Consider two Group IV metal ions X^{2+} and Y | vedclass

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(D) For the precipitation of $YS_{(s)}$,the condition is $[Y^{2+}][S^{2-}] \geq K_{sp}(YS)$.Given $[Y^{2+}] = 0.01 \ M$ and $K_{sp}(YS) = 4 	imes 10^

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$4$

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(D) For the precipitation of $YS_{(s)}$,the condition is $[Y^{2+}][S^{2-}] \geq K_{sp}(YS)$.Given $[Y^{2+}] = 0.01 \ M$ and $K_{sp}(YS) = 4 	imes 10^{-16}$,we have $[S^{2-}] \geq \frac{4 	imes 10^{-16}}{0.01} = 4 	imes 10^{-14} \ M$.For the dissociation of $H_2S$,the equilibrium is $H_2S_{(aq)} ightleftharpoons 2H^{+}_{(aq)} + S^{2-}_{(aq)}$.The equilibrium constant expression is $\frac{[S^{2-}][H^{+}]^2}{[H_2S]} = K_{a1} 	imes K_{a2} = 1.0 	imes 10^{-21}$.Substituting the values: $\frac{(4 	imes 10^{-14})[H^{+}]^2}{0.1} = 1.0 	imes 10^{-21}$.$[H^{+}]^2 = \frac{1.0 	imes 10^{-22}}{4 	imes 10^{-14}} = 0.25 	imes 10^{-8} = 25 	imes 10^{-10}$.$[H^{+}] = 5 	imes 10^{-5} \ M$.$pH = -\log[H^{+}] = -\log(5 	imes 10^{-5}) = 5 - \log 5 = 5 - 0.70 = 4.3$.The nearest integer value for the pH is $4$.

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