Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{NaOH}+{\mathrm{CH}}_3\mathrm{COOH}\to {\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$ $\begin{array}{lllllc}\mathrm{m}\mathrm{mol}& 20& \times 0.2& 50\times 0.2& 0& 0\\ \text{ added }& =4& =10& & & \\ & & & & & \\ \text{-eaˊction }& 0& 6& 4& 4& \end{array}$ $\begin{array}{lcc}\mathrm{m}\text{ mol after }& & \\ \text{ reaction }0& 6& 4\\ [{\mathrm{CH}}_3\mathrm{COOH}=\frac{6}{70}& :& \left[{\mathrm{CH}}_3\mathrm{COONa}\right]\end{array}=\frac{4}{70}$ Now since for a basic buffer $$\begin{array}{l}\mathrm{pH}=-{\mathrm{logk}}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{base}]}\\ \begin{array}{rl}\therefore \mathrm{pH}& =-\log \left(1.8\times 10^{-5}\right)+\log \frac{4/70}{6/70}\\ & =4.56\end{array}\end{array}$$