Question

Find the $\left[H^{+}\right]$ that must be maintained in a saturated $H_{2}S\left(0.1M\right)$ to precipitate $CdS$ not $ZnS$ if $\left[C{d}^{2+}\right]=\left[Z{n}^{2+}\right]=\left(0.1M\right)$initially.
$K_{sp}\left(CdS\right)=8\times {10}^{-27}{K}_{sp}\left(ZnS\right)=1\times {10}^{-21}{K}_{sp}\left(H_{2}S\right)=1.1\times {10}^{-21}$

• A. 0.589 M
• B. 0.105 M
• C. 0.500 M
• D. 0.877 M

Answer 1

The correct option is B 0.105 M

In order to prevent precipitation of $ZnS,$
$\left[Z{n}^{2+}\right]\left[S^{2-}\right]<K_{sp}\left(ZnS\right)or0.1\times \left[S^{2-}\right]<1\times {10}^{-21}\left[S^{2-}\right]<1\times {10}^{-20}$

This is the maximum value of $\left[S^{2-}\right]$ before $ZnS$ will precipitate.

$H_{2}S\left(s\right)\rightleftharpoons 2H^{+}\left(aq\right)+S^{2-}\left(aq\right)K_a=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}=\frac{x^2\times 1\times {10}^{-20}}{0.1}=1.1\times {10}^{-21}x=\left[H^{+}\right]=0.105M$
So, No $ZnS$ will precipitate if concentration of $H^{+}$ greater than 0.105 M.
So, this concentration of $H^{+}$ must be maintained.