AP CHEMISTRYCHAPTER 4

SOLUTION STOICHIOMETRY

Electrons aren’t shared evenly (oxygen is more electronegative)Electrons spend more time close to O than to H.

This uneven distribution of charge makes water polar. Because of this, water is a good solvent. The positive end (H) attracts negative ions or the negative end of another polar molecule. The negative end of water (O) attracts the positive ions or the positive end of another polar molecule.

When water surrounds an ionic crystal, the H end attracts the anion and the O end attracts the cation. This process is called hydration.

Hydration causes salts (ionic compounds) to dissolve. H2O also dissolves polar covalent substances such as C2H5OH. H2O doesn’t dissolve nonpolar covalent substances because there is not enough attraction between the water and the nonpolar molecule.

Hydration

Show the association of the ions with some water molecules when 1 formula unit of KCl dissolves in excess water.

K+

Cl-

A solution is a homogeneous mixture.

In a solution, a solute dissolves in the solvent.

If the solute ionizes in the solution, electricity can be conducted and the solute is said to be an electrolyte.

If the solute ionizes 100% or nearly 100%, it is called a strong electrolyte.

Lesser ionization occurs with weak electrolytes.

 

Svante Arrhenius determined that the extent to which a solution can conduct an electrical current depends directly on the number of ions present.

Solubility- is usually shown as g/given volume solvent or moles/given volume solution

Strong electrolytes1.soluble salts2.strong acids –completely ionize HCl(aq), HNO3(aq), H2SO4(aq)

Ex. Show how HCl dissociates when dissolved in water.

HCl H+ + Cl-

Acid (Arrhenius) – a substance that produces H+ ions in water solution

3.strong bases- completely ionize-contain OH

-bitter taste and slippery feel-NaOH, KOH

Weak electrolytes-only ionize slightly (weak

acids and bases) HC2H3O2 H+ + C2H3O2

99% 1%

Ammonia (NH3) -weak baseNH3 + H2O NH4

+ + OH

Molarity (M) = moles of solute liters of solution

Molarity is the most common unit of concentration used in Chemistry.

We may also see mM (millimolar) = moles of solute mL of solution

Ex. Calculate the molarity of a solution made by dissolving 23.4g of sodium sulfate in enough water to form 125 mL of solution.

23.4 g Na2SO4 1 mol Na2SO4 = 0.165 mol Na2SO4

142.06g Na2SO4

0.165 mol = 1.32 M

0.125 L

Ex. How many grams of Na2SO4 are required to make 350 mL of 0.50 M Na2SO4?

0.350L 0.50 mol Na2SO4 142.06g Na2SO4 = 24.9g

1 L 1 mol Na2SO4

Ex. What volume of 1.000 M KNO3 must be diluted with water to prepare 500.0 mL of

0.250 M KNO3? Dilution problem (M1V1 = M2V2)

(1.000M)(V1) = (0.250M)(500.0mL)

V1 = 125 mL

Remember, this formula can only be used for dilution! Never use it for a chemical reaction (stoichiometry)!

Read procedure for using volumetric flasks and types of pipets. We will be using both in several labs this year.

Let’s Review Equation Writing from Chemistry I

Some reactions fit neatly into a certain “category” of reaction type,

some do not.

DECOMPOSITION REACTIONS

Reaction where a compound breaks down into two or more elements or compounds. Heat, electrolysis, or a catalyst is usually necessary.

A compound may break down to produce two elements.

Ex. Molten sodium chloride is electrolyzed.

2NaCl(l) 2Na + Cl2

A compound may break down to produce an element and a compound.

Ex. A solution of hydrogen peroxide is decomposed catalytically.2H2O2 2H2O + O2

A compound may break down to produce two compounds.

Ex. Solid magnesium carbonate is heated.

MgCO3 MgO + CO2

Metallic carbonates break down to yield metallic oxides

and carbon dioxide.

Metallic chlorates break down to yield metallic chlorides and oxygen.

Hydrogen peroxide decomposes into water and oxygen.

Sulfurous acid decomposes into water and sulfur dioxide.

Carbonic acid decomposes into water and carbon dioxide.

Hydrated salts decompose into the salt and water.

Na2CO3H2O Na2CO3 + H2O

ADDITION REACTIONS

Also known as Synthesis, Combination, or Composition

Two or more elements or compounds combine to form a single product.

A Group IA or IIA metal may combine with a nonmetal to make a salt.

A piece of lithium metal is dropped into a container of nitrogen gas.

6Li + N2 2Li3N

Two nonmetals may combine to form a molecular compound.

C + O2 →CO2

When an element combines with a compound, you can usually sum up all of the

elements on the product side.

