Question

Assertion: A solution contains 0.1 M each of $P{b}^{2+},Z{n}^{2+},N{i}^{2+}$ ions. If $H_{2}S$ is passed into this solution at ${25}^{o}C$
$P{b}^{2+},N{i}^{2+},Z{n}^{2+}$ will get precipitated simultaneously.
Reason: $P{b}^{2+}$ and $Z{n}^{2+}$ will get precipited if the solution contains 0.1 M HCl.
$[K_1{H}_{2}S={10}^{-7},K_2{H}_{2}S={10}^{-14},K_{sp}PbS=3\times {10}^{-29}{K}_{sp}NiS=3\times {10}^{-19}.K_{sp}ZnS={10}^{-25}]$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct
• E. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Here 0.1 M $H_{2}S$ is passed into the solution directly. Selectively precipitation of $P{b}^{2+},Z{n}^{2+}andN{i}^{2+}$ is not possible, as the ionic product of their respective sulphides is more than their $K_{sp}$ values and $S^{2-}$ ion is n()t added slowly or dropwise.
If $H_{2}S$ is passed into the solution containing only $H_{2}O$, then
$\left[S^{2-}\right]\approx K_2$ of $H_{2}S={10}^{-14}M$
But in 0.1 M HCl solution.
$K_1\times K_2=\frac{\left[H^{\oplus }{]}^2\right[S^{2-}]}{H_{2}S}$
${10}^{-7}\times {10}^{-14}=\frac{\left(0.1\right)\left[S^{2-}\right]}{0.1}\therefore \left[S^{2-}\right]={10}^{-20}M$
$\therefore Q_{sp}$ of PbS and ZnS $(0.1\times {10}^{-20})>K_{sp}$ of PbS and ZnS (Hence both precipitate) $Q_{sp}$ of $NiS\left({10}^{-21}\right)<K_{sp}$ of NiS (Hence do not precipitate).