Consider two Group IV metal ions and . A solution containing and is saturated with . The pH at which the metal sulphide YS will form as a precipitate is ________. (Nearest integer) (Given: at at ) [2026]

Question

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 sulphide YS will form as a precipitate is ________. (Nearest integer)

(Given: $K_{s p} (X S) = 1 \times 10^{- 22}$ at $25 ° C,$ $K_{s p} (Y S) = 4 \times 10^{- 16}$ at $25 ° C,$

$[H_{2} S] = 0.1 M in solution, K_{a 1} \times K_{a 2} (H_{2} S) = 1.0 \times 10^{- 21},$ $\log 2 = 0.30, \log 3 = 0.48, \log 5 = 0.70$ ) [2026]

Smart Achievers · Accepted Answer

(4)

$XS (s) ⇌ X^{2 +} (aq) + S^{2 -} (aq)$

$For precipitation of XS(s)$

$[X^{2 +}] [S^{2 -}] \geq K_{s p} (XS)$

$[S^{2 -}] \geq \frac{1 \times 10^{- 22}}{0.01} = 10^{- 20}$

$YS (s) ⇌ Y^{2 +} (aq) + S^{2 -} (aq)$

$For precipitation of YS(s)$

$[Y^{2 +}] [S^{2 -}] \geq K_{s p} (YS)$

$[S^{2 -}] \geq \frac{4 \times 10^{- 16}}{10^{- 2}} = 4 \times 10^{- 14}$

$Now, H_{2} S (aq) ⇌ 2 H^{+} (aq) + S^{2 -} (aq)$

$\frac{[S^{2 -}] {[H^{+}]}^{2}}{[H_{2} S]} = K_{a 1} \times K_{a 2} = 1 \times 10^{- 21}$

$[S^{2 -}] = \frac{1 \times 10^{- 21} \times [H_{2} S]}{{[H^{+}]}^{2}} \geq 4 \times 10^{- 14}$

${[H^{+}]}^{2} \leq \frac{1}{4} \times 10^{- 7} \times 10^{- 1}$

$[H^{+}] \leq \frac{1}{2} \times 10^{- 4} \Rightarrow pH \geq 4.3$

Solution

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