I.
Stoichiometry of Precipitation Reactions
A.
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)
 0.100mol Ag  
  0.150mol Ag 
1.50L

L




 58.45 g
  1mol NaCl 

0.150mol Ag

0
.
150
mol
NaCl
 1mol Ag  
 mol





  8.77 g NaCl

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)
 0.0500mol Pb2
1.25L
L

0.0500mol SO 
2
4

  0.0625mol Pb2


 1mol PbSO
4

 1mol SO 2
4

 0.0250mol SO 2
4
2.00L
L


  0.0500 mol PbSO  303.3g
4

 mol


  0.0500mol SO 2
4



  15.2 g PbSO 4

II. Acid-Base Reactions
A.
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+
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)
 0.350mol OH - 
  8.75 x 10 -3 mol OH 
0.0250L 

L


 


1L
-3
  1mol H
  8.75 x 10 -3 mol H  
8.75 x 10 mol OH
 0.100mol   0.0875L HCl  87.5 ml HCl
 1mol OH  






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
 0.250mol H 
0.0280 L 
L

7.00 x10
3

  7.00 x10 3 mol H 


 0.320mol OH 
0.0530 L 
L


  1.70 x10 2 mol OH 



 1mol H 2 O 
  7.00 x10 3 mol mol H 2 O
mol H  

 1mol H 
3. H+ was limiting, so calculate how much OH- is left over and concentration
1.70x10
2
 

mol OH - initially - 7.00x10 -3 mol OH - reacted  1.00x10 -2 mol OH - left
28.0ml  53.0ml  81.0ml  0.081L
1.00x10 -2 mol OH  0.123M OH 0.081L
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
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. The
buret 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)
 1mol OH - 
 1mol KHP 

3
  6.3701x10 3 mol OH  6.3701x10 mol KHP
1.3009 g KHP

 1mol KHP 
 204.22 g KHP 


6.3701x10 3 mol OH  0.1546M OH 0.04120 L
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)
0.01059L NaOH 0.1546mol NaOH   1.637 x10 3 mol NaOH 1mol HC7 H 5 O 2   1.637 x10 3 mol HC7 H 5 O 2

1.637 x10
3
L


 122.12 g HC 7 H 5 O 2
mol HC 7 H 5 O 2 
 mol HC H O
7 5 2


1mol NaOH 

  0.1999 g HC7 H 5 O 2


0.1999 g HC 7 H 5 O 2
0.3518 g sample
 56.82%
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
Ammonium salts
Equation for Formation of the gas
NH4+ + OH-  NH3 + H2O
Gas produced
NH3
III. Oxidation-Reduction Reactions
A.
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. NO3a. 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:
a.
b.
2Al(s) + 3I2(s) -----> 2AlI3(s)
(0)
(0)
(+3)(-1)
Al is oxidized = reducing agent
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+ + 1eBalanced!
3. Equalize the number of electrons in each half-reaction and add reactions
a. 5(Fe2+ -----> Fe3+ + 1e-) = 5Fe2+ -----> 5Fe3+ + 5eb. 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- + 4eAdd 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 + 4OH4Ag + 8CN- + 4H2O + O2 -----> 4Ag(CN)2- + 2H2O + 4OHg. Cancel water molecules appearing on both sides of the equation
4Ag + 8CN- + 2H2O + O2 -----> 4Ag(CN)2- + 4OHh. Check that everything balances REALLY DONE!