Calculate the pH of a $0.050\text{M}$ aqueous solution of aniline (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2$) at $25^{\circ }\mathrm{C}$ given $K_b=4.3\times 10^{-10}$.

Step-by-step solution

1. Write the base‑dissociation equilibrium: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_3^{+}+\mathrm{O}{\mathrm{H}}^{-}$.
2. Set up the expression $K_b=\frac{[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_3^{+}][\mathrm{O}{\mathrm{H}}^{-}]}{[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2]}$ and assume $x$ mol L${}^{-1}$ of aniline dissociates.
3. Because $K_b$ is very small, $[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2]\approx 0.050\text{M}$ and $[\mathrm{O}{\mathrm{H}}^{-}]=[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_3^{+}]=x$.
4. Thus $K_b=\frac{x^2}{0.050}$, giving $x=\sqrt{K_b\times 0.050}=\sqrt{4.3\times 10^{-10}\times 0.050}=4.64\times 10^{-6}\text{M}$.
5. The hydroxide concentration is $[\mathrm{O}{\mathrm{H}}^{-}]=4.64\times 10^{-6}\text{M}$, so $pOH=-{\log }_{10}(4.64\times 10^{-6})=5.33$.
6. Finally $pH=14-pOH=14-5.33=8.67$.

Concept tested: pH calculation of weak base solution

Published 9 September 2026 · How this solution was prepared and checked