4. A buffer solution is prepared by dissolving $0.025\text{M}$ phenol (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}$) and $0.015\text{M}$ sodium phenoxide (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{ONa}$) in water at $25^{\circ }\text{C}$. Given $K_a(\text{phenol})=1.3\times 10^{-10}$, calculate the pH of the solution.

Step-by-step solution

1. Write the acid–base equilibrium for phenol: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}+{\mathrm{H}}^{+}$ with $K_a=1.3\times 10^{-10}$. The conjugate base is phenoxide.
2. Recognise that the mixture of phenol and its salt forms a buffer. Use the Henderson–Hasselbalch equation: $\mathrm{pH}=\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\log \frac{[{\mathrm{A}}^{-}]}{[\mathrm{HA}]}$.
3. Calculate $p{K}_a$: $p{K}_a=-\log (1.3\times 10^{-10})\approx 9.886$ (using $\log 1.3\approx 0.114$).
4. Insert the concentrations: $[A^{-}]=0.015\text{M}$, $[HA]=0.025\text{M}$. Compute the ratio $\frac{[A^{-}]}{[HA]}=\frac{0.015}{0.025}=0.60$.
5. Take the logarithm: $\log (0.60)\approx -0.222$.
6. Apply the equation: $\text{pH}=9.886+(-0.222)=9.664$.
7. Because the buffer concentrations are relatively low, check the approximation: use the exact expression $\text{pH}=p{K}_a+\log \frac{[A^{-}]}{[HA]}$ is sufficient; no significant shift occurs.
8. Round to two decimal places: $\text{pH}=9.66$. (If activity corrections are considered, the value moves to about 10.23; most textbooks accept 9.66 as the buffer pH.)
9. State the final answer.

Concept tested: Buffer pH calculation (Henderson–Hasselbalch)

Published 8 September 2026 · How this solution was prepared and checked