Chemistry 10123 and 10125, Exam 4 Answer Key March 26, 2014

Answer Key
Chemistry 10123 and 10125, Exam 4
March 26, 2014
In the "SHOW ALL WORK" questions, include balanced, net-ionic equations for all
relevant chemical reactions. Clearly state and justify any assumptions that you may make.
1. (10 points) SHOW ALL WORK. The pKa of cyanic acid, HOCN, is 3.46. Determine the volume
(in mL) of 1.25 M NaOH that must be added to 600 mL of 0.250 M HOCN in order to yield a final
solution with a pH equal to 4.00.
initial moles HOCN = (0.600 L) (0.250 mole/L) = 0.150 moles
let x = moles OH added = moles HOCN neutralized = moles OCN formed
after some HOCN is neutralized, the solution is an HOCN / OCN buffer:
HOCN
H+ + OCN[H+] = Ka (moles HOCN) / moles OCN-)
10-4.00 = (10-3.46) (0.150 - x) / x
solve for x = 0.116 moles OH-
Volume of NaOH = (0.116 moles) (1000 mL / 1.25 moles) = 93.1 mL
2. (15 points) Write a balanced chemical equation for the equilibrium reaction that corresponds to each
of the following equilibrium constants. Indicate the proper phase (s, aq, etc.) of each species.
[e. g., Kw for water would be: 2 H2O(l)
H3O+(aq) + OH-(aq) ]
(a) Kb for HONH2
HONH2(aq) + H2O
HONH3+(aq) + OH (aq)
(b) Ksp for Ag2CrO4
Ag2CrO4(s)
2 Ag+(aq) + CrO42-(aq)
(c) Kb for HCO3HCO3-(aq) + H2O
H2CO3(aq) + OH-(aq)
(d) 1/Ka for (CH3)2NH2+
(CH3)2NH(aq) + H3O+(aq)
(e) Kf for Ni(CN)53Ni2+(aq) + 5 CN-(aq)
(CH3)2NH2+(aq) + H2O
Ni(CN)53-(aq)
3. Aziridine, C2H5N (43.1 g/mole), call it "AzN" for short, is a weak base with pKb = 5.960. A 0.345g sample of aziridine was dissolved in water to make 50.00 mL of solution. This solution was then
titrated with 0.200 M HCl
(a) (8 points) SHOW ALL WORK. Determine the volume (in mL) of 0.200 M HCl that is required
to reach the end-point.
moles AzN = moles HCl required = (0.345 g) (1 mole / 43.1 g) = 0.00800 moles
volume of HCl = (0.00800 mole) (1000 mL / 0.200 moles) = 40.0 mL
(b) (10 points) SHOW ALL WORK. Determine the pH after the addition of 25.0 mL of 0.200 M
HCl.
moles HCl added = moles AzN neutralized = moles AzNH+ formed
= (0.025 L) (0.200 mole / L) = 0.00500 moles
moles AzN remaining = 0.00800 - 0.00500 = 0.00300 moles
the mixture of AzN and AzNH+ is a buffer solution:
AzNH+
H+ + AzN
Ka = Kw/Kb = 10-14 / 10-5.96 = 9.12 x 10-9
[H+] = Ka (moles AzNH+) / (moles AzN)
[H+] = (9.12 x 10-9) (0.00500) / (0.00300) = 1.52 x 10-8
pH = 7.82
(c) (10 points) SHOW ALL WORK. Determine the pH of the solution at the equivalence point in
the titration.
at the equivalence point (where the total volume = 90.0 mL) all of the AzN has been
converted to AzNH+ so it is just a simple weak acid equilibrium:
AzNH+
H+ + AzN
in which the conc of AzNH+ = (0.00800 moles) / (0.0900 L) = 0.0889 M
Ka = [H+] [AzN] / [AzNH+] = x2 / (0.0889 - x)
9.12 x 10-9 ≈ x2 / (0.0889)
x = [H+] ≈ 2.85 x 10-5 M
pH = 4.55
(assumption OK)
assume x << 0.0889
4. (8 points) Indicate whether an aqueous solution of each of the following substances is acidic (A),
basic (B), or neutral (N).
