Consider two Group IV metal ious $$X^{2+}\text{ and }Y^{2+}$$
A solution containing $$0.01 M X^{2+}\text{ and }0.01MY^{2+}$$ is satmated with $$H_{2}S$$. The pH at which the metal sulphide YS will form as a precipitate is __ . (Nearest integer)
$$(Given:K_{sp}(XS)=1\times 10^{-22} \text{ at } 25^{\circ}C,K_{sp}(YS)=4\times 10^{-16} \text{ at } 25^{\circ}C,[H_{2}S]=0.1M\text{ in solution },K_{a1}\times K_{a2}(H_{2}S)=1.0\times 10^{-21},\log{2}=0.30,\log{3}=0.48,\log{5}=0.70)$$

We need to find the pH at which YS precipitates from a solution saturated with $$H_2S$$.

The concentration $$[S^{2-}]$$ required to precipitate YS follows from the solubility product: $$[S^{2-}] = \frac{K_{sp}(YS)}{[Y^{2+}]} = \frac{4 \times 10^{-16}}{0.01} = 4 \times 10^{-14} \text{ M}$$.

The dissociation constants of $$H_2S$$ are related by $$K_{a1} \times K_{a2} = \frac{[H^+]^2[S^{2-}]}{[H_2S]}$$, so with $$1.0 \times 10^{-21} = \frac{[H^+]^2 \times 4 \times 10^{-14}}{0.1}$$ one finds $$[H^+]^2 = \frac{10^{-22}}{4 \times 10^{-14}} = 2.5 \times 10^{-9}$$ and hence $$[H^+] = \sqrt{25 \times 10^{-10}} = 5 \times 10^{-5} \text{ M}$$.

Therefore, $$\text{pH} = -\log(5 \times 10^{-5}) = 5 - \log 5 = 5 - 0.70 = 4.30$$. Rounding to the nearest integer gives pH = 4.