentration goes from0.10M to 0.01 M, a change of 0.09M H3O+ In going from pH = 2 to pH = 3, thehydronium concentration goes from 0.01 M to 0.001 M, a change of 0.009 MH30+. The change in hydronium concentration is ten times less when the pH isone unit closer to 7.0.

Titration Curves



The mixtures listed in Table 5.2 can be studied quantitatively with the help ofgraphs. Weakacid and weak base titration curves have distinct features. Case2a, shown in Figure5-7, shows the titrationof a weak acid by a strong base.

Equivalence point (pH > 7)

mL strong base addedCase 2a: Weak acid titrated by a strong base

Figure 5-7

Case2b,showninFigure 5-8, shows the titrationof a weakbase by a strong acid.

pHLip-o-weakness

Because [A"] = [HA]

PH = pKa

Note that strong curves share half-equivalence point at pH = pKa.

,Equivalence point (pH < 7)

mL strong acid addedCase 2b: Weak base titrated by a strong acid

Figure 5-8

Titration curves exhibit an initial cusp when the reagent being titrated is weak.This is referred to as a lip-o-zveakness, and may be used to distinguish the natureof the reagent from its titration curve. The lip-o-weakness is due to the fact thatthe equilibrium betweenweak acid and conjugatebase heavily favors one of thetwo species,so the pH changessignificantly. This can be seen the experiment inFigure5-1. ThepH at half-equivalence point is always equal to the pKa, whichcan be inferred from the Henderson-Hasselbalch equation.In Case 2a, the equivalence pH is above 7.0, because the neutralized product is aweak base (the conjugate base of the weak acid). The weaker the acid titrated,the stronger the conjugate base formed at equivalence, and consequently thehigher the pH at theequivalence point. ThepH of the conjugate basedepends onboth the concentration and the Kb of the base at equivalence.In Case 2b, the equivalence pH is below 7.0, because the neutralized product is aweak acid (the conjugate acid of the weak base). The weaker the base titrated,the stronger the conjugate acid formed at equivalence, and consequently thelower the pH at the equivalence point. ThepH of the conjugate acid depends onboth the concentration and the Ka of the acid at equivalence.



Example 5.9All of the following are a feature of a titration curve in which both reactants arestrong EXCEPT:A. an equivalence pH of 7.0.B. a sigmoidal graph shape.C. the most rapid pH change near pH = 7.D. a largechange in pH as the first fewdrops of titrant are added.

SolutionThis question tests your knowledge about strong-by-strong titration curves.When both reagents are strong, they neutralizeeachother and leavebehind a pH= 7aqueous salt solution. Thismakes choice A valid. Because the reagent fullydissociates (in the case of a strong acid) or hydrolyzes (in the case of a strongbase), the highest concentrations are initially observed. The result is a slowchange in pH until justbefore equivalence. This causes thecurve tobesigmoidaland makes choice Bvalid. The pH change is always rapid near the equivalencepoint, so for strong-by-strong curve, the pH change near pH = 7 is the mostdrastic. Choice C is valid. The pH of the titration mixture is nearly constant atthe start of the titration. This is to say that a strong-by-strong curve is "lip-free"in the beginning. ChoiceD is invalid, and is thus the best answer.

Example 5.10All of the following are features of a titration curve inwhich one reactant isweakand the titrant is strong EXCEPT:A. a half-equivalence pH equal to pKa.B. a sinusoidal graph shape.C. a pH at equivalence that is not equal to 7.D. a large change in pH as the first few dropsof titrant areadded.

SolutionThis question tests your knowledge about weak-by-strong titration curves.When a weak reagent is half-titrated by a strong titrant, half of the originalspecies in converted to its conjugate. This leaves half of the original reagentunreacted in solution. Because the two components of the conjugate pair are inequal concentration, the pH of the solution is equal to the pKa. This makeschoice A a valid statement and eliminates it. A sinusoidal graph implies a sinewave, which is not observed with weak-by-strong titration curves. This makeschoice B invalid, and thus the correct choice. When a weak acid is fullyneutralized bya strong base, it forms itsconjugate base. This results in a solutionwith pH greater than 7. The equivalence pH is not equal to7. When aweak baseis fully neutralized bya strong acid, it forms itsconjugate acid. This results in asolution with pH less than 7. The equivalence pH is not equal to 7. The pH atequivalence for a weak-by-strong titration curve isnot7, sochoice C isvalid andthus eliminated. Early in a weak-by-strong titration curve, the equilibriumbetween conjugates favors the one initially in solution. As a titrant is added, theequilibrium shifts drastically, resulting in a big change in the hydroniumconcentration. Abigchange in the hydronium concentration causes a significantchange in the pH of the solution. This makes choice D a valid statement andeliminates it.

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To understand titration curves better, Table 5.3 shows how the pH is calculatedat different points along a titration. The equations used to calculate pH are all inTable5.2. The titration represents the titration of a hypothetical weak acid with apKa of 5.0 and an initial concentration of 0.10 M. The weak acid is titrated by0.10M KOH(aq). The pH represents the solution pH after the titrant base hasreacted with the weak acid in solution. The value points in Table 5.3 are plottedin Figure 5-9, and by connecting the dots (yes, that's right, studying for theMCAT involves a little connect-the-dots), a titration curve can be generated.

mLKOH pH calculation pH

0.00 pH -pKa -1 log [HA] -5 -1log (0.10) - 2.5 +0.5 - 3.02 2 2 2

3.00

1.00 pH=pKa +log mLOH' -5.0 +log10 -5.0-log4950-mLOH" 49.0

3.31

2.00 pH=pKa +log mLOH" -5.0 +log20 -5.0-log2450-mLOH" 48.0

3.62

5.00 pH=pKa +log mLOH" -5.0+log5-0 -5.0-log950 - mLOH" 45.0

4.05

10.00 PH=pKa +log mLOH" -5.0 +log100 -5.0-log450-mLOH" 40.0

4.40

12.50 pH=pKa +log mLOH- -5.0 +log125 -5.0-log3Y 50-mLOH" 37.5 4.52

25.00 pH-pKa +log mLOH" -5.0 +log25 -5.0 +log1r r 50-mLOH" 25 5.00

37.50 pH=pKa+log mLOH" -5.0 +log37-5 -5.0 +log350-mLOH" 12.5

5.48

40.00 pH-pKa+log mLOH" -5.0 +log40-0 -5.0 +log450-mLOH" 10.0

5.60

45.00 pH=pKa+log mLOH" -5.0+log45-0 -5.0 +log950 - mLOH" 5.0

5.95

48.00 pH=pKa +log mLOH" -5.0 +log480 -5.0 +log2450-mLOH" 2.0

6.38

49.00 pH-pKa +log mLOH" -5.0 +log49-0 -5.0 +log4950-mLOH" 1.0

6.69

50.00 pH _ PKa +pHtitrant _5.0 +13 _182 2 2

8.85

51.00 pOH=-log [OHIexcess =-log(o.l0 x-L-) => pH =14 -pOH 11.00

55.00 pOH=-log [OH'lexcess =-log(o.l0 x-M => pH =14 -pOHV 105'

11.68

Table 5.3

Initially (at 0.00 mLadded), the pH is found using the shortcut equation. Beforeequivalence (from 1.00mL added to 49.00mL added), the pH is found using amodified version of the Henderson-Hasselbalch equation, where the moles OH"added are substituted for moles A" (given that the OH"converts to A-) and 50 -mL OH" is substituted for moles HA, because that describes the leftover HA. Atequivalence (at50.00 mLadded), the pH is approximated by averaging pH of thetitrant base and the pKa of the acid. The approximation is off by 0.15, so the pHcolumn shows the actual pH at equivalence. After equivalence (beyond 50.00mLadded), the pOH is found by taking the negative log mL OH" - 50 (for what isreacted) over the total volume (50 + mL OH").



Plotting a Titration CurveThe points from Table 5.3 are represented in Figure 5-9as circles. The curves isdrawn to fit the circles. The shape of the curve for a weak acid is distinct. Thebuffer region is not perfectly flat, showing that pH changes slightly in the bufferregion. The pH values at 1.00 mL and 2.00 mL are not exact, because theHenderson-Hasselbalch equation does not hold as well outside the pKa ± 1range. Nevertheless, the values are close enough to generate a reasonabletitration curve.

10 15 20 25 30

mL 0.10 M KOH added

11.68

35

Figure 5-9

The titration curve should become familiar with enough examples. What makescurves usefulis that they summarize a greatdealof information. Ifyou thinkoftitrationcurves in terms of equivalentsand regions, you canextracta substantialamount of information from them. For instance, when 21.27 mL of 0.1 M KOHhas been added, the pH is roughly 4.7 to 4.8. That range is small enough that aneducated guess can be made on a multiple-choice question. Weshall emphasizeusing titration curves in lieuofcalculating thepH,when it comes tobuffers andother mixtures.

Titration Curves

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Example 5.11What is the pH after 30 mL of 1.00M NaOH(aq) has been added to 100mL 0.50MHOAc(aq)? HOAc has a pKa = 4.74.A. 3.51B. 4.56C. 4.92D. 5.97

SolutionBecause the strong base is twice as concentrated as the weak acid, only half thevolume of strong base (relative to the weak acid) is required to reach theequivalence point. This means that 50 mL of 1.00 M NaOH(aq) fully neutralizesthe 100 mL of 0.50 M HOAc(aq). The halfway point of the titration is reachedwhen exactly 25 mL of 1.00M NaOHfaq) has been added. At the halfway point,the pH of the solution equals the pKa of the weak acid. The additional strongbase beyond the 25 mL makes the pH of the solution slightly greater than thepKa of the acid, 4.74. The best choice is answer C. The titration curve belowshows a summary of the intuitive approach:

pH

7"30 mL NaOH added

pKa- 1< pH < pKjZ 4

pKa-<pH<pKa + lA

25mL 1.00 M NaOH(padded

50

EquivalencePoint

When 30 mL has been added, the mixture is just beyond the half-titrated point onthe titration curve (as shown by the arrow). This makes the pH fall into therange of pKa < pH < pKa + 1, according to the titration curve. According to theHenderson-Hasselbalch equation, the pH equals the pKa + log (conjugate baseover acid). Past the half-titrated point, the concentration of the conjugate baseexceeds the concentration of the acid, so the ratio of conjugate base to acid isgreater than one. The log of a number greater than 1.0 is a positive value. Whena positive value is added to the pKa, the final value is greater than the pKa,confirming that pH > pKa. These questions should be answered quickly, usingeither a titration curve or the Henderson-Hasselbalch equation in a purelyconceptual manner.



Example 5.12What is the pH after 70 mL of 0.20 M HCl has been added to 50 mL 0.60 MH3CNH2? H3CNH2 has a pKb = 3.42.A. 9.44B. 10.51C 10.65D. 11.72

SolutionThe strong acid is one-third as concentrated as the weak base, so three times thevolume of HCl (relative to the H3CNH2) is needed to reach the equivalencepoint. Thismeans that 150 mL of 0.20 MHCl fullyneutralizes the 50mLof 0.60MH3CNH2. The halfway point of the titration is reached when 75mL of 0.20 MHCl is added. At the halfway point, the pH of the solution equals the pKa of theweak acid. Less than the 75 mL has been added, so the pH of the solution isslightly greater than the pKa of the conjugate acid, 10.58. The best choice isanswer C. The titration curve below shows how to estimate the value.

mL 0.20 M HCl(aq) added

EquivalencePoint

Example 5.13What is the pH ofa solutionmadebymixing 10.0mL0.10MHC02H(aq) with4.0mL 0.10M KOH(aq)? The pKa for HCO2H is 3.64.A. 1.34B. 3.46C. 3.82D. 9.36

SolutionThebest way to solve this question is to think in termsof equivalents. Theweakacid and titrant strong base are of equal concentration, so 10 mL KOH is oneequivalent (the amount needed to reach the equivalence point.) If 5.0mL areadded, then the acid is half-titrated, so pH = pKa. However,less than 5.0mLhasbeenadded, so pH is a little less than pKa. The pKa is 3.64, so the best answer ischoiceB. Choice A is too much less than the pKa (more than 1.0 is beyond the 10: 1 ratio, which would be when less than 1.0 mL of KOH had been added.)

Titration Curves



Concentration Effect on Titration Curve ShapeThe pH of a solution depends on the strength and concentration of the reagents,so strength and concentration also affect titration curves. The concentration ofthe reagents affects the dimensions of the titration curve, but not its basic shape.A strong-by-strong curve maintains the same fundamental shape (sigmoidal),but with varying concentrations, the curve may skew and elongate. Theequivalence point is always at pH = 7. A weak-by-strong curve also maintainsits same fundamental shape (i.e., has a lip-o-weakness), but with varyingconcentrations, the curve also skews and elongates. The half-equivalence point isalways at pH = pKa.

For a strong acid titrated by a strong base, as the concentrations of both reagentsincrease proportionally, the respective curves start lower and finish higher, butthey have the same distance in the x-direction (the mL axis). If the acidconcentration is increased but the base remains the same, then the curve startslower and stretches to a point farther from the origin along the x-axis for theequivalence point. If the base concentration is increased but the acid remains thesame, then the curve finishes higher and contracts to a point closer to the originalong the x-axis for the equivalence point. This is shown in Figure 5-10.

pH1.0 M HCl(aq) + l.OMKOH(aq)O.lMHCl(aq) + 0.1 M KOH(aq)

.01M HCl(aq) + .01 M KOH(aq)

Note that strong curves sharean equivalence point at pH = 7.

mL titrant strong base addedFigure 5-10

In each titration, the concentration of the strong acid is equal to the concentrationof the strong base, so the volume of base required is the same in each case. Thisis why the curves are similar in the x-direction. In the y-direction, the curvedepends on concentration. As the concentration lessens, the curve contracts withrespect to a line through pH = 7. Note that the initial pH is 0, 1, and 2respectively for the three titration curves. The variation in the concentration ofbase also causes the ends of the curves to vary (in the region of excess titrant).But at equivalence, the pH is 7, no matter what the concentrations are.

For a weak acid titrated by a strong base, as the concentrations of both reagentsincrease proportionally, the curve starts lower and finishes higher but advancesthe same distance in the x-direction (the mL axis) and has the same buffer region.If the acid concentration is increased but the base remains the same, then thecurve starts lower and stretches to a point farther from the origin along the x-axisfor its equivalence point. If the base concentration is increased but the acidremains the same, then the curve finishes higher and contracts to a point closer tothe origin along the x-axis for its equivalence point. This is shown in Figure 5-11.


General Chemistry

On each curve, because

[A-] = [H],pH = p£

Buffers and Titration

LOM HOAc(aq) + 1.0 M KOH(aq)0.1M HOAc(aq) + 0.1 M KOH(aq)01 M HOAc(aq) + .01 M KOH(aq)

The equivalence points vary, because higherinitial HOAC concentration leads to a higherOAc"concentration at the equivalence point.The approximate pH at equivalence is anaverage of the pl^ and pH of the titrant base.

As acid concentration lessens, initial pHincreases, and the size of the lip lessens.

mL titrant strong base added

Figure 5-11

In each titration, the concentration of the weak acid is equal to the concentrationof the strong base, so the volume of base required is the same in each case. Thisis why the curves are similar in the x-direction. In the y-direction, the curvedepends on concentration. As the concentration lessens, the curve contracts withrespect to a line through pH=pKa. The initial pH is)- pKa, i- pKa +0.5, and A-pKa + 1.0respectively for the three curves. The variation in the concentrationofbase also causes the ends of the curves to vary (in the region of excess titrant).But at half-equivalence, the pH is pKa, no matter what the concentrations are.Strength Effect on Titration Curve ShapeThe examples in Figure 5-10 and Figure 5-11 show the effect of concentration ontitration curves. The shape of a titration curve is also a reflection of the strengthof the reagents. When the concentrations are uniform between titration curves,then the shape of the curve and location of key points give clues as to the natureof the acidbeing titrated. As the acidbecomes weaker, thereis a largerinitiallip,a greater midpoint (where pH = pKa), and a greater equivalence point. This isshown in Figure 5-12. As acid strength increases, initial pH decreases, and thesize of the lip-o-weakness lessens.

0.iMH22^, N4-0lMKOH(aq}

0lMHNO3(aq^^™mL 0.10 M KOH (aq) added

Figure 5-12

CIO-

nAMKOH(*q |̂|OAc->N03'

Titration Curves



Polyprotic Acid Titration CurvesPolyprotic acids are acids which yield multiple equivalents of hydronium ion(H30+). Three common polyprotic acids with which every test-taker must befamiliar are carbonic acid (H2CO3), phosphoric acid (H3PO4), and sulfuric acid(H2SO4). Polyprotic titration curves have multiple equivalence points, one foreachdissociable proton. Butthe protons are removed one at a time, so the curvesfor each proton are separate. They should be treated as separate titration curvesthat happen to overlapon the samegraph. Figure5-13 shows the titration curvefor a typical diproticacid,whereboth protons are weak. Examples that share thesame basic curve shape include carbonic acid, glycine, or a mixture of two weakmonoprotic acids in the samesolution. Figure5-14 shows the titration curve fora typical diprotic acid, where the first proton is strong and the second proton isweak. An examplethat shares the samebasiccurve shape is sulfuric acid. Figure5-15shows the titration curve for a typical triprotic acid, where all three protonsare weak. Examples that share the same basic curve shape include phosphoricacid, citric acid, and glutamic acid.

Weak firstproton

pH


mL strong base added

Figure 5-13

Second equivalence point

PH = pKaSecond equivalence point

First equivalence point


Figure 5-14


Figure 5-15

312

Third equivalence point

Second equivalence point

The Berkeley Review


IndicatorsIndicator Composition and FunctionAn indicator is an organic compound with extended conjugation. Its weak acid(protonated) form and conjugate weak base (deprotonated) form have twodistinctcolors. The colors of both speciesare reflected colors. The energy statetransition that produces color involves the rc-bonding and rc-antibondingorbitalsand depends on the electron donating and withdrawing nature of substituents onthe rc-system. The lone pair formed upon deprotonation affects this conjugation.When the pH of the solution is less than the pKa of the indicator (pHsoiution <pKa (indicator))/ tne indicator exists predominantly in its protonated form (H—Ind > Ind"). If this is true, the solution assumes the hue of the protonated form ofthe indicator (H—Ind). When the pH of the solution exceeds the pKa of theindicator (pHsoiuti0n > pKa (indicator))/tne indicator exists predominantly in itsdeprotonated form (Ind" > H—Ind). If this is true, the solution assumes the hueof the deprotonated form of the indicator (Ind'). When the pH of the solution isequalto the pKa of the indicator (pHsoiution =pKa (indicator))/ me indicator existsequally in its protonated and deprotonated forms ([H—Ind] = [Ind']). If this istrue, the solution assumes a hue that is the result of a mixture of the protonatedand deprotonated forms of the indicator. When the pH of the solution is near thepKa of the indicator, the color varies with small changes in pH. Reaction 5.3represents the dissociation of an indicator in water.

