Question

If $H_{2}S$ is passed into the above mixture in the presence of $HCl$, which ion will be precipitated first?

• A. $SnS$
• B. $CdS$
• C. $NiS$
• D. $SnS$ and $CdS$ (both together)

Answer 1

The correct option is C $NiS$

$[S^{2-}{]}_{min}$ for $SnS=\frac{K_{sp}\left(SnS\right)}{\left[S{n}^{2+}\right]}$
$[S^{2-}{]}_{min}$ for $CdS=\frac{K_{sp}\left(CdS\right)}{\left[C{d}^{2+}\right]}$
$[S^{2-}{]}_{min}$ for $NiS=\frac{K_{sp}\left(NiS\right)}{\left[N{i}^{2+}\right]}$
and $\left(NiS\right)<K_{sp}\left(SnS\right)<K_{sp}\left(CdS\right)$
$\therefore$ $NiS$ will precipitate first.