Chem 112
Question 1
8 Points
Spring 2007
Exam III
How many moles of potassium fluoride should be added to 2.5 L of a 0.197 M hydrofluoric
acid (HF) solution to prepare a buffer with a pH of 2.210
pH = pKa + log
2.210 = 3.143 + log
[F-]
[HF]
[F-] = 0.023
0.023 mol.L-1 x 2.5 L = 0.058
-
[F ]
[HF]
[F-]
[HF]
= 0.117
[Base]/[Acid] =
Question 2
8 Points
Whelan
Moles of KF =
An aqueous solution contains 0.393 M ammonia. How many moles of hydroiodic acid would
have to be added to 250 mL of this solution to prepare a buffer with a pH of 9.020?
Ka NH4+ = 5.56x10-10
pH = pKa + log
9.020 = 9.255 + log
[NH3]
[NH4+]
[NH3]
[NH4+]
0.392 - x
= 0.582
x
x = 0.248
0.248 mol.L-1 x 0.250 L = 0.062 mol NH4+
0.062 mol HI
[NH3]
= 0.582
[NH4+]
[Base]/[Acid] =
Question 3
8 Points
Moles of HI =
Assuming 100Ka < 0.1 then the area where
the pH can be determine by:
1)
–log10{(Ka x 0.1)0.5}
A
2)
–log10(Ka)
C
3)
–log10{(Kb x 0.05)0.5}
B
4) Circle the area on the curve in
which the pH can be determined by:
pH = pKa + log10([A-]/[HA])
Question 4
6 Points
Determine the equilibrium concentration of hydroxide ion in a saturated manganese(II)
hydroxide solution.
I
C
E
Mn(OH)2(s)
Ù
-s
-s
Mn2+
0
s
s
2 OH0
2s
2s
+
s(2s)2 = 4.6x10-14
s = 2.26x10-5
Ksp = [Mn2+][OH-]2
M
Question 5
5 Points
The maximum amount of barium phosphate that will dissolve in a 0.279 M ammonium
phosphate solution is
I
C
E
A
Ba2(PO4)3(s)
Ù
-s
-s
-s
3 Ba2+
0
3s
3s
3s
+
2 PO430.279
2s
0.279 + 2s
0.279
(3s)3(0.279)2 = 1.3x10-29
s = 1.84x10-10
Ksp = [Ba2+]3[PO43-]2
M
Question 6
8 Points
Question 7
10 Points
Each of the insoluble salts below are put into 0.10 M hydroiodic acid solution. Do you expect
their solubility to be: more, less, or about the same as in a pure water solution ?
1. Lead iodide
B
2. Barium fluoride
A
3. Calcium sulfite
A
4. Silver bromide
C
A. More
B. Less
C. Same
Predict whether ∆S for each reaction would be greater than zero, less than zero, or too
close to zero to decide.
A. 2H2O2(l) = 2H2O(l) + O2(g)
1
B. 2HBr(g) + Cl2(g) = 2HCl(g) + Br2(g)
3
1. Greater than 0
C. I2(g) + Cl2(g) = 2ICl(g)
3
2. Less than 0
D. 2CO(g) + 2NO(g) = 2CO2(g) + N2(g)
2
3. Too close to tell
E. PCl5(g) = PCl3(g) + Cl2(g)
1
Question 8
8 Points
When 15.0 mL of a 2.52x10-4 M silver acetate solution is combined with 15.0 mL
of a 2.70x10-4 M potassium hydroxide solution does a precipitate form ?
Yes
For these conditions the Reaction Quotient, Q, is equal to
AgOH(s) Ù Ag+ + OH-
Q = [Ag+][OH-] = 1.70x10-8
2.52x10-4(0.015) = 3.78x10-6 mol Ag+
[Ag+] = 1.26x10-4
Q < Ksp
No
2.70x0-4(0.015) = 4.05x10-6 mol OH[OH-] = 1.35x10-4
Q =
Question 9
4 Points
Using standard absolute entropies at 298K, calculate the entropy change for the following
reaction:
H2S(g) + 2H2O(l) = 3H2(g) + SO2(g)
∆So = 3SoH2(g) + SoSO2(g) – SoH2S(g) – 2SoH2O(l)
∆So = 3(131) + 248 – 206 – 2(70)
∆So =
Question 10
6 Points
J.K-1
Consider the reaction:
2N2(g) + O2(g) = 2N2O(g)
Using standard thermodynamic data at 298K, calculate the entropy change for the
surroundings when 1 mole of O2(g) reacts at standard conditions:
∆HoRxn = 2∆HofN2O(g) – 2∆HofN2(g) – ∆HofO2(g)
∆HoRxn = 2(82) – 2(0) – 0 = 164 kJ/mol
∆SSurr = -164,000/298
∆SSurroundings =
Question 11
6 Points
J.mol-1.K-1
For the reaction:
2H2S(g) + 3O2(g) = 2H2O(g) + 2SO2(g)
o
∆H = -1036 kJ and ∆So = -153 J/K.
Assuming that ∆Ho and ∆So are independent of temperature, determine the maximum
amount of work that could be done when 1.8 moles of H2S(g) react at 326 K and 1 atm.
∆GRxn = ∆H - T∆S
∆GRxn = -1036 –326(-0.153)
∆GRxn = -986 kJ
Work = 986 kJ/2 mol H2S
kJ
Question
12
12 Points
Without doing any calculations, match the following thermodynamic properties with their
appropriate numerical sign for the following reactions.
A. N2(g) + 3H2(g) = 2NH3(g)
∆SRxn
2
∆GRxn
5
∆SUniverse
4
Exothermic
1.
2.
3.
4.
5.
B. 2H2O(g) + 2SO2(g) = 2H2S(g) + 3O2(g)
Question
13
5 Points
∆SRxn
1
∆GRxn
4
∆SUniverse
5
>0
<0
=0
> 0 low T, < 0 high T
< 0 low T, > 0 high T
Endothermic
1.
2.
3.
4.
5.
>0
<0
=0
> 0 low T, < 0 high T
< 0 low T, > 0 high T
∆Ho = -683kJ
∆So= -366J/K
2NH3(g) + 2O2(g) = N2O(g) + 3H2O(l)
Above or below what temperature would the above reactions ∆Go be less than zero?
∆Ho – T∆So = 0
-683,000 – T(-366) = 0
366T = 683,000
Above/Below
Question
14
6 Points
K
Below
Consider the reaction:
N2(g) + 3H2(g) = 2NH3(g)
Calculate ∆G for this reaction at 298.15K if the pressure of NH3(g) is reduced to
24.64 mm Hg while the pressures of N2(g) and H2(g) remain at 1 atm.
Q=
PNH32
PN2PH23
=
(0.0324)2
= 0.00105
1(1)3
∆GRxn
∆GRxn = ∆GoRxn + RTLn(0.00105)
= -33 + 0.008314(298.15)Ln(0.00105)
∆GoRxn = 2∆GofNH3(g) – ∆GofN2(g) – 3∆GofH2(g)
∆GoRxn = 2(-16.5) – 0 – 0 = -33kJ
∆G =
Exam III Score
kJ