Question 16.21: Calculate the pH of a 0.10-M solution of ammonium chloride (...
Calculate the \text{pH} of a 0.10-M solution of ammonium chloride (\text{NH}_4\text{Cl}) at 25°\text{C}.
Strategy A solution of \text{NH}_4\text{Cl} contains \text{NH}_4^+ cations and \text{Cl}^– anions. The \text{NH}_4^+ ion is the conjugate acid of the weak base \text{NH}_3. Use the K_b value for \text{NH}_3 (1.8 × 10^{–5} from Table 16.8) and Equation 16.8 to determine K_a for \text{NH}_4^+.
K_a = \frac{K_w}{K_b} = \frac{1.0 × 10^{–14}}{1.8 × 10^{–5}} = 5.6 × 10^{–10}
Setup Again, we write the balanced chemical equation and the equilibrium expression:
\text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{NH}_3(aq) + \text{H}_3\text{O}^+(aq) K_a = \frac{[\text{NH}_3][\text{H}_3\text{O}^+]}{[\text{NH}_4^+]}
Next, construct a table to determine the equilibrium concentrations of the species in the equilibrium expression:
\text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{NH}_3(aq) + \text{H}_3\text{O}^+(aq)
| Initial concentration (M): | 0.10 | 0 | 0 | |
| Change in concentration (M): | –x | +x | +x | |
| Equilibrium concentration (M): | 0.10 − x | x | x |
Equation 16.8 K_a × K_b = K_w
| TA B L E 1 6 . 8 Ionization Constants of Some Weak Acids at 25°C | |||
| Name of acid | Formula | Structure | K_b |
| \text{Ethylamine} | \text{C}_2\text{H}_5\text{NH}_2 |  |
5.6 × 10^{–4} |
| \text{Methylamine} | \text{CH}_3\text{NH}_2 |  |
4.4 × 10^{–4} |
| \text{Ammonia} | \text{NH}_3 |  |
1.8 × 10^{–5} |
| \text{Pyridine} | \text{C}_5\text{H}_5\text{N} |  |
1.7 × 10^{–9} |
| \text{Aniline} | \text{C}_6\text{H}_5\text{NH}_2 |  |
3.8 × 10^{–10} |
| \text{Urea} | \text{H}_2\text{NCONH}_2 |  |
1.5 × 10^{–14} |
Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve for x, we get
5.6 × 10^{–10} = \frac{x^2}{0.10 − x} ≈ \frac{x^2}{0.10}
x = \sqrt{(5.6 × 10^{–10})(0.10)} = 7.5 × 10^{–6} M
According to the equilibrium table, x = [\text{H}_3\text{O}^+]. The \text{pH} can be calculated as follows:
\text{pH} = –\text{log} (7.5 × 10^{–6}) = 5.12
The \text{pH} of a 0.10-M solution of ammonium chloride (at 25°\text{C}) is 5.12.