Why does CdS dissolve in HCl, when CuS and Bi2S3 will not?
The Correct Answer and Explanation is :
The differing solubility of CdS, CuS, and Bi₂S₃ in HCl can be explained by their distinct chemical properties, solubility products (Ksp), and interactions with acids.
Correct Answer:
CdS dissolves in HCl because it reacts with the acid to form soluble cadmium chloride (CdCl₂) and releases H₂S gas. CuS and Bi₂S₃, however, have much lower solubility products (Ksp) and do not react significantly with HCl under similar conditions.
Explanation (300 words):
The solubility of metal sulfides in HCl depends on their Ksp values and the extent to which they can react with the acid. CdS, CuS, and Bi₂S₃ are all sulfides of transition metals, but their reactivity with HCl varies significantly:
- CdS Dissolution in HCl:
CdS is a moderately soluble sulfide compared to CuS and Bi₂S₃. In the presence of HCl, CdS reacts as follows:
[
\text{CdS (s) + 2 HCl (aq)} \rightarrow \text{CdCl}_2 \text{(aq) + H}_2\text{S (g)}.
]
The formation of the soluble salt CdCl₂ and the removal of H₂S gas drive the reaction forward, allowing CdS to dissolve readily in HCl. - CuS and Bi₂S₃ Resistance to HCl:
CuS and Bi₂S₃ have significantly lower Ksp values, making them less soluble. CuS, for instance, forms strong Cu-S bonds that resist reaction with HCl. Similarly, Bi₂S₃ has a highly stable lattice structure and is not easily attacked by H⁺ ions. These compounds are more resistant to protonation and thus remain insoluble in HCl. - Role of Sulfide Ion Hydrolysis:
The solubility of metal sulfides is influenced by the hydrolysis of the sulfide ion (S²⁻). CdS is more prone to release S²⁻ into solution, which then reacts with H⁺ from HCl to form H₂S gas. In contrast, CuS and Bi₂S₃ do not release sufficient sulfide ions to sustain the reaction with HCl.
Thus, the dissolution of CdS in HCl is a result of its moderate solubility, weaker metal-sulfur bonds, and the favorable reaction that forms soluble products and gaseous H₂S, unlike CuS and Bi₂S₃.