Final Exam, Spring 2005
Final Exam, Spring 20051. Complete and balance the following chemical equations.
a. aqueous perchloric acid plus aqueous potassium hydroxide
b. aqueous phosphoric acid plus aqueous magnesium hydroxide
c. HCN(aq) + NH3(aq)
d. Co(CH3CO2)2(aq) + H2S(aq)
e. Fe3+(aq) + EDTA4–(aq)
f. aqueous bicarbonate ion plus aqueous cyanide ion
g. Cu(s) + HNO3(aq) → Cu2+(aq) + NO2(g)
h. Al(s) + MnO4–(aq) → Al(OH)3(s) + MnO2(s)
g. H2C2O4(aq) + CrO42–(aq) → Cr3+(aq) + CO2(g)
a. HClO4(aq) + KOHaq(aq) → H2O(l) + K+(aq) + ClO4–(aq)
b. 2 H3PO4(aq) + 3 Mg(OH)2(aq) → 6 H2O(l) + Mg3(PO4)2(s)
c. HCN(aq) + NH3(aq) → ← NH4+(aq) + CN–(aq) (Kc = 6.2×10–10/ 5.6×10–10 = 1.1 < 100)
d. Co(CH3CO2)2(aq) + H2S(aq) → CoS(s) + 2 CH3CO2H(aq)
e. Fe3+(aq) + EDTA4–(aq) → ← [FeEDTA]–(aq)
f. HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO32–(aq)
g. Cu(s) + HNO3(aq) → Cu2+(aq) + NO2(g)
Oxidation: Cu(s) → Cu2+(aq) + 2 e–
Reduction: HNO3(aq) + H+(aq) + e– → NO2(g) + H2O(l)
Net: Cu(s) + 2 HNO3(aq) + 2 H+(aq) → Cu2+(aq) + 2 NO2(g) + 2 H2O(l)
h. Al(s) + MnO4–(aq) → Al(OH)3(s) + MnO2(s)
Oxidation: Al(s) + 3 OH–(aq) → Al(OH)3(s) + 3 e–
Reduction: MnO4–(aq) + 2 H2O(l) + 3 e– → MnO2(s) + 4 OH–(aq)
Net: Al(s) + MnO4–(aq) + 2 H2O(l) → Al(OH)3(s) + MnO2(s) + OH–(aq)
g. H2C2O4(aq) + CrO42–(aq) → Cr3+(aq) + CO2(g)
Oxidation: H2C2O4(aq) + → 2 CO2(g) + 2 H+(aq) + 2 e–
Reduction: CrO42–(aq) + 8 H+(aq) + 3 e– → Cr3+(aq) + 4 H2O(l)
Net: 3 H2C2O4(aq) + 2 CrO42–(aq) + 10 H+(aq) → 2 Cr3+(aq) + 6 CO2(g) + 8 H2O(l)
2. For each of the following solutions or mixtures, estimate the molar concentrations of all dissolved ions and molecular species (except water) and then estimate the pH of the solution.
a. 0.1 M hydrobromic acid
b. 0.1 M sodium hydroxide
c. 0.1 M Na2CO3
d. 0.1 M Fe(NO3)2
e. 0.1 mol Al(OH)3 placed in 1 L of water
a. HBr(aq) + H2Ol) → H3O+(aq) + Br–(aq)
[H3O+] = 0.1 M
[Br–] = 0.1 M
[OH–] = 1.0×10–14/0.1 = 1.×10–13 M
pH = –log(0.1) = 1.0
b. NaOH(aq) → Na+(aq) + OH–(aq)
[Na+] = 0.1 M
[OH–] = 0.1 M
[H3O+] = 1.0×10–14/0.1 = 1.×10–13 M
pH = –log(1×10–13) = 13.0
c. Na2CO3 → 2 Na+(aq) + CO32–(aq)
[Na+] = 2×0.1 = 0.2 M
CO32–(aq) + H2O(l) → ← HCO3–(aq) + OH–(aq)
Kb = [HCO3–]e[OH–]e/ [CO32–]e = 1.0×10–14/ 4.7×10–11 = 2.1×10–4
Initial0.100
Change–x+x+x
Equil.0.1 – xxx
0.1/2.1×10–4 = 500 > 100 so approximation is allowed
2.1×10–4 = x2/0.1
x = 5×10–3
[OH–] = 5×10–3 M
[H3O+] = 1.0×10–14/5×10–3 = 2×10–12
[CO32–] = 0.1 M
[HCO3–] = 5×10–3 M
pH = –log(2×10–12) = 11.7
d. Fe(NO3)2 → Fe2+(aq) + 2 NO3–(aq)
[NO3–] = 2×0.1 = 0.2 M
Fe2+(aq) + 2 H2O(l) → ← FeOH+(aq) + H3O+(aq)
Ka = [FeOH+]e[H3O+]e/ [Fe2+]e = 3.2×10–10
Initial0.100
Change–x+x+x
Equil.0.1 – xxx
0.1/3.2×10–10 = 3×108 > 100 so approximation is allowed
3.2×10–10 = x2/0.1
x = 6×10–6
[H3O+] = 6×10–6 M
[OH–] = 1.0×10–14/6×10–6 = 2×10–9
[Fe2+] = 0.1 M
[FeOH+] = 6×10–6 M
pH = –log(6×10–6) = 5.2
e. 0.1 mol Al(OH)3 placed in 1 L of water
Al(OH)3(s) → ← Al3+(aq) + 3 OH–(aq)