H—Indyellow

H+

Reaction 5.3

Ind"blue

Table 5.4 shows the pH effects on the generic indicator in Reaction 5.3, alongwith a ratio of protonated-to-deprotonated species, and the color of the indicator.

pHsoiution Ratio of H—Ind to Ind" Mixture of colors to form solution color

pKa + 3.0 1:1000 1 yellow : 1000blue .-. bluepKa + 2.0 1:100 1 yellow : 100blue /. bluepKa +1.0 1:10 1 yellow: 10 blue .*. greenish bluepKa + 0.7 1:5 1 yellow: 5 blue .*. blue-greenpKa + 0.3 1:2 1 yellow : 2 blue /. bluish greenpKa 1:1 1 yellow: 1 blue .*. green

pKa-0.3 2:1 2 yellow : 1 blue .\ yellowish greenpKa-0.7 5:1 5 yellow : 1 blue ,\ yellow-greenpKa-1.0 10:1 10 yellow : 1 blue /. greenish greenpKa-2.0 100:1 100 yellow with 1 blue .\ yellowpKa - 3.0 1000:1 1000 yellow with 1 blue /. yellow

Table 5.4

When the solution is bluish green, it can be concluded that the pH of the solutionis slightly (about 0.2 to 0.4 times) greater than the pKa of the indicator. Thismeans that the pH of the solution can be approximated from the color of thesolution. The color change range (and thus the useful range) of an indicator ispKa(Indicator) ± 1- An indicator is generally used for one of two purposes. Thefirst is to detect the endpoint of a titration, and the second is to approximate thepH of a solution by observing the color of the indicator in the solution.

Indicators



Detecting the EquivalencePoint (Using Indicators)Indicators make the equivalence point of a titration visible, either by convertingfrom an uncolored species to the colored conjugate or by converting from onecolored species to its conjugate, which has a different color. An indicatorchanges color at a pH near theequivalence pointof the titration. The change incolor can thus be used to determine the equivalence point. Indicators are addedin small quantities, so they do not interfere with the titration. The pH of thesolution thus dictates the color of the indicator. As the titration approachesequivalence, the pH changes rapidly, so the color changes rapidly. A colorchange indicates that equivalence has been reached, if the correct indicator ischosen.

The ideal scenario for an indicator is when the pH at equivalence equals the pKaof the indicator. The range of a color change for an indicator is centered on itspKa. For the titration of a weak acid with a strong base, the pH at theequivalence point may not be known exactly, so a best approximation of theequivalence pH must be made. When approximating the equivalence pH,consider that the pH at equivalence is greater than the pKa of the weak acidbeing titrated, but less than the pH of the strong base being added. The pH atequivalence lies roughly halfwaybetween the pKa of the acid and the pH of thebase being added. A good approximation of the pH at equivalence is an averageof the pKa of theweakacid and the pH of the titrant strongbase. ThepKa of theindicator should be somewhere around (within ± 1 unit) the average of the pKaof the acid being titrated and the pH of the strong base. For the titration of aweak base with a strong acid, the pH at the equivalence point is less than the pKaof the conjugate acid of the weak base being titrated. The pKa of the indicatorshould be somewhere around (within ± 1 unit) of the average of the pKa of theconjugate acid of the weakbase being titrated and the pH of the strong acid. Theactive range for an indicator is represented by the two titration curves in Figure5-16 and Figure 5-17.

pH = pK^(acid)

Indicatorregion

pH at equivalence point > 7.0


Best scenario: pHequivaience =pKa(indicator) ±1PKa(acid) +PH(titrant base) v . ,

Best approximation: = P^a(indicator) ± 1

Choosing an indicator for the titration of a weak acid by a strong base

Figure 5-16



pH

pH =pl^(acid) " N.

7"\ Equivalence\ point

V ipH at equivalencepoint < 7.0 %^

mL strong acid added

Best scenario: pHequivalence =pKa(indicator) ±1

Indicatorregion

Best approximation: pKa(conjugate acid) + P^(titrant acid) = pK a(indicator) ±1

Choosing an indicator for the titration of a weak base by a strong acid

Figure 5-17

You should make note of the fact that the pH at equivalence for the titration of aweak acid by a strong base can be approximated quite closely by averaging thepKa of the weak acid with the pH of the titrant strong base. This value will bewithin 0.5 of the actual value, meaning that for the titration of acetic acid bysodium hydroxide, the pKa of the indicator should be greater than seven(somewhere around nine.) As the titration is carried out, the pH of solutionincreases; thus, the indicator starts out in its protonated form and eventuallybecome deprotonated. For the sample indicator in Reaction 5.2, the solutionwould go from yellow to green to blue.For the titration of a weak base by a strong acid, the pH at equivalence can beapproximated by averaging the pKa of the conjugate acid of the weak base withthe pH of the titrant strong acid. This means that for the titration of ammonia byhydrochloric acid, the pKa of the indicator should be less than 7.0 (somewherearound 5.0.) As the titration is carried out, the pH of solution decreases; thus, theindicator starts out in its deprotonated form and eventually becomes protonated.For the sample indicator in Reaction 5.3, the solution would go from blue togreen to yellow.

Indicators



Estimating Solution pH (using Indicators)For a solution of unknown pH, if the pH is within one unit of the pKa of theindicator, then the hue of the indicator can be used to estimate the pH of thesolution. Consider an indicator with a pKaof 6.83 that is used to test pool water.It isyellow when protonated and red when deprotonated. Table 5.5 can be usedto estimate the pH of the pool water.

pHsoiutionpH > 7.83pH = 7.83pH = 7.53pH = 7.13pH = 6.83pH = 6.53pH = 6.13pH = 5.83pH<5.83

Ratio of H—Ind to Ind"

1:10+1:10

1:51:2

1:1

2:1

5:1

10:1

10+:1

Solution Color

Red

Orangish redRed-orange

Reddish orangeorange

Yellowish orangeYellow-orangeOrangish yellow

Yellow

Table 5.5

If the results of the pool water test is red, the water isbasic; andbecause red isoutside the color change range, the exact pH cannot be approximated. If theresults of the pool water test is yellow, then someone has left some acid in it.Most people are aware that yellow pool water has some acid in it, and it shouldnot be swum in. Theidealcolor is orangewith a slighthint of red. Indicators areused inpHtest kits for swirrrming pools and fish tanks. The kits usually containmore than oneindicator, increasing theaccuracy of theapproximation. All oftheindicators in sucha testkit shouldhave pKa valuesbetween6.0 and 8.0, becausethe pHof the water should be around 7.0. Indicators are also used in pH teststicks, where a series of threeor four indicatorsare on the stick. The pKavaluesoftheindicators differ byroughly two units. This allows for a wider range ofpHvalues from which one can correlate the color to the solution pH. For instance, apHstick with three indicators, with pKa values of5.05,6.98, and 9.11, has a rangeofroughly 4.05 to10.11. This isbecause each indicator hasa two-pH-unit range.

Example 5.14Given the following indicators on a pHstick, what is thepH ofa solution thatyieldsX: red, Y: blue,and Z: red?

Indicator X: pKa =4.96; when deprotonated, it goes from yellow toredIndicator Y: pKa =7.01; when deprotonated, it goes from yellow toblueIndicator Z: pKa =8.98; when deprotonated, it goes from red toblue

A. 5B. 6

C. 7D. 8

SolutionBecause Indicator Yis blue, the pH must be at least one unit greater than 7.01(the pKa of Indicator Y). Because Indicator Zisred, the pHmust beat least oneunit less than 8.98 (the pKa ofIndicator Z). The only value greater than 7.01 andless than 8.98 is 8, choice D.


Buffersand

TitrationPassages

15 Passages

100 Questions

Suggested Buffers and Titration Passage Schedule:I: After reading this section and attending lecture: Passages I, IV, VII, VIII & XI


Following Task I: Passages II, III, V, VI, IX, & XIV (41 questions in 53 minutes)Time yourself accurately, grade your answers, and review mistakes.

Review: Passages X, XII, XIII, & XVFocus on reviewing the concepts. Do not worry about timing

II

III

R-E-V-I.E'WSpecializing in MCAT Preparation

isipiii^

I. Buffer pH and Weak Acids

II. Buffer Composition

III. Buffer Chart and pKa Chart Passage

IV. Molecular Weight from neutralization of an Organic

V. Conjugate Pair Titration Curve

VI. Titration Curves and Concentration Effects

VII. Titration Curves and Strength Effects

VIII. normality and neutralization

IX. Titration Curve of a Polyprotic Acid

X. Carbonate Titration Curve

XI. Indicator Selection

XII. Indicator Color and Solution pH

XIII. Indicator Table

XIV. pH Sticks and Indicators

XV. Acidity and Electronic Influences

Buffers and Titration Scoring Scale


84 - 100 13- 15

66-83 10 - 12

47 -65 7 -9

34-46 4-6

1 -33 1 -3

(1 -7)

(8- 14)

(15 -21)

Acid (22 - 29)

(30 - 36)

(37 - 42)

(43 - 49)

(50 - 56)

(57 - 63)

(64 - 70)

(71 -76)

(77 - 83)

(84 - 89)

(90 - 96)

(97 - 100)


An integral part of any biological study conducted invivo is the accurate simulation of body conditions with asmuch precision as possible. As important as any factor tobiological systems is pH. To accommodate the need for aconstant and accurate pH, organisms of all types use buffersto maintain a relatively constant internal pH range. A bufferexists when there is both a weak acid and its weak conjugatebase present in solution in roughly equal molarconcentrations. It is important that both the acid and the basebe water-soluble and exhibit no side reactions. Organic acidsand their conjugate bases are best for this purpose.

The Henderson-Hasselbalch equation, Equation 1, is usedto calculate the value of the pH for a buffer.

PH =pKa +log£**l[Acid]

Equation 1

In human blood, a buffer of bicarbonate and carbonic acidexists. Inorganic phosphates also play a role in bufferingwithin the body. The buffer must have a pH that is relativelyclose to 7.4, known as biological pH. This can be simulatedin the lab by starting with a weak acid whose pKa is as closeto 7.4 as possible. After the acid has been added to water, thesolution is titrated with strong base until a pH of 7.4 isobtained. This method ensures an accurate value. Buffers canalso be made by mixing equal molar portions of the weakacid and its conjugate base (the conjugate base may come inits salt form). The pH equals the pKa of the acid when abuffer is made with equal molar portions of acid andconjugate base.

1. What is the pH of a solution made by adding 0.839grams NaHC03(s) (MW = 83.9 grams/mole) to 100 mL0.10MH2CO3(aq)?

pKaj=6.4 pKa2=10.8

A. 3.20B. 6.40C. 8.60D. 10.80

2. Which of the following pH values is the BEST choicefor the pH of a buffer initially at pH = 7.21 after HClhas been added?

A. 7.14B. 7.21C. 7.28D. 8.31


3. What should be mixed to make a pH = 4.2 buffer?pKa(benzoic acid) = 4.2A. 10 grams C6H5CC>2Na + 10 grams C6H5C02HB. 10 mL 0.10 M C6H5CO2H + 5 mL 0.10 M NaOHC. 10 mL 0.10 M C6H5CO2H + 10 mL 0.10 M

NaOH

D. 10 mL 0.10 M C6H5C02H + 15 mL 0.10 MNaOH

4. How many mL of 0.20 M NaOH must you add to 50mL of 0.10 M HF to produce a solution with a pH of3.3? (pKa = 3.3)

A. 10.0 mLB. 12.5 mLC. 16.7 mLD. 25.0 mL

5. What is the acetate anion concentration in a solutionmade by mixing 20 mL 0.30 M HOAc with 10 mL 0.30MNaOH?

A. 0.07MH3CCO2-B. O.IOMH3CCO2-C. 0.15MH3CCO2-D. 0.20MH3CCO2-

6. Which of the following combinations produces a buffer?A. 10 mL 0.25 M NH3(aq) + 20 mL 0.25 M HCl(aq)B. 20 mL 0.25 M NH4Cl(aq) + 10 mL 0.25 M

HCl(aq)C. 10 mL 0.25 M NH3(aq) + 20 mL 0.25 M NaOH(aq)D. 20 mL 0.25 M NH4Cl(aq) + 10 mL 0.25 M

NaOH(aq)

What is the pH of 100 mL of 0.10 M propanoic acid,pKa(propanoic acid) = 5.0?A. 2.5B. 3.0

C. 5.0D. 6.0



A buffer is composed of a weak acid and its conjugatebase. In an effective buffer, the ratio of the conjugate pairmust be less than 10:1, in favor of either component.According to Equation 1, when the two components of theconjugate pair are roughly equal in concentration, the pH ofthe solution is approximately the pKa of the weak acid.

pH = pKa + log [Base][Acid]

Equation 1

Abuffer iscapable ofconsuming any acid or base that isadded to solution. The pH of an aqueous buffer solutionchanges only slightly after an acid or base is added. This isbecause there is an equilibrium between the conjugates.Table 1 lists a series of weak acids and conjugate bases, alongwith the pKa values for each acid. The pKa value of theweak acid and the acid-base ratio can be used to determine thepHof a bufferfrom Equation 1.

Weak Acid Conjugate Base pKa value

CIH2CCO2H CIH2CC02K 2.82

HF NaF 3.19

HC02H HC02Na 3.64

H3CCOCO2H H3CC0C02Na 3.86

C6H5C02H C6H5C02K 4.19

C6H5NH3C1 C6H5NH2 4.62

H3CCO2H H3CC02Na 4.74

C5H5NHCI C5H5N 5.16

4-N02C6H4OH 4-N02C6H40K 7.15

HCIO KCIO 7.49

HBrO KBrO 8.67

NH4CI NH3 9.26

C6H50H C6H5OK 10.01

H3CNH3CI H3CNH2 10.56

Table 1

To make a buffer, an acid must be chosen that has a pKavalue within one unit of the target pH. The closer the pKavalue is to the pH, the better the buffer. Buffers are madeeither by mixing the weak acid with itsconjugate base, or bypartially titrating either the weak acid with strong base, or theweak conjugate base with strong acid. When the pH exceedsthe pKa, there ismore conjugate base present in solution.

8. Which mixture does NOT produce a buffer?A. H3CCO2H with 2 equivalents of H3CCO2KB. NH3 with 2 equivalents of NH4CIC. H2CO3 1.5equivalents of KOHD. H3CNH2 with 1.5 equivalents of HCl


9. What is the pH of a solution that contains two partsweakacid and one part conjugatebase?A . pKa(weak acid) + 2B. pKa(weak acid) +'°g 2C . pKa(weak acid)" 2D. pKa(weak acid) - 'og 2

10. At which of the following pH values would it beMOST difficult to establish a pH buffer?A. 1.0

B. 3.0

C. 5.0

D. 7.0

11. A buffered solution is BEST described as an aqueoussolution where the:

A . hydronium and hydroxide concentrations are equal.B. hydronium-to-hydroxide concentration ratio never

exceeds 10:1 or is less than 1:10.C. hydronium and hydroxide concentrations are

relatively constant.D. hydronium-to-hydroxide concentration ratio is

within one unit of the pKa value for the weak acid.

12. Addition of 1.00 mL of 0.10 M KOH(aq) to a solutionmade by mixing 15.00 mL0.10 M H3CC02H^/J with10.00 mL 0.10 M U^CC02^n(aq) results in:

A . a solution with a pH less than 3.74.B. a solution with a pH between 3.74 and 4.74.C. a solution with a pH between 4.74 and 5.74.D. a solution with a pH greater than 5.74.

13. In which solution is there the LARGEST ratio otconjugatebase to conjugate acid?A . A solution of HC02H and HC02" with pH = 4.00B. A solution of HCIO and CIO" with pH = 7.00C . A solution of HBrO and BrO" with pH = 8.50D. A solution of NH3 and NH4+with pH = 9.50

14. After 5.00 mL H20f7J has been added to 50.00 mL of abuffered solution with pH initially at 5.0, the pH:A. drops slightly.B. remains constant.

C. increases slightly.D. increases drastically.



Buffers are aqueous solutions of weak acids and theirconjugate base. The pH of solution is dictated by theHenderson-Hasselbalch equation: pH = pKa + log base/acid.This means that a buffer solution should be mixed in amanner where the pKa of the acid is close to the pH desired.Table 1 lists the Ka and pKa values for some commonmonoprotic weak acids:

Acid Ka value pKa valueH2NCONH3+ 6.6 x 10"1 0.18

HF 6.8 x 10"4 3.17

HCNO 3.5 x 10"4 3.49

HC02H 1.7 x 10-4 3.78

H3CCOC02H 1.4 x 10"4 3.89

C6H5C02H 6.5 x 10"5 4.19

C6H5NH3+ 2.3 x 10"5 4.64

H3CCO2H 1.8 x 10'5 4.74

C5H5NH+ 7.1 x lO'6 5.16

HCIO 3.5 x lO'8 7.49

B(OH)3 5.9 x 10-10 9.22

NH4+ 5.6 x 10"10 9.26

HCN 4.9 x lO'10 9.32

H3CNH3+ 2.2 x lO"11 10.66

(H3C)2NH2+ 1.9 x lO'11 10.77

Table 1

Polyprotic acids can also be used in making buffers. Adifficulty that arises with polyprotic acids involves the two ormore pKa values. For any given polyprotic acid, theconjugate pair will buffer at the respective pKa for the acid ofthe conjugate pair. For instance, carbonate/bicarbonate willbuffer at a pH around 10.8 because pKa2 of carbonic acid is10.81. Carbonic acid/bicarbonate will buffer at a pH around6.4 because pKaj of carbonic acid is 6.37. Table 2 lists theKa values for some common polyprotic acids:

Acid Kal value Ka2 value Ka3 value

H2C204 5.6 x lO'2 5.1 x lO'5H2SO3 1.3 x 10"2 6.3 x 10"8H3P04 6.9 x lO"3 6.2 x 10"8 4.8 x lO'13H2C03 4.3 x lO'7 1.5 x 10"nH2S 8.9 x 10"8 3.8 x lO'13

Table 2

As a point of interest, carbonates and phosphates arebelieved to be the major contributors to buffering in humanblood. For years it was believed that carbonate played themajor role, but recent research indicates that phosphate mayplay a more significant role than carbonate in the overallbuffering. Within the kidneys, phosphates are known to playa significant role.


