The goal of the first ten questions is to compare the solubility of manganese(II) hydroxide at 25 °C and 60 °C.
1. Calculate the molar solubility of Mn(OH)2 at 25 °C.
Mn(OH)2(s) → ← Mn2+(aq) + 2 OH–(aq)
Ksp = [Mn2+]e[OH–]e2 = 1.9×10–13
Initial00
Change+x+2x
Equilibriumx2x
Ksp = [x][2x]2 = 1.9×10–13
4x3 = 1.9×10–13
x = 3.6×10–5
The molar solubility of manganese(II) hydroxide is 3.6×10–5 mol/L at 25 °C.
2. Calculate the pH of a saturated solution of Mn(OH)2 at 25 °C.
[OH–] = 2× 3.6×10–5 = 7.2×10–5
pOH = –log(7.2×10–5) = 4.14
pH = 14.00 – 4.14 = 9.86
3. Calculate ΔG° for the solubilization reaction of Mn(OH)2 at 25 °C.
ΔG° = –RTlnKsp = –(8.314)(298)ln(1.9×10–13) = 72600 J/mol = 72.6 kJ/mol
4. Calculate ΔG°f for Mn(OH)2 at 25 °C.
ΔG°(rxn) = [ΔG°f(Mn2+) + 2(ΔG°f(OH–))] – [ΔG°f(Mn(OH)2)]
72.6 = [(–228.1) + 2(–157.2)] – [ΔG°f(Mn(OH)2)]
[ΔG°f(Mn(OH)2)] = –615.1 kJ/mol
5. Calculate ΔH° for the solubilization reaction of Mn(OH)2 at 25 °C. ΔH°f(Mn(OH)2(s)) = –693.7 kJ/mol
ΔH° = [–220.8 + 2(230.0)] – [–693.7] = +12.9 kJ/mol
6. Calculate ΔS° for the solubilization reaction of Mn(OH)2 at 25 °C.
ΔS° = (ΔH° – ΔG°)/T = (12.9 – 72.6)/298 = –0.200 kJ/mol·K = –200. J/mol·K
7. Calculate S° for Mn(OH)2 at 25 °C.
S°(Mn(OH)2) = [&ndashh;73.6 + 2(–10.75)] – (–200.) = 105 J/mol·K
8. Calculate ΔG° for the solubilization reaction of Mn(OH)2 at 60 °C.
ΔG° = 12.9 – (333)(–0.200) = 79.5 kJ/mol
9. Calculate Ksp for Mn(OH)2 at 60 °C.
ΔG° = –RTlnKsp
79500 = –(8.314)(333)lnKsp
Ksp = e–79500/(8.314×333) = 3.4×10–13
10. Calculate the molar solubility of Mn(OH)2 at 60 °C.
Mn(OH)2(s) → ← Mn2+(aq) + 2 OH–(aq)
Ksp = [Mn2+]e[OH–]e2 = 3.4×10–13
Initial00
Change+x+2x
Equilibriumx2x
Ksp = [x][2x]2 = 3.4×10–13
4x3 = 1.9×10–13
x = 4.4×10–5
The molar solubility of manganese(II) hydroxide is 4.4×10–5 mol/L at 60 °C.
The next eight questions look at the solubility of manganese(II) hydroxide at 60 °C in terms of pH.
11. Calculate ΔH° for the autoionization of water at 25 °C.
2 H2O(l) → ← H3O+(aq) + OH–(aq)
ΔH° = [(–285.8) + (–230.0)] – [2(–285.8)] = +55.8 kJ/mol
12. Calculate ΔS° for the autoionization of water at 25 °C.
ΔS° = [(69.91) + (–10.75)] – [2(69.91)] = –80.66 J/mol·K
13. Calculate ΔG° for the autoionization of water at 25 °C.
ΔG° = 55.8 – (298)(–0.08066) = +79.8 kJ/mol
14. Calculate ΔG° for the autoionization of water at 60 °C.
ΔG° = 55.8 – (333)(–0.08066) = +82.7 kJ/mol
15. Calculate Kw at 60 °C.
82700 = –(8.314)(333)ln Kw
Kw = e–82700/(8.314×333) = 1.1×10–13
16. Calculate [H3O+]e and [OH–]e of pure water at 60 °C.
Kw = [H3O+]e[OH–]e
1.1×10–13 = x2
x = [H3O+]e = [OH–]e = 3.3×10–7 M
17. Calculate the pH of pure water at 60 °C.
pH = –log[H3O+] = –log(3.3×10–7) = 6.48
18. Calculate the pH of a saturated solution of Mn(OH)2 at 60 °C.
[OH–] (at 60 °C) = 2(4.4×10–5) = 8.8×10–5
[H3O+] (at 60 °C) = 1.1×10–13/[OH–] (at 60 °C)
[H3O+] (at 60 °C) = 1.1×10–13/8.8×10–5 = 1.3×10–9 M
pH = –log(1.3×10–9) = 8.89
The next six questions look at the acid/base properties of manganese(II) and manganese(II) hydroxide.
