If to a 100 mL of solution of Q. 100, 0.002$\mathrm{mol}$ of $\mathrm{HCl}$ is added, its pH changes to

In the solution of $\text{ Q.100, }\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=\left[{\mathrm{CH}}_3\mathrm{COONa}\right]=0.01\mathrm{M}$

$n\left({\mathrm{CH}}_3\mathrm{COOH}\right)=n\left({\mathrm{CH}}_3\mathrm{COONa}\right)=\left(0.01\mathrm{M}\right)\left(0.1{\mathrm{dm}}^3\right)=0.001\mathrm{mol}$

On adding 0.002l, $\mathrm{mol} \mathrm{HCl}$,the entire CH3COONa is replaced by CH3COOH. Thus, solution contains 0.002 $\mathrm{mol}$ acetic acid and 0.001$\mathrm{mol} \mathrm{HCl}$. The solution concentrations are

$\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=0.02\mathrm{M}\text{ and }\left[\mathrm{HCl}\right]=0.01\mathrm{M}$

The major source of ${\mathrm{H}}^{+}$ in the solution will be $\mathrm{HCl}$.

Hence, $\mathrm{pH}=-\log \left(0.01\right)=2$