Calculate the pH of a $0.1\text{ M}$ solution of a weak acid $\text{HA}$ if its dissociation constant $K_a$ is $10^{-5}$.

Solution

1. The dissociation is $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$.
2. The equilibrium constant is $$\begin{array}{rl}{K}_a& =\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}\\ & =\frac{x^2}{0.1-x}\approx \frac{x^2}{0.1}\end{array}$$.
3. Solving for $x=[{\text{H}}^{+}]$ gives $$\begin{array}{rl}{x}^2& =10^{-5}\times 0.1\\ & =10^{-6}\end{array}$$.
4. Thus, $[{\text{H}}^{+}]=10^{-3}\text{ M}$.
5. The pH is $-\log (10^{-3})=3$.
6. Hence the answer is (C).