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: Recall solubility product (K\(_\text{sp}\)) principle.
The lower the solubility product, the less soluble the sulphide, so it precipitates first.

Step 2: Compare K\(_\text{sp}\) values.
\[ K_\text{sp} (\text{HgS}) = 4\times10^{-53} \lt K_\text{sp} (\text{NiS}) = 4.7\times10^{-5} \] Hence, Hg\(^{2+}\) will precipitate first and Ni\(^{2+}\) last.

Step 3: Conclusion.
The first and last ions to precipitate are \[ \boxed{\text{Hg}^{2+}, \text{Ni}^{2+}}. \]