Final Exam A, Spring 2018

1. Complete and balance the following in aqueous solutin:

a. chlorite ion plus hydrobromic acid

b. calcium hydroxide plus perchloric acid

c. H3PO4(aq) + Zn(C2H3O2)2(aq)

d. Sc3+(aq) + H2O(l)

e. cyanate ion plus water

f. Na2S(aq) + Fe(NO3)2(aq)

g. chromate ion plus nitrite ion giving chromium 3+ ion and nitric acid

h. MnO4–(aq) + C2O42–(aq) → Mn(OH)2(s) + CO2(g)

Answer

a. ClO2–(aq) + HBr(aq) → HClO2(aq) + Br–(aq)

b. Ca(OH)2(s) + 2 HClO4(aq) → 2 H2O(l) + Ca2+(aq) + 2 ClO4–(aq)

c. 2 H3PO4(aq) + 3 Zn(C2H3O2)2(aq) → Zn3(PO4)2(s) + 6 HC2H3O2(aq)

d. Sc3+(aq) + 2 H2O(l) → ← ScOH2+(aq) + H3O+(aq)

e. OCN–(aq) + H2O(l) → ← HOCN(aq) + OH–(aq)

f. Na2S(aq) + Fe(NO3)2(aq) → FeS(s) + 2 Na+(aq) + 2 NO3–(aq)

g. chromate ion plus nitrite ion giving chromium 3+ ion and nitric acid

Oxidation: NO2–(aq) + H2O(l) → HNO3(aq) + H+(aq) + 2 e–

Reduction: CrO42–(aq) + 8 H+(aq) + 3 e– → Cr3+(aq) + 4 H2O(l)

Net: 2 CrO42–(aq) + 3 NO2–(aq) + 13 H+(aq )→ 2 Cr3+(aq) + 3 HNO3(aq) + 5 H2O(l)

h. MnO4–(aq) + C2O42–(aq) → Mn(OH)2(s) + CO2(g)

Oxidation: C2O42–(aq) → 2 CO2(g) + 2 e–

Reduction: MnO4–(aq) + 6 H+(aq) + 5 e– → Mn(OH)2(s) + 2 H2O(l)

Net in acid: 2 MnO4–(aq) + 5 C2O42–(aq) + 12 H+(aq) → 2 Mn(OH)2(s) + 10 CO2(g) + 4 H2O(l)

Neutralization: 12 H2O(l) → 12 H+(aq) + 12 OH–(aq)

Net in base: 2 MnO4–(aq) + 5 C2O42–(aq) + 8 H2O(l) → 2 Mn(OH)2(s) + 10 CO2(g) + 12 OH–(aq)

2. The data given below was found for the reaction

[Cr(NH3)5(H2O)]3+(aq) + Cl–(aq) → [Cr(NH3)5Cl]2+(aq) + H2O(l)

[[Cr(NH3)5(H2O)]3+] (M) [Cl–] (M)Initial Rate (M·s–1)


0.00100.00107.0×10–3


0.00200.00101.4×10–2


0.00200.00202.8×10–2


Find the rate law and the rate constant (including units).

Answer

Rate = –k[[Cr(NH3)5(H2O)]3+]p [Cl–]q

From experiments 1 and 2, doubling [Cr(NH3)5(H2O)]3+] while keeping [Cl–] constant doubles the rate, which means that p = 1.

From experiments 2 and 3, doubling [Cl–] while keeping [Cr(NH3)5(H2O)]3+] constant doubles the rate, which means that q = 1.

Thus, k = Rate/[Cr(NH3)5(H2O)]3+][Cl–]

k = 7.0×103 M–1s–1 for all three experiments.

3. Find the pH of a 0.10 M solution of butyric acid (C3H7CO2H).

Answer

C3H7CO2H(aq) + H2O(l) → ← H3O+(aq) + C3H7CO2–(aq)

Ka = [H3O+]e[C3H7CO2–]e/ [C3H7CO2H]e = 1.5×10–5

Initial0.1000

Change–x+x+x

Equilibrium0.10 – xxx

Approximate? 0.10/1.5×10–5 = 6700 > 100, so yes.

1.5×10–5 = x2/0.10

x = [H3O+] = 1.2×10–3 M

pH = –log(1.2×10–3) = 2.91

4. Find the molar solubility of PbBr2 in water.

Answer

PbBr2(aq) → ← Pb2+(aq) + 2 Br–(aq)

Ksp = [Pb2+]e[Br–]e2 = 4.0×10–5

Initial00

Change+x+2x

Equilibriumx2x

4.0×10–5 = (x)(2x)2 = 4x3

x = molar solubility = 2.2×10–2 M

5. Consider the reaction between Pb(OH)2(s) and HCl(g) at 25 °C.

a. Find ΔG° for the solubilization of Pb(OH)2.

b. Find ΔG° for the autoionization of water.

c. Find ΔG° for the solubilization of PbCl2.

d. Find ΔG° for the reaction of HCl(g) with water.

e. Find ΔG° for the reaction of Pb(OH)2(s) with HCl(g).

