The pH of an aqueous solution containing 1 M benzoic acid pK a = 4 . 20 and 1 M sodium benzoate is 4 . 5 .…

$\begin{array}{ccccc}& \mathrm{Benzoic} \mathrm{acid}& +& \mathrm{Sodium} \mathrm{benzoate}& \\ \mathrm{Molarity}& 1\mathrm{M}& & 1\mathrm{M}& \\ \mathrm{Volume}& {\mathrm{V}}_{\mathrm{a}}\mathrm{ml}& & {\mathrm{V}}_{\mathrm{s}}\mathrm{ml}& \\ \mathrm{milli} \mathrm{moles}& {\mathrm{V}}_{\mathrm{a}}\times 1& & {\mathrm{V}}_{\mathrm{s}}\times 1& \\ & & & & \end{array}$

$\mathrm{pH}=4.5$

${\mathrm{pK}}_{\mathrm{a}}=4.20$

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\text{ salt }\right]}{\left[\text{ acid }\right]}$

$$\begin{gathered} 4.5=4.2+\log \left(\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{a}}}\right) \\ \mathrm{Where},{\mathrm{V}}_{\mathrm{s}}- \mathrm{Volume} \mathrm{of} \mathrm{salt} \\ {\mathrm{V}}_{\mathrm{a}}- \mathrm{Volume} \mathrm{of} \mathrm{acid} \end{gathered}$$

$\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{a}}}=2$..........(i)

${\mathrm{V}}_{\mathrm{s}}+{\mathrm{V}}_{\mathrm{a}}=300$........(ii)

${\mathrm{V}}_{\mathrm{a}}=100\mathrm{ml}$