chm1045 chapter 4b

1

Chapter 4Chemical

Quantities and Aqueous Reactions

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Tro, Chemistry: A Molecular Approach 2

Reaction Stoichiometry• the numerical relationships between chemical

amounts in a reaction is called stoichiometry• the coefficients in a balanced chemical equation

specify the relative amounts in moles of each of the substances involved in the reaction

2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O

We can come up with lots of unit factors

8 18(l) 8 18(l) 2(g)

2(g) 2(g) 2 (g)

2 C H 2 C H 16 CO 1

25 O 16 CO 18 H O= = =


Predicting Amounts from Stoichiometry• How much CO2 can be made from 22.0 moles of

C8H18 in the combustion of C8H18?2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

2 moles C8H18 : 16 moles CO2

2188

2188 CO moles 176

HC mol 2CO mol 16HC moles 22.0 =×

2


Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline

• assuming that gasoline is octane, C8H18, the equation for the reaction is:

2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline

since 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2

3.4 x 1015 g C8H18g CO2

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

g 114.22mol 1

216

2

2

188

2

188

188188

15

CO g 101.0

CO mol 1CO g 44.01

HC mol 2CO mol 16

HC g 114.22HC mol 1HC g 10.43

×=

××××

188

2HC mol 2

CO mol 61

g C8H18 mol CO2 g CO2mol C8H18

mol 1g 44.01


Practice• According to the following equation, how

many milliliters of water are made in the combustion of 9.0 g of glucose?

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

density of water = 1.00 g/mL

3


Practice

OH mL 5.4

OH g 1.00OH mL 1 x

OH mole 1OH g 18.0 x

OHC mole 1 OH mole 6 x

g 10 x 80.1OHC mole 1 x OHC g 0.9

2

2

2

2

2

6126

22

61266126

=

According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose?

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)


• Limiting Reagents: The extent to which a reaction takes place depends on the reactant that is present in limiting amounts—the limiting reagent.

The limiting reagent is the one that runs out first. It is the one that limits the extent of the

reaction.

•the limiting reactant gets completely consumed

•reactants not completely consumed are called excess reactants


Limiting Reactant• the reactant that limits the amount of product is

called the limiting reactantsometimes called the limiting reagentthe limiting reactant gets completely consumed

• reactants not completely consumed are called excess reactants

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Limiting Reactant• the amount of product that can be made from

the limiting reactant is called the theoretical yield

• the amount of product that is made in a reaction is called the actual yield

• the efficiency of product recovery is generally given as the percent yield

%100yield ltheoretica

yield actualYieldPercent ×=


• Limiting Reagent Calculation: Lithium oxide is a drying agent used on the space shuttle. If 80.0 kg of water is to be removed and 65 kg of lithium oxide is available, which reactant is limiting?

Li2O(s) + H2O(l) → 2 LiOH(s)

• MM(Li2O) = 29.88 g/mol

• MM(H2O) = 18.02 g/mol

•The best way to calculate this is to convert all reactants to moles.

•Divide by the coefficient of that reagent in the balanced chemical equation

•The smallest quotient is the limiting reagent.

All the calculations are then based on the limiting reagent.


• 25.0 g of Lead (II) nitrate and 18.2 g of sodium chloride are mixed in 250.00 mL of solution. How many grams of Lead (II) chloride are made.? We first need a balanced chemical equation.

Pb(NO3)2 (aq) + 2NaCl (aq) PbCl2 (s) + 2 NaNO3 (aq)

Which reactant do we base the calculation on?

The Limiting Reactant.

mass Limiting Reactant

moles limiting reactant

moles Desired product

mass Desired product

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mass Limiting Reactant

moles limiting reactant

moles Desired product

mass Desired product

1. Make sure have a balanced chemical reaction

2. Determine limiting reactant

3. Every subsequent calculation is dependent on the limiting reactant


• Limiting Reagent Calculation: Cisplatin is an anti-cancer agent prepared as follows:

•K2PtCl4 + 2 NH3 → Pt(NH3)2Cl2 + 2 KCl

• If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react:

(a) which is the limiting reagent? (b) How many grams of cisplatin are formed?