Answer

a. Pb(OH)2(s) → ← Pb2+(aq) + 2 OH–(aq)

ΔG° = –RTlnKsp = –(8.314)(298)ln(1.2×10–15) = 85100 J/mol = 85.1 kJ/mol

b. 2 H2O(l) → ← H3O+(aq) + OH–(aq)

ΔG° = –RTlnKw = –(8.314)(298)ln(1.0×10–14) = 79900 J/mol = 79.9 kJ/mol

OR ΔG° = [(–237.1) + (–157.2)] – [2(–237.2)] = 80.0 kJ/mol

a. PbCl2(s) → ← Pb2+(aq) + 2 Cl–(aq)

ΔG° = –RTlnKsp = –(8.314)(298)ln(1.6×10–5) = 27400 J/mol = 27.4 kJ/mol

d. HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq)

ΔG° = [(–237.2) + (–131.2)] – [(–95.30) + (–237.2)] = –35.9 kJ/mol

e. Pb(OH)2(s) + 2 HCl(g) → PbCl2(s) + 2 H2O(l)

The reactions in parts a through d can be used to attain the desired reaction:

Pb(OH)2(s) → ← Pb2+(aq) + 2 OH–(aq) ΔG° = 85.1 kJ/mol

HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq) ΔG° = –35.9 kJ/mol

HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq) ΔG° = –35.9 kJ/mol

H3O+(aq) + OH–(aq) → ← 2 H2O(l) ΔG° = –80.0 kJ/mol

H3O+(aq) + OH–(aq) → ← 2 H2O(l) ΔG° = –80.0 kJ/mol

Pb2+(aq) + 2 Cl–(aq) → ← PbCl2(s) ΔG° = –27.4 kJ/mol

Summing the reactions means summing the individual ΔG° values, keeping track of reversed reactions, which changes the sign of ΔG°.

ΔG° = 85.1 – 35.9 –35.9 –80.0 –80.0 – 27.4 = –174.1 kJ/mol

6. Consider the reaction F2(g) + H2O(l) → OF2(g) + F–(aq)

a. Find the oxidation numbers in all of the atoms in the reactants and products.

F in F2

H in H2O

O in H2O

O in OF2

F in OF2

F in F–

b. Balance the reaction under acidic conditions.

c. Find the standard potential for the reaction.

d. Find ΔG° for the reaction.

e. Is the reaction spontaneous or nonspontaneous? Why?

Answer

a. Oxidation numbers

F in F2 : 0

H in H2O : +1

O in H2O : –2

O in OF2 : +2

F in OF2 : –1

F in F– : –1

b. Oxidation: H2O(l) + 2 F–(aq) → OF2(g) + 2 H+(aq) + 4 e–

Reduction: F2(g + 2 e– → 2 F–(aq)

Net: 2 F2(g) + H2O(l) → OF2(g) + 2 H+(aq) + 2 F–(aq)

c. E° = 2.86 + (–2.1) = 0.8 V

d. ΔG° = –nFE° = –(4)(96485)(0.8) = –300000 J/mol = –300 kJ/mol

e. Spontaneous because E° > 0 and ΔG° < 0.

7. Consider the cell:

Pt(s) | PbSO4(s) | PbO2(s) | SO42–(aq) | pH = 0 || PbO2(s) | Pb2+(aq) | pH = 0 | Pt(s)

a. Write the oxidation half-reaction.

b. Write the reduction half-reaction.

c. Write the net reaction.

d. Find the standard potential at 25 °C.

e. Find the equilibrium constant for the reaction.

f. What is the appropriate label (Ka, Kb, etc.) for the equilibrium constant?

Answer

a. PbSO4(s) + 2 H2O(l) → PbO2(s) + SO42–(aq) + 4 H+(aq) + 2 e–

b. PbO2(s) + 4 H+(aq) + 2 e– → Pb2+(aq) + 2 H2O(l)

c. PbSO4(s) → ← Pb2+(aq) + SO42–(aq)

d. E° = 1.453 + (–1.69) = –0.24 V

e. E° = (RT/nF)ln Keq

Keq = enFE°/RT = e(2)(96485)(–0.24)/(8.314)(298) = 7.6×10–9

f. Ksp - the net reaction is the solubilization of lead sulfate.