15. Combining all of the following results in a buffer withpH = 9.5 EXCEPT:

A . 1.8 equivalents NH3 with 1.0 equivalents NH4+.B. 1.6 equivalents NaCN with 1.0 equivalents HCN.C. 0.7 equivalents HCl with 1.0 equivalents NH3.D. 0.65 equivalents NaOH with 1.0 equivalents HCN.

16. To make a buffer at pH = 10.83, which of the followingshould be mixed?

A. One-half equivalent of NaOH with one equivalentof H2CO3

B. One and one-half equivalents of NaOH with oneequivalent of H2CO3

C. One and one-half equivalents of NaOH with oneequivalent of H3PO4.

D. Two and one-half equivalents of NaOH with oneequivalent of H3P04

17. Which of the following solutions forms a buffer with apH greater than 7.0?A. One and one-half equivalents of KOH with one

equivalent of H2C204B. One-half equivalent of KOH with one equivalent of

H2CO3C. One-half equivalent of KOH with one equivalent of

HCIOD. One-half equivalent of HCl with one equivalent of

C5H5N

18. Which of the following mixtures would NOT result in asolution with pH = 4.00?

A. Excess HCO2" mixed with HCO2HB. Excess C6H5NH3+ mixed with C6H5NH2C. Excess C6H5C02- mixed with C6H5CO2HD. Excess H3CCOC02_ mixed with H3CCOCO2H

19. Carbon dioxide when dissolved into the blood formscarbonic acid. What is observed in C02-enriched blood?

A. The pH is less than 7.4.B. The [P043_] increases.C. The [HCO3"] decreases.D. The [H30+] decreases.


20. In which of the following solutions is the conjugatebase in GREATER concentration than the acid?

A . HF(aq) with F"(aq) at pH = 3.00B. H2C03(aq) withHC03"(aq) at pH = 6.00C. HC204-(aq) with C2042-(aq) atpH =4.00D. H2P04"(aq) with HP042-(aq) atpH =8.00

21. Biological pH is approximately 7.4. Which of thefollowing is NOT true about theconcentration of bufferspecies at this pH?A. [HCO3-] > [H2C03]B. [HCO3I > [C0321C. [H2P04-] > [HPO42-]D. [HP0421 > [PO43-]



The formula weight of an acid can be determined bytitration, using a strong base of known concentration. Theprocess involves the titration of an exact mass of someunknown acid. Once enough base has been added to reachequivalence, the moles ofbase added are used todetermine themoles of acid that were present in the solution initially. Themass of the acid divided by the moles of the acid gives theformula weight for the acid, not themolecular weight. Oftenboth weights are the same, but a difference arises if theunknown is a polyprotic acid. Only when the acid ismonoprotic is the formula weight equal to the molecularweight. Table 1 shows some phenols with their respectivepKa values. All of the phenols are solids at roomtemperature and aremonoprotic acids.

Structure Formula pKa

02N—U^ \- OH P-O2NC6H4OH 7.2

"VQ-oH P-H3CCOC6H4OH 8.4

Qkoh C6H5OH 10.0

^-o^ p-H3CC6H4OH 10.4

h3co-^ y~m p-H3COC6H5OH 11.2

Table 1

If the molecular mass of an acid is known, the pH of asolution can be calculated from its gram concentration insolution. To do this requires converting from grams intomoles. Once the concentration is known, the shortcutequation, Equation 1, may be used toquickly calculate the pHof the aqueous solution of the acid:

pH =1 pKa -1 log [HA]2 2

Equation 1

22. 1.0 gram of which of the above acids requires exactly30.0 mL of 0.20 M NaOH to reach equivalence?A . Acetic acid (H3CC02H) MW = 60B. Trichloroacetic acid (CI3CCO2H) MW = 151.5C. p-nitro benzoic acid (02NC6H4C02H) MW=167D. Benzoic acid (C6H5C02H) MW = 122


23. If 1.0 gram of an unknown acid requires exactly 40.00mL of 0.25 M NaOH to reach the equivalence point,what is the formula weight for the unknown acid?A. 50 grams per moleB. 80 grams per moleC. 100 grams per moleD. 125 grams per mole

2 4. Consider this graph:

mL of titrant

The graph represents the titration of:A. H3CCO2H by NaOH.B. HCl by NaOH.C. NH3 by H3CCO2H.D. NH3byHCl.

25. Which of the following indicators would be BEST forthe titration of p-nitrophenol(02NC6H40H) by NaOH?

A. Thymol blue (pH range of color change is 1.2 to2.8)

B. Methyl red (pH range of color change is 4.6 to 5.8)C. Bromthymol Blue (pH range of color change is 6.0

to 7.6)D. Phenolphthalein (pH range of color change is 8.0

to 9.6)

26. Which of the following mixtures results in a bufferedsolution?

A. 10 mL 0.25 M NaOH + 10 mL 0.25 M H3CC02HB. 20 mL 0.25 M NaOH + 10 mL 0.25 M H3CC02HC. 10 mL 0.25 M NaOH + 20 mL 0.25 M H3CC02HD. 10 mL 0.25 M HCl + 10 mL 0.25 M H3CCO2H


2 7. The BEST choice for a pH = 8.5 buffer would be whichof the following?A. O2NC6H4OH with less than one full equivalent of

NaOHB. H3COC6H4OH with less than one full equivalent

ofNaOHC. H3CC6H4OH with less than one full equivalent of

NaOHD. H3CCOC6H4OH with less than one full equivalent

of NaOH

28. What is the formula weight (equivalent weight) ofoxalic acid (HO2CCO2H)?

A. 45 grams per moleB. 90 grams per moleC. 135 grams per moleD. 180 grams per mole

29. 25.0 mL of an unknown acid when titrated by exactly30.0 mL of 0.100 M KOH(aq), requires seven drops of0.100 M HCl(aq) to return to equivalence. What is theconcentration of the unknown acid?

A. 0.1217Macid(aq)B. 0.1183Macid(aq)C. 0.0849 M acid(aq)D. 0.0817 Macid(aq)



Conjugate acid/base pairs areconnected bya relationshipbetween pKa and pKb- Equation 1 shows the relationshipwithin a conjugatepair at 25°Cin aqueous solution.

pKa (conjugate acid) +P^b(conjugate base) = 14Equation 1

When titrating a weak acid or weak base, theportion ofthe curve following the initial drops of titrant up until justbefore theequivalence point is an equilibrium mixture of theconjugate pair. As such, the titration curve of onecomponent in a conjugate pair have similarities to thetitration curve of the other component.

The titration curves for conjugatepairs are inversegraphsthat intersect at the half-titrated point. At this point, the pHof the solution equals the pKaof the conjugate acid. At thissame point, the pOH of the solution equals the pKb of theconjugate base. Figure 1 shows the titration of acetic acid(H3CCO2H) with strong base (NaOH) overlaid onto thetitration of sodium acetate (H3CC02Na) with strong acid(HCl). In bothtitration curves, all species are in equal molarconcentrations. The pKa for carboxylic acids is generallybetween 3 and 5.

T r12.5 25.0

mL titrant solution added •

25mL0.10MHjCCO2Htitrated by 0.10 M NaOH

25mL0.10MH$CCO2Natitrated by 0.10 M HCl

Figure 1

Figure 2 shows the titration of methyl ammoniumchloride (CH3NH3CI) with strong base(NaOH) overlaid ontothe titration of methyl amine (CH3NH2) with strong acid(HCl). In both titrations, all species are in equal molarconcentrations. The pKa for amines is generally between 9and 11.


12.5mL titrant solution added

25 mL 0.10 M CH>NH3C1titrated by 0.10 M NaOH

25mL0.10MCHjNH2titrated by 0.10 M HCl

Figure 2

Both a weak acid and weak conjugate base titration curveof a conjugate pair show the same pH at the half-titrationpoint (indicated by the empty circle on both graphs),regardless of the initial concentration of the conjugatespecies. The pH at the half-titration point inFigure 1 is lessthan the pH at the half-titration point in Figure 2. This isbecause the pKa value of acetic acid is less than the pKavalue of methyl ammonium cation. At this point, theconjugate base concentration equals the acid concentration;therefore, according to theHenderson-Hasselbalch equation,pH = pKa.

30. By roughly how much do the two equivalents points inthe first graph differ?A. Fewer than 2.0 pH unitsB. Fewer than 4.0 pH units, but more than 2.0 pH

unitsC. Fewer than 8.0 pH units, but more than 4.0 pH

unitsD. More than 8.0 pH units

31. ThepHat equivalence is GREATEST forwhich of thefollowing titrations?A. The titration of 0.10 M H3CCO2H by NaOHB. The titration of 0.10 M H3CC02Na by HClC. The titration of 0.10 M CH3NH3CI by NaOHD. The titration of 0.10 M CH3NH2 by HCl


32. What is true in the titration of ammonia byhydrochloric acid, when the pH of the solution is greaterthan the pKa for ammonium chloride?A. [NH4+] > [NH3I; Ka(ammonium chloride) > [H+]B. [NH4+1 > [NH3]; Ka(ammonium chloride)< [H+]C. [NH4+1 < [NH3]; Ka(ammonium chloride) > [H+]D. [NH4+] < [NH3I; Ka(ammonium chloride) < [H+]

3 3. How does the pH at point a in Figure 1 compare to thepH at point d in Figure 1?A. The pH at point a is more than 1.0 pH unit greater

than the pH at point d.B. The pH at point a is greater than the pH at point d,

but the difference is less than one pH unit.C. The pH at point a is less than the pH at point d,

but the difference is less than one pH unit.D. The pH at point a is more than 1.0 pH unit lower

than the pH at point d.

34. Given that H3CCO2H has a lower pKa value thanCH3NH3CI, which of the following statements is true?

A. H3CCO2H buffers at a higher pH value thanCH3NH3+

B. H3CCO2H has a conjugate base with a lower pKbvalue than the conjugate base of CH3NH3+.

C. H3CCO2H dissociates less than CH3NH3+D. H3CCO2H is a better electron pair acceptor than

CH3NH3+

35. Given that H3CCO2H is a stronger acid thanCH3NH3+, which of the following statements is NOTtrue?

A. H3CCO2H yields a lower pH value thanCH3NH3+of equimolar concentration.

B. H3CCO2H has a conjugate base with a higher pKbvalue than the conjugate base of CH3NH3+.

C. H3CCO2H produces more conjugate base thanCH3NH34" when added to water.

D. H3CCO2H is a worse proton donor thanCH3NH3+


36. If a similar experiment with identical concentrations andvolumes were conducted using hydrofluoric acid andfluoride anion, what would be true, knowing that thepKa of hydrofluoric acid is lower than the pKa of aceticacid?

A. The initial pH for the titration of HF would begreater than the equivalence point for the titrationofH3CC02Na.

B. When both HF and H3CCO2H are one-third titratedby equimolar NaOH, [F"] > [H3CC02"! andKa(acetic acid) < I" ]•

C. When both HF and H3CCO2H are one-third titratedby equimolar NaOH, [HF] > [H3CC02H] andKa(acetic acid) < [H ].

D. When both HF and H3CCO2H are one-half titratedby equimolar NaOH, Ka(acetic acid) < tH+] in tneHF titration.



Titration involves the quantitative additionof one reagentto another, where the concentration is known for only one ofthe species. Acids are often titrated by strong bases, so thatthe concentration of the acid may be ascertained. Once theequivalence point has been reached, the volume of titrant ismeasured. Using Equation 1, it is possible to solve for themolarity of the acid, if the molarity of the titrantstrong baseis known.

M(acid)V(acid) =M(baSe)V(base)Equation 1

Besides the quantitative aspects, qualitative features ofthe acid may also be determined. The shape of the titrationcurve varies with the strength of the acid. For strong acids,the shape is sigmoidal, with a nearly horizontal initialregion. For weakacids, the shape is not sigmoidal, with aninitial vertical ascent before leveling off into a horizontalbuffering region. As the degree of the initialascent increases,it can be observed that the acid being titrated is weaker. Theconcentration of the acid also affects the titration curve. Thesame fundamental shape is observed, but the pH values aredifferent.

A researcher conducts two experiments studying theeffect of acid concentration. In the first experiment, shetitrates the same strong acid, HCl, at three differentconcentrations, keeping the acid and titrant base in the sameconcentration as each other. Figure 1 shows the threetitrations overlaid onto one graph. She finds that all threetitrations generate the same pH at their equivalence point,regardlessof the initial concentrationof strong acid.

pH7H

Titrations of HCl with NaOH

12.5

mL NaOH solution added

25.0

25 mL 0.01 M HCl titrated by 0.01 M NaOH



Figure 1

Copyright © by TheBerkeley Review®

In the secondexperiment, the researcher titrates the sameweak acid, HOAc, at three different concentrations, keepingthe acid and titrant base in the same concentration as eachother. Figure 2 shows the three titrations overlaid onto onegraph. It is found that all three titrations generated the samepH at theirhalf-equivalence points.

A

pH7H

Titrations of HOAc with NaOH

112.5

mL NaOH solution added

25.0

25 mL 0.01 M HOAc titrated by 0.01 M NaOH



Figure 2

Weak acid titration curves show the same equivalencepoint, regardless of the initial concentration of weak acid.The pH at the half-equivalence point in all three titrationcurves in Figure2 is equal to the pKa for acetic acid. At thehalf-equivalence point, the concentration of the conjugatebase equals theconcentration of the acid; therefore, accordingto the Henderson-Hasselbalch equation, pH = pKa.

37. The initial pH is GREATEST in which titration?A. The titration of 0.01 M H3CCO2H by NaOHB. The titration of 1.00 M H3CCO2H by NaOHC. The titration of 0.01 M HCl by NaOHD. The titration of 1.00 M HCl by NaOH

38. What is true in the titration of acetic acid by sodiumhydroxide when the pH of the solution is greater thanthe pKa for acetic acid?

A. [H3CC02-1 > [H3CCO2H]; Ka(acetic acid) > [H+lB. [H3CCO2-] > [H3CCO2H]; Ka(acetic acid) < [H+lC. [H3CCO2I < [H3CCO2HI; Ka(acetic acid) > [H+lD. [H3CCO2I < [H3CCO2H]; Ka(acetic acid) < [H+l

326 GO ON TO THE NEXT PAGE

39. The BEST explanation for the greater pH at theequivalence point observed with the higher initialconcentration of weak acid can be attributed to:

A. the greater number of mL of base solution added toreach the equivalence point.

B. the lower number of mL of base solution added toreach the equivalence point.

C. the greater conjugate base concentration at theequivalence point.

D. the lower conjugate base concentration at theequivalence point.

40. How would the titration curves in Figure 1 be affected ifthe base concentrations were all doubled, while the acidconcentrations remained the same?

A. Equivalence would be achieved with half thevolume of titrant base, and the shape of thetitration curves would change.

B. Equivalence would be achieved with twice thevolume of titrant base, and the shape of thetitration curves would change.

C. Equivalence would be achieved with half thevolume of titrant base, and the shape of thetitration curves would not change.

D. Equivalence would be achieved with twice thevolume of titrant base, and the shape of thetitration curves would not change.

41. Even though the NaOH concentration in the third trialis 100 times greater than the NaOH concentration in thefirst trial, the two graphs follow a similar slope. Thisis BEST explained by which of the followingstatements?

A. The solution is a buffered solution, so the pHchange is minimal.

B. The NaOH is a weak base and does not fully reactwith the HCl.

C. The pH is a log scale, so as the pH increases up to7.0, the amount of base necessary to increase thepH becomes less.

D. The pH changes only at the equivalence point.

Copyright © by The BerkeleyReview® 327

42. Which of the following graphs represents the resultsthat would be observed if the experiment described inthe passage were carried out with ammonia (NH3) andhydrochloric acid (HCl)?A.

pH

B.

A

pH

C.

A

pH

D.

A

pH

25mL0.10MNH3!with 0.10 M HCl

25mL1.00MNH3with LOOM HCl

1 1-12.5 25.0

mL HCl solution added

25 mL 0.01 MNHjlwith 0.01 MHCl \125mL0.10MNH3'»Vwith 0.10 MHCl V^^^

—

25 mL 1.00 MNH 3 ^^-^.^^with LOOM HCl

1 112.5 25.0

mL HCl solution added

25mL0.10MNH3|with 0.10 M HCl

25mL1.00MNH3with 1.00 M HCl

12.5 25.0mL HCl solution added

25 mL 0.01 MNHwith 0.01 M HCl

25mL0.10MNH3,with 0.10 M HCl

25 mL 1.00 MNH 3with LOOM HCl

12.5 25.0mL HCl solution added



The four titrations curves shown in Figure 1, representthe titration of three weak acids (HCN, HCIO, and HOAc)and the titration of the strong acid HCl. The equivalencepoint is represented by thedotat thevertical inflection pointof each curve. The respective conjugate base is shown nextto its equivalence point. At the startof each titration, 25mLof 0.10M acid are present. All of the acids are titrated byO.lOMKOH(aq).

mL 0.10 M KOH(aq) solution added •

Figure 1

The initial and equivalence pH values were collectedandrecorded for some of the titrations. Table 1 shows datacollected during the experiment.

Titration Initial pH Equivalence pH

HCl by KOH 1.00 7.00

HOAc by KOH 2.87 8.72

HCIO by KOH 4.23

HCN by KOH 11.01

Table 1

The pKa for each weak acid can be found by measuringthe pH of the aqueous solution at the half-equivalence pointin its respective titration. For instance, the pKa for HCN is9.32, which is the pH at the half-equivalence point of thetitration curve. The values are approximated, because thepoints on the curve cannot be read that accurately. The pHvalues are listed in Table 1 were recorded from a pH meter, sothey are considered to be reliable.

43. The greatest Ka value is found with which of thefollowing acids?A. HCN

B. HCIOC. HOAcD. HCl


44. If pKa of HOAc is 4.74, the pH at the equivalence inthe titration of HCN is 11.01, and the initial pH in thetitration of HCIO is 4.23, then what else must be true?

A. Initial pH in the titration of HCN is 3.86; the pHat equivalence in the titrationof HCIOis 9.61.

B. Initial pH in the titration of HCN is 5.16; the pHat equivalence in the titrationof HCIOis 10.08.

C. Initial pH in the titration of HCN is 6.12; the pHat equivalence in the titrationof HCIOis 11.42.

D. Initial pH in the titration of HCN is 7.42; the pHat equivalence in the titrationof HCIOis 10.34.

45. Which of the following statements are valid whencomparing the titration curve associated with a weakacid by strong base titration to the titration curveassociatedwith a strong acid by strong base titration?I. Strong acid titration curves have an initial drop inpH due to the dissociation of the protons, whileweak acid titration curves start with a plateau.