• MM(K2PtCl4) = 415.08 g/mol MM(NH3) = 17.04 g/mol


Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

6


Solutions• when table salt is mixed with water, it seems to disappear,

or become a liquid – the mixture is homogeneousthe salt is still there, as you can tell from the taste, or simply boiling away the water

• homogeneous mixtures are called solutions• the component of the solution that changes state is called

the solute• the component that keeps its state is called the solvent

if both components start in the same state, the major component is the solvent


Describing Solutions• since solutions are mixtures, the composition can

vary from one sample to anotherpure substances have constant compositionsalt water samples from different seas or lakes have different amounts of salt

• so to describe solutions accurately, we must describe how much of each component is present

we saw that with pure substances, we can describe them with a single name because all samples identical


Solution Concentration• qualitatively, solutions are often

described as dilute or concentrated

• dilute solutions have a small amount of solute compared to solvent

• concentrated solutions have a large amount of solute compared to solvent

• quantitatively, the relative amount of solute in the solution is called the concentration

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Solution ConcentrationMolarity

• moles of solute per 1 liter of solution• used because it describes how many molecules

of solute in each liter of solution

L)(in solution ofamount moles) (in solute ofamount M molarity, =


Preparing 1 L of a 1.00 M NaCl Solution

Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution

since most solutions are between 0 and 18 M, the answer makes sense

1 mol KBr = 119.00 g, M = moles/L

25.5 g KBr, 1.75 L solutionMolarity, M

Check:• Check

Solution:• Follow the Concept Plan to Solve the problem

Concept Plan:

Relationships:

• Strategize

Given:Find:

• Sort Information

g 119.00mol 1

M 0.122L 1.75

KBr mol 29421.0solution L

KBr moles M molarity,

KBr mol 2940.21 KBr g 119.00

KBr mol 1KBr g 5.52

===

=×

g KBr mol KBr

L sol’nML

mol M =

8


Using Molarity in Calculations• molarity shows the relationship between the

moles of solute and liters of solution• If a sugar solution concentration is 2.0 M, then

1 liter of solution contains 2.0 moles of sugar 2 liters = 4.0 moles sugar 0.5 liters = 1.0 mole sugar

• 1 L solution : 2 moles sugar

solution L 1sugar mol 2

sugar mol 2solution L 1

Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH?

since each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should

require a little more than 2 L

0.125 mol NaOH = 1 L solution

0.125 M NaOH, 0.255 mol NaOHliters, L

Check:• Check


Concept Plan:

Relationships:

• Strategize

Given:Find:


NaOH mol 0.125solution L 1

solution L 2.04 NaOH mol 0.125

solution L 1NaOH mol 552.0 =×

mol NaOH L sol’n


Dilution• often, solutions are stored as concentrated stock

solutions• to make solutions of lower concentrations from these

stock solutions, more solvent is addedthe amount of solute doesn’t change, just the volume of solution

moles solute in solution 1 = moles solute in solution 2• the concentrations and volumes of the stock and new

solutions are inversely proportionalM1·V1 = M2·V2

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Example 4.7 – To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH?

since the solution is diluted by a factor of 5, the volume should increase by a

factor of 5, and it does

M1V1 = M2V2

V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 MV2, L

Check:• Check


Concept Plan:

Relationships:

• Strategize

Given:Find:


22

11 VM

VM=

•

( )L 1.00

Lmol3.00

L 200.0L

mol15.0=

⎟⎠⎞

⎜⎝⎛

•⎟⎠⎞

⎜⎝⎛

V1, M1, M2 V2


Solution Stoichiometry

• since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

Example 4.8 – What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the

reaction 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq)

since need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x volume Pb(NO3)2

1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 = 2 mol KCl

0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2

L KCl

Check:• Check


Concept Plan:

