If 50 mL of 0.2 M KOH is added to 40 mL of 0.5 M ${\mathrm{CH}}_3\mathrm{COOH}\text{, }$the pH of the resulting solution is :  

$\left(K_{\alpha }=1.8\times {10}^{-4},\log 18=1.26\right)$

$\begin{array}{l}{\mathrm{n}}_{\mathrm{KOH}}=\frac{50\times 0.2}{1000}=0.01\mathrm{mol}\\ {\mathrm{n}}_{\mathrm{HCOOH}}=\frac{40\times 0.5}{1000}=0.02\mathrm{mol}\end{array}$

Moles of salt form HCOOK $=\frac{50\times 0.2}{1000}=0.01\mathrm{mol}$

Moles of left acid  $=0.02-0.01=0.01\mathrm{mol}$

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\text{ Salt }}{\text{ Acid }}\\ \mathrm{pH}=-\log {\mathrm{K}}_{\mathrm{a}}+\log \frac{0.01}{0.01}\\ \mathrm{pH}=3.744\end{array}$