100 mL of 0.02 M benzoic acid (pK_a = 4.2) is titrated by using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been added are

Millimoles of benzoic acid = 2

Millimole of NaOH in first addition = 1

Millimoles of NaOH in second addition = 2

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{NaOH}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Initial} 2 0 0 0 \\ \mathrm{After} \mathrm{first} \\ \mathrm{addition} \mathrm{of} \mathrm{NaOH} (2-2)=0 (2-2) = 0 2 \end{gathered}$$

After first addition, mixture is a buffer containing 1 millimole each of ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}\text{ and }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$

$\therefore \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}^{\ast }+\log \frac{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right]}\Rightarrow {\mathrm{pK}}_{\mathrm{a}}=4.2$

After second addition, there are only 2 millimoles of, C₆H₅COONa in 200 mL solution and due to hydrolysis, solution is basic.

$\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\right]=\frac{2\times {10}^{-3}}{02}=1\times {10}^{-2}\mathrm{M}$

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+{\mathrm{OH}}^{-} \\ \therefore \mathrm{pH}=7+\frac{{\mathrm{pK}}_{\mathrm{a}}}{2}+\frac{\mathrm{logC}}{2} \\ =7+2.1-1=8.1. \end{gathered}$$