An amount of $0.1$ millimole of $\mathrm{CdSO}_{4}$ is present in $10 \mathrm{ml}$ acid solution of $0.08 \mathrm{M}$ $-\mathrm{HCl}$. Now $\mathrm{H}_{2} \mathrm{~S}$ is passed to precipitate all the $\mathrm{Cd}^{2+}$ ions. What would be the $\mathrm{pH}$ of solution after filtering off precipitate, boiling off $\mathrm{H}_{2} \mathrm{~S}$ and making the solution $100 \mathrm{ml}$ by adding water?
(a) $3.0$
(b) $2.0$
(c) $4.0$
(d) $2.22$