20.0 mL of 0.10 M $H X_{2} A$ ( $p K_{a 1} = 3.0$ , $p K_{a 2} = 7.0$ ) is titrated with 0.10 M NaOH. Use $lo g 3 = 0.477$ and treat each stage as a buffer. The pH after 25.0 mL of NaOH has been added is

Answer: A

H X_{2} A = 2.0

mmol; NaOH

= 2.5

mmol. The first 2.0 mmol convert

H X_{2} A

HA X^{-}

; the next 0.5 mmol give 0.5 mmol

A X^{2 -}

, leaving 1.5 mmol

HA X^{-}

. pH

= p K_{a 2} + lo g (0.5/1.5) = 7.0 - 0.477 = 6.52

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