$100 mL$ of $0.1 M$ acetic acid ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) is mixed with $100 mL$ of $0.1 M$ $HCl$ . The degree of ionisation of acetic acid in the mixture is nearest to

Answer: A

After mixing,

[HA] \approx 0.05 M

and

[H X^{+}] \approx 0.05 M

(from

HCl

K_{a} = [H X^{+}] [A X^{-}] / [HA]

gives

α = [A X^{-}] / [HA] = K_{a} / [H X^{+}] = 1.8 \times 1 0^{- 5} /0.05 = 3.6 e - 04

. Ignoring the common ion gives

K_{a} /0.05 = 9.5 \times 1 0^{- 3}

, which is wrong.

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