$25.0 mL$ of a solution that is $0.100 M$ in HCl and $0.100 M$ in acetic acid ( $p K_{a} = 4.74$ ) is titrated with $0.100 M$ NaOH. The pH after $30.0 mL$ of NaOH have been added is

Answer: D

The strong acid is neutralised first:

2.5 mmol

HCl uses

2.5 mmol

NaOH. The remaining

0.5 mmol

NaOH converts

0.5 mmol

of acetic acid into acetate, leaving

2.0 mmol

acid.

pH = 4.74 + lo g (0.5/2.0) = 4.14