Consider two Group IV metal ions X^2+ and Y^2+. A solution containing 0.01 M X^2+ and 0.01 M Y^2+ is saturated with H_2 S. The pH at which the metal…

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