Relationships:

• Strategize

Given:

Find:


23)Pb(NO L 1mol 0.175

KCl L .3500 mol .1500

KCl L 1)Pb(NO mol 1

KCl mol 2)Pb(NO L 1

mol 0.175)Pb(NO L .15002323

23

=

×××

23)Pb(NO mol 1KCl mol 2

L Pb(NO3)2 mol KCl L KClmol Pb(NO3)2

mol0.150KCl L 1

10

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What Happens When a Solute Dissolves?• there are attractive forces between the solute particles

holding them together• there are also attractive forces between the solvent

molecules• when we mix the solute with the solvent, there are

attractive forces between the solute particles and the solvent molecules

• if the attractions between solute and solvent are strong enough, the solute will dissolve


Table Salt Dissolving in WaterEach ion is attracted to the surrounding water molecules and pulled off and away from the crystalWhen it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity


Electrolytes and Nonelectrolytes• materials that dissolve

in water to form a solution that will conduct electricity are called electrolytes

• materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes

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Molecular View of Electrolytes and Nonelectrolytes

• in order to conduct electricity, a material must have charged particles that are able to flow

• electrolyte solutions all contain ions dissolved in the water

ionic compounds are electrolytes because they all dissociate into their ions when they dissolve

• nonelectrolyte solutions contain whole molecules dissolved in the water

generally, molecular compounds do not ionize when they dissolve in water

the notable exception being molecular acids


Salt vs. Sugar Dissolved in Water

ionic compounds dissociate into ions when they dissolve

molecular compounds do not dissociate when they dissolve


Acids• acids are molecular compounds that ionize when they

dissolve in waterthe molecules are pulled apart by their attraction for the waterwhen acids ionize, they form H+ cations and anions

• the percentage of molecules that ionize varies from one acid to another

• acids that ionize virtually 100% are called strong acidsHCl(aq) → H+(aq) + Cl-(aq)

• acids that only ionize a small percentage are called weak acids

HF(aq) ⇔ H+(aq) + F-(aq)

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Strong and Weak Electrolytes• strong electrolytes are materials that dissolve

completely as ionsionic compounds and strong acidstheir solutions conduct electricity well

• weak electrolytes are materials that dissolve mostly as molecules, but partially as ions

weak acidstheir solutions conduct electricity, but not well

• when compounds containing a polyatomic ion dissolve, the polyatomic ion stays together

Na2SO4(aq) → 2 Na+(aq) + SO42-(aq)

HC2H3O2(aq) ⇔ H+(aq) + C2H3O2-(aq)


Classes of Dissolved Materials


Solubility of Ionic Compounds• some ionic compounds, like NaCl, dissolve very well in

water at room temperature• other ionic compounds, like AgCl, dissolve hardly at all

in water at room temperature• compounds that dissolve in a solvent are said to be

soluble, while those that do not are said to be insolubleNaCl is soluble in water, AgCl is insoluble in waterthe degree of solubility depends on the temperatureeven insoluble compounds dissolve, just not enough to be meaningful

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When Will a Salt Dissolve?

• Predicting whether a compound will dissolve in water is not easy

• The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results

we call this method the empirical method


Ag+, Ca2+, Sr2+, Ba2+, Pb2+SO42–

Ag+, Hg22+, Pb2+Cl–, Br–, I–

noneNO3–, C2H3O2

–

noneLi+, Na+, K+, NH4+

Exceptions(when combined with ions on the left the compound is insoluble)

Compounds Containing the Following Ions are Generally Soluble

Solubility RulesCompounds that Are Generally Soluble in Water


Li+, Na+, K+, NH4+CO3

2–, PO43–

Li+, Na+, K+, NH4+,

Ca2+, Sr2+, Ba2+

S2–

Li+, Na+, K+, NH4+,

Ca2+, Sr2+, Ba2+

OH–

Exceptions(when combined with ions on the left the compound is soluble or slightly soluble)

