Question

$\frac{x}{20}M$ concentration of $H^{+}$ ion must be maintained in a saturate $H_{2}S\left(0.1M\right)$ to precipitate $CdS$ but not $ZnS$ , if $[C{d}^{2+}]=[Z{n}^{2+}]=0.1M$ initially. $K_{sp}(CdS)=8\times 10^{-27},K_{sp}\left(ZnS\right)=1\times {\left(10\right)}^{-21}{K}_a\left(H_{2}S\right)=1\times {\left(10\right)}^{-21}$ . $ZnS$ will not precipitate at a concentration of $H^{+}$ greater than $\frac{x}{20}M$ . The value of $x$ is.

Answer 1

Answer: 2

$CdS\rightleftharpoons \underset{0.1}{C{d}^{2+}}+\underset{s}{S^{2-}}$
For precipitation of CdS
$[{\left(Cd\right)}^{2+}][S^{2-}]=K_{sp(CdS)}$
$0.1\times [S^{2-}]=8\times 10^{-27}$
$[S^{2-}]=8\times 10^{-26}$
For precipitation of ZnS
$[{\left(Zn\right)}^{2+}][S^{2-}]=K_{sp(ZnS)}$
$0.1\times [S^{2-}]=10^{-21}$
$[S^{2-}]=10^{-20}$
CdS will start precipitate as the conc of $[S^{2-}]=8\times 10^{-26}$ and ZnS will start precipitate as the conc of $[S^{2-}]=10^{-20}$ so we can raise the concentration of $[S^{2-}]$ upto $10^{-20}$
Now
$\underset{\underset{\approx 0.1}{0.1(1-\alpha )}}{H_{2}S}\rightleftharpoons \underset{\underset{=\frac{x}{20}M}{2\times 0.1\alpha }}{2H^{+}}+\underset{0.1\alpha =10^{-20}}{S^{2-}}$
$\frac{{\left(\frac{x}{20}\right)}^2\times 10^{-20}}{0.1}=10^{-21}$
${\left(\frac{x}{20}\right)}^2=10^{-2}$
$\frac{x}{20}=10^{-1}$
$x=2$