K_sp of CdS is 8.0x10^-27 and that of H₂S is 1x 10^-22. 1 x 10^-14 M CdCl₂, solution is precipitated on passing H₂S when pH is about

${\mathrm{H}}_2\mathrm{S}\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{S}}^{2-};$

$$\begin{gathered} \mathrm{CdS}\rightleftharpoons {\mathrm{Cd}}^{2+}+{\mathrm{S}}^{2-} \\ {\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{CdS}\right)=\left[{\mathrm{Cd}}^{2+}\right]\left[{\mathrm{S}}^{2-}\right] \\ 8\times {10}^{-27}=\left(1\times {10}^{-14}\right)\left[{\mathrm{S}}^{2-}\right] \\ \therefore \left[{\mathrm{S}}^{2-}\right]=8\times {10}^{-13}\mathrm{M} \\ \text{ Also, }{\left[{\mathrm{H}}^{+}\right]}^2\left[{\mathrm{S}}^{2-}\right]={\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{H}}_2\mathrm{S}\right) \\ {\left[{\mathrm{H}}^{+}\right]}^2\left(8\times {10}^{-13}\right)=1\times {10}^{-22} \\ {\left[{\mathrm{H}}^{+}\right]}^2=1.25\times {10}^{-10} \\ \therefore \left[{\mathrm{H}}^{+}\right]=1.11\times {10}^{-5},\mathrm{pH}\approx 5 \end{gathered}$$