19. Calculate Kb for Mn(OH)2 at 25 °C. (Hint: use the rule of multiple equilibria. Kb for MnOH+ = 3.6×10–4)
Mn(OH)2(s) → ← MnOH+(aq) + OH–(aq) 1 Kb1
MnOH+(aq) → ← Mn2+(aq) + OH–(aq) 2 Kb2 = 3.6×10–4
Mn(OH)2(s) → ← Mn2+(aq) + 2 OH–(aq) 3 Ksp = 1.9×10–13
From the Law of Multiple Equilibria: Ksp = Kb1×Kb2
Kb1 = Ksp/Kb2 = (1.9×10–13)/(3.6×10–4) = 5.3×10–10
20. Calculate the equilibrium concentration of MnOH+(aq) at 25 °C. Use the usual assumption that only the first ionization is important for a weak base.
Mn(OH)2(s) → ← MnOH+(aq) + OH–(aq)
Kb1 = [MnOH+]e[OH–]e = 5.3×10–10
Initial00
Change+x+x
Equilibriumxx
Kb1 = [x][x] = 5.3×10–10
x2 = 5.3×10–10
x = [MnOH+] = 2.3×10–5
21. Calculate the [OH–]e and pH of a saturated Mn(OH)2 solution at 25 °C using Kb(Mn(OH)2) and Kb(MnOH+).
The contribution from the solid salt is found in question 20: [MnOH+] = [OH–] = 2.3×10–5
MnOH+(aq) → ← Mn2+(aq) + OH–(aq)
Kb2 = [Mn2+]e[OH–]e/[MnOH+]e = 3.6×10–4
Initial2.3×10–50 2.3×10–5
Change–x+x+x
Equilibrium2.3×10–5 – xx 2.3×10–5 + x
Kb2 = [x][x + 2.3×10–5]/ [2.3×10–5 – x] = 3.6×10–4
x2 + 2.3×10–5x = 8.28×10–9 – 3.6×10–4x
x2 + 3.83×10–4x – 8.28×10–9 = 0
x = 2.1×10–5 or x =–4.9×10–4
[OH–] = 2.3×10–5 + 2.1×10–5 = 4.4×10–5 M
pOH = –log(4.4×10–5) = 4.36
pH = 14.00 – 4.36 = 9.64
22. The pH values calculated from question 2 and question 21 should be the same; why are they different?
The method of Kb1 and Kb2 is flawed because, atypically, Kb1 < Kb2. (Normally, the second ionization is less than the first ionization.) Thus, a significant portion of the hydroxide ion generated in the reaction comes from the second step. This also reduces the MnOH+ concentration (note, that in problem 21 the MnOH+ concentration is reduced by 90% from "initial" conditions). To properly treat this as an acid/base problem requires using both equilibria simultaneously, which is mathematically tedious. Note, also, that treatment of the equilibria through the solubility method (Ksp) is also flawed because it does not allow for any concentration of MnOH+.
23. Calculate Ka for Mn2+(aq) at 25 °C.
Mn2+(aq) + 2 H2O(l) → ← MnOH+(aq) + H3O+(aq)
Ka = Kw/Kb = 1.0×10–14/3.6×10–4 = 2.8×10–11
24. Calculate the pH of a 0.10 M solution of Mn(NO3)2 at 25 °C.
Mn2+(aq) + 2 H2O(l) → ← MnOH+(aq) + H3O+(aq)
Ka = [MnOH+]e[H3O+]e/[Mn2+]e = 2.8×10–11
Initial0.1000
Change–x+x+x
Equilibrium0.10 – xxx
Approximate? 0.10/2.8×10–11 = 3.6×109 > 100, so Yes
2.8×10–11 = x2/0.10
x = 1.7×10–6
pH = –log(1.7×10–6) = 5.77
The goal of the next seven questions is to find the standard reduction potential of manganese(II) in a basic environment.
25. Calculate E° for Pt(s) | H2(g) | H+(aq) || Mn2+(aq) | Mn(s) at 25 °C.
Oxidation: H2(g) → 2 H+(aq) + 2 e– E°ox = 0.00 V
Reduction: Mn2+(aq) + 2 e– → Mn(s) E°red = –1.18 V
Net: Mn2+(aq) + H2(g) → 2 H+(aq) + Mn(s) E° = E°red + E°ox = –1.18 + 0.00 = –1.18 V
