$100 mL$ of $0.15 M$ $NaOH$ is added to $250 mL$ of $0.10 M$ $HCOOH$ ( $p K_{a} = 3.75$ ). Using $lo g 2 = 0.301$ and $lo g 3 = 0.477$ , the pH is

Answer: B

OH X^{-}

15 mmol

and

HCOOH

25 mmol

give

HCOO X^{-} = 15

and

HCOOH = 10 mmol

pH = 3.75 + lo g (15/10) = 3.75 + 0.176 = 3.93

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