The concentration of sulphide ion in $0.1 \ M \ HCl$ solution saturated with hydrogen sulphide is $1.0 \times 10^{-19} \ M$. If $10 \ mL$ of this is added to $5 \ mL$ of $0.04 \ M$ solution of the following: $FeSO_4, MnCl_2, ZnCl_2$ and $CdCl_2$,in which of these solutions will precipitation take place? Given $K_{sp}$ for $FeS = 6.3 \times 10^{-18}, MnS = 2.5 \times 10^{-13}, ZnS = 1.6 \times 10^{-24}, CdS = 8.0 \times 10^{-27}$.

(D) For precipitation to occur,the ionic product must exceed the $K_{sp}$ value.
After mixing,the total volume is $15 \ mL$.
The new concentration of sulphide ion is $[S^{2-}] = \frac{1.0 \times 10^{-19} \times 10}{15} = 6.67 \times 10^{-20} \ M$.
The new concentration of metal ion is $[M^{2+}] = \frac{0.04 \times 5}{15} = 1.33 \times 10^{-2} \ M$.
The ionic product is $[M^{2+}][S^{2-}] = (1.33 \times 10^{-2}) \times (6.67 \times 10^{-20}) = 8.87 \times 10^{-22}$.
Comparing this with the given $K_{sp}$ values:
For $FeS$: $8.87 \times 10^{-22} < 6.3 \times 10^{-18}$ (No precipitation)
For $MnS$: $8.87 \times 10^{-22} < 2.5 \times 10^{-13}$ (No precipitation)
For $ZnS$: $8.87 \times 10^{-22} > 1.6 \times 10^{-24}$ (Precipitation occurs)
For $CdS$: $8.87 \times 10^{-22} > 8.0 \times 10^{-27}$ (Precipitation occurs)
Thus,precipitation will take place in $ZnCl_2$ and $CdCl_2$ solutions.