I. Stoichiometry of Precipitation ReactionsA. Procedure

B. Example: ??? g NaCl needed to precipitate all Ag+ from 1.50L of

0.100 M AgNO3?

1. Net Ionic Equation Ag+(aq) + Cl-(aq) -------> AgCl(s)

NaCl 77.845.58

NaCl mol 150.0Ag 1

NaCl 1Ag 150.0

Ag 150.0Ag 100.0

50.1

gmol

g

mol

molmol

molL

molL

C. Example: ??? g PbSO4 form when 1.25L 0.0500M Pb(NO3)2 and 2.00L 0.0250M Na2SO4 are mixed?

Net ionic equation: Pb2+(aq) + SO42-(aq) -------> PbSO4(s)

II. Acid-Base ReactionsA. Acid Base models

1. Arrhenius model: correct, but too limiting

a. Acid = H+ producer in water

b. Base = OH- producer in water

2. Bronsted-Lowry model: includes Arrhenius model, but more general

a. Acid = proton (H+) donor

b. Base = proton acceptor

c. Example: HC2H3O2 + H2O -------> C2H3O3- + H3O+

4424

424

24

242

2

PbSO 2.153.303

PbSO mol 0500.0SO 1

PbSO 1SO 0500.0

SO 0500.0SO 0250.0

00.2 Pb 0625.0Pb 0500.0

25.1

gmol

g

mol

molmol

molL

molLmol

L

molL

acid base

B. Neutralization Reactions

1. Water is a nonelectrolyte (not highly ionized in solution)

2. H+(aq) + OH-(aq) -------> H2O(l)

3. Reaction of an acid and a base (usually to form water) = Neutralization

4. H+ is a strong acid and will completely react with any weak base present

H+(aq) + NH3(aq) -------> NH4+(aq)

5. OH- is a strong base and will completely react with any weak acid present

OH-(aq) + HC2H3O2(aq) -------> H2O(l) + C2H3O2-(aq)

6. Similar to a precipitation reaction, except the product is a liquid (H2O)

C. Example: ??? L of 0.100M HCl to neutralize 25.0 ml 0.350M NaOH?

Net Ionic Equation: H+(aq) + OH-(aq) -------> H2O(l)

HCl ml 87.5HCl 0875.0100.0

1H 10 x 8.75

OH 1

H 1OH 10 x 8.75

OH 10 x 8.75OH 350.0

0250.0

3-3-

3--

Lmol

Lmol

mol

molmol

molL

molL

D. Example: ??? mol H2O is formed when 28.0ml of 0.250M HNO3 and 53.0ml of 0.320M KOH are reacted? What is the concentration of H+/OH-

left?

1. Net Ionic Equation: H+(aq) + OH-(aq) -------> H2O(l)

2. Calculate how much water is formed

3. H+ was limiting, so calculate how much OH- is left over and concentration

E. Acid-Base Titrations

1. Definition: volumetric analysis of the concentration of an unkown

a. Titrant =solution of known concentration whose volume is measured

b. Analyte =solution whose concentration is to be determined

c. Equivalence Point = amount of titrant just reacts with all analyte

d. Indicator = changes color at endpoint of the titration

OH mol 1000.7H 1

OH 1H 1000.7

OH 1070.1OH 320.0

0530.0 H 1000.7 250.0

0280.0

2323

23

molxmol

molmolx

molxL

molLmolx

L

HmolL

-

-2-

--2--3-2

OH M123.00.081L

OH mol x1000.1 0.081L 81.0ml 53.0ml 28.0ml

left OH mol x1000.1reacted OH mol 7.00x10 - initially OH mol1.70x10

2. Acid-Base Titration results in neutralization of all of the analyte

a. H+(aq) + OH-(aq) -------> H2O(l)

b. Phenolphthalein indicator is colorless in acid and pink in base

c. A buret accurately measures the amount of titrant added

The flask contains acid of unknown concentration and phenolphthalein. Theburet contains base of a known concentration

Base is added dropwise. The faint pink color goes away as you stir. The endpoint hasn’t been reached yet.

The endpoint has now been reached as the pink color persists. Measuring the volume of base dispensed aids calculation.

3. Standardizing the base solution: Example:

a. 1.3009g KHP is dissolved in distilled water (doesn’t matter how much) and titrated with 41.20ml unknown NaOH solution.

b. HP-(aq) + OH-(aq) -------> H2O(l) + P2-(aq)

4. Calculating the concentration of the analyte: Example:

a. 0.3518g of sample was titrated with 10.59ml of 0.1546M NaOH. Calculate the mass percent of HC7H5O2 (benzoic acid) in the sample

b. HC7H5O2 (aq) + OH-(aq) -------> H2O(l) + C7H5O2

-(aq)

