Chapter 14 Key

Nicole Bouvier-Brown; Saori Shiraki; J. Ryan Hunt; Emily Jarvis

Answer Keys to Selected Problems

Chapter 14 Key

14.1. One example for NH₃ as a conjugate acid: NH₂⁻ + H⁺ → NH₃; as a conjugate base: NH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l)

14.3.

(a) H₃O⁺(aq) → H⁺(aq) + H₂O(l)

(b) HCl(aq) → H⁺(aq) + Cl⁻(aq)

(c) NH₃(aq) → H⁺(aq) + NH₂⁻(aq)

(d) CH₃CO₂H(aq) → H⁺(aq) + CH₃CO₂⁻(aq)

(e) NH₄⁺(aq) → H⁺(aq) + NH₃(aq)

(f) HSO₄⁻(aq) → H⁺(aq) + SO₄²⁻(aq)

14.5.

(a) H₂O(l) + H⁺(aq) → H₃O⁺(aq)

(b) OH⁻(aq) + H⁺(aq) → H₂O(l)

(c) NH₃(aq) + H⁺(aq) → NH₄⁺(aq)

(d) CN⁻(aq) + H⁺(aq) → HCN(aq)

(e) S²⁻(aq) + H⁺(aq) → HS⁻(aq)

(f) H₂PO₄⁻(aq) + H⁺(aq) → H₃PO₄(aq)

14.7. (a) H₂O, O²⁻ (b) H₃O⁺, OH⁻ (c) H₂CO₃, CO₃²⁻ (d) NH₄⁺, NH₂⁻ (e) H₂SO₄, SO₄²⁻ (f) H₃O₂⁺, HO₂⁻ (g) H₂S, S²⁻ (h) H₆N₂²⁺, H₄N₂

14.9. The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA.

(a) HNO₃(BA), H₂O(BB), H₃O⁺(CA), NO₃⁻(CB)

(b) CN⁻(BB), H₂O(BA), HCN(CA), OH⁻(CB)

(c) H₂SO₄(BA), Cl⁻(BB), HCl(CA), HSO₄⁻(CB)

(d) HSO₄⁻(BA), OH⁻(BB), SO₄²⁻(CB), H₂O(CA)

(e) O²⁻(BB), H₂O(BA), OH⁻(CB and CA)

(f) [Cu(H₂O)₃(OH)]⁺(BB), [Al(H₂O)₆]³⁺(BA), [Cu(H₂O)₄]²⁺(CA), [Al(H₂O)₅(OH)]²⁺(CB)

(g) H₂S(BA), NH₂⁻(BB), HS⁻(CB), NH₃(CA)

14.11. Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H₂O. As an acid: H₂O(aq) + NH₃(aq) ⇌ NH₄⁺(aq) + OH⁻(aq).

As a base: H₂O(aq) + HCl(aq) ⇌ H₃O⁺(aq) + Cl⁻(aq)

14.13. (a) amphiprotic: NH₃ + H₃O⁺ → NH₄OH + H₂O, NH₃ + OCH₃⁻ → NH₂⁻ + CH₃OH; (b) amphiprotic: HPO₄²⁻ + OH⁻ → PO₄³⁻ + H₂O, HPO₄²⁻ + HClO₄ → H₂PO₄⁻ + ClO₄⁻ (c) not amphiprotic: Br⁻ (d) not amphiprotic: NH₄⁺ (e) not amphiprotic: AsO₄³⁻

14.15. In a neutral solution [H₃O⁺] = [OH⁻]. At 40°C, [H₃O⁺] = [OH⁻] = (2.910 × 10⁻¹⁴)^1/2 = 1.7 × 10⁻⁷.

14.17. x = 3.051 × 10⁻⁷ M = [H₃O⁺] = [OH⁻]; pH = −log (3.051 × 10⁻⁷) = −(−6.5156) = 6.5156; pOH = pH = 6.5156

14.19. (a) pH = 3.587; pOH = 10.413 (b) pOH = 0.68; pH = 13.32 (c) pOH = 3.85; pH = 10.15 (d) pOH = −0.40; pH = 14.4

14.21. [H₃O⁺] = 3.0 × 10⁻⁷ M; [OH⁻] = 3.3 × 10⁻⁸ M

14.23. [H₃O⁺] = 1 × 10⁻² M; [OH⁻] = 1 × 10⁻¹² M

14.25. [OH⁻] = 3.1 × 10⁻¹² M

14.27. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH⁻, which causes the solution to be basic.

