A buffer solution is prepared by dissolving $0.03, M$ phenol ( $C_{6} H_{5} OH$ ) and $0.02, M$ sodium phenoxide ( $C_{6} H_{5} ONa$ ) in $200, mL$ of water. Calculate the pH of the resulting solution at $2 5^{\circ} C$ , given $K_{a} (phenol) = 1.3 \times 1 0^{- 10}$ .

hard 5 marks Class 12 · Chemistry · Alcohols Phenols and Ethers

Answer

9.71

The correct answer is 9.71 — Conclusion: The buffer solution has a pH of “ $9.71$ ”.

Step-by-step solution

1. Compute $p K_{a}$ of phenol: $p K_{a} = - l o g_{10} (1.3 \times 1 0^{- 10}) \approx 9.89$ . 2. For a buffer, use the Henderson–Hasselbalch equation: $p H = p K_{a} + l o g_{10} \frac{[ A ^{-} ]}{[ H A ]}$ . 3. Determine concentrations in the solution. The given molarities are already the concentrations in the final volume ( $0.03, M$ phenol, $0.02, M$ phenoxide). 4. Calculate the ratio $\frac{[ A ^{-} ]}{[ H A ]} = \frac{0.02}{0.03} = 0.6667$ . 5. Compute $l o g_{10} (0.6667) \approx - 0.1761$ . 6. Add to $p K_{a}$ : $p H = 9.89 + (- 0.1761) = 9.7139$ . 7. Round to two decimal places: $p H \approx 9.71$ .
Conclusion: The buffer solution has a pH of “ $9.71$ ”.

Common mistake: Using the wrong ratio or forgetting the sign of the logarithm

Concept tested: Buffer pH calculation using Henderson–Hasselbalch

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked