AP CHEMISTRYCHAPTER 4

SOLUTION STOICHIOMETRY

-Covalent bonds-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.

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 #- The number of H2O molecules associated with a particular ion, usually 4 or 6.

Hydration causes salts to dissolve. H2O also dissolves polar covalent substances such as C2H5OH. H2O doesn’t dissolve nonpolar covalent substances.

Hydration

Review:SoluteSolventElectrical conductivityStrong electrolyteWeak electrolyte

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

Solubility- 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

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

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

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)

Precipitation Reaction- A reaction in which two solutions are mixed and an insoluble solid forms.

Review solubility rules

Use solubility rules to determine which product, if any, will ppt in a double replacement rxn.

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---

Insolubles….Hydroxide,

Sulfide and Carbonate and Phosphate,And all of these can be dried!

By Kimberly Chin and Jeannie Chen, class of 2001

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 x 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 x 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.

titrant- solution of known concentration (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

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

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.

Photo by Will Rooney (AP 2008)

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.