Chemistry - Ionic Equilibrium Question with Solution | TestHub

Ionisation of ${\mathrm{CH}}_3\mathrm{COOH}$ is $1\%$ in aqueous solution. So, $\left[{\mathrm{H}}^{+}\right]=\frac{1}{100}\times$ concentration of ${\mathrm{CH}}_3\mathrm{COOH}$ $\begin{array}{rl}& =\frac{1}{100}\times 10^{-2}=10^{-4}\\ \mathrm{pH}& =-\log \left[{\mathrm{H}}^{+}\right]=-\log \left(1\times 10^{-4}\right)\\ \mathrm{pH}& =4\end{array}$