20 mL of benzoic acid (K_a = 6.3x 10^-5) is exactly half-neutralised by NaOH. The pH of the solution is

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$20 mL$ of benzoic acid ( $K_{a} = 6.3 \times 1 0^{- 5}$ ) is exactly half-neutralised by NaOH. The pH of the solution is

At half-neutralisation

[acid] = [salt]

, so by the Henderson equation

pH = p K_{a} = - lo g (6.3 \times 1 0^{- 5}) = 4.20