Ksp = [Al3+]e[OH–]e3 = 1.8×10–5
Initial00
Change+x+3x
Equil.x3x
1.8×10–5 = (x)(3x)3 = 27x3
x = 8.7×10–3
[OH–] = 3×8.7×10–3 = 0.026 M
[H3O+] = 1.0×10–14/0.026 = 3.8×10–13
[Al3+] = 8.7×10–3 M
pH = –log(3.8×10–13) = 12.41
3. A reaction, A(aq) + B(aq) → products, has a rate law that is first order in each reactant at 25 °C.
a. Write the rate law for the reaction.
b. If the concentration of A is doubled, how does the reaction rate change and how does the rate constant change?
c. If the concentration of A is doubled and simultaneously the concentration of B is halved, how does the reaction rate change and how does the rate constant change?
d. If the temperature is doubled, how does the reaction rate change and how does the rate constant change?
a. Rate = –k[A][B]
b. The reaction rate doubles but the rate constant is unchanged.
c. Neither the reaction rate nor the rate constant change.
d. Both the reaction rate and the rate constant increase.
4. Consider the reaction between hydrogen phosphate ion and carbonate ion.
a. Write the balanced reaction.
b. Find ΔH° for the reaction.
c. Find ΔS° for the reaction.
d. Find ΔG° for the reaction at 25 °C.
e. Calculate the equilibrium constant for the reaction using two different methods. Compare the two values.
a. HPO42–(aq) + CO32–(aq) → ← PO43–(aq) + HCO3–(aq)
b. ΔH° = [(–1277) + (–691.11)] – [(–1292.1) + (–677.1)] = 1 kJ/mol
c. ΔS° = [(–222) + (95.0)] – [(–33) + (–56.9)] = –37 J/mol·K
d. ΔG° = 1 – (298)(–0.037) = 12 kJ/mol
e. Use ΔG° and use Ka for the acid and conjusgate acid:
Kc = e–12000/(8.314×298) = 7.9×10–3
Kc = 4.2×10–13/4.7×10–11 = 8.9×10–3
The agreement between the two methods is quite good.
5. Consider the reaction:
[Al(OH)4]–(aq) + Zn(s) → [Zn(OH)4]2–(aq) + Al(s)
a. Write the balanced oxidation half-reaction in base.
b. Write the balanced reduction half-reaction in base.
c. Write the balanced net reaction in base.
d. Write the cell notation for the reaction at standard conditions.
a. Zn(s) + 4 OH–(aq) → [Zn(OH)4]2–(aq) + 2 e–
b. [Al(OH)4]–(aq) + 3 e– → Al(s) + 4 OH–(aq)
c. 2 [Al(OH)4]–(aq) + 3 Zn(s) + 4 OH–(aq) → 3 [Zn(OH)4]2–(aq) + 2 Al(s)
d. Zn(s) | [Zn(OH)4]2–(aq) | pH = 14 || [Al(OH)4]–(aq) | pH = 14 | Al(s)
6. Briefly answer the following questions:
a. Zn(OH)2 is more soluble in a solution that is buffered with pH = 10 than in a solution that is not buffered with pH = 10. Explain.
b. A solution of hydrochloric acid with pH = 4 and a solution of acetic acid with pH = 4 are prepared. If equal volumes of each solution are titrated with 0.1 M sodium hydroxide, which solution requires more base and why?
c. H+ and H3O+ are used interchangeably to represent the cationic autoionization product of water and any associated chemical reactivity. Despite this, the thermodynamic parameters for the two ions are different. Suggest an explanation.
d. What does entropy measure?
a. Zinc(II) hydroxide reacts with the available hydroxide ion to form a complex ion:
Zn(OH)2(s) + 2 OH–(aq) → ← [Zn(OH)4]2–(aq)
In the buffered solution there is sufficient hydroxide present for the complex ion formation to proceed to equilibrium, consuming some of the zinc(II) hydroxide. In the unbuffered solution, the hydroxide is in too low of concentration, essentially a limiting reagent, for the reaction to proceed as far towards products, leaving more undissolved salt.
b. The acetic acid solution requires much more base. This is because the hydrochloric acid, a strong acid, is 100% ionized and the pH represents all of the available acid. In contrast, acetic acid is a weak acid, only a few per cent ionized, so the pH represents only a small fraction of the acid that can react with the base.
c. Hydronium ion is a representation of hydrogen ion bound to a water molecule. By definition, the thermodynamic parameters for hydrogen ion are zero (this arises from the standard potential definition) so the thermodynamic parameters for hydronium ion are the same as those found for liquid water.
d. Entropy is a measure of the degeneracy of a system, i.e. the number of states that have the same energy. This is directly embodied in the Third Law of Thermodynamics (S = RlnW). Entropy can also be thought of as where energy is stored in a system.