Question 16.14: Determine the pH and percent ionization for acetic acid solu...
Determine the \text{pH} and percent ionization for acetic acid solutions at 25°C with concentrations (a) 0.15 M, (b) 0.015 M, and (c) 0.0015 M.
Strategy Using the procedure described in Worked Example 16.13, we construct an equilibrium table and for each concentration of acetic acid, we solve for the equilibrium concentration of \text{H}^+. We use equation 16.2 to find \text{pH}, and Equation 16.7 to find percent ionization.
Setup From Table 16.7, the ionization constant, K_a , for acetic acid is 1.8 × 10^{–5}
\text{pH} = –\text{log} [\text{H}_3\text{O}^+] or \text{pH} = –\text{log} [\text{H}^+] Equation 16.2
Equation 16.7 \text{percent ionization} = \frac{[\text{H}^+]_\text{eq}}{[\text{HA}]_0} × 100\%
| TA B L E 1 6 . 7 Ionization Constants of Some Weak Acids at 25°C | |||
| Name of acid | Formula | Structure | K_a |
| \text{Chloroacetic acid} | \text{CH}_2\text{ClCOOH} |  |
5.6 × 10^{–2} |
| \text{Hydrofluoric acid} | \text{HF} |  |
7.1 × 10^{–4} |
| \text{Nitrous acid} | \text{HNO}_2 |  |
4.5 × 10^{–4} |
| \text{Formic acid} | \text{HCOOH} |  |
1.7 × 10^{–4} |
| \text{Benzoic acid} | \text{C}_6\text{H}_5\text{COOH} |  |
6.5 × 10^{–5} |
| \text{Acetic acid} | \text{CH}_3\text{COOH} |  |
1.8 × 10^{–5} |
| \text{Hydrocyanic acid} | \text{HCN} |  |
4.9 × 10^{–10} |
| \text{Phenol} | \text{C}_6\text{H}_5\text{OH} |  |
1.3 × 10^{–10} |
(a) \text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{H}_3\text{O}^+(aq) + \text{CH}_3\text{COO}^–(aq)
| Initial concentration (M): | 0.15 | 0 | 0 | |
| Change in concentration (M): | –x | +x | +x | |
| Equilibrium concentration (M): | 0.15 − x | x | x |
Solving for x gives [\text{H}_3\text{O}^+] = 0.0016 M and \text{pH} = –\text{log} (0.0016) = 2.78.
\text{percent ionization} = \frac{0.0016 M}{0.15 M} × 100\% = 1.1\%
(b) Solving in the same way as part (a) gives [\text{H}_3\text{O}^+] = 5.2 × 10^{–4} M and \text{pH} = 3.28.
\text{percent ionization} = \frac{5.2 × 10^{–4} M}{0.015 M} × 100\% = 3.5\%
(c) Solving the quadratic equation, or using successive approximation [≫ Appendix 1] gives[\text{H}_3\text{O}^+] = 1.6 × 10^{–4} M and \text{pH} = 3.78.
\text{percent ionization} = \frac{1.6 × 10^{–4} M}{0.0015 M} × 100\% = 11\%