25.0 mL of a weak monoprotic acid HA requires 20.0 mL of 0.100 M NaOH to reach its equivalence point. After adding 8.0 mL of the same NaOH, the pH is 4.45. Calculate $p K_{a}$ of HA to two decimal places, treating the mixture as a buffer.

Numerical answer type, tolerance 0.02.

Answer: 4.63

Total HA

= 20.0 \times 0.100 = 2.0

mmol. After 8.0 mL of NaOH:

[A X^{-}] = 0.80

mmol and

[HA] = 1.20

mmol.

p K_{a} = pH - lo g (0.80/1.20) = 4.45 + 0.176 = 4.63

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