A solution that may contain Cu^2+, Bi^3+, Sn^4+, or Sb^3+ ions is treated with t
ID: 950316 • Letter: A
Question
A solution that may contain Cu^2+, Bi^3+, Sn^4+, 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 show the presence or absence of an ion.)Explanation / Answer
Thioacetamide is generally used as a source of sulphide ion. Thus the metals are preipitated as their sulphides and the ions that are precipitated with thioacetamide are divalent ions Ni, Pb, Cd and Hg trivalent ions like As, Sb and Bi and monovalent ions Ag and Cu+1.
Thus in the given solution the ions that can be precipitated are onlyCu+2, Bi+3 and Sb+3 only. A black precipitate is due to Bi2S3 or CuS.
The balck precipitae soluble in alkali is bismuth/antimony which gives organge precipitate of bismuthat / Sb2S3.
The reaimaing balck precipitate which is soluble in concentrted nitric acid is CuS, which on reaction with excess ammonia gives blue coloration due to the formation of cuprammonium complex.
Thus ion definitely present is Cu+2
in doubt are Bi+3/Sb+3
Absent Sn+4
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