Final Exam, Spring 2001

1. Complete and balance the following reactions.

a. H2SO4(aq) + BaBr2(aq)

b. H2SO3(aq) + H2O(l)

c. HCN(aq) + Sr(OH)2(s)

d. AgBr(s) + NH3(aq)

e. FeCl3(aq) + H2S(aq)

f. Mg(ClO4)2(aq) + Rb2CO3(aq)

g. Mn2+(aq) + H2O(l)

h. HF(aq) + NH3(aq)

i. ClO2(aq) + HOCl(aq) → ClO3–(aq) + Cl–(aq) [hint: ClO2 is named chlorine dioxide]

j. ClO2(aq) → ClO2–(aq) + ClO3–(aq) (pH = 13)

Answer

a. H2SO4(aq) + BaBr2(aq) + 2 H2O(l) → BaSO4(s) + 2 H3O+(aq) + 2 Br–(aq)

b. H2SO3(aq) + H2O(l) → ← H3O+(aq) + HSO3–(aq)

c. 2 HCN(aq) + Sr(OH)2(s) → Sr(CN)2(s) + 2 H2O(l)

d. AgBr(s) + 2 NH3(aq) → ← [Ag(NH3)2]+(aq) + Br–(aq)

e. 2 FeCl3(aq) + 3 H2S(aq) + 6 H2O(l) → Fe2S3(s) + 6 H3O+(aq) + 6 Cl–(aq)

f. Mg(ClO4)2(aq) + Rb2CO3(aq) → MgCO3(s) + 2 Rb+(aq) + 2 ClO4–(aq)

g. Mn2+(aq) + 2 H2O(l) → ← MnOH+(aq) + H3O+(aq)

h. HF(aq) + NH3(aq) → ← NH4+(aq) + F–(aq) (Kc = 6.2×10–/5.6×10– = 1.1 < 100)

i. ClO2(aq) + HOCl(aq) → ClO3–(aq) + Cl–(aq) [hint: ClO2 is named chlorine dioxide]

Oxidation: ClO2(aq) + H2O(l) → ClO3–(aq) + 2 H+(aq) + e–

Reduction: HOCl(aq) + H+(aq) + 2 e– → Cl–(aq) + H2O(l)

Net: 2 ClO2(aq) + HOCl(aq) + H2O(l) → 2 ClO3–(aq) + Cl–(aq) + 3 H+(aq)

j. ClO2(aq) → ClO2–(aq) + ClO3–(aq) (pH = 13)

Oxidation: ClO2(aq) + H2O(l) → ClO3–(aq) + 2 H+(aq) + e–

Reduction: ClO2(aq) + e– → ClO2–(aq)

Net in acid: 2 ClO2(aq) + H2O(l) → ClO2–(aq) + ClO3–(aq) + 2 H+(aq)

Neutralization: 2 H+(aq) + 2 OH–(aq) → 2 H2O(l)

Net in base: 2 ClO2(aq) + 2 OH–(aq) → ClO2–(aq) + ClO3–(aq) + H2O(l)

2. Hydrogen azide reacts with hydrogen peroxide to give products of water and nitrogen gas.

a. Write the balanced reaction.

b. Suppose that the rate of reaction doubles if the concentration of either reactant is doubled but the reaction rate quadruples if both reactant concentrations are doubled. What is the rate law?

Answer

a. 2 HN3(aq) + H2O2(aq) → 2 H2O(l) + 3 N2(g)

b. Rate = –k[HN3][H2O2]

3. Consider the reaction: OCl–(aq) + I–(aq) → OI–(aq) + Cl–(aq)

a. Write the balanced reaction.

b. The following rate data was obtained. Find the order of reaction in OCl–, I–, and OH–.

[OCl–] (M) [I–] (M) pH Initial Rate (M·s–1)

 

0.0040 0.0020 14.00 4.8×10–4

 

0.0020 0.0040 14.00 5.0×10–4

 

0.0020 0.0020 14.00 2.4×10–4

 

0.0020 0.0020 13.70 4.5×10–4

 

c. Find the average rate constant.

Answer

a. Already balanced: OCl–(aq) + I–(aq) → OI–(aq) + Cl–(aq)

b. The following rate data was obtained. Find the order of reaction in OCl–, I–, and OH–.

[OCl–] (M) [I–] (M) pH [OH–] (M) Initial Rate (M·s–1)

 

0.0040 0.0020 14.00 1.0 4.8×10–4

 

0.0020 0.0040 14.00 1.0 5.0×10–4

 

0.0020 0.0020 14.00 1.0 2.4×10–4

 

0.0020 0.0020 13.70 0.50 4.5×10–4

 

Experiments 1 & 3 show that the order in OCl– = 1

Experiments 2 & 3 show that the order in I– = 1

Experiments 3 & 4 show that the order in OH– = –1

c. The rate law is Rate = k[OCl–][I–][OH–]–1 so k = Rate×[OH–1]/[OCl–][I–]

[OCl–] (M) [I–] (M) pH [OH–] (M) Initial Rate (M·s–1)

k (s–1)

 

0.0040 0.0020 14.00 1.0 4.8×10–4

4.8×10–4×(1.0)/(0.0040×0.0020) = 60.