26. Calculate ΔG° for Pt(s) | H2(g) | H+(aq) || Mn2+(aq) | Mn(s) at 25 °C.
ΔG° = –nFE° = –(2)(96485)(–1.18) = +228000 J/mol = +228 kJ/mol
27. Calculate Kc for Pt(s) | H2(g) | H+(aq) || Mn2+(aq) | Mn(s) at 25 °C.
–RTln Kc = ΔG°
Kc = e–ΔG°/RT = e–(228000)/(8.314)(298) = 1.×10–40
28. Calculate Kc for Pt(s) | H2(g) | OH–(aq) || Mn(OH)2(s) | Mn(s) at 25 °C. (Use the rule of multiple equilibria.)
Oxidation: H2(g) + 2 OH–(aq) → 2 H2O(l) + 2 e–
Reduction: Mn(OH)2(s) + 2 e– → Mn(s) + 2 OH–(aq)
Net: H2(g) + Mn(OH)2(s) → 2 H2O(l) + Mn(OH)2(s)
This is the target reaction to find Kc.
Mn2+(aq) + H2(g) → ← 2 H+(aq) + Mn(s) Kc = 1×10–40 (from problem 27)
2 H+(aq) + 2 OH–(aq) → ← 2 H2O(l) Kc = (1/Kw)2 = (1/1.0×10–14)2 = 1.0×1028
Mn(OH)2(s) → ← Mn2+(aq) + 2 OH–(aq) Kc = Ksp = 1.9×10–13
The three reactions sum to the target reaction so Kc(target) = (1×10–40)(1.0×1028)(1.9×10–13) = 2×10–25
29. Calculate ΔG° for Pt(s) | H2(g) | OH–(aq) || Mn(OH)2(s) | Mn(s) at 25 °C.
ΔG° = –RTln Kc = –(8.314)(298)ln(2×10–25) = 141000 J/mol = 141 kJ/mol
30. Calculate E° for Pt(s) | H2(g) | OH–(aq) || Mn(OH)2(s) | Mn(s) at 25 °C.
ΔG° = –nFE°
E° = –ΔG°/nF = –141000/(2)(96485) = –0.731 V
31. Calculate the reduction potential for Mn(OH)2(s) + 2e– → Mn(s) + 2 OH–(aq).
From problem 28, E° = E°red + E°ox(H2/OH–)
–0.731 = E°red + 0.828
E° = –1.559 V
Some thought questions:
32. A common misconception amongst beginning chemistry students is that the pH of a solution tells you if an acid is strong or weak. Why is this statement false?
The pH of a solution only indicates the concentration of hydronium ion, not the degree of ionization. To determine if an acid is strong or weak, the pH and the initial concentration of the acid must be known. For example, a 1 M solution of acetic acid and a 4×10–3 M solution of hydrochloric acid have about the same pH, but acetic acid is a weak acid and hydrochloric acid is a strong acid.
33. It can be argued that the Ka of pure water at 25 °C is 1.8×10–16. Show how this value is arrived at.
In Brønsted/Lowry theory, Ka is defined for a reaction of a weak acid with water to give hydronium ion and hydroxide ion. The mass action expression includes the acid and the ionic products. Thus, thinking of water as a weak acid, the reaction should be written as:
H2O(aq) + H2O(l) → ← H3O+(aq) + OH–(aq)
By definition, then: Ka = [H3O+]e[OH–]e/[H2O]e
[H3O+]e[OH–]e = Kw = 1.0×10–14
[H2O] = 55.6 M
Thus, Ka = 1.0×10–14/55.6 = 1.8×10–16
34. The concentration of a solid does not change as a function of reaction progress and so it is not used in mass action expressions. Despite this, the surface area of a solid strongly affects the reaction rate. Explain.
Reactions in solids can only occur on the surface of the material - reactants cannot easily penetrate into the interior of a solid because of the high density of the solid. As the surface area increases, there are more reaction sites available and the reaction can proceed at a faster rate. In terms of a rate law, this can be written as
Rate = k[surface sites]m
where the order of reaction, m, depends on the specific reaction.
This effect is easily seen in metals. For example, aluminum is a perfectly safe material that we all handle routinely as soda cans or aluminum foil. Yet, the reaction of aluminum with oxygen is highly exothermic (ΔHfo of aluminum oxide is -1676 kJ/mol) and the free-energy is highly spontaneous (ΔGf° of aluminum oxide is -1582 kJ/mol). If aluminum is finely powdered, increasing the surface area by many orders of magnitude, now the material becomes a major fire hazard because the rate of reaction has become quite high.
35. The First Law of Thermodynamics says that energy is conserved. Why, then, is it impossible to make a perpetual motion machine?
To make a perpetual motion machine, one would want to exploit a spontaneous process and convert all of the available energy to work, i.e. ΔU = w and q = 0. The Second Law of Thermodynamics says that for this to occur, the total entropy change must be increasing. Further, some of this entropy will be passed from the system to the surroundings in the form of heat, i.e., q > 0. Entropy always wins, so there is always some heat created, leading to a loss of efficiency and some conversion of energy to heat.