pH of a mixture of 1 M benzoic acid $\left({\mathrm{pK}}_{\mathrm{a}} = 4.20\right)$  and $1 \mathrm{M} {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}$ is 4.5. In 300 ml buffer, benzoic acid is [log 2 = 0.3]

$$\begin{gathered} \mathrm{pH} = {\mathrm{pK}}_{\mathrm{a}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} \\ \therefore 4.5 = 4.2 + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} = \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} = 0.3 \\ \left(\mathrm{since} \mathrm{anti} \log 0.3= 2\right) \therefore \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} = 2 \end{gathered}$$

Let V ml of 1 M $C_6{H}_5$ COOH solution and (300 - V) ml of 1 M $C_6{H}_{5}COONa$ solution be mixed together

$$\begin{gathered} \left[\mathrm{Acid}\right] = \frac{\mathrm{V} \times 1}{1000} \times \frac{1000}{300} ; \left[\mathrm{Salt}\right] = \frac{\left(300 - \mathrm{V}\right)}{1000} \times \frac{1000}{300} \\ \left[\mathrm{Acid}\right] = \frac{\mathrm{V}}{300} ; \left[\mathrm{Salt}\right] = \frac{300 - \mathrm{V}}{300} \\ \therefore \frac{300 - \mathrm{V}/300}{\mathrm{V}/300} = 2=300-\mathrm{V} = 2\mathrm{V} \therefore \mathrm{V}=100 \mathrm{ml} \end{gathered}$$