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Unformatted text preview: Separation of Group II Cations
The group II cations are
separated from all other
cations by forming acid
insoluble sulfide salts. Pb2+
Sb3+/Sb5+ Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq)
(aq) Sn2+/Sn4+ 2Bi3+(aq) + 3HS−(aq) ↔ Bi2S3(s) + 3H+(aq)
Cu2+(aq) + HS−(aq) ↔ CuS(s) + H+(aq)
SnCl62−(aq) + 2HS−(aq) ↔ SnS2(s) + 4H+(aq) + 6Cl−(aq)
2SbCl6−(aq) + 5HS−(aq) ↔ Sb2S5(s) + 5H+(aq) + 12Cl−(aq)
Chemistry 123 – Dr. Woodward Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+
HNO3 acts as an
Cl− acts as a
complexing agent HNO3 + HCl
Pb2+, Bi3+, Cu2+, SbCl61−, SnCl62−
heating to give
~0.10 M H2S(aq) Removes excess
acid, be careful
not to overdo it. Evaporate to
CH3CSNH2, heat PbS (black), Bi2S3 (dark brown), CuS (black),
Sb2S5 (orange), SnS2 (yellow)
Copper subgroup PbS, Bi2S3, CuS SnS2 & Sb2S5 are
Antimony subgroup 3−
SbS43−,CSbO43−,123 – 43−,Woodward
hemistry SnS Dr. SnO4 1 Sulfide precipitation, pitfalls
1. If you overheat while evaporating to a paste some cations can
vaporize (Sn & Sb) or the entire solid can pop out of the crucible.
2. We generate H2S from thioacetamide by heating
CH3CSNH2 + 2H2O → CH3COO− + NH4+ + H2S
If you heat to rapidly or much above 80 °C H2S will bubble out of
solution and there won’t be enough to fully precipitate the cations.
3. The nitrate ion (NO3−) is an oxidizing agent. If the nitrate ion
concentration is too high it can oxidize sulfide to elemental sulfur,
which is a pale yellow to yellow-white solid.
3H2S(aq) + 2NO3−(aq) + 2H+(aq) → 3S(s) + 2NO(g) + 4H2O(l)
Chemistry 123 – Dr. Woodward Copper Subgroup
PbS, Bi2S3, CuS
HNO3 + heat Heat to remove HNO3 and
excess acid (sulfates become
more soluble in strong acid). Sulfur Pb2+, Bi3+, Cu2+ Pale yellow ppt discard H2SO4 + heat PbSO4(s)
White ppt Dense white fumes of SO3 (a
choking toxic gas that forms
from decomposition of SO42ions) start to come off when
you’ve heated long enough Bi3+, Cu2+
At this point there may not be
much liquid left, so you will
have to add water to make sure
the ppt doesn’t dissolve.
Chemistry 123 – Dr. Woodward 2 Copper Subgroup
PbS, Bi2S3, CuS
HNO3 + heat Sulfur
Pale yellow ppt discard The blue-green color
of Cu2+ in solution,
and later the violetblue color of
Cu(NH3)42+ are a clear
giveaway for the
presence of Cu2+ Pb2+, Bi3+, Cu2+
H2SO4 + heat PbSO4(s) Bi3+, Cu2+ White ppt Conc. NH3(aq) NH3 + H2O ↔ NH4+ + OH− Bi(OH)3(s) Cu(NH3)42− White ppt Blue-violet123 – Dr. Woodward
Chemistry solution Antimony Subgroup
SbS43−, SbO43−, SnS43−, SnO43− Colorless
solution Neutralize with 3M HCl &
react with thioacetamide
Sb2S5 - orange SnS2 - yellow Sb2S5(s), SnS2(s)
12M HCl & heat Colorless
solution SbCl6−, SnCl62−
Split into two equal parts for
confirming tests Antimony tests Tin tests
Chemistry 123 – Dr. Woodward 3 Confirmation of Antimony
Add oxalic acid, H2C2O4, the oxalate ion C2O42−
forms a stable complex with Sn4+, which
sequesters it from further reaction
SnCl62−(aq) + 3C2O42− ↔ Sn(C2O4)32−(aq) + 6Cl−
Next add thioacetamide CH3CSNH2 and
heat to reacts with Sb5+
Sb2S5 (orange ppt)
What other precipitates could form?
SnS2 is yellow, SnS is123 – Dr. Woodward Confirmation of Tin
Add iron (as a nail) and NaOH. The iron acts as a reducing agent.
The antimony is taken out of solution by reduction to its elemental
SnCl62−(aq) + Fe(s) + 5OH− ↔ Sn(OH)3−(aq) + Fe(OH)(s) + 6Cl−(aq)
2SbCl6−(aq) + 5Fe(s) ↔ 2Sb(s) + 5Fe2+(aq) + 12Cl−(aq)
Centrifuge to remove the solids, and mix the decantate
with Bi(OH)3, which triggers a redox reaction between Bi3+
2Bi(OH)3 + 3Sn(OH)3− + 3OH− ↔ 2Bi(s) + 3Sn(OH)62−(s)
Observation of black precipitate
Chemistry 123 – Dr. Woodward
confirms the presence of tin. 4 Example Problem
7. The group II pretreatment followed by addition of thioacetamide
and heating forms a dark precipitate.
NaOH solution is added to the precipitate. The dark precipitate
(A) is separated from a colorless solution (B).
Precipitate A is treated with 3 M HNO3 which caused it to
dissolve to form a light blue solution.
H2SO4 was added and the mixture heated. There was no
NH3 was added which led to the formation of a gelatinous
precipitate, while the solution became deep blue in color.
When solution B (from step 2) was neutralized with 3 M HCl an
orange-red precipitate was formed.
The orange-red precipitate was dissolved in 12 M HCl, an iron
nail and NaOH were added. The decantate was separated and
added to Bi(OH)3. No reaction was observed. Which group II ions are present? Which are absent? Which are
Chemistry 123 – Dr. Woodward 5 ...
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This note was uploaded on 06/03/2011 for the course CHM 123 taught by Professor Woodward during the Spring '08 term at Ohio State.
- Spring '08