Ex. PCl3 + Cl2 PCl5

This is a trick that works because the common positive oxidation states of P are + 3 and +5.

Two compounds combine to form a single product.

Ex. Sulfur dioxide gas is passed over solid calcium oxide.

SO2 + CaO CaSO3

A metallic oxide plus carbon dioxide yields a metallic carbonate. (Carbon keeps the same oxidation state)

A metallic oxide plus sulfur dioxide yields a metallic sulfite. (Sulfur keeps the same oxidation state)

A metallic oxide plus water yields a metallic hydroxide.

A nonmetallic oxide plus water yields an acid.

Double Replacement (metathesis)

Two compounds react to form two new compounds. No changes in oxidation numbers occur. All double replacement reactions must have a "driving force" that removes a pair of ions from solution.

Formation of a precipitate: A precipitate is an insoluble substance formed by the reaction of two aqueous substances. Two ions bond together so strongly that water can not pull them apart. You must know your solubility rules to write these net ionic equations!

Simple Rules for Solubility

1. Most nitrate (NO3) salts are soluble.

2. Most alkali (group 1A) salts and NH4+ are soluble.

3. Most Cl, Br, and I salts are soluble (NOT Ag+, Pb2+, Hg22+)

4. Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4)

5. Most OH salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble)

6. Most S2, CO32, CrO4

2, PO43 salts are only slightly soluble.

SOLUBILITY SONGTo the tune of “ My Favorite Things” from “The Sound of Music”

Nitrates and Group One and Ammonium,These are all soluble, a rule of thumb.Then you have chlorides, they’re soluble fun,All except Silver, Lead, Mercury I.Then you have sulfates, except for these three:Barium, Calcium and Lead, you see.Worry not only few left to go still.We will do fine on this test. Yes, we will!Then you have the---InsolublesHydroxide,Sulfide and Carbonate and Phosphate,And all of these can be dried!

Ex. Solutions of silver nitrate and lithium bromide are mixed.

AgNO3(aq) + LiBr(aq) AgBr(s) + LiNO3(aq)

Formation of a gas: Gases may form directly in a double replacement reaction or can form from the decomposition of a product such as H2CO3 or H2SO3.

H2CO3 H2O and CO2

H2SO3 H2O and SO2

NH4OH NH3 and H2O

Ex. Excess hydrochloric acid solution is added to a solution of potassium sulfite.

2HCl(aq) + K2SO3(aq) H2SO3 H2O(l) + SO2(g) + 2KCl(aq)

Remember that sulfurous acid decomposes into water and sulfur dioxide!

Ex. A solution of sodium hydroxide is added to a solution of ammonium chloride.Remember that ammonium hydroxide does not exist.

NaOH(aq) + NH4Cl(aq) NaCl(aq) + NH4OH NH3(g) + H2O(l)

Formation of a molecular substance:

When a molecular substance such as water or acetic acid is formed, ions are removed from solution and the reaction "works".

Ex. Dilute solutions of lithium hydroxide and hydrobromic acid are mixed.

LiOH(aq) + HBr(aq) LiBr(aq) +H2O(l)

Ex. Gaseous hydrofluoric acid reacts with solid silicon dioxide.

4HF(g) + SiO2(s) SiF4(g) + 2H2O(l) This reaction occurs when glass is etched.

Single Replacement

Reaction where one element displaces another in a compound. One element is oxidized and another is reduced. A + BC B + AC

Active nonmetals replace less active nonmetals from their compounds in aqueous solution. Each halogen will displace less electronegative (heavier)

halogens from their binary salts.

Activity Series of Nonmetals

Most Active F2

Cl2

Br2

Least Active I2

Ex. Chlorine gas is bubbled into a solution of potassium iodide.

Cl2(g) + 2KI(aq) 2KCl(aq) + I2(s)

Active metals replace less active metals or hydrogen from their

compounds in aqueous solution. Use an activity series or a reduction potential table to determine activity. The more easily oxidized metal replaces the less easily oxidized metal.

ACTIVITY SERIES OF METALS

ElementLithiumPotassiumBarium

Decreasing CalciumActivity Sodium

MagnesiumAluminumZincIronCadmiumNickelTinLeadHydrogen ( a nonmetal)CopperMercurySilverGoldPlatinum

*Metals from Li to Na will replace H from water andacids; metals from Mg to Pb will replace H fromacids only.

Group I,II,+III

Transition Metals

Hydrogen

Jewelry Metals

Ex. Magnesium turnings are added to a solution of iron(III) chloride.

3Mg(s) + 2FeCl3(aq) 2Fe(s)+3MgCl2(aq)

Ex. Sodium is added to water.

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

Alkali metal demo

COMBUSTION REACTIONS

-Elements or compounds combine with oxygen to produce the oxides of each element.

