Calculate the pH of a $0.025\text{M}$ aqueous solution of pyridine at $25^{\circ }\text{C}$ given $K_b=1.7\times 10^{-9}$.

Step-by-step solution

1. Write the base‑hydrolysis equilibrium: ${\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}^{+}+\mathrm{O}{\mathrm{H}}^{-}$.
2. Express the base‑dissociation constant: $K_b=\frac{[C_5{H}_{5}N{H}^{+}][O{H}^{-}]}{[C_5{H}_{5}N]}$.
3. Assume $x=[O{H}^{-}]$ produced; then $[C_5{H}_{5}N{H}^{+}]=x$ and $[C_5{H}_{5}N]\approx 0.025-x\approx 0.025$ (since $x\ll C$).
4. Set up $K_b=x^2/0.025$, giving $x=\sqrt{K_b\times 0.025}=\sqrt{1.7\times 10^{-9}\times 0.025}=6.5\times 10^{-6}\text{M}$.
5. Calculate $pOH=-\log [O{H}^{-}]=-\log (6.5\times 10^{-6})=5.187$.
6. Finally, $pH=14-pOH=14-5.187=8.813\approx 8.81$.

Concept tested: pH calculation of weak base solutions

Published 11 September 2026 · How this solution was prepared and checked