14.29. [H₂O] > [CH₃CO₂H] > [H₃O⁺] ≈ [CH₃CO₂⁻] > [OH⁻]

14.31. The oxidation state of the sulfur in H₂SO₄ is greater than the oxidation state of the sulfur in H₂SO₃.

14.33. Mg(OH)₂(s) + 2 HCl(aq) → Mg²⁺(aq) + 2 Cl⁻(aq) + 2 H₂O(l)
BB BA CB CA

14.35. K_a = 2.3 × 10⁻¹¹

14.37. The stronger base or stronger acid is the one with the larger K_b or K_a, respectively. In these two examples, they are (CH₃)₂NH and CH₃NH₃⁺.

14.39. triethylamine

14.41. (a) HSO₄⁻; higher electronegativity of the central ion. (b) H₂O; NH₃ is a base and water is neutral, or decide on the basis of K_a values. (c) HI; PH₃ is weaker than HCl; HCl is weaker than HI. Thus, PH₃ is weaker than HI. (d) PH₃; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.

14.43.

(a) NaHSeO₃ < NaHSO₃ < NaHSO₄; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner.

(b) ClO₂⁻ < BrO₂⁻ < IO₂⁻; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.

(c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.

(d) HOCl < HOClO < HOClO₂ < HOClO₃; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases).

(e) HTe⁻ < HS⁻ << PH₂⁻ < NH₂⁻; PH₂⁻and NH₂⁻are anions of weak bases, so they act as strong bases toward H⁺. HTe⁻and HS⁻ are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion.

(f) BrO₄⁻ < BrO₂⁻ < BrO₂⁻ < BrO⁻; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.

14.45. [H₂O] >[C₆H₄OH(CO₂H)] > [H⁺] 0>[C₆H₄OH(CO₂)⁻] ≫ [C₆H₄O(CO₂H)⁻] > [OH⁻]

14.47. 1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H₃O⁺.

14.48. (b) The addition of HCl

14.50.

(a) Adding HCl will add H₃O⁺ ions, which will then react with the OH⁻ ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO₂, and decreasing the concentration of NO₂⁻ ions.

(b) Adding HNO₂ increases the concentration of HNO₂ and shifts the equilibrium to the left, increasing the concentration of NO₂⁻ ions and decreasing the concentration of OH⁻ ions.

(c) Adding NaOH adds OH⁻ ions, which shifts the equilibrium to the left, increasing the concentration of NO₂⁻ ions and decreasing the concentrations of HNO₂.

(d) Adding NaCl has no effect on the concentrations of the ions.

(e) Adding KNO₂ adds NO₂⁻ions and shifts the equilibrium to the right, increasing the HNO₂ and OH⁻ ion concentrations.

14.52. This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO₂H exists primarily as HCO₂H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO₂H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H₃O⁺] produced by the stronger acid.

14.54. (a) K_b = 1.8 × 10⁻⁵ (b) K_a = 4.5 × 10⁻⁴ (c) K_b = 6.4 × 10⁻⁵ (d) K_a = 5.6 × 10⁻¹⁰

14.56. K_a = 1.2 × 10⁻²

14.58. (a) K_b = 4.3 × 10⁻¹² (b) K_a = 1.6 × 10¹⁰ (c) K_b = 5.9 × 10⁸ (d) K_b = 4.2 × 10⁻³ (e) K_b = 2.3 × 10⁵ (f) K_b = 6.3 × 10⁻¹³

14.60.

(a)

\frac{[H_{3} O^{+}] [C l O^{-}]}{[H C l O]} = \frac{x \cdot x}{0.0092 - x} \approx \frac{x^{2}}{0.0092} = 2.9 \times 10^{- 8}

Solving for x gives 1.63 × 10⁻⁵ M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H₃O⁺] = [ClO^–] = 1.6 × 10⁻⁵ M; [HClO^–] = 0.0092 M; [OH⁻] = 6.1 × 10⁻¹⁰ M;

(b)

\frac{[C_{6} H_{5} N H_{3}^{+}] [O H^{-}]}{[C_{6} H_{5} N H_{2}]} = \frac{x \cdot x}{0.0784 - x} \approx \frac{x^{2}}{0.0784} = 4.3 \times 10^{- 10}

Solving for x gives 5.81 × 10⁻⁶ M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [C₆H₅NH₃⁺] = [OH⁻] = 5.8 × 10⁻⁶ M; [C₆H₅NH₂] = 0.0784 M; [H₃O⁺] = 1.7 × 10⁻⁹ M;

(c)

\frac{[H_{3} O^{+}] [C N^{-}]}{[H C N]} = \frac{x \cdot x}{0.0810 - x} \approx \frac{x^{2}}{0.0810} = 4.9 \times 10^{- 10}

Solving for x gives 6.30 × 10⁻⁶ M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H₃O⁺] = [CN⁻] = 6.3 × 10⁻⁶ M; [HCN] = 0.0810 M; [OH⁻] = 1.6 × 10⁻⁹ M;

