Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$.
A solution containing $0.01 \mathrm{MX}^{2+}$ and $0.01 \mathrm{MY}^{2+}$ is saturated with $\mathrm{H}_2 \mathrm{~S}$. The pH at which the metal sulphide YS will form as a precipitate is $\_\_\_\_$ . (Nearest integer)
(Given: $\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}$ at $25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}$ at $25^{\circ} \mathrm{C},\left[\mathrm{H}_2 \mathrm{~S}\right]=0.1 \mathrm{M}$ in solution, $\left.\mathrm{K}_{a 1} \times \mathrm{K}_{a 2}\left(\mathrm{H}_2 \mathrm{~S}\right)=1.0 \times 10^{-21}, \log 2=0.30, \log 3=0.48, \log 5=0.70\right)$