Compounds Containing the Following Ions are Generally Insoluble

Solubility RulesCompounds that Are Generally Insoluble

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Precipitation Reactions• reactions between aqueous solutions of ionic

compounds that produce an ionic compound that is insoluble in water are called precipitation reactions and the insoluble product is called a precipitate

41

2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)


No Precipitate Formation = No Reaction

KI(aq) + NaCl(aq) → KCl(aq) + NaI(aq)all ions still present, ∴ no reaction

15


Process for Predicting the Products ofa Precipitation Reaction

1. Determine what ions each aqueous reactant has2. Determine formulas of possible products

Exchange ions(+) ion from one reactant with (-) ion from other

Balance charges of combined ions to get formula of each product

3. Determine Solubility of Each Product in WaterUse the solubility rulesIf product is insoluble or slightly soluble, it will precipitate

4. If neither product will precipitate, write no reaction after the arrow


Process for Predicting the Products ofa Precipitation Reaction

5. If either product is insoluble, write the formulas for the products after the arrow – writing (s) after the product that is insoluble and will precipitate, and (aq) after products that are soluble and will not precipitate

6. Balance the equation

Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of

nickel(II) chloride1. Write the formulas of the reactants

K2CO3(aq) + NiCl2(aq) →2. Determine the possible products

a) Determine the ions present

(K+ + CO32-) + (Ni2+ + Cl-) →

b) Exchange the Ions(K+ + CO3

2-) + (Ni2+ + Cl-) → (K+ + Cl-) + (Ni2+ + CO32-)

c) Write the formulas of the productscross charges and reduce

K2CO3(aq) + NiCl2(aq) → KCl + NiCO3

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3. Determine the solubility of each productKCl is soluble

NiCO3 is insoluble4. If both products soluble, write no reaction

does not apply since NiCO3 is insoluble


nickel(II) chloride


5. Write (aq) next to soluble products and (s) next to insoluble products

K2CO3(aq) + NiCl2(aq) → KCl(aq) + NiCO3(s)6. Balance the EquationK2CO3(aq) + NiCl2(aq) → 2 KCl(aq) + NiCO3(s)


nickel(II) chloride


Ionic Equations• equations which describe the chemicals put into the water

and the product molecules are called molecular equations2 KOH(aq) + Mg(NO3)2(aq) → 2 KNO3(aq) + Mg(OH)2(s)

• equations which describe the actual dissolved species are called complete ionic equations

aqueous strong electrolytes are written as ionssoluble salts, strong acids, strong bases

insoluble substances, weak electrolytes, and nonelectrolyteswritten in molecule form

solids, liquids, and gases are not dissolved, therefore molecule form2K+1

(aq) + 2OH-1(aq) + Mg+2

(aq) + 2NO3-1

(aq) → 2K+1(aq) + 2NO3

-1(aq) + Mg(OH)2(s)

17


Ionic Equations• ions that are both reactants and products are called

spectator ions

2K+1(aq) + 2OH-1

(aq) + Mg+2(aq) + 2NO3

-1(aq) → 2K+1

(aq) + 2NO3-1

(aq) + Mg(OH)2(s)

• an ionic equation in which the spectator ions are removed is called a net ionic equation

2OH-1(aq) + Mg+2

(aq) → Mg(OH)2(s)


Acid-Base Reactions• also called neutralization reactions because the

acid and base neutralize each other’s properties2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)

• the net ionic equation for an acid-base reaction isH+(aq) + OH−(aq) → H2O(l)

as long as the salt that forms is soluble in water


Acids and Bases in Solution• acids ionize in water to form H+ ions

more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+

most chemists use H+ and H3O+ interchangeably• bases dissociate in water to form OH− ions

bases, like NH3, that do not contain OH− ions, produce OH− by pulling H off water molecules

• in the reaction of an acid with a base, the H+ from the acid combines with the OH− from the base to make water

• the cation from the base combines with the anion from the acid to make the salt

acid + base → salt + water

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Common AcidsChemical Name Formula Uses Strength Perchloric Acid HClO4 explosives, catalyst Strong

