Question

Why does CdS dissolve in HCl, when CuS and Bi2S3 will not? please explain!!

Answer 1

The solubility can be explained in terms of the solubility product constant, K_sp.

1) CdS + 2 HCl ------> CdCl₂ + H₂S

CdCl₂ (aq) -----> Cd²⁺ (aq) + 2 Cl^- ; K_sp = [Cd²⁺][Cl^-]²

2) CuS + 2 HCl -----> CuCl₂ + H₂S

CuCl₂ (aq) -----> Cu²⁺ (aq) + 2 Cl^-; K_sp = [Cu²⁺][Cl^-]²

3) 2 Bi₂S₃ + 6 HCl -----> 2 BiCl₃ + 3 H₂S

BiCl₃ (aq) -----> Bi³⁺ (aq) + 3 Cl^- (aq); K_sp = [Bi³⁺][Cl^-]³

A compound is soluble when the product on the ionic concentrations is less than the solubility product of the compound. CdS is soluble because the ionic product Q = [Cu²⁺][Cl^-] is less than the solubility product and hence CdS is soluble in HCl. Moreover, CdCl₂ dissolves in excess HCl forming a complex [CdCl₄]^2- anion.

The solubility product values for CuCl₂ and BiCl₃ are much easily reached in HCl and hence the solids precipitates out.