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A solution that may contain Cu^2+, Bi^3+, Sn^4+, and/or Sb^3+ ions is treated with thioacetamide in an acid medium. The black precipitate that forms is partly soluble in strongly alkaline solution. The precipitate that remains is soluble in 6 M HNO_3 and gives only a blue solution on treatment with excess NH_3. The alkaline solution, when acidified, produces an orange precipitate. On the basis of this information, which ions are present, which are absent, and which are still in doubt? (As with Group I, evidence other than confirmatory tests may indicate the presence or absence of an ion.) Present ______ Absent ______ In doubt ______
Expert Answer
In the solution may 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.
