Question

Maximum $pH$ that must be maintained in a saturated $H_{2}S$ (0.1 M) to precipitate $CdS$ but not $ZnS$ is $x$ and concentration of $C{d}^{2+}$ ion at that $pH$ is $y\times {10}^{-7}$ then $x+y=$
Given $\left[C{d}^{2+}\right]=\left[{Zn}^{2+}\right]=0.1M$ initially
$K_{sp}\left(CdS\right)=8\times {10}^{-27}$
$K_{sp}\left(ZnS\right)=1\times {10}^{-21}$
$K_{a1}\left(H_{2}S\right)=1.0\times {10}^{-8}$
$K_{a2}\left(H_{2}S\right)=1.0\times {10}^{-13}$

Answer 1

To present the principle of $Zns$

$\left[Z{n}^{+2}\right]\left[S^{-2}\right]\times K_{SP}\left(ZnS\right)$

$\Rightarrow \left[Z{n}^{+2}\right[S^{-2}]\times {10}^{-21}]$

$\Rightarrow 0.1\times [S^{-2}\times {10}^{-21}]$

$\Rightarrow [S^{-2}\times {10}^{-20}]$

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

Now,

$H_{2}S\leftrightharpoons 2H^{+}+S^{2-};Ka=K{a}_1\times K{a}_2$

$\Rightarrow Ka=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}=\frac{x^2[1\times {10}^{-20}]}{0.1}$

$\Rightarrow 1\times {10}^{-21}=x^2\times {10}^{-19}$

$\Rightarrow x^2={10}^{-2}$

$\therefore x=\left[H^{+}\right]=0.1$

$\Rightarrow pH=-\log \left[H^{+}\right]=1=X$

Now, at $\left[S^{2-}\right]={10}^{-20}$

$K_{SP}\left(Cds\right)=\left[C{d}^{+2}\right]\left[S^{2-}\right]$

$\Rightarrow \left[C{d}^{+2}\right]=8\times {10}^{-27}/{10}^{-20}=8\times {10}^{-7}$

$\therefore Y=8$

$\therefore X+Y=9$