The oxide of H is H2O, oxide of S is usually SO2, oxide of C is CO2, etc.

Hydrocarbons or alcohols combine with oxygen to form carbon dioxide and water.

Nonmetallic hydrides combine with oxygen to form oxides and water.

Nonmetallic sulfides combine with oxygen to form oxides and sulfur dioxide.

Ex. Carbon disulfide vapor is burned in excess oxygen.

CS2 + 3O2 CO2 + 2SO2

Ex. Ethanol (C2H5OH) is burned completely in air.

C2H5OH + 3O2 2CO2 + 3H2O

Diethyl ether (C2H5OC2H5) is burned in air.

C2H5OC2H5 + 6O2 4CO2 + 5H2O

Selective precipitation- process by which ions are caused to ppt one by one in sequence to separate mixtures of ions.

Qualitative analysis- process of separating and identifying ions

Ex. Separate Ag+, Ba2+, Fe3+

1. Add Cl to remove Ag+ as AgCl.

2. Add SO42- to remove Ba2+ as BaSO4.

3. Add OH or S2- to remove Fe3+ as Fe(OH)3

or Fe2S3.

Ex. Separate Pb2+, Ba2+, Ni2+

1. Add Cl to remove Pb2+ as PbCl2.

2. Add SO42 to remove Ba2+ as BaSO4.

3. Add OH or S2 to remove Ni2+ as Ni(OH)2

or NiS.

Quantitative analysis- determines how much of a component is present.

Gravimetric analysis- quantitative procedure where a ppt containing the substance is formed, filtered, dried & weighed.

Ex. The zinc in a 1.2000g sample of foot powder was precipitated as ZnNH4PO4. Strong heating of the ppt yielded 0.4089 g of Zn2P2O7. Calculate the mass percent of zinc in the sample of the foot powder.

0.4089gZn2P2O7 1 mol Zn2P2O7 2 mol Zn 65.37g = 304.7 g 1 mol Zn2P2O7 1 mol Zn

0.1754g Zn × 100 = 14.62% Zn

1.200g sample

Ex. A mixture contains only NaCl and Fe(NO3)3. A 0.456g sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of Fe(OH)3. The ppt is filtered, dried, & weighed. Its mass is 0.128g.Calculate: a. the mass of the iron

b. the mass of Fe(NO3)3

c. the mass percent of Fe(NO3)3 in the sample

0.128g Fe(OH)3 1 mol Fe(OH)3 1 mol Fe 55.85g Fe= 0.0669g Fe 106.9g Fe(OH)3 1 mol Fe(OH)3 1 mol Fe

0.0669g Fe 1 mol Fe 1 mol Fe(NO3)3 241.9g Fe(NO3)3= 0.290g Fe(NO3)3

55.85g Fe 1 mol Fe 1 mol Fe(NO3)3

0.290g × 100 = 63.6% Fe(NO3)3

0.456g

Acid-Base Reactions

Bronsted-Lowry acid-base definitions:

acid- proton donor base- proton acceptor

When a strong acid reacts with a strong base the net ionic rxn is:H+(aq) + OH(aq) H2O(l)

When a strong acid reacts with a weak base or a weak acid reacts with a strong base, the reaction is complete (the weak substance ionizes completely.)HC2H3O2(aq) + OH(aq) H2O(l) + C2H3O2

(aq)

neutralization reaction - acid-base rxn

When just enough base is added to react exactly with the acid in a solution, the acid is said to be neutralized.

Volumetric Analysis

titration- process in which a solution of known concentration (standard solution) is added to analyze another solution (analyte).

Titrations are most often used for acids and bases, but can be used for other types of reactions, also.

titrant- solution of known concentration (usually in buret)

equivalence point or stoichiometric point-point where just enough titrant has been added to react with the substance being analyzed

Indicator - chemical which changes color at or near the equivalence point

End point- point at which the indicator changes color

Ex. 54.6 mL of 0.100 M HClO4 solution is required to neutralize 25.0 mL of an NaOH solution of unknown molarity. What is the concentration of the NaOH solution?

HClO4 + NaOH H2O + NaClO4

0.0546 L HClO4 0.100 mol HClO4 1 mol NaOH = 1 L HClO4 1 mol HClO4

0.00546 mol NaOH

0.00546 mol NaOH = 0.218 M NaOH0.025L

Writing net-ionic equations

1. molecular equation

-overall reaction stoichiometry

2. complete ionic equation

-all strong electrolytes are represented as ions

3. net ionic equation

-spectator ions are not included

1. NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)

2. Na+ (aq) + Cl(aq) + Ag+(aq) + NO3(aq)

Na+(aq) + NO3(aq) + AgCl(s)

3. Cl(aq) + Ag+(aq) AgCl(s)

Oxidation-Reduction Reactions

Redox Rxns - reactions in which one or more electrons are transferred.

