Chemistry - Ionic Equilibrium Question with Solution | TestHub

Key Idea ${\mathrm{CH}}_3\mathrm{COOH}$ (weak acid) and ${\mathrm{CH}}_3\mathrm{COONa}$ (conjugated salt) form acidic buffer and for acidic buffer, $$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{[\text{ salt }]}{[\text{ acid }]}$$ and $$\begin{array}{rl}\left[{\mathrm{H}}^{+}\right]& =-\mathrm{antilog}\mathrm{pH}\\ \mathrm{pH}& =-\log {\mathrm{K}}_{\mathrm{a}}+\log \frac{[\text{ salt }]}{[\text{ acid }]}\\ & \because {\mathrm{pK}}_{\mathrm{a}}=-\log {\mathrm{K}}_{\mathrm{a}}]\\ & =-\log \left(1.8\times 10^{-5}\right)+\log \frac{(0.20)}{(0.10)}\\ & =4.74+\log 2\\ & =4.74+0.3010=5.041\end{array}$$ Now, $$\begin{array}{c}\left[{\mathrm{H}}^{+}\right]=\mathrm{antilog}(-5.045)\\ =9.0\times 10^{-6}\mathrm{mol}/\mathrm{L}\end{array}$$