II. Weak acid titration curves have a buffer regionwhile strong acid titration curves do not.

HI. All points beforetheequivalence pointare less than7 for both the titration of and weak acid and thetitration of a strong acid.

A. I onlyB. II onlyC. I and II onlyD. II and D3only

46. Given that 0.10 M HIO has a pH greater than that of0.10 M HCN, we can conclude that:A. thepKaofHIOis8.61.B. the pH at equivalence in the titration of HIO by

KOH is 10.06.C. the pH of 0.10 M HlO(aq) is 7.21.D. the difference between the pKa of HIO and the pH

at the equivalence point in the titration of 0.10 MHlO(aq) by 0.10 M KOH(aq) is less than 3 pHunits.

47. When 10 mL 0.10 M HCIO is mixed with 10 mL 0.15M KCIO, the pH is 7.64. What is the pH after 30 mLof water is added to raise the volume to 50 mL?

A. 7.51B. 7.64C. 7.77D. 8.26


48. Which of the following curves accurately representsthe titrations of 0.10 M NaCN(aq) by 0.10 M HCl(aq)and 0.10 M NaClO(aq) by 0.10 M HCl(aq)?A.

D.

CIO

CN"

pH

7H

mL0.10 MHCl(aq) added 25.0

mL 0.10 MHCl(aq) added 25.0

W

HCN

LHCIO

mL0.10MHCl(aq) added 25.0


4 9. Which of these sequences relates the pKa values of theindicated acids in descending order?

A • pKa(HCN) >pKa(HC10) > pKa(HOAc) > pKa(HCl)B• pKa(HOAc) > pKa(HC10) > PKa(HCN) > pKa(HCl)C. pKa(HCl) > pKa(HCN) > pKa(HC10) > pKa(HOAc)D • PK3(HC1) > pKa(HOAc) > pKa(HC10) > pKa(HCN)



A polyprotic acid is an acid that contains more thanoneacidicproton. The secondproton lost by the acid is neverasstrongly acidic as the first proton lost by the acid. Thesecond proton can be removed using strong base, once thefirst proton has been completely removed from the acid.Concentrations of polyprotic acids are measured in terms ofnormality. Normality measures molar equivalence. Thenormality of an acid equals the molarity of the equivalentbase required to neutralizeall of the acidic protons.

A researcher sets out to determine the effect of mixingsolutions together, by studying their pH before and aftermixing. Into a flask (Flask 1) he places exactly 25.0 mL of0.20 N H2SO4 to be titrated by a solution of NaOH ofunknown concentration. Into a second flask (Flask 2) heplacesexactly 40.0 mL of 0.30 N H3PO4, also to be titratedby the same solution of NaOH. Both solutions are titratedinsuccession to a visual endpoint, determined by the colorchange of an indicator. The quantity of base needed toachieve this is recorded accurately to the second decimal place.

5 0. If Flask 1 requires exactly 20.0 mL NaOH solution tobe neutralized, what must be the concentration of theNaOH solution?

A. 0.125 MNaOH(aq)B. 0.200 M NaOH(aq)C. 0.250 M NaOH(aq)D. 0.500 M NaOH(aq)

51. Which of the following statements is true?A. pKai is always larger than pKa2-B. pKa2 is always larger than pKaj.C. There is no rule for pKa2 or pKai.D. pKa2 is greater than pKai only for the oxyacids.

52. A solution of 0.30 M H3P04(aq) has which of thefollowing values for normality?A. 0.10NH3PO4(aq)B. 0.30 N H3P04(aq)C. 0.60 N H3P04(aq)D. 0.90 N H3P04(aq)

53. When 0.1 moles NaH2P04 and 0.2 moles Na2HP04are mixed in 100 mL, what is the pH of the solution?

A. pH<PKal+PKa22

B. pKa2>pH>Pi^L±PKa22

c# pKa2 +pKa3>pH>pKa22

D. pH>PKa2 +PKa32


54. Howmany mL of 0.40 M H2S04(aq) would require thesame amount of base to reach full neutralization aswould 25 mL 0.60 M H3P04(aq)?

A. 25.00 mL 0.40 M H2S04(aq)B. 37.50 mL 0.40 M H2S04(aq)C. 50.00 mL 0.40 M H2S04(aq)D. 56.25 mL 0.40 M H2S04(aq)

55. Which titration curve represents the complete titrationof phosphoric acid?A.


56. What is the phosphate concentration in Flask 2 after 40mL of 0.30 N NaOH has been added?

A. 0.050 MP043-(aq)B. 0.100 MP043-(aq)C. 0.450 MP043"(aq)D. 0.900 MP043"(aq)



Polyprotic acids are acids that contain more than oneacidic hydrogen. A typical example is sulfuric acid (H2SO4).The first proton is easily removed, while the second proton isharder to remove. Sulfuric acid is a strong acid with respectto its first proton, but weak with respect to its second proton.The titration curve for a polyprotic acid looks like twoseparate titration curves that have been connected. Theendpoint of the curve for the first proton is the starting pointof the curve for the second proton. A student titrates anunknown diprotic acid with strong base. The titration curvefor the experiment is shown in Figure 1.

mL Titrant added

Figure 1

The exact pH and exact volume of titrant used are notknown, but the graph is proportional throughout the durationof the titration. The student repeats the experiment threetimes,and the graph in Figure 1 represents the best resultsofthe three trials.

5 7. The unknown acid can BEST be categorized as which ofthe following?

A. A strong monoprotic acidB. A diprotic acid with both protons strongC. A diprotic acid with one strong proton and one

weak protonD. A diprotic acid with both protons weak

58. Which of the following relationships is NOT true aboutthe unknown acid?

A. pKai < pKa2B. pH at first equivalence point > pKaiC. pH at second equivalence point > pKa2D. pH at first equivalence point > pKa2

59. In Figure 1, the titrant can best be described as a:A. strong acid.B. strong base.C. weak acid.D. weak base.


6 0. Which change requires that the MOST titrant be addedto the solution?

A. Going from a pH < pKai to a pH > pKa2B. Going from the first equivalence point to the

second equivalence pointC. Going from a pH > pKai to a pH < pKa2D. Going from the initial point to the first equivalence

point

61. If, after the addition of 5 mL of titrant, the pH of thesolution is less than the pKai of a diprotic acid, whatvolume of titrant is required to reach the secondequivalencepoint from the fully protonated state?A. Between 5 and 10 mLB. Between 10 and 15 mLC. Between 15 and 20 mLD. More than 20 mL

62. What is true the predominant species and any otherspecies present between points d and e on the titrationcurve?

A. A2" (the fully deprotonated form) is all that ispresent.

B. A2- is present along with some HA- (the partiallyprotonated form).

C. HA- (the partially protonated form) is all that ispresent.

D. HA" is present along with some H2A (the fullyprotonated form).

63. Which of these points in Figure 1 is NOT describedcorrectly below?A . Point b is where [H2A] = [HA-].B. Point c is the first equivalence point.C. Point d is where pH = pKa2-D. Point f iswhere [HA"] = [A2"].



Polyequivalent bases are bases that can neutralize morethan one acidic hydrogen per base molecule. A typicalexample of a diequivalent base is carbonate (C032")- Thefirst proton added corresponds to the second proton removed.Carbonate is a stronger base than its conjugate acidbicarbonate (HCO3), although both are considered weakbases, because the pKb values are 3.67 and 7.63 respectively.The titration curve in Figure 1 is obtained when calciumcarbonate (CaC03) is titrated with a strong acid, such ashydrochloric acid.

0.5 1.0 1.5Equivalentsstrong acid added

Figure 1

The exact pH and exact volume of titrant used in thistitration are not provided, but the graph is proportionalthroughout its duration. The nine points marked along thetitration curve note some of the key transitional points.Despite the fact that the pH is recorded for the titrationof thebase (rather than the pOH), the titration curve is stillpredictable.

Among the key points recorded are the two equivalencepoints, and the two points at which the pH = pKa (both thefirst and second pKa points are marked). It is interesting tonote that when pH = pKaj, pOH = pKb2- The other pointsof the curve are interesting in that they represent differentpoints at which the pH is predictable from the trend in the pHand the amount of acid added.

6 4. At which point is the pH equal to the value of pKai ?

A. Point c

B. Point d

C. Point fD. Point h

6 5. Which of the following relationships is NOT true?A. pKai +pKb2= 14B. PKa2 + pKb2=14C • pHjnitially > PKa2D. pHat 2nd equivalence <pKai


66. Howcan the pH of solution be determinedat pointd?A. pH = 7.0B. pH - PKal +PKa2

C. PH=V(pKai)2+(pKa2)2D. PH = 2(pKai-pKa2)

67. What is NOT true at point e on the titration curve?A. Fewer than 1.5equivalentsof HCl have been added.B. The pH is less than 6.37.C. The pH is greater than pKaj.D. The pH is less than pKa2-

68. In whatpH range is the concentration of bicarbonate theGREATEST?

A. pH less than 3.67B. pH between 3.67 and 7.00C. pH between pH 7.00 and 10.33D. pH greater than 10.33

69. Between which two points does the pH of the solutionchange by the GREATEST amount?A. Betweenpoint c and point dB. Between point d and point eC. Between point e and point fD. Between point f and point g

7 0. Which of the following graphs shows changing pOH asa function of equivalents strong acid when calciumcarbonate is titrated by hydrochloric acid?A. B.

Equivalents strongacid Equivalents strongacid

Equivalents strongacid Equivalents strongacid



When titrating an acid with a strong base, an indicator isadded to the solution to indicate when the equivalence pointhas been reached. An indicator works by changing color at ornear the equivalence point of the titration. Because indicatorscan lose or gain a proton, they are susceptible to the effectsof varying solution pH during titration. A conjugateacid andits conjugate base have different colors (although in somecases, one of the two is colorless.) With most indicators,both the conjugate acid and conjugate base are coloredspecies. When both species are colored, it is often harder tosee the color change, unless there is a pronounced differencebetween the two colors. An example with the hypotheticalindicator HQ, and its dissociation to conjugate base Q", isshown in Reaction 1. Assume HQ is colorless and Q" isorange.

HQ(aq) -g-^ H30+(aq) + Q"(aq)Clear Orange

Reaction 1

When the conjugate base (Q") is the predominantspeciesin solution, it appears orange. When the conjugate acid (HQ)is the predominant species in solution, the solution appearsclear. When the pH of the solution is equal to the pKaof theindicator, there are equal parts of HQ and Q" in solution, sothe solution has an orange hue. Indicator are used in lowconcentration, so they don't become visibly detectable untilat least one-tenth of the indicator is in its deprotonated(orange conjugate base) form. This occurswhen the solutionpH is approximately one pH unit below the pKa of theindicator. Equation 1 lists the active range of an indicator.

pH(at equivalence) = pKa(indicator) ± 1-Equation 1

Because it is desirable to have the color change near theequivalence point of the species being titrated, Equation 1helps when selecting an indicator. If the equivalence pH isnot known, then the pH at equivalence can be approximatedaspKa(acjd) +3 (where the pKa is for the acid being titrated).This means that often the pKa of the indicator is about threepH units higher than the pKa of the acid being titrated. Thisholds true only if the pKa for the acid is between 4 and 10. Itis not affected by concentration factors. Figure 1 shows atitration curve for the titration of a weak acid by strong baseand the active range of an indicator:

pH

Equivalence point

mL titrant added

Figure 1


The change in shading represents the increase in theintensity of the orange color of the solution. The equivalenceoccurs in the middle of the color change band.

71. The BEST indicator to use for the titration of an acidwith a pKa of 5.0 would be one having a pKa of:

A. 1.5.

B. 5.0.C. 8.5.D. 11.5.

72. An indicator in a concentration that is too high couldhave what effect on a solution?

A. It could interfere with the acid and/or baseproperties of the solution being titrated.

B. It could change the viscosity of the solution beingtitrated, resulting in a non-homogeneous solution.

C. The color change could be too extreme to be useful.D. The color change could be too subtle to detect.

73. What is the ratio of the conjugate base to conjugate acidat two pH units above the pKa?

A. 100 : 1B. 2: 1C. 1 :2D. 1 : 100

74. Agood indicator haswhich of thefollowing properties?A . It should be transparent throughout the titration.B. It should be unreactive with respect to acid-base

reactions.C. It should be a strong acid or a strong base.D. It should be a weak acid or a weak base.

75. For which of these titrations is the pH at equivalencethe same,regardless of concentration?A. The titration of a weak acid by a strong baseB. The titration of a strong acid by a weak baseC. The titration of the first proton of a weak diprotic

acid by a strong baseD. The titration of the second proton of a weak

diprotic acid by a strong base

76. Which BEST explains why the indicator pKamay beone unit off from the equivalence pH?A. A difference of one pH unit has an insignificant

effect on color at the pH of the indicator's pKa.B. The pH changes rapidly at equivalence.C. At theequivalence pointfor theacid, pH= pKa.D. An indicator's color change occurs at only one very

specific pH value.



Indicators have two main purposes in chemistry. Thefirst use is as an aid in determining the equivalence point in atitration experiment. If the pH at equivalence is known, thenan indicator that changes color at or near that pH valuecan beused to detect the equivalence point. The ideal indicatorhasits pKavaluewithin± 1 of the pH at equivalence.

The second use of an indicator is to approximate the pHof an unknown aqueous solution by examining the color ofthe indicator in a sample of the solution. The indicatorchanges color at a pH value approximately one unit from itspKavalue, so if the pH of solution is within one pH unitofthe indicator's pKa value, then the pH can be estimated.Table 1 shows the relationship between pH and the colorassociated with three indicators.

pH Indicator I Indicator II Indicator in

1 Red Yellow Purple2 Red Yellow Purple3 Red Yellow Violet

4 Red Yellow Fuchsia

5 Red Yellow Clear

6 Red-orange Yellow Clear

7 Orange Yellow Clear

8 Mango Yellow Clear

9 Yellow Chartreuse Clear

10 Yellow Green Clear

11 Yellow Aquamarine Clear

12 Yellow Blue Clear

13 Yellow Blue Clear

Table 1

For the titration of a strong acid by strong base, the pHat the equivalence point is 7.0, while for the titration of aweak acid by a strong base, the pH at equivalence is greaterthan 7.0. Because the pH at equivalence is different, adifferent indicator is required for the two titrations, althoughfor a weak acid with a pKa of 3 or less, the indicator used inthe titration of a strong acid may work.

The equilibrium distribution of the deprotonated andprotonated forms of an indicator obeys Equation 1, where Ind"is the deprotonated form and H-Ind is the protonated form.

PHsoiution = pKa(indicator) + lo8 77^7H-Ind

Equation 1

77. Which indicator could be used to determine theequivalence point in the titration of 1.0 M benzoic acidby l.OMKOH(aq)?The pKa of H5C6COOH is 4.21.A. Indicator II onlyB. Indicator D3onlyC. Both Indicator I and Indicator nD. Both Indicator II and Indicator III


78. Which combination of colors is NOT possible for asolution?

A. I:red H: yellow DI: purpleB. I: mango Di: yellow ITI: violet

C. I: red II: yellow HI: fuchsia

D. I: yellow II: aquamarine DI: clear

7 9. If the ratio of blue species to clear species within anindicator equilibrium is 1000:1 at pH = 6, what is thepKa of the indicator, given that the deprotonated formabsorbs visible light?

A. 3B. 4C. 9D. 10

80. What is the pH of a solution that is clear when IndicatorIII is added, faint chartreuse when Indicator II is added,and faint mango when Indicator I is added?A. 7.0B. 7.5C. 8.0D. 8.5

81. Which of the following accurately describes the pKavalues for the three indicators?

A. 1:6.86 II: 9.87 ni: 4.22

B. 1:8.94 D: 7.21 III: 5.98

C. 1:7.21 II: 8.28 ffl: 4.11

D. 1:8.34 II: 9.66 III: 3.85

82. The active range for Indicator III is which of thefollowing?

A. pH2topH6B. pH3topH4C. pH3topH5D. pH4topH6

83. The best description of the absorbance of lightassociated with Indicator DI is:

A. The protonated form absorbs light with a ^max of426 nm, while the deprotonated form absorbs lightwith a ?lmax of 339 nm.

B. The protonated form absorbs light with a /kmax of339 nm, while the deprotonated form absorbs lightwith a X,max of 426 nm.

C. The protonated form absorbs light with a A,max of598 nm, while the deprotonated form absorbs lightwith a Xmaxof 339 nm.

D. The protonated form absorbs light with a Xmax of339 nm, while the deprotonated form absorbs lightwith a X,max of 598 nm.



In titration, it is common to use indicators to signify acertain pH for the solution. Indicators are made, mostcommonly, from organic dyes that gradually change colorwithin a given pH range. A color change is observed,because the indicator in its protonated state is one color andin its deprotonated state is a different color. Reaction 1shows the equilibrium for a generic indicator.

Hlnd(aq) + H2OO) ^ *• H30+(aq) + Ind'(aq)Reaction 1

Table 1 below lists a series of indicators, the pKa of theindicator, the active range for visual detection, and therespective colors of the protonated form (acid) anddeprotonated form (base). The Henderson-Hasselbalchequation describes the relationship between the members of aconjugate pair. When the concentrations of the acid and baseare equal, the pH equals the pKa. When this occurs, the colorof the solution is an average of the colors listed in Table 1.

Indicator pKa pH range Acid Base

Alizarin Yellow 10.8 10.0- 11.6 Yellow Red

Phenolphthalein 8.8 8.0 - 9.6 Clear Pink

Thymol Blue 8.4 7.6 - 9.2 Yellow Blue

Cresol Red 8.0 7.2 - 8.8 Yellow Red

Bromthymol Blue 6.8 6.0 - 7.6 Yellow Blue

Chlorophenol Blue 5.6 4.8 - 6.4 Yellow Red

Bromcresol Green 4.4 3.7-5.1 Yellow Blue

Methyl Orange 3.9 3.2 - 4.6 Orange Yellow

Bromphenol Blue 3.7 3.0 - 4.4 Yellow Blue

Erythrosin B 2.8 2.2 - 3.6 Orange Yellow

Thymol Blue 2.0 1.2-2.8 Red Yellow

Cresol Red 1.6 1.0-2.2 Red Yellow

Methyl Violet 0.8 0.0 - 1.6 Yellow Violet

Table 1

Depending on the concentrations and species beingtitrated, there is one ideal indicator. To detect the equivalencepoint for a titration, a small portion of indicator is added tosolution. Ideally, the equivalence point of the titrationshould be equal to the pKa of the indicator. At the very least,the equivalence pH must fall within the indicator's activerange. When a strong acid or strong base is titrated with astrong titrant, the pH at the equivalence point is always equalto 7.0. For a weak acid titration, the pH at equivalence canbe estimated by taking an average of the pKa of the acidbeing titrated and the pH of the titrant base.