Electronegativity - attraction for shared electrons

mostelectronegative F>O>N=Cl elements “Phone Call”

These are most likely to have negative oxidation numbers.

Rules for Assigning Oxidation States

1. Oxidation state of an atom in an element = 0

2. Oxidation state of monatomic element = charge

3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1)

4. H = +1 in covalent compounds

5. Fluorine = 1 in compounds

6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

Review oxidation state rules on page 167.

N2O PBr3 HPO32-

P4O6 NH2-

+1-2 +3-1 +1+3-2

+3 -2 -3 +1

Noninteger states are rare, but possible.

Fe3O4 8/3 -2

O = 4(-2) = -8

Fe = 8/3 = 2 2/3 or Fe2+, Fe3+, Fe3+

Oxidation - loss of electrons - increase in oxidation number

Reduction - gain of electrons- decrease in oxidation number

OIL RIG Oxidation Is Loss (of e), Reduction Is Gain (of e)

LEO the lion goes GER Lose Electrons = Oxidation, Gain Electrons = Reduction

Oxidizing agent - electron acceptor- substance that is reduced

Reducing agent - electron donor - substance that is oxidized

The terms oxidizing agent and reducing agent are not tested on the AP test.

2KI + F2 2KF + I2

+1-1 0 +1-1 0

oxidized

I

reduced

F

OA

F2

RA

KI

2PbO2 2PbO + O2

+4 -2 +2 -2 0

oxidized:

O

reduced:

Pb

OA:

PbO2

RA:

PbO2

Balancing redox reactions by the half-reaction method1.Write skeleton half-reactions.

2.Balance all elements other than O and H.

3.Balance O by adding H2O.

4.Balance H by adding H+.

5.Balance charge by adding e- to the more positive side.

6.Make the # of e lost = # of e gained by multiplying each half-rxn by a factor.

7.Add half-reactions together.

8.Cancel out anything that is the same on both sides.

9.If the reaction occurs in basic solution, add an equal number of hydroxide ions to both sides to cancel out the hydrogen ions. Make water on the side with the hydrogen ions. Cancel water if necessary.

10.Check to see that charge and mass are both balanced.

Practice:

Sn2+ + Cr2O72 Sn4+ + Cr3+

(acidic solution)Sn2+ Sn4+ Cr2O7

2 Cr3+

Sn2+ Sn4+ Cr2O72 2Cr3+

Sn2+ Sn4+ Cr2O72 2Cr3+ +7H2O

Sn2+ Sn4+ Cr2O72 + 14H+ 2Cr3+ +7H2O

Sn2+ Sn4++2e Cr2O72 + 14H+ +6e 2Cr3+ +7H2O

3(Sn2+ Sn4++2e ) Cr2O72 + 14H+ +6e 2Cr3+ +7H2O

3Sn2+ + Cr2O72 + 14H+ +6e- 3Sn4++6e + 2Cr3+ +7H2O

3Sn2+ + Cr2O72- + 14H+ +6e- 3Sn4++6e- + 2Cr3+ +7H2O

3Sn2+ + Cr2O72 + 14H+ 3Sn4++ 2Cr3+ +7H2O

MnO42- + I- MnO2 + I2 (basic solution)

MnO42- MnO2 I- I2

MnO42- MnO2 2I- I2

MnO42- MnO2 + 2H2O 2I- I2

MnO42- +4H+ MnO2 + 2H2O 2I- I2

MnO42- +4H+ + 2e- MnO2 + 2H2O 2I- I2+2e-

MnO42- +4H+ + 2e- +2I- MnO2 + 2H2O +I2+2e-

MnO42- +4H+ + 2e- +2I- MnO2 + 2H2O +I2+2e-

MnO42- +4H+ +2I- MnO2 + 2H2O +I2

MnO42- +4H+ + 4OH- +2I- MnO2 + 2H2O +I2 + 4OH-

MnO42- + 4H2O +2I- MnO2 + 2H2O +I2 + 4OH-

MnO42- + 2H2O +2I- MnO2 + I2 + 4OH-

OXIDATION-REDUCTION TITRATIONS

Most common oxidizing agents: KMnO4 & K2Cr2O7

Potassium permanganate used to disinfect ponds and fish in Egypt.

MnO4- in acidic solution:

MnO4- + 8H+ + 5e Mn2+ + 4H2O

Purple colorless

When you titrate with MnO4 , the

solution is colorless until you use up all of the reducing agent (substance being oxidized).

In calculations, work redox titrations like acid-base titrations. You must have a balanced reaction to know the mole ratio.