$25.0 mL$ of $0.100 M$ $Na X_{2} CO X_{3}$ is titrated with $0.100 M$ HCl. For carbonic acid $p K_{a 1} = 6.35$ and $p K_{a 2} = 10.33$ . The pH after $35.0 mL$ of HCl have been added is

Answer: A

Na X_{2} CO X_{3}

2.5 mmol

; HCl:

3.5 mmol

. The first

2.5 mmol

convert carbonate to bicarbonate; the remaining

1.0 mmol

converts

1.0 mmol

HCO X_{3} X^{-}

H X_{2} CO X_{3}

, leaving

1.5 mmol

HCO X_{3} X^{-}

. The pair

HCO X_{3} X^{-} / H X_{2} CO X_{3}

buffers:

pH = 6.35 + lo g (1.5/1.0) = 6.53

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