84. Which of the following does NOT form a bluesolution?

A. Bromphenol blue in a pH = 7.0 bufferB. Bromcresol green in an aqueous ammonia solutionC. Thymol blue in an acetic acid solutionD. Bromthymol blue in a hydroxide solution


8 5. Which of the following indicators CANNOT be used forthe titration of ammonia by hydrochloric acid?

A. PhenolphthaleinB. Bromcresol greenC. Methyl orangeD. Bromphenol blue

86. If KCN has pKb = 4.68 and KF has pKb = 10.83,which indicator is NOT the ideal choice for thefollowing proposed titrations?A. 1.00 M KF(aq) titrated by 1.00 M HCl(aq) with an

indicator of cresol redB. 0.01 M KF(aq) by 0.01 M HCl(aq) with an

indicator of methyl violetC. 1.00 M KCN(aq) by 1.00 M HCl(aq) with an

indicator of bromcresol greenD. 0.01 M KCN(aq) by 0.01 M HCl(aq) with an

indicator of chlorophenol blue

87. Which of the following indicator : color correlations isNOT correct for biological pH (pH = 7.4)?A. Methyl violet: VioletB. Methyl orange : YellowC. Thymol blue : BlueD. Bromcresol green : Blue

88. Which of the following titrations requires thymol blueindicator?

A. A low concentration of strong base titrated by astrong acid

B. A low concentration of weak base titrated by astrong acid

C. A highly concentrated strong base titrated by astrong acid

D. A highly concentrated weak base titrated by astrong acid

89. Which of the following indicators should be chosen toidentify theequivalence pointof a strong acidtitrated bya strong base?A. Methyl violetB. Methyl orangeC. Bromthymol blueD. Alizarin yellow



ApHstick is a device that can approximate the pHof anaqueous solution by reference to a color blend band. It is aplastic stickwith indicators attached to specific segments ofthe stick. Each indicator has a range of color change thatextends to either side of a central pKavalue. When the pHofthe solution is less than the pKa of the indicator, theindicator exists predominantly in its protonated state. Whenthe pH of the solution is greater than the pKa of theindicator, the indicator exists predominantly in itsdeprotonated state. An indicator is ideal when both theprotonated and deprotonated species are colored. When thespecies are primary colors, thechange is easierto observe.

Typical pH sticks come with three to four indicators,such as those listed in Table 1. Each solution must be testedwitha separatepH stick. A pH stick cannot be used formorethan one test, because the acidity of the first test solutionaffects other solutions that are added to the pH stick.

Indicator pKaProtonatedColor

DeprotonatedColor

Bromcresol Green 4.37 Yellow Blue

Methyl Red 5.21 Red Yellow

Bromthymol Blue 6.78 Yellow Blue

Phenolphthalein 8.79 Clear Magenta

Table 1

A student uses a pH stick with four indicator markers toapproximate the pH of five separate solutions. Table 2 liststhe results for the five separate solutions. Each sequence ofcolors represents what is observed for the four indicators inorder of increasing pKa.

Solution BromcresolGreen

MethylRed

BromthymolBlue

Phenolphthalein

1 Yellow Red Yellow Clear

2 Blue Orange Yellow Clear

3 Blue YellowGreenish-blue

Clear

4 Blue Yellow Blue Magenta

5Greenish-blue

Reddish-orange

Yellow Clear

Table 2

An indicator is used to detect the endpoint in a titration.An ideal indicator for a titration has its pKa equal to theequivalence pH of the titration. This is the ideal condition,but in practice, the exact pH at equivalence often cannot bedetermined. The general rule for titration is that the pH atequivalence should be within one of the pKa of the indicator.

9 0. What is the pH range of the pH stick?

A. 3.4 to 8.8B. 3.4 to 9.8C. 4.4 to 8.8D. 4.4 to 9.8


91. It is NOTpossible to estimate pH for:A. Solution 1 only.B. Solution 1 and Solution 2 only.C. Solution 2 and Solution 3 only.D. Solution 1 and Solution 4 only.

92. An aqueous solution that has a hydroxide concentrationof 1.0 x 10"6M would show what colors?A. Yellow with bromcresol greenB. Red with methyl redC. Blue with bromthymol blueD. Clear with phenolphthalein

9 3. What is the approximate pH of Solution 5?A. 4

B. 5

C. 6

D. 7

94. When HCl is added to Solution 2, the colors on the pHstick do not change. How can this be explained?A. The hydronium concentration is too high for the

pH to be affected by the addition of HCl.B. The hydroniumconcentration is too low for the pH

to be affected by the addition of HCl.C. The solution is a buffer made from a carboxylic

acid and its carboxylate conjugate base.D. The solution is a buffer made from an amine and its

ammonium conjugate acid.

9 5. Which indicator could be used in the titration of a weakacid by a strong base?A. Bromcresol greenB. Methyl redC. Bromthymol blueD. Phenolphthalein

9 6. Which of the following statements is INVALID?A. A solution that turns bromthymol blue to blue

would turn Phenolphthalein to magenta.B. A solution cannot show two green marks on the

pH stick.C. A solution that turns methyl red to yellow would

turn bromcresol green to blue.D. The pH stick can estimate pH best when that value

falls between 4.21 and 5.37.


Passage XV (Questions 97 - 100)

A common class of acids in organic chemistry is thecarboxylic acids. Their acidity is attributable to the electron-withdrawing nature of the carbonyl group through resonance.Short-chain carboxylic acids are water-soluble. As the alkylchain length increases, the hydrophilicity decreases, makingcarboxylic acids of five carbons or more rather insoluble.Long-chain fatty acids are often used as surfactants because oftheir insolubility.

Phenols constitute another common class of acids inorganic chemistry. Their acidity is attributable to theelectron-withdrawing nature of the benzene ring throughresonance. Because aromatic rings are less electron-withdrawing than carbonyl groups, phenols are weaker acidsthan their carboxylic acid counterparts. Most phenols areinsoluble in water in their protonated state. Table 1 lists afew examples of carboxylic acids and phenols, along withtheir pKa values.

Organic Acid 2L>0

C13C^ OH0.64

CN-fW0\=/ OH

3.40

x=/ OH4.21

02N—4^-OH 7.18

O^0h 10.01

Table 1

Because of solubility constraints, carboxylic acids andamines are more often involved in aqueous buffering thanphenols. However, it is possible to titrate phenols, becausetheir conjugate base is water-soluble, allowing for thereaction to be monitored.

9 7. Which of these mixtures produces the MOST effectivebuffer for pH 4.0?A. H3CCO2H + H3CC02NaB. CI3CCO2H + Cl3CC02NaC. 02NC6H4C02H + 02NC6H4C02NaD. C6H5C02H + C6H5C02Na


98. For the titration of p-nitrophenol, which of thefollowing reagents would be best?

A. KOHB. Potassium p-nitrophenoxideC. PhenolD. HCl

9 9. What is the pH of a solution made by mixing 20 mL0.10 M phenol with 10 mL 0.10 M KOH(aq)?

A. 10.0

B. 7.0C. 5.8D. 5.5

100. Which of the following graphs represents the titrationof 50 mL 0.1 M p-nitrophenol by 0.20 M KOH(aq)?A. B.

mL 0.20 M KOH!aq) mL 0.20 M KOtfaq)

pH7H

D.

pH7H

T"25

mL 0.20 M KOIfaq)

1. B6. D11. C16. B21. C26. C31. C36. D41. C46. D51. B56. A61. D66. B71. C76. B81. A86. B91. D96. A

2. A7. B12. B17. C22. C27. D32.37.

CA

42. D47. B52. D57. D62. B67. B72. A77.82.87.92.

ACCc

97. D

3. B8. D13. A18. C23. C28. A33. C38. A43. D48. A53. C58. D63. D68. C73. A78. B83. C88. D93. B98. A

50mL 0.20 M KORaq)

4. B9. D14. B19. A24. D29. B34. D39. C44. B49. A54. D59. B64. C69. A74. D79. A84. C89. C94. C99. A

5. B10. A15. C20. D25. D30. C35. D40. C45. B50. C55. B60. A65. B70. A75. C80. D85. A90. B95. D100. A

THAT'S ENOUGH CHEM FOR NOW.

Buffers and Titration Passage Answers

Passage I (Questions 1-7) Buffer pH and Weak Acids

Choice B is correct. The mixture is composed of NaHCC>3 and H2CO3. Without considering how much of eachcomponent is present in solution, recognize first that they are a conjugate pair, meaning that the solution is abuffer. This means that the pH is close to the pKa. Carbonic acid is diprotic, so be sure you understand that thefirst proton is involved in this conjugate pair, and that the pH should be close to pKai- This makes choice Bthe best candidate. If you wish to solve for the exact value, the first step is to convert 0.839 grams NaHCC»3(s)into moles: (0.839 grams)( 1mole )=0.01 moles NaHC03. The number of moles of H2C03 = (0.10 L)(0.10 M) =

84 grams0.01 moles H2CO3. The pH can be found using the Henderson-Hasselbalch equation:

_, . Moles conjugate base u „ , Moles HCO3" ,. , , , oi c a , i„~ 1 capH = pKa + log L_2 _ ... pH = pKai + log *- = 6.4 + log^- = 6.4 + log 1 = 6.4Moles conjugate acid Moles H2CO3 .01

Since the log of 1 is 0, the pH of the solution is equal to pKai, 6.4. The best answer is choice B.

Choice A is correct. Even though the solution is a buffer, the addition of HCl decreases the pH slightly. Abuffer resists extreme pH changes, but a small change is often observed. The addition of an acid to the bufferedsolution lowers the pH. Since the pH is initially 7.21, the final pH value must be lower than 7.21. The onlyanswer choice less than 7.21 is choice A, 7.14.

Choice B is correct. To make a buffer, a weak acid and its conjugate base must be mixed. Benzoic acid is a weakacid, so it must be mixed with benzoate, its conjugate base. The desired pH is equal to the pKa of benzoic acid,so according to the Henderson-Hasselbalch equation, equal parts of benzoate and benzoic acid must be mixed.This can be accomplished either by adding one-half equivalent (in terms of moles) of strong base (NaOH) toconvert half of the benzoic acid into benzoate, or by adding an equivalent amount (in terms of moles) of benzoateto the benzoic acid solution. Choice A has the conjugate pair added together in an equal gram ratio, not equalmole ratio. This does generate equal mole portions, so the pH is not equal to the pKa. This eliminates choice A.The remaining three choices involve the mixture of benzoic acid with a strong base, so they must be mixed in amanner that half-titrates the benzoic acid. The only answer with half as much strong base as weak acid(benzoic acid) is choice B. This mixture yields equal molar portions of the two components of the conjugate pair(benzoic acid and benzoate), so the best answer is choice B.

Choice B is correct. The concentration of the titrant strong base (0.2 M NaOH) is twice that of the weak acidHF (0.1 M), so to reach equivalence, only half the volume of strong base is required. There are initially 50 mLof 0.1 M HF present, so only 25 mL of 0.2 M NaOH are required to reach the equivalence point. The target pHfor the solution is 3.3, which happens to be the pKa of HF. This means that the pH of solution is equal to pKaof HF, which is true when [HF] = [F"]. This occurs when the HF(aq) is half-titrated. If 25 mL 0.2 M NaOH(aq) isrequired for full titration, then 12.5 mL is required for half titration. Choice B is the best answer.

Choice B is correct. This question takes more than the usual amount of effort to answer. There are two factors toconsider: dilution and reactivity. Addition of sodium hydroxide solution converts some of the acetic acid toacetate, and it dilutes the solution. After the completion of the reaction, 3.0 mmole of H3CCO2" are present in30 mL of aqueous solution. From here on, it's strictly a matter ofcalculation, beginning with 3mmole _ q.i y^

30 mLH3CCO2". Choice B is the best answer. The reaction chart below shows how the moles were determined.

HOAc + NaOH

init: 6 mmols 3 mmolsA: -3 mmols -3 mmols

final: 3 mmols 0

H20XXX

XXX

XXX

H3CC02"0

+3 mmols

3 mmols

6. Choice D is correct. A buffer is prepared by mixing (in this case) a weak acid with a half-molar equivalent ofstrong base. This is also referred to as half-titrating a weak acid. Choice A is eliminated, because twice asmuch acid as weak base has been added. Choice B is eliminated, because both species are acids. Choice C iseliminated, because both species are bases. In choice D, half of an equivalent of strong base is added to a weakacid. This results in a buffer, so the best answer is therefore answer choice D.

®Copyright © by The Berkeley Review 338 Section V Detailed Explanations

7. Choice B is correct. The solution is acidic, so the pH is less than 7.0. All of the answer choices are less than 7,so nothing is eliminated. Fora weakacid with pKa between 2 and 12in an aqueous solutionwhere [HAJjnJtial isgreater than Ka, use the shortcut equation to determine the pH. The pKa is 5.0, and [HA] is 0.1M.

pH =IpKa -llog [HA] =1(5.0) - llog (0.10) =2.5 - I(-l) =2.5 +0.5 =3.02 2 2 2 2

The pH is 3.0, so the best answer is choice B.

Passage II (Questions 8 -14) Buffer Composition

8. Choice D is correct. A buffer results when a solution contains roughly equal concentrations of weak acid and itsconjugate base. This can be achieved by mixing the components of the conjugate pair in a roughly one-to-oneratio, as is observed in choices A and B. In choice C, the acid is diprotic, so the 1.5 equivalents of strong basecompletely remove the first proton (to form HCO3") and then pull off a second proton from half of thebicarbonate ions. The result is a solution with equal parts HCO3" and CO32". Because these are a conjugatepair, the solution forms a buffer. This eliminates choice C. In choice D, methyl amine is capable of gainingonly one proton, so the 1.5 equivalents of HCl completely converts the weak base (H3CNH2) into its conjugateacid (H3CNH3+), with a leftover of 0.5 equivalents of HCl. The solution is a mixture of weak acid and strongacid, which does not result in a buffer. The best answer is thus choice D.

9. Choice D is correct. Because the solution contains more weak acid than conjugate base, the pH is less than thepKa of the acid. This eliminates choices A and B. According to Equation 1, it is a log relationship, so the bestanswer is choice D. Plugging into Equation 1 would yield pH = pKa + log 1/2 = pKa - log 2.

10. Choice A is correct. The pH can be regulated (maintained at a relatively constant value) by a buffer. A bufferis composed of a weak acid and its conjugate base. Because the pH of the buffermust be close to (within one unitof) the pKa of the acid, the pH becomesharder to regulate as the weak acid becomes stronger. To regulate thepH at 1.0 requires an acidwith a pKa no greater than 2.0. An acid with a pKa of 2.0 or less would be one of thestrongest weak acids. Increasing acid strength diminishes its ability to buffer. You should note that Table 1does not contain any acids that could be used to buffer at 1.0. The best answer is thus choice A.

11. Choice C is correct. A buffer serves to maintain relatively constant pH. Choice A is eliminated, because anaqueous solution where the hydronium and hydroxide concentrations areequal hasa pHof7.0, notnecessarily aconstant pH. A buffer solution is one in which the ratio of the acid and basein the conjugate pair neverexceeds10: 1, but that does not address the hydronium-to-hydroxide ratio. Consider a pH = 4.0buffer, for instance. AtpH =4.0, the hydronium concentration is 10"4 M, and the hydroxide concentration is 10"10 M. This results in aratio of 1,000,000 : 1. This is far greater than 10:1, so choice B is eliminated. Choice C is the best, because arelatively constant pH implies that the hydronium concentration is relatively constant. If hydroniumconcentration is relatively constant, so is hydroxide concentration. This makes choice C the best answer.Choice D is eliminated, because the pH should be within one unit of the pKa value for the weak acid, not thehydronium-to-hydroxide ratio. The best answer is thus choice C.

12. Choice B is correct. The solution before adding the hydroxide has a 15 : 10mole ratio of weak acid to conjugatebase. Addition of 1.00mL of KOH converts some of the weak acid into conjugate base so that the ratio of 15 : 10becomes 14 : 11, still in favor of the weak acid. Because the weak acid is slightly more concentrated than itsconjugate base, the pH of the solution is slightly lower than the pKa. The pKa is 4.74, so the best answer ischoice B.

13. Choice A is correct. The largest ratio of conjugate base to weakacid is found in the solutionwith a pH greaterthan the pKa of the weak acid by the largest amount. Formic acid (HCO2H) has a pKa of 3.64. In choice A,when the solution pH is 4.00, the pH is 0.36 greater than the pKa. Hypochlorous acid (HCIO) has a pKa of7.49. In choice B, when the solution pH is 7.00, the pH is 0.49 less than the pKa. There is more weak acid thanconjugate base, which eliminates choice B. Hypobromous acid (HBrO) has a pKa of8.67. In choice C, when thesolution pH is 8.50, the pH is 0.17 less than the pKa. There is more weak acid than conjugate base, whicheliminates choice C. Ammonium ion (NH4+) has a pKa of 9.26. In choice D,when the solution pH is 9.50, thepH is 0.24 greater than the pKa. 0.36 is greater than 0.24, so the best answer is choice A.

Copyright © by The Berkeley Review*- 339 Section V Detailed Explanations

14. Choice B is correct. Addition of water to a buffer equally dilutes the weak acid and its conjugate base. Theresult is that the ratio of base to acid has not changed, so the buffer does not shift. The equilibrium has notbeen disturbed, so the reaction exhibits no net shift. The pH remains constant. The best answer is choice B.

Passage III (Questions 15 - 21) Buffer Chart and pKa Chart

15. Choice C is correct. Ammonium (NH4+) has a pKa value of 9.26 as given in Table 1. Thismeans that when theacid and conjugate base arepresent in equal molar concentration, the pH of thesolution is 9.26. Tomake thepHof the solution equal 9.5, the conjugate base must be in excess to such a degree that the log of the base-to-acidratio is 0.24. Choice A lists the conjugate base (ammonia) in excess, so choice A is valid. Hydrocyanic acid(HCN) has a pKa value of 9.32 as given in Table 1. This means that when the acid and conjugate base arepresent in equalmolar concentration, the pH of the solution is 9.32. Tomake the pH of the solution equal 9.5,the conjugate basemust be in excess to sucha degree that the logof the base-to-acid ratio is 0.18. Choice Bliststhe conjugate base (sodium cyanide) in excess, so choice Bis valid. From choice A,we know that an ammoniumand ammonia solution must have excess ammonia for the pH to equal 9.5. If one-half equivalent of the strongacid HCl were added to one equivalent of ammonia, then the pH would equal the pKa (9.26). The excess HCladded lowers the pH, so the value is less than 9.5. This makes choice C invalid and thus the answer youshould choose. From choice D, we know that an hydrocyanic acid and cyanide solution must have excesscyanide for the pH to equal 9.5. If one-half equivalent of thestrong base NaOH were added to one equivalentof hydrocyanic acid, then the pH would equal the pKa (9.32). The excess NaOH added raises the pH to avalue greater than 9.32 and approximately equal to 9.5, making choice D valid. Choose answer C.

