Why does CdS dissolve in HCl, when CuS and Bi2S3 will not?
The Correct Answer and Explanation is :
Cadmium sulfide (CdS) dissolves in hydrochloric acid (HCl), while copper sulfide (CuS) and bismuth sulfide (Bi₂S₃) do not, due to differences in the solubility product constants (K(_{sp})) of these sulfides and their ability to form soluble complexes in acidic conditions.
Explanation:
- Solubility Product (K(_{sp})):
- CdS has a relatively higher K(_{sp}) compared to CuS and Bi₂S₃, meaning it is less insoluble in water. Thus, CdS can partially dissociate into Cd²⁺ and S²⁻ ions, making it more reactive in acidic conditions.
- CuS and Bi₂S₃ have much lower K(_{sp}) values, making them highly insoluble and less reactive to acids.
- Reaction with HCl:
- When HCl is added to CdS, the hydrogen ions (H⁺) react with the sulfide ions (S²⁻) to form hydrogen sulfide gas (H₂S):
[
CdS + 2H⁺ \rightarrow Cd^{2+} + H_2S
] - This reaction is favored because H₂S is a weak acid and easily escapes as a gas, shifting the equilibrium and enhancing CdS dissolution.
- Complex Formation:
- Cadmium ions (Cd²⁺) can also form soluble chloride complexes such as ([CdCl_4]^{2-}), further driving the dissolution of CdS in HCl.
- Resistance of CuS and Bi₂S₃:
- CuS and Bi₂S₃ do not dissolve because their sulfide ions are more tightly bound in the solid lattice, and the strong H⁺ from HCl cannot displace them easily.
- Additionally, Cu²⁺ and Bi³⁺ ions do not form highly stable chloride complexes in acidic solutions, further reducing their solubility.
In summary, CdS dissolves in HCl due to its higher K(_{sp}), the release of H₂S gas, and the formation of soluble Cd²⁺ complexes, while the low K(_{sp}) and chemical stability of CuS and Bi₂S₃ prevent their dissolution.