--3

-3-

3

OH 1546.004120.0

OH mol 103701.6

OH mol 103701.6KHP 1

OH 1KHP mol 103701.6

KHP 22.204

KHP 1KHP 3009.1

ML

x

xmol

molx

g

molg

%82.56sample 3518.0

OHHC 1999.0 OHHC 1999.0

OHHC

OHHC 12.122OHHC mol 10637.1

OHHC mol 10637.1NaOH 1

OHHC 1NaOH mol 10637.1

NaOH 1546.0NaOH 01059.0

257257

257

257257

3

25732573

g

gg

mol

gx

xmol

molx

L

molL

F. Reactions that give off gases

1. Sometimes the product of a reaction is not a solid, but a gas

2. We can still observe that something happened: bubbles form

Compound that reacts with acid

Equation for Formation of the gas Gas produced

Sulfides 2H+ + S2- H2S H2S

Carbonates 2H+ + CO32- H2CO3 H2O + CO2 CO2

Bicarbonates H+ + HCO3- H2CO3 H2O + CO2 CO2

Sulfites 2H+ + SO32- H2O + SO2 SO2

Bisulfites H+ + HSO32- H2O + SO2 SO2

Cyanides H+ + CN- HCN HCN

Compound that reacts with base

Equation for Formation of the gas Gas produced

Ammonium salts NH4+ + OH- NH3 + H2O NH3

III. Oxidation-Reduction ReactionsA. Definition = Reactions in which electrons are transferred (aka Redox Rxns)

1. Example: 2Na(s) + Cl2(g) -------> 2NaCl(s)

a. Each Na loses one electron to become Na+ (Oxidation)

b. Each Cl gains one electron to become Cl- (Reduction)

2. Oxidation States = tool for keeping track of electrons in Redox reactions

a. The total charge on all atoms must match the molecule or ion charge

b. Oxidation states of a few elements help us calculate for all others

B. Example:

1. CO2

a. O = -2 so we have a total of -4 coming from the 2 oxygens

b. C must be +4 to balance the negative charge

2. SF6

a. F = -1 so we have a total of -6 coming from the 6 fluorines

b. S must be +6 to balance the negative charge

3. NO3-

a. O = -2 so we have -6 coming from the 3 oxygens

b. N must be +5 in order to give us an overall 1- charge

C. Recognizing what is happening in Redox Reactions

Radius decreases Radius increases

1. CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g)

(-4)(+1) (0) (+4)(-2) (+1)(-2)

a. CH4 -----> CO2 + 8e- Carbon = oxidized, CH4 = reducing agent

1(-4) 1(+4)

b. 2O2 + 8e- -----> CO2 + 2H2O Oxygen = reduced = oxidizing agent

2(0) 2(-2) 2(-2)

2. Example: 2Al(s) + 3I2(s) -----> 2AlI3(s)

(0) (0) (+3)(-1)

a. Al is oxidized = reducing agent

b. I is reduced; I2 is the oxidizing agent

3. Example: Oxidized? Reduced? Oxidizing agent? Reducing agent?

a. 2PbS(s) + 3O2(g) -----> 2PbO(s) + 2SO2(g)

b. 2PbO(s) + CO(g) -----> Pb(s) + CO2(g)

D. Balancing Redox Equations: Half-Reaction Method in Acidic Solution

MnO4-(aq) + Fe2+(aq) -----> Fe3+(aq) + Mn2+(aq) (acidic solution)

1. Identify and write equations for the two half-reactions

a. MnO4- -----> Mn2+ (this is the reduction half-reaction)

(+7)(-2) (+2)

b. Fe2+ -----> Fe3+ (this is the oxidation half-reaction) (+2) (+3)

2. Balance each half-reaction

a. Add water if you need oxygen

b. Add H+ if you need hydrogen (since we are in acidic solution)

c. Balance the charge by adding electrons

d. MnO4- -----> Mn2+ + 4H2O

e. 8H+ + MnO4- -----> Mn2+ + 4H2O

(+7) (+2)

f. 5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O Balanced!

g. Fe2+ -----> Fe3+ + 1e- Balanced!

3. Equalize the number of electrons in each half-reaction and add reactions

a. 5(Fe2+ -----> Fe3+ + 1e-) = 5Fe2+ -----> 5Fe3+ + 5e-

b. 5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O

c. 5Fe2+ + 8H+ + MnO4- ------> 5Fe3+ + Mn2+ + 4H2O

d. Species (including e-) on each side cancel out (algebra)

e. Check that the charges and elements all balance: DONE!

4. Example: H+ + Cr2O72- + C2H5OH -----> Cr3+ + CO2 + H2O

E. Balancing Redox Equations: Half-Reaction Method in Basic Solution

1. Follow the Acidic Solution Method until you have the final balance eqn

2. H+ can’t exist in basic solution, so add enough OH- to both sides to turn all of the H+ in H2O

3. Example: Ag + CN- + O2 -----> Ag(CN)2- (basic solution)

a. Ag + CN- -----> Ag(CN)2- (oxidation half-reaction)

b. Becomes: Ag + 2CN- -----> Ag(CN)2- + 1e- Balanced

c. O2 -----> (reduction half-reaction)

d. Becomes: 4e- + 4H+ + O2 -----> 2H2O Balanced

e. Equalize the electrons in each half-reaction and add reactions

4(Ag + 2CN- -----> Ag(CN)2- + 1e-)

Becomes: 4Ag + 8CN- -----> 4Ag(CN)2- + 4e-

Add to: 4e- + 4H+ + O2 -----> 2H2O

Gives: 4Ag + 8CN- + 4H+ + O2 -----> 4Ag(CN)2- + 2H2O DONE!

f. Add OH- ions to both sides to remove H+ ions

4Ag + 8CN- + 4H+ + O2 + 4OH- -----> 4Ag(CN)2- + 2H2O + 4OH-

4Ag + 8CN- + 4H2O + O2 -----> 4Ag(CN)2- + 2H2O + 4OH-

g. Cancel water molecules appearing on both sides of the equation

4Ag + 8CN- + 2H2O + O2 -----> 4Ag(CN)2- + 4OH-

h. Check that everything balances REALLY DONE!