 

0.0020 0.0040 14.00 1.0 5.0×10–4

5.0×10–4×(1.0)/(0.0020×0.0040) = 63.

 

0.0020 0.0020 14.00 1.0 2.4×10–4

2.4×10–4×(1.0)/(0.0020×0.0020) = 60.

 

0.0020 0.0020 13.70 0.50 4.5×10–4

4.5×10–4×(0.50)/(0.0020×0.0020) = 56.

 

kave = 60. s–1

4. Consider Fe(OH)3:

a. Write the solubilization reaction in pure water.

b. Find ΔH° for the reaction in part a in units of kJ.

c. Find ΔS° for the reaction in part a in units of J/K.

d. Find ΔG° at 25 °C for the reaction in part a in units of kJ.

e. Find ΔG° at 80 °C for the reaction in part a in units of kJ.

f. Find Ksp for Fe(OH)3 at 25 °C.

g. Find the molar solubility of Fe(OH)3 at 25 °C.

h. Find the pH of a saturated solution of Fe(OH)3 at 25 °C.

Answer

a. Fe(OH)3(s) → ← Fe3+(aq) + 3 OH–(aq)

b. ΔH° = [(–48.5) + 3(–230.0)] – [–823.0] = +84.5 kJ/mol

c. ΔS° = [(–315.9) + 3(–10.75)] – [107] = –455 J/mol·K

d. ΔG° = 84.5 – (298)(–0.455) = +220. kJ/mol

e. ΔG° = 84.5 – (353)(–0.455) = +245. kJ/mol

f. Ksp = e–220000/(8.314×298) = 2.7×10–39

g. Fe(OH)3(s) → ← Fe3+(aq) + 3 OH–(aq)

Ksp = [Fe3+]e[OH–]e3 = 2.7×10–39

Initial00

Change+x+3x

Equilibriumx3x

2.7×10–39 = (x)(3x)3 = 27x4

x = 1.0×10–10 M = molar solubility

h. [OH–] from Fe(OH)3 = 3×1.0×10–10 = 3.0×10–10

This is much less than the contribution from water, thus the pH = 7.00

5. 1.0 L of a 0.10 M solution of Fe2+ solution is prepared at 25 °C. Find the pH of the solution at 25 °C.

Answer

Fe2+(aq) + 2 H2O(l) → ← FeOH+(aq) + H3O+(aq)

Ka = [FeOH+]e[H3O+]e/ [Fe2+]e = 3.2×10–10

Initial0.1000

Change–x+x+x

Equilibrium0.10 – xxx

Approximation is acceptable, so:

3.2×10–10 = x2/0.10

x = 5.7×10–6

pH = –log(5.7×10–6) = 5.24

6. 1.0 L of a 0.0025 M solution of ammonia is prepared at 25 °C. Find the pH of the solution at 25 °C.

Answer

NH3(aq) + H2O(l) → ← NH4+(aq) + OH–(aq)

Kb = [NH4+]e[OH–]e/ [NH3]e = 1.0×10–14/5.6×10–10 = 1.8×10–5

Initial0.0.002500

Change–x+x+x

Equilibrium0.0.0025 – xxx

Approximation is not acceptable, so:

1.8×10–5 = x2/(0.0025 – x)

x2 + 1.8×10–5x – 4.5×10–8 = 0

x = 2.0×10–4 or –2.2×10–4

pOH = –log(2.0×10–4) = 3.70

pH = 14.00 – 3.70 = 10.30

7. Consider the cell Ag(s) | Ag+(aq, 0.10 M) || H2O2(aq, 0.50 M) | H+(aq, pH = 2.00) | Pt(s) at 25 °C.

a. Write the oxidation half-reaction.

b. Write the reduction half-reaction.

c. Write the net reaction.

d. Find the standard cell potential.

e. Find the cell potential.

Answer

a. Ag(s) → Ag+(aq + e–

b. H2O2(aq) + 2 H+(aq + 2 e– → 2 H2O(l)

c. 2 Ag(s) + H2O2(aq) + 2 H+(aq → 2 Ag+(aq + 2 H2O(l)

d. E° = 1.763 + (–0.800) = +0.963 V

e. E = E° – (RT/nF)lnQ, where Q = [Ag+]2/[H2O2][H+]2

[Ag+] = 0.10 M, [H2O2] = 0.50 M, [H+] = 1.0×10–2 M

Q = (0.10)2/(0.50)(1.0×10–2)2 = 2.0×102

E = 0.963 – (8.314×298/2×96485)ln(200) = +0.895 V

8. Answer each of the following questions in 10 words or less.

a. Which is the stronger acid, Co2+ ion or Co3+ ion? Why?

b. Solutions are prepared of two different acids. The pH of the Acid A solution is 3.0 and the pH of the acid B solution is 4.0. Which is the stronger acid?

c. Without using mathematics, describe the First Law of Thermodynamics.

d. Distinguish between a Lewis acid and a Brønsted-Lowry acid.

e. What is the difference between solubility and Ksp?

Answer

a. Co3+ ion because of the higher charge.

b. Not enough information to tell - also need the initial concentrations.

c. Conservation of energy.

d. Lewis acid: electron pair acceptor; Brønsted-Lowry acid: hydrogen ion donor.

e. Solubility: the amount in solution; Ksp: an equilibrium constant.