Hard chemistry question...precitates..?

Question

A solution that may containn Cu^2+, Bi^2+,Sn^4+, or Sb^2+ is treated with Thioacetamide in an acidic medium. The black precipitate which forms is partly soluble in strongly basic solution. The precipitate which remains is soluble in 6M HNO3 and gives only a white precipitate upon the addition oh NH4OH. The basic solution, when acidified, produces an orange precipitate.

(1) Which group II ions are present?

(2) Which are absent?

(3)Which are in doubt?

(4) HOW WOULD YOU REMOVE ALL DOUBT??

Thank you very much for your help!

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.