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

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

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Answer #1

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

1) CdS + 2 HCl ------> CdCl2 + H2S

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

2) CuS + 2 HCl -----> CuCl2 + H2S

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

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

BiCl3 (aq) -----> Bi3+ (aq) + 3 Cl- (aq); Ksp = [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.

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