(d)

\frac{[(C H_{3})_{3} N H^{+}] [O H^{-}]}{[(C H_{3})_{3} N]} = \frac{x \cdot x}{0.11 - x} = \frac{x^{2}}{0.11} = 6.3 \times 10^{- 5}

Solving for x gives 2.63 × 10⁻³ M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [(CH₃)₃NH⁺] = [OH⁻] = 2.6 × 10⁻³ M; [(CH₃)₃N] = 0.11 M; [H₃O⁺] = 3.8 × 10⁻¹² M;

(e)

\frac{[F e (H_{2} O)_{5} (O H)^{+}] [H_{3} O^{+}]}{[F e (H_{2} O)_{6}^{2 +}]} = \frac{x \cdot x}{0.120 - x} = \frac{x^{2}}{0.120} = 1.6 \times 10^{- 7}

Solving for x gives 1.39 × 10⁻⁴ M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [Fe(H₂O)₅(OH)⁺] = [H₃O⁺] = 1.4 × 10⁻⁴ M; [Fe(H₂O)₆²⁺] = 0.120 M; [OH⁻] = 7.2 × 10⁻¹¹ M

14.62. pH = 2.41

14.64. [C₁₀H₁₄N₂] = 0.049 M; [C₁₀H₁₄N₂H⁺] = 1.9 × 10⁻⁴ M; [C₁₀H₁₄N₂H₂²⁺] = 1.4 × 10⁻¹¹ M; [OH⁻] = 1.9 × 10⁻⁴ M; [H₃O⁺] = 5.3 × 10⁻¹¹ M

14.66. K_a = 1.2 × 10⁻²

14.68. K_b = 1.77 × 10⁻⁵

14.70. (a) acidic (b) basic (c) acidic (d) neutral

14.72. [H₃O⁺] and [HCO₃⁻] are practically equal

14.74. [C₆H₄(CO₂H)₂] 7.2 × 10⁻³ M, [C₆H₄(CO₂H)(CO₂)⁻] = [H₃O⁺] 2.8 × 10⁻³ M, [C₆H₄(CO₂)₂²⁻] 3.9 × 10⁻⁶ M, [OH⁻] 3.6 × 10⁻¹² M

14.76. (a) K_ba2 = 1.5 × 10⁻¹¹ (b) K_b = 4.3 × 10⁻¹²

(c)

\frac{[T e^{2 -}] [H_{3} O^{+}]}{[H T e^{-}]} = \frac{x (0.0141 + x)}{0.0141 - x} = \frac{x \cdot 0.0141}{0.0141} = 1.5 \times 10^{- 11}

Solving for x gives 1.5 × 10⁻¹¹ M. Therefore, compared with 0.014 M, this value is negligible (1.1 × 10⁻⁷%).

14.78.

Excess H₃O⁺ is removed primarily by the reaction: H₃O⁺(aq) + H₂PO₄⁻(aq) → H₃PO₄(aq) + H₂O(l)

Excess base is removed by the reaction: OH⁻(aq) + H₃PO₄(aq) → H₂PO₄⁻(aq) + H₂O(l)

14.80. [H₃O⁺] = 1.5 × 10⁻⁴ M

14.82. [OH⁻] = 4.2 × 10⁻⁴ M

14.84.

(a) The added HCl will increase the concentration of H₃O⁺ slightly, which will react with CH₃CO₂⁻ and produce CH₃CO₂H in the process. Thus, [CH₃CO₂⁻] decreases and [CH₃CO₂H] increases.

(b) The added KCH₃CO₂ will increase the concentration of [CH₃CO₂⁻] which will react with H₃O⁺ and produce CH₃CO₂ H in the process. Thus, [H₃O⁺] decreases slightly and [CH₃CO₂H] increases.

(c) The added NaCl will have no effect on the concentration of the ions.

(d) The added KOH will produce OH⁻ ions, which will react with the H₃O⁺, thus reducing [H₃O⁺]. Some additional CH₃CO₂H will dissociate, producing [CH₃CO₂⁻] ions in the process. Thus, [CH₃CO₂H] decreases slightly and [CH₃CO₂⁻] increases.

(e) The added CH₃CO₂H will increase its concentration, causing more of it to dissociate and producing more [CH₃CO₂⁻] and H₃O⁺ in the process. Thus, [H₃O⁺] increases slightly and [CH₃CO₂⁻] increases.

14.86. pH = 8.95

14.88. 37 g (0.27 mol)

14.90. (a) pH = 5.222 (b) The solution is acidic. (c) pH = 5.220

14.92. At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.

14.94. (a) pH = 2.50 (b) pH = 4.01 (c) pH = 5.60 (d) pH = 8.35 (e) pH = 11.08

Chapter 14 Key

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