CHM 136 General Chemistry II
Name
SOLUTIONS
Exam 3, Spring 2009 – Dr. Steel
1. (4 points) Using 1 – 4, organize the species in this group from the strongest acid (1)
to weakest acid (4).
H2S
[Pb(H2O)6]2+
2
H2PO41-
HNO2
4
1
3
2. (4 points) Using 1 – 4, organize the species in this group from which produces the
strongest conjugate base (1) to which produces the weakest conjugate base (4).
HSO41-
4
H2PO41-
[Al(H2O)6]3+
HClO
1
3
2
3. (6 points) Will each of these salts form an acidic, basic, or neutral solution when
dissolved in water?
BASIC
Na2S
ACIDIC
BASIC
NaF
NEUTRAL
ACIDIC
AlCl3
BASIC
NaH2PO4
K2SO4
LiC6H5CO2
4. (3 points) Circle the species in each group of acids that will produce the weakest
conjugate base.
HCl
HBr
HI
H2O
H2Se
H2S
HMnO4
HMnO3
HMnO2
5. (4 points) Write the appropriate equilibrium constant expression for each of the
reactions below.
a. CH4(g) + 2 H2O(g) ↔ 4 H2(g) + CO2(g)
(write an expression for Kp)
b. C(s) + 2 H2(g) ↔ CH4(g)
(write an expression for Kc)
Kp =
Kc =
PH4 2 × PCO 2
PCH 4 × PH2 2 O
[CH 4 (g)]
[H 2 (g)]2
6. (8 points) Select the answer from the column on the right that best matches each
description from the column on the left. Each answer can be used, at most, only once.
_ H___
This equilibrium constant describes a system that is
A. exothermic
heavily product-favored
__ F __
The equilibrium constant for an
reaction
will increase as you raise the temperature.
B. concentration
__G __
Removing some of the reactants of a reaction at
equilibrium will cause it to shift in this direction
C. K << 1
__C __
This equilibrium constant describes a system that is
D. reactant
strongly reverse-reaction favored
__B __
will not change the size of
A change in
equilibrium constant of a reaction.
E. temperature
__ I___
A reaction proceeds in this direction when its
current value of Q is less than its Keq
F. endothermic
__A___
The equilibrium constant for an
reaction
will increase as you lower the temperature.
G. reverse
__D___
When the reaction quotient is greater than the
equilibrium constant a system is
-favored
H. K >> 1
I. forward
J. product
7. (5 points) Consider the equilibrium below. Indicate whether each of the stresses
listed will cause the system to shift to the right (favor the products), left (favor the
reactants), or have no effect.
2 SO2(s) + O2(g) ↔ 2SO3(g)
ΔHrxn = -198 kJ/mol
left
increase the temperature
right
add more oxygen
right
remove some of the SO3 by condensing it
right
increase the volume of the reaction chamber
no effect
add more sulfur dioxide
8. (8 points) What is the pKa of the conjugate acid of a weak base whose Kb = 7.7×10-6?
pK b = − log ( K b )
= − log (7.7 × 10 −6 )
= 5.11
pK a + pK b = 14.00
pK a = 14.00 − 5.11 = 8.89
9. (8 points) The reaction below has a concentration-based equilibrium constant equal
to 279 M-1 at 1000 K. Use this information to determine the value of the pressurebased equilibrium constant for the system.
2 SO2(g) + O2(g) ↔ 2 SO3(g)
K P = K c ( RT )
Δngas
where Δngas = −1
-1
atm ⋅L
= (279)(0.08 21 mol
⋅K ⋅ 1000 K)
= 3.40
10. (10 points) The pH of a 0.133-M solution of an unknown weak acid is 3.652.
Determine the pKa of the acid.
HA
+ H2 O →
A- +
[ ]0
0.133
0
Δ[]
-x
+x
[ ]eq
0.133
2.23×10
H3O+
0
+x
-4
2.23×10-4
p H = 3.652 ⇒ [H 3 O + ] = 10 - p H = 2.23 × 10 −4 M
Ka =
[A − ][H 3 O + ] (2.23 × 10 − 4 ) 2
=
= 3.74 × 10 −7
[HA]
0.133
pK a = −log K a = 6.427
11. (10 points) The reaction below is prepared at 12 °C with an initial pressure of NO of
0.824 atm and an initial pressure of Br2 of 0.658 atm. At equilibrium the pressure of
the NOBr is measured to be 0.506 atm. Determine Kp for the reaction.
2 NO(g)
+ Br2(g)
↔
2 NOBr(g)
P0
0.824
0.658
0
ΔP
- 2x
+x
+ 2x
Peq
0.824 – 2x
0.658 - x
2x
Peq
0.318 atm
0.405 atm
0.506 atm
2x = 0.506
x = 0.253
The equilibrium constant is found:
0.506
0.318 0.405
·
6.25
12. (10 points) Determine the pH of a solution prepared by dissolving 8.82 grams of
lithium hydrogen carbonate, LiHCO3, in enough water to make 1.00 L of solution.
[LiCO3 ] =
1 mol
8.82 g LiHCO ×
3 67.96 g
1.00 L
∴ ⎡HCO 1 − ⎤ = 0.130 M
⎢⎣
3 ⎥⎦
= 0.130 M
1−
HCO 3 + H 2 O ←
⎯→ H 2 CO 3 + OH 1−
[ ]0
0.130
0
0
Δ[]
-x
+x
+x
[ ]eq
0.130
x
x
Kb =
[H 2 CO 3 ][OH1− ] =
[ HCO ]
1−
3
x2
= 2.4 × 10 −8
0.130
x = 5.59 × 10 −5 = [OH1− ]
p OH = −log[OH1− ] = 4.25
pH = 9.75
13. (10 points) For the following reaction, Kc = 0.592 at 200 K.
COCl2(g) ↔ CO(g) + Cl2(g)
If the reaction mixture initially contains a COCl2 concentration of 0.400 M, a CO
concentration of 0.500 M, and a Cl2 concentration of 0.600 M, what will be the
equilibrium concentration of each species?
COCl2(g) ↔ CO(g) +
Cl2(g)
[ ]0
0.400
0.500
0.600
Δ[]
-x
+x
+x
[ ]eq
0.400-x
0.500+x
0.600+x
[ ]eq
0.438 M
0.462 M
0.562 M
0.500
0.592
0.600
0.400
0.300 1.1
0.400
0.237 0.592
0.300 1.1
0
1.692
0.063
x = -0.0381, -1.654 (reject -1.654)
0.592
14. (10 points) Determine the [OH-], [H3O+], pH, and pOH for a solution that has a
potassium hydroxide, KOH, concentration of 4.8×10-4 M.
[OH-] = [ KOH] = 4.8×10-4 M
pOH = -log[OH-] = 3.32
1.0 10
4.8 10
2.1
pH = -log[H3O+] = 10.68
(or 14 – pOH = 14 – 3.32 = 10.68)
10