Question 16.12: Calculate the pH of a 0.20 M solution of HCN. (Refer to Tabl...
Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for the value of K_{a}.)
| TABLE 16.2 Some Weak Acids in Water at 25 °C | |||
| Acid | Structural Formula* | Conjugate Base | K_{a} |
| Chlorous (HClO_{2}) | H—O—Cl—O | ClO^{ -}_{2} | 1.0 × 10^{-2} |
| Hydrofluoric (HF) | H—F | F^{-} | 6.8 × 10^{-4} |
| Nitrous (HNO_{2}) | H—O—N=O | NO^{ -}_{2} | 4.5 × 10^{-4} |
| Benzoic (C_{6}H_{5}COOH) |  |
C_{6}H_{5}COO^{-} | 6.3 × 10^{-5} |
| Acetic (CH_{3}COOH) | H-O-\overset{\overset{O}{||} }{C} -\overset{\overset{H}{|} }{\underset{\underset{H}{|} }{C}} -H | CH_{3}COO^{-} | 1.8 × 10^{-5} |
| Hypochlorous (HOCl) | H—O—Cl | OCl^{-} | 3.0 × 10^{-8} |
| Hydrocyanic (HCN) | H—C≡N | CN^{-} | 4.9 × 10^{-10} |
| Phenol (HOC_{6}H_{5}) |  |
C_{6}H_{5}O^{-} | 1.3 × 10^{-10} |
| *The proton that ionizes is shown in red. | |||
Analyze We are given the molarity of a weak acid and are asked for the pH. From Table 16.2, K_{a} for HCN is 4.9 × 10^{-10}.
Plan We proceed as in the example just worked in the text, writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H^{+} is our unknown.
Solve Writing both the chemical equation for the ionization reaction that forms H^{+}(aq) and the equilibrium expression for the reaction:
HCN (aq) ⇌ H^{+} (aq) + CN^{-} (aq)K_{a} =\frac{[H^{+}][CN^{-}]}{[HCN]}= 4.9 × 10^{-10}
Next, we tabulate the concentrations of the species involved in the equilibrium reaction, letting x = [H^{+}] at equilibrium:
Substituting the equilibrium concentrations into the equilibrium expression yields:
K_{a} =\frac{(x)(x)}{0.20 – x }= 4.9 × 10^{-10}We next make the simplifying approximation that x, the amount of acid that dissociates, is small compared with the initial concentration of acid, 0.20 -x \simeq 0.20. Thus,
\frac{x²}{0.20}= 4.9 × 10^{-10}Solving for x, we have:
x² = (0.20)(4.9 × 10^{-10}) = 0.98 × 10^{-10}
x = \sqrt{0.98 × 10^{-10}} = 9.9 × 10^{-6} M = [H^{+}]
A concentration of 9.9 × 10^{-6} M is much smaller than 5% of 0.20 M, the initial HCN concentration. Our simplifying approximation is therefore appropriate. We now calculate the pH of the solution:
pH = – log [H^{+}] = -log (9.9 × 10^{-6}) = 5.00