Nitric Acid HNO3 explosives, fertilizer, dye, glue Strong

Sulfuric Acid H2SO4 explosives, fertilizer, dye, glue,

batteries Strong

Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong

Phosphoric Acid H3PO4 fertilizer, plastics & rubber,

food preservation Moderate

Chloric Acid HClO3 explosives Moderate

Acetic Acid HC2H3O2 plastics & rubber, food preservation, vinegar Weak

Hydrofluoric Acid HF metal cleaning, glass etching Weak Carbonic Acid H2CO3 soda water Weak

Hypochlorous Acid HClO sanitizer Weak Boric Acid H3BO3 eye wash Weak


Common BasesChemical

Name Formula Common Name Uses Strength

sodium hydroxide NaOH lye,

caustic soda soap, plastic,

petrol refining Strong

potassium hydroxide KOH caustic potash soap, cotton,

electroplating Strong

calcium hydroxide Ca(OH)2 slaked lime cement Strong

sodium bicarbonate NaHCO3 baking soda cooking, antacid Weak

magnesium hydroxide Mg(OH)2

milk of magnesia antacid Weak

ammonium hydroxide

NH4OH, {NH3(aq)}

ammonia water

detergent, fertilizer,

explosives, fibers Weak


HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

19


Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric

acid with aqueous calcium hydroxide1. Write the formulas of the reactants

HNO3(aq) + Ca(OH)2(aq) →2. Determine the possible products

a) Determine the ions present when each reactant dissociates(H+ + NO3

-) + (Ca+2 + OH-) →b) Exchange the ions, H+1 combines with OH-1 to make H2O(l)

(H+ + NO3-) + (Ca+2 + OH-) → (Ca+2 + NO3

-) + H2O(l)c) Write the formula of the salt

cross the charges (H+ + NO3

-) + (Ca+2 + OH-) → Ca(NO3)2 + H2O(l)


3. Determine the solubility of the saltCa(NO3)2 is soluble

4. Write an (s) after the insoluble products and a (aq) after the soluble products

HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + H2O(l)5. Balance the equation2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)


acid with aqueous calcium hydroxide



acid with aqueous calcium hydroxide6. Dissociate all aqueous strong electrolytes to

get complete ionic equationnot H2O

2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq) →

Ca+2(aq) + 2 NO3-(aq) + H2O(l)

7. Eliminate spectator ions to get net-ionic equation

2 H+1(aq) + 2 OH-1(aq) → 2 H2O(l)H+1(aq) + OH-1(aq) → H2O(l)

20


Titration• often in the lab, a solution’s concentration is

determined by reacting it with another material and using stoichiometry – this process is called titration

• in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio

the unknown solution is added slowly from an instrument called a burette

a long glass tube with precise volume markings that allows small additions of solution


Acid-Base Titrations• the difficulty is determining when there has been just

enough titrant added to complete the reactionthe titrant is the solution in the burette

• in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity

the chemical is called an indicator

• at the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH−

aka the equivalence point


Titration

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TitrationThe base solution is thetitrant in the burette.As the base is added tothe acid, the H+ reacts withthe OH– to form water. But there is still excess acid present so the colordoes not change.

At the titration’s endpoint,just enough base has been added to neutralize all theacid. At this point the indicator changes color.


Example 4.14:The titration of 10.00 mL of HClsolution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

• Write down the given quantity and its units. Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH


• Write down the quantity to find, and/or its units.Find: concentration HCl, M

InformationGiven: 10.00 mL HCl

12.54 mL of 0.100 M NaOH


22


• Collect Needed Equations and Conversion Factors:HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

∴ 1 mole HCl = 1 mole NaOH0.100 M NaOH ∴0.100 mol NaOH ≡ 1 L sol’n


12.54 mL of 0.100 M NaOHFind: M HCl


solution literssolute molesMolarity =


• Write a Concept Plan:

mLNaOH

LNaOH

molNaOH

NaOH L 1NaOH mol 1000.

mL 1L 0010.

molHCl

NaOH mol 1HCl mol 1


12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH

0.100 mol NaOH = 1 LM = mol/L


mLHCl

LHCl

mL 1L 0010. HClliters

HCl molesMolarity =


NaOH mole 1HCl mol 1

L 1NaOH mol 1000

mL 1L 0.001NaOH mL 2.541 ×××

.

