Equilibrium · Selective sulphide precipitation ksp

Consider two Group IV metal ions $\ce{X^{2+}}$ and $\ce{Y^{2+}}$.

A solution containing 0.01 M $\ce{X^{2+}}$ and 0.01 M $\ce{Y^{2+}}$ is saturated with $\ce{H2S}$. The pH at which the metal sulphide YS will form as a precipitate is (Nearest integer)

(Given :

$K_\mathrm{sp}(\mathrm{XS}) = 1\times10^{-22}$ at 25$^\circ$C, $K_\mathrm{sp}(\mathrm{YS}) = 4\times10^{-16}$ at 25$^\circ$C,

[$\ce{H2S}$] = 0.1 M in solution, $K_\mathrm{a1}\times K_\mathrm{a2}$ ($\ce{H2S}$) $= 1.0\times10^{-21}$,

$\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$)