Q2 / 120 · VITEEE 2015 · Chemistry
A saturated solution of $\mathrm{H}_{2} \mathrm{~S}$ in $0.1 \mathrm{M} \mathrm{HCl}$ at $25^{\circ} \mathrm{C}$ contains $\mathrm{S}^{2-}$ ion concentration of $10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}$. The solubility product of some sulphides are $\mathrm{CuS}=10^{-44}, \mathrm{FeS}=10^{-14}, \mathrm{MnS}=10^{-15}, \mathrm{CdS}=$ $10^{-25}$. If $0.01 \mathrm{M}$ solution of these salts in $1 \mathrm{M}$ $\mathrm{HCl}$ are saturated with $\mathrm{H}_{2} \mathrm{~S}$, which of these will be precipitated?
- AAll
- BAll except MnS
- Correct answerC$\mathrm{AU}$ except $\mathrm{MnS}$ and $\mathrm{FeS}$
- DOnly-CuS
Solution
$\left[\mathrm{S}^{2-}\right]=10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}$
$\left[\mathrm{M}^{2+}\right]=10^{-2} \mathrm{M}$
Ionic product, $\mathrm{K}_{1 \mathrm{P}}=\left[\mathrm{M}^{2+}\right]\left[\mathrm{S}^{2-}\right]=10^{-25}$
$\therefore \quad$ ionic product is greater than $\mathrm{K}_{\mathrm{sp}}$ of CuS and $\mathrm{CdS}$,
Therefore, all except $\mathrm{MnS}$ and $\mathrm{FeS}$ are precipitated.
Hence, the correct answer is option: 'C'.