Question

$Cu$ is precipitated as $CuS$ while $Zn$ is not precipitated when $H_{2}S$ is passed through an acidic solution of $CU(N{O}_3{)}_2$ and $Zn(N{O}_3{)}_2$ respectively.

Answer 1

This is because $K_{sp}$ of $CuS$ is very low as compared to that of $ZnS$. Therefore, a low concentration of $S^{2-}$ ions can precipitate $C{u}^{2+}$ ions as $CuS$, whereas, a high concentration of $S^{2-}$ ions is required to precipitate $Z{n}^{2+}$ ions as $ZnS$.
Now, in presence of an acid i.e., $HCl$ dissociation of $H_{2}S$ is suppressed due to common ion effect.
$HCl\rightleftharpoons H^{+}+C{l}^{-}$ (strong electrolyte)
$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$ (weak electrolyte)
As a result, in acidic solution copper will precipitate as $CuS$ but not the Zinc as $ZnS$.