Question

a flow scheme with the minimum number of steps for the of only these cations Pb Cu SN 4 Devise and diasram and identification

Answer 1

Answer.

1. Among the metal ions Pb2+, Cu2+, Sb3+ will on reaction with thioacetamide form insoluble sulphides whereas Ni2+ will remain in solution due to high Ksp value of its sulphide. Thus, separate three ions from the solution by filtration and keep mother liquid for Ni2+ (See chart).

2. Dissolve the precipitate of step 1 in NaOH, the Sb2S3 will form a soluble complex Sb(OH)4 and PbS and CuS will remain insoluble, Filter them,

3. Dissolve the precipitate of step 2 in NH3, Cu2+ will form a soluble complex [Cu9NH3)4]2+ but PbS will remain insoluble, so filter the solution, precipitate will contain Pb2+ and Cu2+ will remain in solution makin color of solution blue. Now check for their confirmatory test as shown in chart. [Cu2+ and Pb2+ over]

4. Take the filtrate of step 2, acidify this with HCl slowly, so solid Sb2S3 precipitae, now add excess of HCl to dissolve this. In the resultant solution add the reagent shown in chart. So [Sb3+ is also confirm]

5. Now take filtrate of step one, add NH3 solution followed by addition Dimethylglyoxime. It will formprecipitae of Ni(DMG)2 having strawberry red color. Thus, Ni2+ is also confirmed.