• Apply the Solution Map:

= 1.25 x 10-3 mol HCl


12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH


CP: mL NaOH → L NaOH → mol NaOH → mol HCl;mL HCl → L HCl & mol ⇒ M

Example:The titration of 10.00 mL of HClsolution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

23


HCl L 010000mL 1

L 0.001NaOH mL 0.001 .=×

• Apply the Concept Plan:


12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH




M 1250HCl L 0.01000

HCl moles 10 x 1.25Molarity-3

.==


• Check the Solution:HCl solution = 0.125 M

The units of the answer, M, are correct.The magnitude of the answer makes sense since

the neutralization takes less HCl solution thanNaOH solution, so the HCl should be more concentrated.


12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH





Gas Evolving Reactions• Some reactions form a gas directly from the ion

exchangeK2S(aq) + H2SO4(aq) → K2SO4(aq) + H2S(g)

• Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water

K2SO3(aq) + H2SO4(aq) → K2SO4(aq) + H2SO3(aq)H2SO3 → H2O(l) + SO2(g)

24


NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)


Compounds that UndergoGas Evolving Reactions

K2SO3(aq) + 2HCl(aq) →2KCl(aq) + SO2(g) + H2O(l)

SO2yesH2SO3acidmetalnSO3

metal HSO3

K2CO3(aq) + 2HCl(aq) →2KCl(aq) + CO2(g) + H2O(l)

CO2yesH2CO3acidmetalnCO3,metal HCO3

KOH(aq) + NH4Cl(aq) →KCl(aq) + NH3(g) + H2O(l)

NH3yesNH4OHbase(NH4)nanion

no

Decom-pose?

K2S(aq) + 2HCl(aq) →2KCl(aq) + H2S(g)

H2SH2SacidmetalnS,metal HS

ExampleGasFormed

Ion ExchangeProduct

ReactingWith

ReactantType


Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution

of nitric acid, a gas evolves1. Write the formulas of the reactants

Na2CO3(aq) + HNO3(aq) →2. Determine the possible products

a) Determine the ions present when each reactant dissociates(Na+1 + CO3

-2) + (H+1 + NO3-1) →

b) Exchange the anions(Na+1 + CO3

-2) + (H+1 + NO3-1) → (Na+1 + NO3

-1) + (H+1 + CO3-2)

c) Write the formula of compoundscross the charges

Na2CO3(aq) + HNO3(aq) → NaNO3 + H2CO3

25


3. Check to see either product H2S - No4. Check to see if either product decomposes –

YesH2CO3 decomposes into CO2(g) + H2O(l)

Na2CO3(aq) + HNO3(aq) → NaNO3 + CO2(g) + H2O(l)


of nitric acid, a gas evolves


5. Determine the solubility of other productNaNO3 is soluble

6. Write an (s) after the insoluble products and a (aq) after the soluble products

Na2CO3(aq) + 2 HNO3(aq) → 2 NaNO3(aq) + CO2(g) + H2O(l)

7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) → 2 NaNO3 + CO2(g) + H2O(l)


of nitric acid, a gas evolves


Other Patterns in Reactions• the precipitation, acid-base, and gas evolving

reactions all involved exchanging the ions in the solution

• other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions

also known as redox reactionsmany involve the reaction of a substance with O2(g)

4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

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Combustion as Redox2 H2(g) + O2(g) → 2 H2O(g)


Redox without Combustion2 Na(s) + Cl2(g) → 2 NaCl(s)