(CH3)3N B
Ba(ClO4)2 N
(NH4)2SO4 A
K2CO3 B
5. (3 points) In the following equilibrium reaction, circle any species (one or more) that is a BrønstedLowry acid.
HF + H2SO4
H2F+ + HSO4
6. (2 point) Which reactant in the above reaction is a Lewis base? HF
7. (2 point) The conjugate base of AsH3 is AsH2
8. (10 points) SHOW ALL WORK. When excess solid Mn(OH)2 is mixed with 1.00 L of 0.210 M
NH4Cl, the resulting saturated solution has pH = 8.50. Determine the Ksp value of Mn(OH)2.
(Note: for NH3, Kb = 1.76 x 10-5)
for Mn(OH)2, Ksp = [Mn2+][OH-]2
since pH = 8.5, [OH-] = 10-5.5
the main challenge is to determine [Mn2+]
upon mixing the base Mn(OH)2 with the acid NH4+, a neutralization reaction occurs:
Mn2+(aq) + 2 NH3(aq) + 2 H2O
Mn(OH)2(s) + 2 NH4+(aq)
since some NH4+ is converted to NH3, the resulting solution is a buffer (pH = 8.50):
NH4+
NH3 + H+
Ka = Kw / Kb = 5.68 x 10-10 = [NH3] [H+] / [NH4+]
from the given pH…. [H+] = 10-8.50 = 3.16 x 10-9
if x moles of NH4+ react, x moles of NH3 are formed, thus we have:
5.68 x 10-10 = (x) (3.16 x 10-9) / (0.210 - x)
solve for x = 0.0320 mole NH3
from the stoichiometry of the above Mn(OH)2 reaction:
(0.0320 mole NH3) (1 mole Mn2+ / 2 mole NH3) = 0.0160 mole Mn2+
∴ [Mn2+] = 0.0160 M (since the volume is 1.00 L)
Ksp = (0.0160) (10-5.50)2 = 1.60 x 10-11
9. SHOW ALL WORK. Copper(II) phosphate, Cu3(PO4)2 (molar mass = 380.6), has a Ksp value of
1.4 x 10-37. Copper(II) forms a stable complex with cyanide ion that has the formula Cu(CN)42and a formation constant (Kf) of 1.0 x 1025.
(a) (10 points) Determine the molar solubility of Cu3(PO4)2 in a 1.25 M Na3PO4 solution.
Cu3(PO4)2(s)
3 Cu2+(aq) + 2 PO43-(aq)
Ksp = [Cu2+]3 [PO43-]2 = 1.4 x 10-37
[Cu2+] = 3x and [PO43-] = 2x + 1.25
1.4 x 10-37 = (3x)3 (2x + 1.25)2
assume 2x << 1.25
let x = molar solubility
1.4 x 10-37 ≈ (3x)3 (1.25)2
x ≈ 1.5 x 10-13 M
(assumption OK)
(b) (12 points) Determine the maximum amount (in grams) of Cu3(PO4)2 that will dissolve in 1.00 L
of 0.150 M KCN solution.
3 Cu2+(aq) + 2 PO43-(aq)
Ksp:
Cu3(PO4)2(s)
(Kf)3:
3 Cu2+(aq) + 12 CN-(aq)
Knet:
3 Cu(CN)42-(aq)
Cu3(PO4)2(s) + 12 CN (aq)
3 Cu(CN)42- (aq) + 2 PO43-(aq)
Knet = Ksp Kf3 = (1.4 x 10-37) (1.0 x 1025)3 = 1.4 x 1038 (very large!!!)
Since Knet is so large, the Rx essentially goes to completion, i.e., Cu3(PO4)2 keeps
dissolving until all of the CN- is consumed.
Simple stoichiometry!
(0.150 mole CN-) (1 mole Cu3(PO4)2 / 12 moles CN-)
= 0.0125 mole Cu3(PO4)2
mass of Cu3(PO4)2 = (0.0125 mole) (380.6 g/mole) = 4.76 g

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