$20 \mathrm{mL}$ of $0.1\mathrm{MNaOH}$ is added to $50 \mathrm{mL}$ of $0.1\mathrm{M}$ acetic acid solution. The $\mathrm{pH}$ of the resulting solution is $\times {10}^{-2}$. (Nearest integer) Given : $\mathrm{pK}a\left({\mathrm{CH}}_3\mathrm{COOH}\right)=4.76$

$\log 2 = 0.30$

$\log 3 = 0.48$

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula:

$\begin{array}{ccc}\mathrm{pH}& =& \mathrm{pKa}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}\end{array}$

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}⟶{\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O} \\ 5 2 — — \\ 3 0 2 \end{gathered}$$

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{2}{3}\right)$

$\mathrm{pH}=4.76+0.30-0.48$

$= 4.76 – .18$

$= 4.58$