A solution contains $0.10 M$ $HA$ ( $K_{a} = 2.0 \times 1 0^{- 5}$ ) and $0.050 M$ $NaA$ . Without volume change, $0.020 mol L^{- 1}$ of $HCl$ is then added. Using $lo g 2 = 0.301$ , the pH is

Answer: D

H X^{+}

converts

A X^{-}

HA

HA = 0.10 + 0.020 = 0.12

and

A X^{-} = 0.050 - 0.020 = 0.030 M

p K_{a} = 5 - 0.301 = 4.699

, so

pH = 4.699 + lo g (0.030/0.12) = 4.699 - 0.602 = 4.10

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