$100 mL$ of $0.10 M$ $H X_{2} A$ ( $p K_{a 1} = 3.00$ , $p K_{a 2} = 7.00$ ) is titrated with $0.10 M$ $NaOH$ . The volume of base needed to bring the pH to $7.00$ is closest to

Answer: D

pH = 7.00 = p K_{a 2}

[HA X^{-}] = [A X^{2 -}]

(and

H X_{2} A

is negligible), so the average number of protons removed is

0.5 \times 1 + 0.5 \times 2 = 1.5

. Base needed

= 1.5 \times 10.0 mmol = 15 mmol

, i.e.

15/0.10 = 150 mL