Question

A solution that may contain Cu²⁺, Bi³⁺, Sn⁴⁺, or Sb³⁺ ions is treated with thioacetamide in an acid medium. The black precipitate that forms is partly soluble in strongly basic solution. The precipitate that remains is soluble in 6 M HNO₃ and gives only a white precipitate upon the addition of NH4OH. The basic solution, when acidified, produces, an orange precipitate.. which group 2 ions are present? which are absent? and which are still in doubt? How would you remove all doubt?

Answer 1

I say only Sb3+ is definite. Its IIB sulfide is the only one that's orange when re-precipitated from the OH- extract. However, the orange color could mask the yellow As2S3 and SnS2 precipitates, but not the black HgS.

As for the original precipitate being black, a IIA cation (Pb2+, Cu2+, Cd2+, Bi3+) must be present. The only IIB cation with a black sulfide is Hg2+ which we ruled out above. Of these four, when the acid extract is made alkaline with NH3, Cu2+ and Cd2+ will form soluble amine complexes, which could happen. Bi3+ would form the white ppt. Bi(OH)3 and Pb2+, if not removed in Group I, will form solid Pb(OH)2, also white. (NOTE: Pb(OH)2 will dissolve in excess NaOH to form Pb(OH)42-, but not in NH3 solution).

There are many ways you could remove doubts. For instance, when adding NH3 above to the IA extract, if the solution turns blue, Cu2+ is confirmed. If not, Cu2+ is absent or in very low concentration. If you treated the precipitate with NaOH first, if it dissolves, that confirms Pb2+. Bi(OH)3 will not dissolve in NaOH solution.