Transcript Pb 2+ (aq) - Chemistry

Separation of Group II Cations
Pb2+
Bi3+
The group II cations are
separated from all other
cations by forming acid
insoluble sulfide salts.
Cu2+
Sb3+/Sb5+
Sn2+/Sn4+
Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq)
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
oxidizing agent
Cl− acts as a
complexing agent
HNO3 + HCl
(Aqua regia)
Pb2+, Bi3+, Cu2+, SbCl61−, SnCl62−
CH3CSNH2
(thioacetamide)
decomposes on
heating to give
~0.10 M H2S(aq)
Removes excess
acid, be careful
not to overdo it.
Evaporate to
a paste
HNO3
CH3CSNH2, heat
PbS (black), Bi2S3 (dark brown), CuS (black),
Sb2S5 (orange), SnS2 (yellow)
NaOH
Copper subgroup
PbS, Bi2S3, CuS
SnS2 & Sb2S5 are
amphoteric
Antimony subgroup
3− SnO 3−
SbS43−,Chemistry
SbO43−,123
SnS
4
– 4Dr. ,Woodward
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
Copper Subgroup
PbS, Bi2S3, CuS
HNO3 + heat
Sulfur
Pb2+, Bi3+, Cu2+
Pale yellow ppt discard
H2SO4 + heat
PbSO4(s)
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+
Bi3+, Cu2+
White ppt
Conc. NH3(aq)
NH3 + H2O ↔ NH4+ + OH−
Bi(OH)3(s)
White ppt
Cu(NH3)42−
Blue-violet
solution
Chemistry 123
– Dr. Woodward
Antimony Subgroup
SbS43−, SbO43−, SnS43−, SnO43−
Colorless
solution
Neutralize with 3M HCl &
react with thioacetamide
Sb2S5 - orange
Sb2S5(s), SnS2(s)
SnS2 - yellow
12M HCl & heat
SbCl6
−,
Colorless
solution
SnCl62−
Split into two equal parts for
confirming tests
Antimony tests
Tin tests
Chemistry 123 – Dr. Woodward
Confirmation of Antimony
SbCl6−, SnCl62−
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?
Chemistry
– Dr. Woodward
SnS2 is yellow,
SnS is123
gray-brown
Confirmation of Tin
SbCl6−, SnCl62−
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
form.
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+
and Sn2+
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.
Example Problem
1.
2.
3.
4.
5.
6.
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
precipitate.
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
undetermined?
Chemistry 123 – Dr. Woodward