16. Choice B is correct. One half-equivalent of base when added to one equivalent of carbonic acid removes thefirst proton from half of the carbonic acid molecules present in solution. This results in a solution of halfcarbonic acid and half bicarbonate ([H2CO3] = [HCO3"]). When the concentration of acid equals theconcentration of conjugate base, pH equals pKai for carbonic acid, which is 6.37, as given in the passage. Whenone and one-half equivalents of base are added to one equivalent of carbonic acid, the first proton is completelyremoved from all of the carbonic acid molecules present in solution. The remaining half of an equivalentcontinues to remove the second proton from half of the bicarbonatemolecules present in solution. This results ina solution where [HCO3"] = [CO32"]. When the concentration of bicarbonate equals the concentration ofcarbonate, pH equals pKa2 for carbonic acid, which is 10.83, as given in the passage. This makes choice Bcorrect. When one and one-half equivalents of base are added to one equivalent of phosphoric acid, the firstproton is completely removed from all of the phosphoric acid molecules present in solution. The remaining halfof an equivalent continues to remove the second proton from half of the dihydrophosphate present in solution.This results in a solution where [H2PO4"] = [HPO42"]. When this equality holds true, the pH equals pKa2 forphosphoric acid. From Table 2, we know that Ka2 for phosphoric acid is 6.2 x 10"8, which equates to a pKa2value of 8 - log 6.2, which is a little over 7. This does not make a pH of 10.83, so choice C is eliminated. Whentwo and one-half equivalents of base are added to one equivalent of phosphoric acid, the first and secondprotons are completely removed from all of the phosphoric acid molecules present in solution. The remaininghalf of an equivalent removes the third proton from half of the hydrophosphate in solution. This results in asolution where [HPO42"] = [PO43"]. When this equality holds true, the pH equals pKa3 for phosphoric acid.From Table 2, we know thatKa3 for phosphoric acid is4.8 x 10"13, which equates toa pKa3 value of13 - log 4.8,which is a little over 12. This does not make a solution with a pH of 10.83, so choice D is eliminated.

17. Choice C is correct. One and one-half equivalents of base when added to one equivalent of oxalic acidcompletely removes the first proton from all of the oxalic acid molecules present in solution and continues on toremove the second proton from half of the bioxalate present in solution. This results in a solution where[HC2O4"] = [C2O42"]. When this equality holds true, the pH equals pKa2 for oxalic acid. From Table 2, weknow that Ka2 for oxalic acid is 5.1 x 10"^, which equates to a pKa2 value of5 - log 5.1, which is less than 7, sochoice A is eliminated. One-half equivalent of base when added to one equivalent of carbonic acid removes thefirst proton from half of the carbonic acid present in solution. This produces a solution where [H2CO3] =[HCO3"]. When this equality holds true, the pH equals pKai for carbonic acid, which is 6.37, as given in thepassage. Choice B is eliminated. One-half equivalent of base when added to one equivalent of hypochlorousacid removes the proton from half of the hypochlorous acid present in solution. This results in a solution where[HCIO] = [CIO"]. When the concentration of acid equals the concentration of conjugate base, the pH equals pKafor HCIO, which is 7.49 as given in the table. Choice C is therefore correct. One-half equivalent of acid whenadded to one equivalent of pyridine adds a proton to half of the pyridine present in solution. This results in asolution where [C5H5N] = [Cs^NH+J. When this is true, the pH equals pKa for C5H5N, which is 5.16, asgiven in Table 1. Choice D is eliminated.

Copyright © by The Berkeley Review® 340 Section V Detailed Explanations

18. Choice C is correct. ThepKa value for formic acid (HCO2H) is 3.78. In order to have a pH =4.00, the solutionmust be mixed with excess conjugate base (HCO2"). Thismakes choice A valid. The pKa value for aniliniumcation (C6H5NH3+) is 4.64. In order tohave a pH=4.00, thesolution must bemixed withexcess conjugate acid(C6H5NH3+). This makes choice B valid. The pKa value for benzoic acid (C6H5CO2H) is 4.19. In order tohave a pH = 4.00, the solution must be mixed with excess conjugate acid (C.6H5CO2H). This makes choice Cinvalid. If excess C6H5CO2" were mixed with C6H5CO2H, then the pH of the solution would be greater than4.19, which is greater than 4.00. This makes choice C invalid. You should select choice C. The pKa value forpyruvic acid (H3CCOCO2H) is 3.89. In order to have a pH = 4.00, the solution must be mixed with excessconjugate base (H3CCOC02-)- This makes choice D valid.

19. Choice A is correct. Blood that is rich in carbon dioxide is also rich in H2CO3. Since carbonic acid is acidic, thepH decreases as carbonic acid is added to solution. The final solution thus has a pH lower than 7.4. This makeschoice A the correct choice. Because the amount of acid is increasing, the phosphate buffer system is affected.The additional hydronium ion present is absorbed by the phosphate buffer, thus increasing the amount ofH2PO4" present and decreasing the amount ofHPO42" and PO43" present in solution. This eliminates choice B.IfH2CO3 isaddedto solution, it equilibrates bymaking hydronium ion(H3O4") andbicarbonate anion (CO32").This results in increases in the concentrations of both bicarbonate and hydronium ions. This eliminates choicesCandD.

20. Choice D is correct. The concentration of the conjugate base is greater than the concentration of the acid whenthe pH is greater than the pKa. When the pH is greater than the pKa, the log of the ratio of base to acid is apositive value, indicating that the ratio of conjugate base to acid is greater than 1.0. The pKa for HF is 3.17, soat pH = 3.00, the pH is less than the pKa. This eliminates choice A. The value for pKai for H2CO3 is 6.37, soat pH = 6.00, the pH is less than the pKai. This eliminates choice B. The value for pKa2 for H2C2O4 is 5 - log5.1 (which is roughly 4.3), so at pH = 4.00, the pH is less than the pKa2- This eliminates choice C. The valuefor pKa2 forH3PO4 is 8 - log6.2 (which is roughly 7.2), so at pH =8.00, the pH is greater than the pKa2- Thisresults in the base beingmore concentrated than the acid,makingchoice D the correctchoice.

21. Choice C is correct. The value of pKai is 6.37 and pKa2 is 10.33 for carbonic acid (H2CO3). The pH is greaterthan pKai, so the conjugate base (bicarbonate) is in greater concentration than the acid (carbonic acid). Thismakes choice A valid. ThepH is less than pKa2, so the conjugate acid (bicarbonate) is in greater concentrationthan the base (carbonate). This makes choice B valid. The value for pKa2 is 7.20 and pKa3 is 12.32 forphosphoric acid. The pH is greater than pKa2, so the conjugate base (HPO42") is in greater concentration thanthe acid (H2PO4"). Thismeans that choice C is invalid, making it the correctchoice for the question. The pHis less than pKa3, so the conjugate acid (HP042-) is in greater concentration than the base (PO43"). This makeschoice D valid.

Passage IV (Questions 22 - 29) Molecular Weight from Neutralization of an Organic Acid

22.

23.

24.

Choice C is correct. To reach equivalence, it takes (0.03 L)(0.20 M) = 0.006 moles ofNaOH. Formula weight isgrams per mole equivalent, so the unknown acid has aformula weight of 1-0 8/0.OO6 moles =1°7 grams per mole.From the choices given, the only acid that is close to 167 grams per mole is p-nitro benzoic acid(O2NC6H4CO2H). Toverify your answer, add up the atomic weights of themolecule, O = 12, N = 14, C = 12and H = 1 (all in grams per mole). The sum is 46 (for NO2) + 76 (for C6H4) + 45 (for CO2H) = 167 grams permole, so the answer is definitely C.

Choice C is correct. The formula weight of a compound is measured in grams per mole. The moles aredetermined from 0.040 liters x 0.25 M=0.010 moles. The formula weight is 1-0 g/n.010 moles = 10°grams permole. Pick choice C for best results.

Choice D is correct. Since the graph begins at a high pH, the initial solution must be basic. This rules outchoices A and B. The equivalence point is at pH = 4.6, which is < 7. This indicates that the neutralizedproduct is a weak acid, so the titrant must be a strong acid. (Because the equivalence point is not 7.0, thecompound mustbea weak base). The graph represents D, thetitration ofa weakbase bystrong acid.


25. Choice Dis correct. To predict the pHnear the equivalence point, which iswhat you're doing when you choosean indicator, you should use the Henderson-Hasselbalch equation. At 2pH units beyond the pKa ofthe acid,the ratio ofconjugate base to acid is 100:1, which means that the reaction is almost at equivalence. This is thepoint at which the indicator should start to show some color change. In this question, the pKa for the weakacid is 7.20. The best indicator is active around pH = 9.20. This is choiceD, phenolphthalein, which is activebetween 8.0 and 9.6. You may remember usingphenolphthalein in titration labsfromgeneralchemistry.

26. ChoiceC is correct. A buffer is formed whena conjugate pair (acid and base) are present in solution in roughlyequal molar amounts. To achieve this, you can: (1) mix the conjugate acid and base evenly, (2) add the acid andhalf-titrate it (add a half molar equivalent of strong base), or (3) add the base and titrate it halfway (addhalf an equivalent of strong base). Choice A is a case of full titration, which leads to the complete conversionof the weak acid to its conjugate base, so that is out. Choice Bis a case of over-titrating by adding double theamount of base needed. In choice C, the weak acid (H3CCO2H) is added to a half-equivalent of strong base,which results in a buffered solution. Pick C. Choice D is an acid with an acid, which doesn't react.

27. Choice D is correct. The best choice for a buffer of pH = 8.5 is a weak acid with a pKa close to 8.5 mixed insolution with its conjugate base. The acid 4-ethanoylphenol (H3CCOC6H4OH) has a pKa of 8.4, which isclosest to the 8.5 value, so choose D. To make the buffer equal to 8.5 (slightly to the basic side of 8.4), thereshould be a slight excess of the conjugate base relative to 4-ethanoylphenol. The log of the ratio of conjugatebase to weak acid should be 0.1, a positivenumber, so the ratio of conjugate base to weak acid must be greaterthan 1.

28. Choice A is correct. The formula weight, also known as the empirical weight, is derived from the mass ofcompound per mole of equivalent. Adding up the molecular mass of the compound yields 90 grams/mole.Because there are two equivalents of protons per oxalic acid, the formula weight is 45 grams/mole. PickA. Besure not to pick Bby mistake.

29. Choice B is correct. If the titration required 30.0 mLof 0.10 MKOH(aq) to neutralize 25.0 mL of 0.10 M HA(aq),then according the relationship MacidVacjd =MbaSeVbase/ ^e m°larity °f tne ac"*must be 0.12M. However,seven drops (although the exact number is irrelevant) of HCl must be added to back titrate to the equivalencepoint. This means that the molarity of the unknown weak acid is actually a little less than 0.12M, makingchoice B the best answer.

Passage V (Questions 30 - 36) Conjugate Pair Titration Curve

30. Choice C is correct. As a general rule, the pH at the equivalence point is either three units above or below thepKa for the acid depending on the type of titration. This would mean that the pH at the two equivalencepoints differs by approximately 6 pH units. This makes choice C the best choice.To calculate the pH exactly for the two titrations, one can plug into the shortcut equation for both the conjugatebase (formed at point b) and conjugate acid (formedat point d). At the equivalence point, the volume is doubleits original value, so the concentration is half of its initial value. At point b, there is 0.05M OAc", and at pointd, there is 0.05 M HOAc. From Figure 1, the pKa for HOAc is roughly 5.0. This means that pKb for OAc" isroughly 9.0 The two calculations are shown below:

For HOAc atpoint d: pH = lpKa - llog [HA] = 1(5.0) - llog (0.05) = 2.5 - 1(-1.3) = 2.5 + 0.65 = 3.15

For OAc- atpoint b: pOH =1 pKb -1 log [A"] =1(9.0) -1 log (0.05) =4.5 -1(-1.3) =4.5 +0.65 =5.15 .-. pH=8.85The difference in pH is 5.7, so the best answer is choice C.

31. Choice C is correct. The pH of a solution depends on the concentration and strength of the reagents in solution.The highest pH at equivalence results from the presence of the strongest conjugate base (since all of theconjugate bases are in equal concentration). At equivalence, choices B and D are both acids, so they are botheliminated (pH at equivalence is less than 7.0). Methyl ammonium is less acidic than acetic acid based on therelative pKa general values given in the passage. The stronger base is the conjugate base of the weaker acid,making methylamine the more basic compound between acetate and the amine. The greater hydroxideconcentration results from the titration of 0.10 M H3CNH3CI, so choice A is eliminated, and thus choice C isthe best answer.


32.

33.

34.

35.

36.

Choice C is correct. When the pH is greater than the pKa, then according to the Henderson-Hasselbalchequation, the conjugate base is in greater concentration than the acid. This eliminates choices A and B.Because thepH and pKa are negative logvalues, if pH >pKa, then [H+] <Ka. Choice C is correct.

Choice C is correct. Initially in the titration of H3CCO2H by NaOH (at point a), the concentration ofH3CCO2H is 0.10 M. At the equivalence point in the titration of H3CC02Na by HCl (point d), theconcentration ofH3CCO2H is 0.05 M, due to dilution that occurs when the aqueousbasesolution is added. Notein the graph below that the final volume is 50mL, not 25 mL. The concentrations of H3CCO2H between the twopoints in question differ by a factor of 2. This means that the pH of the two solutions differs by -logV2. Withthe H3CCO2H concentration being greater at the initial point of the titration of H3CCO2H by NaOH than theequivalence point of the titration of H3CC02Na by HCl, the pH is lower at point a (the initial point of thetitration of H3CCO2H by NaOH). The best answer is therefore choice C. The graph belowshows the data:

pH

Start with 25 mL

O.IOMH3CCO2"

Start with 25 mL0.10 M H3CCO2H

mL titrant solution added

Finish with 50 mL

0.05 M H3CCO2"

Start with 50 mL0.05 M H3CC02H

25.0

Choice D is correct. By having a lower pKa value, H3CCO2H is more acidic than CH3NH3+. Because it ismore acidic, H3CCO2H buffers at a lower pH than CH3NH34", so choice A is eliminated. Because it is moreacidic, H3CCO2H has a conjugate base that is weaker (and thus has a higher pKD value) than the conjugatebase of CH3NH3+. Choice B is thus eliminated. Because it is more acidic, H3CCO2H dissociates more thanCH3NH3+, so choice C is eliminated. Because it is more acidic, H3CCO2H is a better electron pair acceptorthan CH3NH3"4" making choice D correct. The point of this question is to view the many different ways inwhich relative acidity can be expressed.

Choice D is correct. Because it is more acidic, H3CCO2H yields a lower pH value than CH3NH34" of equimolarconcentration, so choice A is valid and consequently eliminated. Because it is more acidic, H3CCO2H has aconjugate base with a higher pKb value than the conjugate base of CH3NH3+. Choice B is valid and thuseliminated. Because it is more acidic, H3CCO2H producesmore conjugatebase than CH3NH34" when added towater, so choice C is valid and consequently eliminated. Because it is more acidic, H3CCO2H is a better protondonor than CH3NH3+, so choice D is not true. The correct answer is choice D.

Choice D is correct. Given that the pKa value for HF is lower than that of H3CCO2H, HF is a stronger acidthan H3CCO2H. This means that equimolar HF and H3CCO2H solutions result in a lower pH for the HFsolution. This makes choice A invalid and thus eliminated. When equimolar HF and H3CCO2H are titratedequally, they both have converted into the same amount of conjugate base in the buffer region, meaning that [F"]equals [H3CCO2']. This eliminates choices Band C. When both HF and H3CCO2H are half-titrated, the pHof solution is lower for the HF solution, because the pH equals the pKa of the acid and the pKa value for HF islower than that of H3CCO2H. This means that pH(HF solution) <pKa(acetic acid) which means that [H+](HFsolution) > Ka(acetic acid)- This makes choice D the best choice.

Passage VI (Questions 37 - 42) Titration Curves and Concentration Effects

37. Choice A is correct. The initial pH is greatest in the solution with the lowest concentration of the weakestacid. H3CCO2H is the weaker acid of the two choices, and 0.010 M is the lowest of the concentrations. Thismeans that choice A is correct.


38.

39.

40.

41.

42.

Choice A is correct. When the pH of the solution exceeds the pKa for the acid, then there must be an excess ofconjugate base relative to the conjugate acid, according to the Henderson-Hasselbalch equation. Thiseliminates choices C and D. The pH is the negative log of the [H30+] and the pKa is the negative log of theKa. If pH > pKa, then the [H30+] must be less than Ka. This makes choice A correct. To make this clearer,when pH =pKa', then [H30+] =Ka. As the pH increases, the [H30+] decreases, while both pKa and Ka remainconstant. The[H30+] is thus less than Ka. The derivation from theequation for Ka is shown below:

Ka =[PW+][A- K

[HA] [H30+] [HA][A], WhenpH>pKa, [

[HA]•> 1.0, so K

[H3O4> 1 and Ka> [H30+]

Choice C is correct. In all three weak acid titrations, the volume of sodium hydroxide solution added to reachthe equivalence point was 25.0 milliliters, so choices A and B are eliminated. The higher the acidconcentration initially, the more conjugate base that forms at the equivalence point. The higher theconcentration ofconjugate base, thehigher pH for thesolution at theequivalence point. Pick choice C.

Choice C is correct. If the base concentration is doubled while the acid concentration remains constant, then thebase is twice as concentrated as the acid. Half as much titrant (base) is required, so the equivalence point isreached in half the base volume. This eliminates choices B and D. The acid is still HCl (a strong acid), andthe titrant base is still NaOH (a strong base), so the pH is still 7.0 at equivalence, and the curve still has thesame shape (lip-free and sigmoidal). The best answer is choice C.

