pH of a mixture of 1 M benzoic acid (pK_a, = 4.20) and 1 M sodium benzoate is 4.5. In 300 mL buffer, benzoic acid is

If benzoic acid = x mL x millimoles sodium benzoate

$$\begin{gathered} =(300-\mathrm{x})\mathrm{mL}=(300-\mathrm{x})\text{ millimoles } \\ \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\text{ [sodium benzoate] }}{\text{ [benzoic acid] }} \\ 4.5=4.2+\log \left(\frac{300-\mathrm{x}}{\mathrm{x}}\right) \\ \frac{300-\mathrm{x}}{\mathrm{x}}=2 \\ \mathrm{x} = 100 \mathrm{mL} \end{gathered}$$