Question

$K_{sp}$ of $CdS=8\times {10}^{-27}$, $K_{sp}$ of $ZnS=1\times {10}^{-21}$, $K_a$ of $H_{2}S=1\times {10}^{-21}$.
What $\left[H^{+}\right]$ must be maintained in saturated $H_{2}S\left(0.1\text{M}\right)$ to precipitate $CdS$ but not $ZnS$, if $\left[C{d}^{2+}\right]=\left[Z{n}^{2+}\right]=0.1\text{M}$ initially?

• A. $>0.1\text{M}$
• B. $>0.01$
• C. $>0.2\text{M}$
• D. $>0.02\text{M}$

Answer 1

The correct option is A $>0.1\text{M}$

$K_{sp}$ of $CdS=8\times {10}^{-27}=\left[C{d}^{2+}\right]\left[S^{2-}\right]=0.1\times \left[S^{2-}\right]$
Now for $\left[S^{2-}\right]>8\times {10}^{-26},CdSwillprecipitate$
Similarly for $\left[S^{2-}\right]>{10}^{-20},ZnSwillprecipitate$
now $K_a{H}_{2}S={10}^{-21}=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}$
$\therefore [H^{+}{]}^2=0.01$
so for $\left[H^{+}\right]>0.1M$ $CdS$ will precipitate but $ZnS$ will not precipitate