Question

2. A solution that may contain Sn.Bi. Cu and Cd²+ ions is treated with thioacetamide in an acidic medium. The black precipitate that forms is partly soluble in a strongly alkaline solution. The precipitate that remains is soluble in 6M HNO, and gives a blue solution when treated with excess NH. Based on this information, which ions are present, which are absent, and which are still in doubt? 3. Explain how controlling the solution pH allows separation of Group II cations from those of Group III.

Answer 1

The group II ions are Sn2+ , Cd2+, Bi3+ , Cu2+

The ions of Group II are precipitated as their sulfides from an acidic solution of hydrogen sulfide. These sulfides are insoluble at a pH of 0.5 but not the ions of Group III, the sulfides of which are more soluble.

In the laboratory, hydrogen sulfide is generated by the hydrolysis of thioacetamide.

+ CH2CNH, + H2O thioacetamide CHZÖNH2 + H,S acetamide

In the given unknown mixture of ions, the ions that precipitated are Bi3+ and Cu2+.

The black precipitate is due to the formation of Bi2S3 which is partly soluble in alkaline solution.

2Bi+3 + 3H2S 2 Bi2S3(s) + 6H+

                The precipitate left insoluble is CuS.

Cu++ H2S = Cus(s) + 2H+

CuS precipitate is soluble in 6M HNO3 and gives a deep blue complex when treated with excess of NH3.

Cu?+ + 4NH, + Cu(NH3)* (Cu(NH3)* is deep blue)

The ions that are absent in the solution are Sn2+ , Cd2+

3. Effect of pH on the separation of group II ions and group III ions

Group II ions form insoluble sulfides with H2S at a pH of 0.5 . The sulfides of the ions of Group III are more soluble and cannot precipitated in the acidic solution of hydrogen sulfide of Group II .

Hence, the ions of Group III are precipitated as their sulfides, or as their hydroxides, from a basic solution of hydrogen sulfide. This is done by carrying out the reaction in the presence of ammonia-ammonium chloride buffer.