A 25.0 mL sample of $0.10 M$ acetic acid (CH $_{3}$ COOH) is titrated with $0.10 M$ NaOH. Calculate the pH of the solution at the half‑equivalence point. (Acetic acid $K_{a} = 1.8 \times 1 0^{- 5}$ .) Give your answer correct to two decimal places.

hard 5 marks Class 10 · Science · Acids, Bases and Salts

Answer

4.74

The correct answer is 4.74 — Therefore the pH at the half‑equivalence point is $4.74$ .

Step-by-step solution

At the half‑equivalence point the amounts of acid and its conjugate base are equal, so $[CH_{3} COOH] = [CH_{3} COO^{-}]$ .
When the concentrations are equal, the Henderson–Hasselbalch equation reduces to $p H = p K_{a}$ .
Calculate $p K_{a} = - lo g_{10} (K_{a}) = - lo g_{10} (1.8 \times 1 0^{- 5}) \approx 4.74$ .
Therefore the pH at the half‑equivalence point is $4.74$ .

Common mistake: Using the Ka value directly in the pH formula instead of recognising that pH = pKa at half‑equivalence.

Concept tested: Half‑equivalence point in weak acid–base titration