Choice C is correct. Choice A is eliminated, because a strong acid and strong base do not make a buffer whenmixed together. Choice B is eliminated, because NaOH is not a weak base. Choice D is eliminated, becausethe pH changes throughout the titration; but it is at the equivalence point that the pH changes drastically.Choice C is the best answer, because the log scale means that as long as the concentrations are 100 timesdifferent, then the linear difference is 2.0 on the log scale. This in turn means that the slopes are equal throughmost of the titration, except near the equivalence point. Perhaps the answer would be better if it mentioned thedifferent concentrations of the titrant bases in each trial. The best answer is not always perfect.

Choice D is correct. Ammonia is a weak base, so choices A and C are eliminated, because the curves do not showthe initial dip in pH (lip-o-weakness). The greater the concentration of ammonia initially, the greater thepH initially, so the 1.00M NH3 has a higher starting pH than the 0.010 M NH3. The correct choice is D.

PassageVII (Questions 43 - 49) Titration Curves and Strength Effects

43. Choice D is correct. The greatest Ka value correlates with the strongest acid. Of the acids in the experiment,only HCl shows a strong acid titration curve (lip-free) and an equivalence pH of 7, so choice D is the answer.

44. Choice B is correct. The initial pH in the titration of HCN is greater than the initial pH in the titration ofHCIO, according to the titration curve in the passage. This means that the initial pH in the titration of HCNmust be greater than 4.23, which eliminates choice A. Because HCN is acidic, the initial pH must be below 7.0,which eliminates choice D. The pH at the equivalence point in the titration of HCIO is less than the pH atthe equivalence point in the titration of HCN, according to the titration curve in the passage. This means thatthe pH at the equivalence point in the titration of HCIOmust be less than 11.01, which eliminates choice C.The only choice that remains is choice B, so choice B is the best answer. The numbers are reasonable, because5.16 is greater than 4.23 (and 4.74 if you look at the titration curve), and 10.08 fits between 8.72 and 11.01 in theequivalence pH data.

45. Choice B is correct. Statement is invalid, as the curves in Figure 1 show. Strong acids fully dissociate uponaddition to water, so the pH is low in the beginning, It remains relatively constant, because pH is a log scale.It is the strong acid curve that has a plateau in the beginning. A buffer is defined as a roughly equal molarmixture of a weak acid and its conjugate base. Basedon the definition, a strong acid when mixed with a strongbase does not form a buffer. Do not be fooled by the flat region of the strong acid titration curve, which isattributed to mathematics, not buffering. Statement II is valid. Statement III holds true for strong acids, butnot for weak acids. For instance, a weak acid with a pKa greater than 7, has a pH greater than seven for mostof the buffer region, which is observed before the equivalence point. For a weak acid titration curve, the pH isgreater than 7 at equivalence, so there must be a point on the curve before the equivalence point that is greaterthan 7. Only statement II is valid, so choice B is correct.


46.

47.

48.

49.

Choice D is correct. Because 0.10 MHIO has a greater pH than0.10 MHCN, HIO must be a weaker acid thanHCN, and the titration curve associated with HIO would have all of the points greater than the titrationcurve ofHCN. The pKa ofHCN is greater than 8.72 (the half-equivalence pH is greater than the equivalencepH in the HOAc titration curve). Thus, the pKa of HIO must be greater than 8.72, which makes choice Aincorrect. The pH at equivalence in the titration of HCN is 11.01, so the pH at equivalence in the titration ofHIO must be greater than 11.01, and choice Bis thus eliminated. The pH of0.10 MHIO must have a pH lessthan 7.0, because although it is a weak acid, it is still acidic. The pH cannot be 7.21, so choice C is eliminated.The pKa ofHCN is9.32, so the pKa ofHIO must be greater than 9.32. The pH at equivalence mustbe at least apH unit less than 13 (the pH of0.10 MKOH), so the difference between the pKa and the pH at equivalence isless than three pH units. This makes choice D a true statement. The otherwayof confirming choice D is to lookat the trend whereby as the acid gets weaker, the difference between pKa and the pH at equivalence getssmaller. The difference for HCN is 11.01 - 9.32 = 1.69. For HIO, the difference should be less than 1.69.

Choice B is correct. When 10 mL 0.10 M HCIO are mixed with 10 mL 0.15 M KCIO, a buffer is formed. Whenwater is added to a buffer, theacidand conjugate base are diluted equally, so the pH does not change. The pHconsequently remains at 7.64. The best answer is choice B.

Choice A is correct. Because CN" is a stronger base than CIO" (HCN is a weaker acid than HCIO), the titrationcurve for NaCN should have a higher initial pH, higher pH at the half-titrated point (because the pKa forHCN is greater than the pKa for HCIO), and higher pH at the equivalence point. This can be seen only in thetitration curve in choice A. Choices C and D should have been eliminated immediately, because theirrespective graphs show that the two acids have the same pKa, which according to Figure 1 in the passage,they don't have.

Choice A is correct. The stronger the acid, the lower its pKa value. HCl is the strongest acid, so it has thelowest (and only negative) pKa value. Choices C and D are eliminated. From the titration data, it can be seenthat the next strongest acid is HOAc, meaning that the pKa for HOAc should be just ahead of the pKa for HClin the sequence. This makes choice A the best answer.

Passage VIII (Questions 50 - 56) Normality and Neutralization

50. C is correct. The moles of NaOH can be calculated from the moles of the acid (H+) that were neutralized, sincethe amount of base added is to the point of neutralization, where the moles acid equals the moles base. Keep inmind that sulfuric acid is diprotic and that the units given are units of normality. Normality already takesinto account the two equivalents of hydronium ion per sulfuric acid. Given in the passage, the number of molesof acid (H+) is (.025 L)(0.20 M) = 0.005 moles. To reach equivalence, this must be the moles of NaOH as well.The question asks for concentration, so the moles NaOH must be divided by the volume of solution.

Molarity = moles = 0-005 mole = 0.25 Mliter 0.02 L

Thus, the concentration of NaOH(aq) is 0.25M. Answer C is the best answer for you to choose. Do what is best!

51. Choice B is correct. The first proton is always the easiest to remove (that is, pKai is smaller or more acidicthan pKa2). Reversing this wording gives the correct answer B. The first proton must be easiest to remove,because it comes off first. This also a rare case where the correct choice contains the word "always."

52. Choice D is correct. Normality is based on the molarity of equivalent base required to neutralize all of theacidic protons. Since there are three (3) acidic protons on H3PO4, the normality is three times the molarity.

Normality = 3(0.3 M) = 0.90 N H3PO4.Pick choice D if you want to be a supernova of chemistry wisdom. Or pick it because it's the right answer.

53. Choice C is correct. The conversion from H3PO4 to H2PO4" involves the loss of the first proton, so the acidity iscalculated using pKa-|. The conversion from H2PO4" to HPO42" involves the loss of the second proton, so theacidity is calculated using pKa2- Mixing NaH2P04 and Na2HP04 yields H2PO4" and HPO42", so the pH ofthe solution is close to pKa2- Because there ismore of HPO42" in solution than H2PO4", the pH is greater thanpKa2- This eliminates choices A and B. If thesolution contained pure HPO42", the pH would be an average ofpKa2 and pKa3. The presence of H2PO4" lowers the pH, making the best answer choice C. Using theHenderson-Hasselbalch equation would give a pH = pKa2 + log 2 = pKa2 + 0.3.


54. Choice D is correct. Sulfuric acid has only two (2) acidic protons, whereas phosphoric acid has three (3) acidicprotons to neutralize. Thus, phosphoric acid requires only two-thirds the amount of base that is required forsulfuric acid. The number of moles of phosphoric acid is (25 mL)(0.6 M) = 15 mmoles H3PO4 . This means thatthere are 45 mmoles of H+ toneutralize. The question asks how many mL of 0.40 MH2S04(aq) contains thisamount of H+. 0.40 MH2S04(aq) =0.8 MH+. This means that the number of mL is: fcfflmflk, which equals

0.8 M

56.25 mL 0.40M H2SO4, given as answer choice D.

55. Choice Bis correct. Phosphoric acid is a triprotic acid, so choices Aand Ccan be eliminated. Because the firstproton of phosphoric acid is weak, the start of the curve should have a cusp (lip-o-weakness). It is only inchoice B that the cusp is present, so the best answer is choice B.

56. Choice Ais correct. Flask 2 is initially filled with 40.0 mL of 0.30 NH3PO4. The concentration is equivalent to0.10 MH3PO4. Adding 40 mL of 0.30 NNaOH completely neutralizes all three protons on phosphoric acid,leaving only the conjugate base P043"(aq) in solution. All of the moles of phosphoric acid(H3P04) presentinitially are converted into phosphate (PO43-). The addition of 40 mL of NaOH solution also doubles thevolume ofthe solution, increasing it from 40 mL to 80 mL. This cuts the concentration in half. Had the volumeremained 40 mL, and if all of the H3PO4 were converted into PO43", then the concentration of P043"(aq) wouldhave been 0.10 M. However, the finalconcentration after the extra 40mL of solution is accounted for is 0.5x 0.10MP043"(aq). That equals 0.050 MP043"(aq), which ischoice A.

PassageIX (Questions 57- 63) Titration Curve of a Polyprotic Acid

57. Choice D is correct. Because there are two vertical inflection points on the titration curve, the acid has twoequivalence points and thus is diprotic. Because of the lip-o-weakness (initial cusp in the titration curve), thefirst proton is associated with a weak acid. The second proton is always weaker than the first proton, so bothprotons are weak. Choice D is the best answer.

58. Choice D is correct. The first pKa is always lower than thesecond pKa by definition, meaning that choice A isvalid for all polyprotic acids. The first equivalence point (point c) is greater than the point where pH = pKai(point b). This makes choice Ba valid statement and thus eliminates it. The second equivalence point (point e)is greater than the point where pH = pKa2 (point d). This makes choice C a valid statement and thuseliminates it. The first equivalence point (point c) is less than the point where pH = pKa2 (point d). Thismakes choice D an invalid statement and thus makes it the best answer. To go from the first equivalence pointto the point at which the pH equals pKa2 (where [HA"] = [A2"]), base must be added to the solution. Thismeans that pKa2 is greater than the pH at the first equivalence point (addition of base increases the pH),confirming choice D.

Choice Bis correct. To carry out the titration, a strong base must be added toreact with the unknown weak acid.The pH increases as base is added, and according to the curve in Figure 1, the pH does increase when movingleft to right. The bestanswer ischoice B. Aweak base isnot strong enough tocarry out the titration.

Choice A is correct. To gofrom one equivalence point to another requires that one equivalent (whatever exactquantity that may be) ofbase titrant be added. Choices Band Dcan both be eliminated, because they requirean equal amount ofbase. It takes one equivalent oftitrant to go from pH =pKai topH=pKa2, so inchoice A,more than one equivalent is required. In choice C, less than one equivalent is required. This makes the bestanswer choice A. On the graph, to go from one labeled point to the next labeled point, requires one-half of anequivalent. So starting to before point band finishing after point d clearly is more than one equivalent.

61. Choice D is correct. When pH = pKai (true at point a), exactly one-half of an equivalent of titrant has beenadded tosolution. To reach the second equivalence point, two equivalents must be added. If5mL does not reachthe point at which pH equals pKa], then one-half of an equivalent must be greater than 5mL. This means thatone equivalent is greater than 10 mL and thus, two equivalents are greater than 20 mL. More than 20 mL isrequired to reach the second equivalence point, so the best answer is choice D.

62. Choice Bis correct. Point d is the point at which pH =pKa2 (where [HA"] = [A2"]). Point e is the secondequivalence point (at which all of the species present is A2"). Between the two points (d and e), it is safe toassume thatA2" is the predominant species with some HA" present. This isbest described by choice B.

59.

60.


63. Choice D is correct. Point b is the point at which pH =pKal, so [H2A] = [HA"]. This makes choice A valid.Point c is the first vertical inflection point, so it is the first equivalence point. Choice Bis valid. Point d is thepoint at which pH =pKa2, because [HA"] =[A2"]. Choice Cis valid. Point f represents a point where excessbase is being added and all of the species exists in the fully deprotonated form (A2"). Pick Dto feel good.


64.

65.

Carbonate Titration Curve

The passage is easier when all of the points in Figure 1 are labeled.

a (All exists asC032")

H2COs(aq) P*aUpKb2

d (First equivalence point, all exists as HC03")

f(pH=pKai):••• >*g (pH<PKal)

H (aq) + H CO-^(aq)

h (Second equivalence point,all exists as H2CO3)

i (excess acid added)

2.0 Equivalents strong acid added

Choice C is correct. The pH equals the pKa of the acid when the titration is half way to the equivalence point.The two points marked by x represent the pointsat which pH = pKa. The value of pKai is lower than pKa2, sothe correct answer is point f, choice C. Point c is where the pH = pKa2, point d is the first equivalence point(where the species exists as HCO3"), and point h is the second equivalence point (where the species exists asH2CO3).

Choice B is correct. Conjugate pairs have pK values that sum to 14. In the case of a diprotic acid, the firstproton to be lost in an acid dissociation reaction corresponds to the second (last) proton to be gained in a basehydrolysis reaction. This means that the following conjugate pair relationships hold true.

HC03-(aq) P*a2,P*M

H+(aq) + C032"(aq)

The conjugate pair relationship yields pKa] + pKb2 = 14 and pKa2 + pKbi = 14. This makes choice A valid andchoice B invalid. The question asks you to seek a statement that is not true, so choice B is best. According toFigure 1, the initial pH is the highest point, and it is greater than pKa2, making choice C a valid statement.According to Figure 1, the second equivalence point is lower than all pKa values, including pKai. This makeschoice D a valid statement.

66. Choice B is correct. In a fashion similar to the way in which we determine the isoelectric point of an aminoacid, the middle equivalence point can be determined by knowing that it lies exactly in the middle of the twopKa values. The pH of the first equivalence point is thus found by averaging pKa] and pKa2- The pH cannot be7.0 for this titration, and the Ka when dealing with the acidity of bicarbonate would involve pKa2 (because itis the second proton to be lost). The best answer for this question is choice B.

67. Choice B is correct. At point e of the titration curve, the amount of HCl added is greater than one equivalentbut less than one and one-half equivalents. This makes choice A a valid statement. The pH should be lowerthan the pH of point c (the point at which pH = pKa2), so choice D is a valid statement. The value of pKai is6.37, so choices B and C are contradictory answers. The pH is greater than the pH at point f (the point atwhich pH = pKal)/ which makes choice B invalid, and thus the best answer. It is critical in this passage thatyou recognize where pKai and pKa2 lie.


68 Choice Cis correct. Bicarbonate is in its greatest concentration at the first equivalence point. This point liesmidway between pKai and pKa2- From the passage, the values for pKbi and pKb2 are 3.67 and 7.63respectively. The values for pKai and pKa2 are therefore 6.37 and 10.33, respectively. The bicarbonateconcentration is greatest midway between 6.37 and 10.33. The best answer is choice C.

69 Choice Ais correct. The pH change between points cand dcorresponds to the change between pKa2 and the firstequivalence point. This is greater than the incremental change from point d to e (first equivalence to a pointwith pH greater than pKai), greater than the incremental change from point e to f (a point with pH barelygreater than pKai to apoint where pH =pKal), and greater than the incremental change from point f to g(apoint with pH =pKal to apoint where pH is just less than pKal). The best answer is choice A. It may be easierto answer this question by observing the titration curve and noting the change in pH (y-axis) between the twopoints in question.

70. Choice A is correct. Changing the labeling of the y-axis from pH to pOH does not change the shape of thegraph. The base is still weak, so choice Bcan be eliminated. The pOH starts lower than 7(because pH startsgreater than 7), so choices Cand Dare eliminated. The best answer is choice A, because the graph flipsvertically about the pH orpOH =7line when pH and pOH are interchanged.

Indicator SelectionPassage XI (Questions 71- 76)

71. Choice C is correct. Because the indicator has to be active at the equivalence point, the pKa of the indicatorhas to be greater then the pKa of the acid being titrated. This is because the equivalence pH is higher thanthe pKa for the acid being titrated. Because the acid being titrated is weak, the pH at equivalence is greaterthan 7.6. It is stated in the passage that the pKa values should differ by about 3. This makes the best choice C(which differs by 3.5 pH units) the best. Choice C is roughly three units above the pKa for the acid and isgreater than 7.0.

72.

73.

74.

75.

Choice A is correct. An indicator concentration that is too high results in a second competing acid that affectsthe titration and a faint color in the solution that is permanent. If the indicator were in concentration equal tothe acid being titrated, then the curve would look like a diprotic acid. While the color is affected by theexcess indicator, the color change is certainly not going to be too minimal, so choice Dgets eliminated. It is acolor change, which involves frequency of light. Ahigher indicator concentration affects the intensity of thelight, so the color change would just be more intense, but every bit as detectable. This eliminates choice C.While the viscosity may change with the addition of the indicator, this should not affect the reactionequilibrium, only the rate. The best answer is choice A, because the indicator can react with the titrant base.You may recall from titration experiments in general chemistry lab that you add only a few drops of indicatorto the solution.

Choice A is correct. When the pH is greater than the pKa, the solution has more conjugate base than acid, sochoices C and D are eliminated. The difference between the concentrations is 2 on the log scale, so on a linearscale, it can't be 2 : 1 (itmust be 100 : 1). This eliminates choice B and confirms that choice A is the correctanswer. Using the Henderson-Hasselbalch equation, the value can be determined.

pH =pKa +log-^-L =pKa +2=> log-^-LV F 5[HA] F [HA]=2 => i^-L = 102 = 100

[HA]

Choice Discorrect. Agood indicator changes color (thus itcannot be transparent) and itmust be present inlowconcentration in solution. This eliminateschoice A. If it is not affected by acid-base reactions, it can't react in away that indicates a pH change. Choice Bis eliminated. Because the indicator must undergo some acid-basechemical reaction, it is best when the indicator is a weakacid or weak base, so it does not affect the pH much.If it were a strong acid orstrong base would alter the pH and affect the titration. The best answer ischoice D.

Choice C is correct. The question calls for a titration in which the pH at equivalence is the same, regardless ofthe concentration of reactants. Choices A and B can be eliminated, because the pH of weak acids and basesvaries with the concentration. The pH at equivalence when the first proton of a weak diprotic acid has beenremoved is the average of pKai and pKa2- You most likely can recall this from determining the isoelectric pHof an amino acid. It applies to any polyprotic acid, including the amino acids. Because the pKa values do notchange, the pH at equivalence does not change, so the same pH at equivalence is observed. The best answer ischoice C. The second proton ofa diprotic acid isaffected by the concentration, sochoice Diseliminated.


76. Choice Bis correct. Choice Cshould be eliminated first, because the pH at equivalence is greater than the pKaof the acid. Choice Dshould also be eliminated, because the color changes over a range, not at one exact value.At the equivalence pH (when the indicator is changing color), a change of one pH unit is substantial. Thiseliminates choice A. The best answer is choice B, because the pH changes so rapidly that one drop of titrantcan change the pH bymore than one pH unit at theequivalence point. The indicator candetermine this valueto the accuracy of the one drop.

PassageXII (Questions 77 - 83)

77.

78.

Indicator Color and Solution pH

Choice A is correct. The titration of 1.0 Mbenzoic acid by 1.0 Mpotassium hydroxide has an equivalence pHvalue greater than 7.0, because the acid is weak and the base is strong, so the conjugate base that is formed atequivalence makes the pH higher than 7. The conjugate base, benzoate, has a pKb value of 9.79. Thisconfirmsthat benzoate is a weak base. The best indicator must have a color change range greater than 7.0. Indicator Igoes from reddish orange to mango as pH increases from 6 to 8, so Indicator I has a range that is too low.Indicator I is eliminated. Indicator I may seem close, but as the benzoic acid titration reaches equivalence, theindicator would have already turned yellow, so no color change occurs at equivalence. Indicator II goes fromchartreuse to aquamarine as pH increases from 9 to 11, so Indicator II is a good choice. The pH at theequivalence point falls between 9 and 11, in all likelihood. Indicator III goes from violet to clear as pHincreases from 3 to 5, so Indicator III is eliminated. Only Indicator II will work, so the best answer is choice A.If you wish to solve for the pH at equivalence, keep in mind that the conjugate base has formed and that thesolution's volume has increased, diluting all components. The pOH at the equivalence point is found using theshortcut equation.

pOH =IpKb-llog [A-] =1(9.79) - llog (0.5) =4.89 -l(-0.3) =4.89 +0.15 =5.04 .-. pH =8.962 2 2 2 2

This means that the pH at equivalence is about 9.0. The 0.5 M conjugate base is determined, because when thestrong base was added to the benzoic acid, it not only converted the benzoic acid to conjugate base, it alsodiluted the solution. With an equivalence pH of 9.0, only Indicator II can work. A faster way to get theapproximate pH at equivalence is to average the pKa of the weak acid and the pH of the titrant. This yieldsan approximate value. In this case, the average of 4.21 and 14 is 9.1. The approximate value is usually off byabout 0.15, which is close enough for our purposes.

Choice B is correct. The combination of colors that is not possible for a solution is the one that does not have apH value at which the three colors may exist. In choice A, Indicator I is red at pH values of 5 or less, IndicatorII is yellow at pH values of 8 or less, and Indicator III is purple at pH values of 2 or less. At a pH value of 2 orless, the color combination in choice A is possible. In choice B, Indicator I is mango at a pH value of 8, IndicatorII is yellow at pH values of 8 or less, and Indicator III is violet at a pH value of 3. The solution cannotsimultaneously be pH = 3 and pH = 8, so the color combination in choice B is notpossible. In choice C, Indicator Iis red at pH values of 5 or less, Indicator II is yellow at pH values of 8 or less, and Indicator III is fuchsia at apH value of 4. At a pH value of 4, the color combination in choice C is possible. In choice D, Indicator I isyellow at pH values of 9 or greater, Indicator II is aquamarine at a pH value of 11, and Indicator III is clear atpH values of 5 or greater. At a pH value of 11, the color combination in choice D is possible. Choice B is thebest answer choice.

79. Choice A is correct. If the ratio of blue species to clear species within the indicator equilibrium is 1000 : 1 at pH= 6, and the deprotonated form absorbs visible light (which makes the deprotonated form blue), then at a pHof 6, the indicator is predominantly in the deprotonated form. This means that the pKa value is less than 6.0.This eliminates choices C and D. Because the ratio is 1000:1, the log value is 3, making the pKa 3 units lessthan the pH. The equation is as follows:

pH =pKa(indicator) +log ^Pro ona g - pKa(indicator) +log 100°Protonated

6.0 = pKa(indicator) + log1000 = pKa(indicator) + 3Thus, the pKa(indicator) = 3-0' so tne Dest answer is choice A.

80. Choice D is correct. The key colors are faint mango with Indicator I (which implies that the pH is just greaterthan 8) and faint chartreuse with Indicator II (which implies that the pH is just less than 9). The best optionfor pH is thus choice D, 8.5, which falls between 8 and 9. Pick D to feel a rainbow of happy colors shine down.


81 Choice Ais correct. The color change range for Indicator I is from 6 to 8, so the pKa value for Indicator I isroughly 7. The color change range for Indicator II is from 9to 11, so the pKa value for Indicator II is roughly 10.The color change range for Indicator III is from 3to 5, so the pKa value for Indicator III is roughly 4. The pKavalue for Indicator I is incorrect in choices Band D, the pKa value for Indicator II is incorrect in choices Band C,and the pKa value for Indicator III is incorrect in choice B. The best answer is thus choice A.

82. Choice C is correct. In Indicator III the color change of detectability is from violet (at pH of 3) to the first signof clear (at a pH near 5). The pH range is thus best described as from 3 to 5, which makes choice Cthe bestanswer.

83. Choice C is correct. Indicator III is purple when protonated (at low pH) and clear when deprotonated (at highpH). This means that the absorbance by the deprotonated form is beyond the visible range, which eliminateschoices Band D. The visible range is 400 nm to 700 nm, so only the 339 nm absorbance can apply to thedeprotonated species. Because the protonated form is purple, the highest absorbance (kmax) must be in therange of the complementary color of purple, which is yellow. Yellow light lies in the middle of the visiblespectrum, so it must have avalue somewhere in the middle between 400 and 700 nm. Avalue of 426 nm is lowenough to be associated with violet light, so the value of 598 nm must be the best choice for the absorbance ofthe protonated species. The best answer is choice C. As apoint of interest, there is no known indicator whichgoes from colored to clear when deprotonated, because the negative charge usually lowers the transition energybetween the ground state and first excited state in a highly conjugated organic molecule. Ahighly conjugatedorganicmolecule is typical for an indicator.

PassageXIII (Questions 84- 89) Indicator Table

84. Choice C is correct. Bromphenol blue at pH =7.0 has a pHvalue greater than the pKa, so the species is presentinits anionic (conjugate base) form. The base form of bromphenol blue according to Table 1isblue, sochoice Aiseliminated. Bromcresol green in ammonia solution has a pH > 7.0, so the pH value is greater than the pKa.This means that the species is present in its anionic (conjugate base) form. The base form ofbromcresol greenaccording to Table 1 is blue, so choice Bis eliminated. Thymol blue in acetic acid solution has a pH <7.0, so thepH value is less than the pKa, and the species is present in its cationic (conjugate acid) form. The acid form ofthymol blue according to Table 1 is also yellow, so choice C is correct. Bromthymol blue in hydroxide solutionhas a pH > 7.0, so the pH value is greater than the pKa. This means that the species is present in its anionic(conjugate base) form. The base form of bromthymol blue according to Table 1 is blue, so choice Dis eliminated.

85. Choice A is correct. The titration of ammonia with hydrochloric acid forms an acid at the equivalence point.This means that the pH at the equivalence point is less than 7.0. Of the choices, only phenolphthalein has apKa greater than 7.0, so phenolphthalein cannot be used as the indicator. The pH at the equivalence pointshould equal the pKa of the indicator, so bromcresol green (pKa = 4.4), methyl orange (pKa = 4.0), andbromphenol blue (pKa = 3.7) should all be close enough together that they can all work. The better of theindicators depends on the initial concentration of ammonia. Choice A is correct.

86. Choice B is correct. The ideal indicator has a pKa close to the pH at equivalence. To solve this question, thepH atequivalence must be determined for each answer choice. In choices Aand B, HF is formed at equivalence.The pKa for HF is 3.17. In choice A, the pHof1.00 MHCl is 0. The average of the pKa for HF and the titrantpH leads to a pH at equivalence of approximately 1.6. Cresol red has a pKa of 1.6, so choice Ahas a validindicator for the titration. In choice B, the pH of 0.01 M HCl is 2. The average of the pKa for HF and thetitrant pH leads to a pH at equivalence of approximately 2.6. Methyl violet has a pKa of0.8, so choice Bhasan invalid indicator for the titration. Choice B is the best answer.

If you are highly observant, you'll note that as the concentration of the species decreases, the equivalence pHgets closer to 7. This means that choice Brequires an indicator with a higher pKa than choice A. This is alsotrue when comparing choices Cand D. Chlorophenol blue has a higher pKa than bromcresol green, so choices Cand D fit the desired trend. To verify this, let's consider the pH at equivalence for the remaining choices.In choices Cand D, HCN is formed at equivalence. The pKa for HCN is9.32. In choice C, the pH of1.00 MHClis 0. The average of the pKa for HCN and the titrant pH leads to a pH at equivalence of approximately 4.7.Bromcresol green has a pKa of 4.4, so choice Chas a valid indicator for the titration. In choice D, the pH of0.01 M HCl is 2. The average of the pKa for HCN and the titrant pH leads to a pH at equivalence ofapproximately 5.7. Chlorophenol blue has a pKa of5.6, sochoice Dhasa valid indicator for the titration.


87.

89.

Choice C is correct. When the pH is greater than the pKa of the indicator, the conjugate base is thepredominant species in solution, so the solution assumes the color of the conjugate base. When the pH is lessthan the pKa of the indicator, the conjugate acid is thepredominant species in solution, so thesolution assumesthe color of the conjugate acid. The pKa values for the four choices are: methyl violet (pKa =0.8), methylorange (pKa =4.0), thymol blue (pKa =8.4), and bromcresol green (pKa =4.4). At a pH of 7.4, methyl violetexists in its violet conjugate base form, so choice A is valid. At a pH of 7.4, methyl orange exists in its yellowconjugate base form, so choice Bis valid. At apH of 7.4, thymol blue exists in its yellow conjugate acid form, sochoice C is invalid. At a pHof7.4, bromcresol green exists in its blue conjugate base form, sochoice Dis valid.The only choice that does not correctly correlate the indicator with the solution color is C.

Choice D is correct. Thymol blue has a pKa value of 2.0, so it changes color in a highly acidic medium. Whentitratinga strong base with a strong acid, the equivalence point is 7.0, so choices A and C are eliminated. Thetitration of a weak base by a strong acid leads to an equivalence pH value less 7.0, so choice thymol blue isappropriate for choices B and D. A pKa value of 2.0 is rather low, so the conjugate acid present at theequivalence pointmust dissociate readily and be in high concentration. The higher the concentration of weakbase initially, the higher the concentration of weak acid at the equivalence point. To ensure that theequivalence pH is as low as 2.0, the weak base and titrant strong acid should both behighly concentrated. Thebase must be very weak with a pKb of10.0 or greater to be a reasonable choice for this problem. Given thelimited information, the best answer is choice D.

Choice C is correct. A strong acid when titrated by a strongbaseshowsan equivalence pH of 7.0. The indicatorchosen should have a pKa value near 7.0. It is okay if the pKa is as high as 8.0, but not any greater than that.The pKa values for the four choices are: methyl violet (pKa = 0.8), methyl orange (pKa = 4.0), bromthymol blue(pKa = 6.8), and alizarin yellow (pKa = 10.8). Only bromthymol blue has a pKa value close to 7.0, so onlybromthymol blue changes color at the equivalence point. The best answer is choice C.

Passage XIV (Questions 90 - 96) pH Sticks and Indicators

90.

91.

92.

93.

Choice B is correct. The pH range of the stick is found by considering the pH range of each indicator attachedto the pHstick. Indicators have a pH range (color change band) of approximately pKa(incjicat0r) ±1/ dependingon the colors and intensity of the indicator. The four indicators have respective ranges of 3.37 to 5.37, 4.21 to6.21, 5.78 to 7.78, and 7.79 to 9.79. This means that the overall range of the pH stick is 3.37 to 9.79, whichmakes choice B, 3.4 to 9.8, the best answer.

Choice D is correct. If bromcresol green appears yellow, the pH of the solution is less than 3.37 (4.37 -1). Atthis pH, the other indicators would be protonated, so the pH could not be determined from any of theindicators. Because Solution 1 appears yellow with bromcresol green, it is not possible to know the exact pH.Solution 1 could have any pH value less than 3.37. If phenolphthalein appears magenta, the pH of thesolution is greater than 9.79 (8.79 +1). At this pH, the other indicators would be deprotonated, so the pH couldnot be determined from any of the indicators. Because Solution 4 appears magenta with phenolphthalein, it isnot possible to know the exact pH. Solution 4 could have any pH value greater than 9.79. The pH of Solution 1and Solution 4 cannot be approximated from the pH stick, so the best answer is choice D.

Choice C is correct. An aqueous solution with a hydroxide concentration of 1.0 x10"6 Mhas a pOHof 6.0 andtherefore a pH of 8.0. The pH of the solution is more than one unit greater than 4.37 (the pKa of bromcresolgreen), so the bromcresol green indicator will turn blue. Choice A can be eliminated. The pH of the solution ismore than one unit greater than 5.21 (the pKa of methyl red), so the methyl red indicator will turn yellow.Choice B can be eliminated. The pH of the solution is more than one unit greater than 6.78 (the pKa ofbromthymol blue), so the bromthymol blue indicator will turn blue. Choice C is the best answer. The pH of thesolution is roughly one unit less than 8.79 (the pKa of phenolphthalein), so the phenolphthalein indicatorappears clear. Choice D can be eliminated.

Choice B is correct. Solution5 is greenish-blue with bromcresol blue, so the pH ofSolution 5 is greater than 4.37and less than 5.37. Solution 5 is reddish-orange with methyl red, so the pH of Solution 5 is less than 5.21 butgreater than 4.21. The pH of Solution 5 falls between4.37 and 5.21, so choice B is the best answer. Only whenthe color of the indicator is a composite of the protonated and deprotonated colors can the pH of the solution beapproximated. Bromthymol blue and phenolphthalein are purely the protonated color, so they were not usefulin approximating the pH of Solution 5.


94 Choice C is correct. When HCl is added to a solution, hydronium ions are released. If the colors of theindicators do not change, that means that the pH does not change significantly, and thus the hydronium ionconcentration does not change significantly. There are three explanations for the pH not changingSignificantly One reason pH does not change is that the hydronium concentration is so high that any change inhydronium is negligible on the log scale. Asecond reason pH does not change is that the hydroxideconcentration is so high that any change in hydroxide concentration is negligible on the log scale. The lastreason is that the solution is a buffer. Because Solution 2 appears orange with methyl red indicator, theapproximate pH is around 5.2. This does not describe asolution rich in hydronium or hydroxide. The bestexplanation is that Solution 2is abuffer with apH around 5. The pKa for acarboxylic acid is around 5, so thebest answer is choice C.

95. Choice D is correct. Remember in choosing an indicator for titration that the pKa of the indicator must bewithin ±1 of the pH at equivalence. When titrating a weak acid with a strong base, the products the weakconjugate base and water. An aqueous weak base solution forms at equivalence, which has apH greater than 7,so the indicator should have a pKa value greater than 7. The best answer is phenolphthalein, choice D.

96 Choice Aiscorrect. Asolution that turns blue with bromthymol blue has a pH greater than 7.78 (pKa =6.68, sothe pure blue color starts at 6.68 +1). ApH greater than 7.78 does not guarantee that the pH is greater than8.79, the pKa for phenolphthalein. This means that the solution may or may not turn magenta withphenolphthalein. Choice Ais invalid. If asolution turns green with bromcresol green, the pH is roughly 4.37.If a solution turns bromthymol blue green, the pH is roughly 6.68. The pH cannot simultaneously be 4.37 and6.68, so the stick cannot simultaneously have two green marks. Choice Bis a valid statement. Asolution thatturns methyl red yellow has apH greater than 6.21 (pKa =5.21, so the pure yellow color starts at 5.21 +1). ApH greater than 6.21 guarantees that the pH is greater than 4.31, the pKa for bromcresol green. This meansthat the solution must turn blue with bromcresol green. Choice C is a valid statement. When the pH of thesolution falls between 4.21 and 5.37, it falls into the color blend range of two separate indicators. As a result,the pH can be approximated with twice the accuracy. Choice Dis a valid statement.

Passage XV (Questions 97-100) Acidity and Electronic Influences

97.

98.

99.

100.

Choice D is correct. The best choice for a buffer ofpH =4.0 is a weak acid with a pKa close to4.0 mixed with itsconjugate base. Benzoic acid has apKa of 4.21 and nitrobenzoic acid has apKa of 3.40. Benzoic acid is closer, sochoose D. If the buffer is to be 4.0 (the acidic side of 4.21), there should be a slight excess of benzoic acidrelative to the conjugate base.

Choice A is correct. To titrate a weak acid (such as p-nitrophenol), you must add a strong base. The only strongbase listed among the choices isKOH. Choose A,and moveon.

Choice A is correct. The addition of10 mL 0.10 MKOH(aq) to 20 mL 0.10 Mphenol results in thehalf-titrationof the phenol. At the halfway point, pH =pKa. The pKa for phenol is 10.0, so choice Ais the best answer.

Choice A is correct. This titration is of a weak acid with a strong base that is twice as concentrated (0.20 Mbase with 0.10 Mweak acid), so one-half thevolume of thestrong base is used. One-half of 50 mL amounts to25 mL at equivalence. This eliminates choice Band D. Since the titrant is a strong base, the neutralizedproduct is aweak base, so the equivalence pH is greater than 7. Only the titration curve in choice Ashows pHat equivalence and pKa for thephenol as being greater than7.

'One book down, one book to go!"


PERKELEYDr*v-i*w"

PERIODIC TABLE OP THE ELEMENTS

1

H2

He1.0 4.0

3 4 5 6 7 8 9 10Li Be B C N O F Ne6.9 9.0 10.8 12.0 14.0 16.0 19.0 20.2

11 12 13 14 15 16 17 18Na Mg Al Si P S CI Ar

23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe85.5 87.6 88.9 91.2 92.9 95.9 (98) 101.1 102,9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3

55 56 571LaT72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Cs Ba Hf la W Re Os Ir Pt Ail Hg TI Pb Bi Po At Rn

132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222)

87 88 89 KAc§

104 105 106 107 108 109 110 HI 112

Fr Ra Rf Db SR Bh Hs Mt Uun Uuu Uub(223) 226.0 227.0 (261) (262) (263) (262) (265) (266) (269) (272) (277)

58 59 60 61 62 63 64 65 66 67 68 69 70 71

t Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tin Yb Lu

140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0

90 91 92 93 94 95 96 97 98 99 100 101 102 103

§ Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

232.0 (231) 238.0 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)


GeneralChemistry

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