A solution containing possibly Bi +3 ,Cd +2 ,Cu 2+, and Sn 4+ ions was treated w
ID: 972201 • Letter: A
Question
A solution containing possibly Bi+3,Cd+2,Cu2+,and Sn4+ ions was treated with H2S at a pH of 0.5, and a black precipitate (#1) formed. Precipitate (#1) was separated from the supernatant liquid and treated with 3M KOH. The remaining precipitate (#2) was separated from its supernatant liquid (#3). Solution (#3) was treated with 6 M HCl and H2S with no resultant precipitate. Residue (#2) was treated with 1 M HCl. The remaining precipitate was separated from the supernatant liquid (#4). Solution (#4) was treated with 6 M NH3 and H2S producing a yellow precipitate.
Put an X in the appropriate column in the Table below for each ion in the Group.
Ion
Definitely Present
Definitely Absent
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Bi3+
Cd2+
Cu2+
Sn+4
Ion
Definitely Present
Definitely Absent
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Bi3+
Cd2+
Cu2+
Sn+4
Explanation / Answer
Ion
Definitely Present
Definitely Absent
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Bi3+
present
Cd2+
present
Cu2+
Present
Sn+4
present
Group II (Cu2+,Bi3+,Sb3+, Sn4+)
cations produce very insoluble sulfides(Kspvalues less than 10-30) so they can be precipitated by low amounts of sulfide ion; this can be
achieved by adding an acidic solution of H2S.
Ion
Definitely Present
Definitely Absent
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Bi3+
present
Cd2+
present
Cu2+
Present
Sn+4
present
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