2 Na → 2 Na+ + 2 e−

Cl2 + 2 e− → 2 Cl−


Reactions of Metals with Nonmetals

• consider the following reactions:4 Na(s) + O2(g) → 2 Na2O(s)2 Na(s) + Cl2(g) → 2 NaCl(s)

• the reaction involves a metal reacting with a nonmetal• in addition, both reactions involve the conversion of

free elements into ions 4 Na(s) + O2(g) → 2 Na+

2O– (s)2 Na(s) + Cl2(g) → 2 Na+Cl–(s)

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Oxidation and Reduction• in order to convert a free element into an ion, the

atoms must gain or lose electronsof course, if one atom loses electrons, another must accept them

• reactions where electrons are transferred from one atom to another are redox reactions

• atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na → Na+ + 1 e– oxidationCl2 + 2 e– → 2 Cl– reduction

Leo

Ger


Electron Bookkeeping• for reactions that are not metal + nonmetal, or do

not involve O2, we need a method for determining how the electrons are transferred

• chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction

even though they look like them, oxidation states are not ion charges!

oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges


Rules for Assigning Oxidation States• rules are in order of priority1. free elements have an oxidation state = 0

Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)2. monatomic ions have an oxidation state equal

to their chargeNa = +1 and Cl = -1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

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Rules for Assigning Oxidation States3. (b) the sum of the oxidation states of all the atoms in

a polyatomic ion equals the charge on the ionN = +5 and O = -2 in NO3

–, (+5) + 3(-2) = -1

4. (a) Group I metals have an oxidation state of +1 in all their compounds

Na = +1 in NaCl

4. (b) Group II metals have an oxidation state of +2 in all their compounds

Mg = +2 in MgCl2


Rules for Assigning Oxidation States5. in their compounds, nonmetals have oxidation

states according to the table belownonmetals higher on the table take priority

NH3-3Group 5ACS2-2Group 6ACCl4-1Group 7ACO2-2OCH4+1HCF4-1F

ExampleOxidation StateNonmetal


Practice – Assign an Oxidation State to Each Element in the following

• Br2

• K+

• LiF

• CO2

• SO42-

• Na2O2

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Practice – Assign an Oxidation State to Each Element in the following

• Br2 Br = 0, (Rule 1)

• K+ K = +1, (Rule 2)

• LiF Li = +1, (Rule 4a) & F = -1, (Rule 5)

• CO2 O = -2, (Rule 5) & C = +4, (Rule 3a)

• SO42- O = -2, (Rule 5) & S = +6, (Rule 3b)

• Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a)


Oxidation and ReductionAnother Definition

• oxidation occurs when an atom’s oxidation state increases during a reaction

• reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2

oxidationreduction


Oxidation–Reduction• oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them • the reactant that reduces an element in another reactant

is called the reducing agentthe reducing agent contains the element that is oxidized

• the reactant that oxidizes an element in another reactant is called the oxidizing agent

the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent

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Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ → 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O


Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ → 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O

+1 -2 +5 -2 +1 0 +2 -2 +1 -2

ox agred ag

+4 -2 +1 -1 +2 -1 0 +1 -2

oxidationreduction

oxidationreduction

red agox ag


Combustion Reactions• Reactions in which O2(g) is a

reactant are called combustion reactions

• Combustion reactions release lots of energy

• Combustion reactions are a subclass of oxidation-reduction reactions

2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

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Combustion Products• to predict the products of a combustion

reaction, combine each element in the other reactant with oxygen

M2On(s)contains metal

NO(g) or NO2(g)contains N

SO2(g)contains S

H2O(g)contains H

CO2(g)contains C

Combustion ProductReactant


Practice – Complete the Reactions

• combustion of C3H7OH(l)

• combustion of CH3NH2(g)


Practice – Complete the Reactions

C3H7OH(l) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

CH3NH2(g) + 3 O2(g) → CO2(g) + 2 H2O(g) + NO2(g)

chm1045 chapter 4b

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