Calculate the pH of a $0.010\text{M}$ solution of phenol (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}$) at $25^{\circ }\text{C}$ given $K_a=1.3\times 10^{-10}$.

Step-by-step solution

1. Write the dissociation equilibrium: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}+{\mathrm{H}}^{+}$.
2. Let $x$ be the concentration of ${\mathrm{H}}^{+}$ formed. Then $[{\mathrm{H}}^{+}]=x$ and $[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}]=x$.
3. The equilibrium expression is $K_a=\frac{x^2}{C-x}$ where $C=0.010\text{M}$.
4. Since $K_a$ is very small, assume $x\ll C$, so $C-x\approx C$.
5. Solve for $x$: $x=\sqrt{K_a\times C}=\sqrt{1.3\times 10^{-10}\times 0.010}=\sqrt{1.3\times 10^{-12}}\approx 1.14\times 10^{-6}\text{M}$.
6. Calculate pH: $\text{pH}=-{\log }_{10}(1.14\times 10^{-6})\approx 5.94$.

Concept tested: Acidity of phenols

Published 6 September 2026 · How this solution was prepared and checked