Question

The solubility products of NiS, ZnS, CdS, and HgS are \(4.7\times10^{-5}\), \(1.6\times10^{-24}\), \(8\times10^{-27}\), and \(4\times10^{-53}\) respectively. An aqueous solution contains Ni\(^{2+}\), Zn\(^{2+}\), Cd\(^{2+}\), and Hg\(^{2+}\) of equal concentration. H\(_2\)S gas is passed into this solution very slowly. The first and the last ions that precipitate as sulphides are respectively:

• A. Ni\(^{2+}\), Hg\(^{2+}\)
• B. Hg\(^{2+}\), Cd\(^{2+}\)
• C. Zn\(^{2+}\), Hg\(^{2+}\)
• D. Hg\(^{2+}\), Ni\(^{2+}\)

Hint:

The ion with the lowest solubility product precipitates first when H\(_2\)S is added slowly; the highest K\(_\text{sp}\) ion precipitates last.

Answer 1

The Correct Option is D

Step 1: Rank Ksp values (lowest precipitates first).
HgS \((4 \times 10^{-53})\) \(<\) CdS \((8 \times 10^{-27})\) \(<\) ZnS \((1.6 \times 10^{-24})\) \(<\) NiS \((4.7 \times 10^{-5})\).

Step 2: Identify ions remaining.
Under the given H\(_2\)S conditions, ZnS, CdS, and HgS precipitate preferentially. NiS has the highest Ksp and resists precipitation. The two ions remaining in solution are Hg\(^{2+}\) (partial precipitation conditions) and Ni\(^{2+}\).
\[ \boxed{Hg^{2+},\ Ni^{2+}} \]