Chemistry 114
Third Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1A. (6 points) Here are two chemical equations, write the equilibrium expression that
corresponds to these equations.
CO(g) + 2H2(g) WCH3OH(g)
MnO2(s) + 2KI(s) + 2H2SO4(aq) WMnSO4(aq) + K2SO4(aq) + 2H2O(l) + I2(s)
1B. (6 points) Here are two equilibrium expressions, write the chemical equation that
corresponds to these expressions. (Hint: Some species in the equation may not appear
in the equilibrium expression you may have to be as tricky as Dr. Z.)
H2O(l) + HA(aq) WH3O+(aq) + A-(aq)
(P refers to the partial pressure of a gas)
C(s) + S2(g) W CS2(g)
2. Ammonia (NH3) is a base so it undergoes the base reaction:
NH3(aq) + H2O(l) W NH4+(aq) + OH-(aq) The K for this reaction is 1.8x10-5.
A. (4 points) What is the K for the reaction NH4+(aq) + OH-(aq) WNH3(aq) +
H2O(l)?
Reverse reaction Knew =1/K Old = 1/1.8x10-5 = 55,600
B. (4 points) I want to combine the reaction given in A with the reaction
H2O(l)WH+(aq) + OH-(aq). What is the net reaction when you combine these two
reactions? NH4+(aq) + OH-(aq) + H2O(l) W H+(aq) + OH-(aq)+NH3(aq) + H2O(l)
NH4+(aq) W H+(aq) + NH3(aq)
C. (4 points) What is the K for the combined reaction?
K=K1×K2
=55,600 × 1x10-14 = 3.36x10-10
2
3. The reaction Al(OH)3(s)WAl3+(aq) + 3 OH-(aq) has a K of 1.3x10-33
If I mix 2 mLs of 1x10-5 M Al3+ with 8 mLs of 2x10-6 M OH- will a precipitate form?
Step 1. (3 points) What is the concentration of Al3+ when you mix the two
solutions?
M1V1 = M2V2
V2 = 2 + 8 = 10 mL
1x10-5 M x 2mL = X M x 10 mL
X = 1x10-5 x (2/10)
=2x10-6 M
Step 2. (3 points) What is the concentration of OH- when you mix the two
solutions?
M1V1 = M2V2
V2 = 2 + 8 = 10 mL
2x10-6 M x 8mL = X M x 10 mL
X = 2x10-6 x (8/10)
=1.6x10-6
Step 3. (3 points) What is the Q for the reaction?
Q = [Al3+][OH-]3
=2x10-6 M (1.6x10-6)3
=8.2x10-24
Step 4. (3 points) Use the answer from step 3 to tell be why a precipitate does
(or does not) form.
Q>K
Too many products
Reaction will go to left
Precipitate will form
4. ( 12 points) Given that the following system is at equilibrium:
2NOCl(g) 62NO(g) + Cl2(g), predict what will happen with the following changes to the
system:
Change
[NOCl] increases
The system will: Circle one
Shift to Right,
Shift to Left,
or remain the same
[Cl2] increases
Shift to Right,
Shift to Left,
or remain the same
[NO] decreases
Shift to Right,
Shift to Left,
or remain the same
He is added to system
Shift to Right,
Shift to Left,
or remain the same
Overall volume decreases Shift to Right,
Shift to Left,
or remain the same
T increases
Shift to Right,
(Assume rxn is Exothermic)
Shift to Left,
or remain the same
3
5A. (4 points)
What is the Arrhenius definition for acids and bases?
Acids produce H+, Bases produce OH-
B. (4 points) What is the Brønstead-Lowry definition for acids and bases?
Acids are proton donors, bases are proton acceptors
C. (4 points) What is the Lewis definition for acids and bases?
Acids are electron pair acceptors, bases are electron pair donors
6. (12 points) Fill in the following table:
[H+]
.01M Hcl
.01
1.58x10-9
Ca(OH)2
3.16x10-6
pH
1x10-12
2
3.16x10-6
3.16x10-6 M
HClO4
Area below left blank for calculations
[OH-]
pOH
12
5.5
3.16x10-9
8.5
5.5
3.16x10-9
8.5
I was hoping somebody would point out my mistake in the second row. I added a base
(Ca(OH)2) yet I had an acidic pH!. I would have given you BONUS points if you had said
something!
4
7. An acid is 1.7% ionized when is has an concentration of .0035M.
A. (6 points) What is the pH of the solution?
% ionized = [A-]/[Total Acid] x 100%
1.7% = [A-]/.0035 x 100%
1.7%/100% = [A-]/.0035
.017= [A-]/.0035
[A-] = .017 x .0035 = 5.95x10-5
[H+]=[A-]
so [H+] = 5.95x10-5
pH = -log(5.95x10-5)
=4.26
B. (6 points) What is the KA of the acid?
KA = [H+][A-]/[HA]
[HA] + [A-] = Total Acid = .0035M
[HA] = .0035 - [A-]
[HA] = .0035 - 5.95x10-5
[HA] = .00344
KA =
(5.95x10-5)2/ .00344
=1.03x10-6
8. (12 points) What is the pH of a 3.5x10-4 M solution of propanoic acid (CH3CH2COOH)
with a KA of 1.4x10-5?
KA = [H+][A-]/[HA]
1.4x10-5 = X2/(3.5x10-4-X)
Because KA is < 10-4 I will assume 3.5x10-4-X . 3.5x10-4
1.4x10-5 = X2/3.5x10-4
1.4x10-5 x 3.5x10-4 = X2
4.9x10-9 = X2
X = SQRT(4.9x10-10 )
X = 7x10-5 = [H+]
pH = -log(7x10-5)
=4.15