To a $10 \mathrm{mL}$ buffer solution of $0.1 \mathrm{M}$ acetic acid and $0.1 \mathrm{M}$ sodium acetate at ${25}^{\circ }\mathrm{C}, 0.04 \mathrm{g}$ of $\mathrm{NaOH} \left(\mathrm{s}\right)$ is added.

The pH of the resultant solution will be (Given: $\mathrm{p}{K}_{\mathrm{a}}^{\circ }$ (acetic acid) 4.74 and log 2 = 0.30)

Amount of acetic acid in the buffer solution is $n_1=V{M}_1=\left(10\times {10}^{-3}\mathrm{L}\right)\left(0.1 {\mathrm{molL}}^{-1}\right)={10}^{-3}\mathrm{mol}$

Amount of sodium acetate in the buffer solution is $n_2=V{M}_2=\left(10\times {10}^{-3}\mathrm{L}\right)\left(0.1{\mathrm{molL}}^{-1}\right)={10}^{-3}\mathrm{mol}$

Amount of $\mathrm{NaOH}$ added,$n_3=\frac{m}{M}=\frac{0.04 \mathrm{g}}{40 \mathrm{g} {\mathrm{mol}}^{-1}}={10}^{-3} \mathrm{mol}$

The addition of NaOH causes the replacement of equivalent amount of acid into conjugate base. Thus, the resultant solution will not contain acetic acid and it has only sodium acetate whose amount is

 ${10}^{-3}\mathrm{mol} + {10}^{-3}=2\times {10}^{-3}\mathrm{mol}$

(already present)          (from acid)

The pH of the solution will be

$\begin{array}{l}\mathrm{pH}=\frac{1}{2}\left[\mathrm{p}{K}_{\mathrm{w}}^{\circ }+\mathrm{p}{K}_{\mathrm{a}}^{\circ }+\log \left(\mathrm{c}/\mathrm{mol}{\mathrm{dm}}^{-3}\right)\right]=\frac{1}{2}\left[14+4.74+\log \left(2\times {10}^{-3}\right)\right]\\ =\frac{1}{2}[14+4.74-2.7]=8.02\end{array}$