Calculate the pH of a $0.020\text{M}$ aqueous solution of pyridine at $25^{\circ }\text{C}$, given that $K_b=1.7\times 10^{-9}$.

Step-by-step solution

1. Write the base dissociation equation: ${\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}^{+}+\mathrm{O}{\mathrm{H}}^{-}$
2. Let $x$ be the concentration of $\mathrm{O}{\mathrm{H}}^{-}$ produced. Then $x=[\mathrm{O}{\mathrm{H}}^{-}]$ and $[C_5{H}_{5}N{H}^{+}]=x$ and $[C_5{H}_{5}N]=0.020-x$
3. Set up the expression: $K_b=\frac{x^2}{0.020-x}$
4. Solve the quadratic $x^2+1.7\times 10^{-9}x-3.4\times 10^{-11}=0$
5. Take the positive root: $x\approx 1.65\times 10^{-5}\text{M}$
6. Compute pOH: $pOH=-log(1.65\times 10^{-5})=4.78$
7. Finally, $pH=14-pOH=14-4.78=9.22$
8. The pH is 9.22 (to two decimal places).

Concept tested: pH calculation for weak base solution

Published 15 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked