Calculate the pH of a $0.030\text{M}$ aqueous solution of pyridine at 25°C, given $K_b=1.7\times 10^{-9}$.

Step-by-step solution

1. Write the equilibrium: ${\text{C}}_5{\text{H}}_5\text{N}+{\text{H}}_2\text{O}\rightleftharpoons {\text{C}}_5{\text{H}}_5\text{N}\cdot {\text{H}}^{+}+{\text{OH}}^{-}$.
2. Define $x=[{\text{OH}}^{-}]$.$K_b=[{\text{OH}}^{-}][{\text{C}}_5{\text{H}}_5\text{N}\cdot {\text{H}}^{+}]/[{\text{C}}_5{\text{H}}_5\text{N}]=x^2/(0.030-x)$.
3. Assume $x\ll 0.030$ so $K_b\approx x^2/0.030$.
4. Solve for $x$: $x=\sqrt{K_b\times 0.030}=\sqrt{1.7\times 10^{-9}\times 0.030}=\sqrt{5.1\times 10^{-11}}=7.14\times 10^{-6}\text{M}$.
5. Calculate $pOH=-\log (7.14\times 10^{-6})=5.146$.
6. Determine $pH=14-pOH=8.854$.
7. Round to two decimal places: pH≈8.85.

Concept tested: Basicity of pyridine and pH calculation

Published 16 September 2026 · How this solution was prepared and checked