Chemistry 123 – Dr. Woodward

Separation of Group II CationsSeparation of Group II Cations

PbPb2+2+

BiBi3+3+

CuCu2+2+

SbSb3+3+/Sb/Sb5+5+

SnSn2+2+/Sn/Sn4+4+

The group II cations are The group II cations are separated from all other separated from all other cations by forming acid cations by forming acid insoluble sulfide salts.insoluble sulfide salts.

PbPb2+2+(aq)(aq) + 2HS + 2HS−−(aq)(aq) ↔ PbS ↔ PbS(s)(s) + H+ H++(aq)(aq)

2Bi2Bi3+3+(aq)(aq) + 3HS + 3HS−−(aq)(aq) ↔ Bi ↔ Bi22SS33(s)(s) + + 3H3H++(aq)(aq)

CuCu2+2+(aq)(aq) + HS + HS−−(aq)(aq) ↔ CuS ↔ CuS(s)(s) + H+ H++(aq)(aq)

SnClSnCl662−2−(aq)(aq) + 2HS + 2HS−−(aq)(aq) ↔ SnS ↔ SnS22(s)(s) + 4H + 4H++(aq)(aq)

+ 6Cl+ 6Cl−−(aq) (aq) 2SbCl2SbCl66

−−(aq)(aq) + 5HS + 5HS−−(aq)(aq) ↔ Sb ↔ Sb22SS55(s)(s) + 5H + 5H++(aq)(aq) + + 12Cl12Cl−−(aq) (aq)

Chemistry 123 – Dr. Woodward

Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+

HNO3 + HCl(Aqua regia)

Pb2+, Bi3+, Cu2+, SbCl61−, SnCl62−

HNO3 acts as an oxidizing agent

Cl− acts as a complexing agent

Removes excess acid, be careful not to overdo it.

Evaporate to a paste

HNO3 CH3CSNH2,

heat

CH3CSNH2 (thioacetamide) decomposes on

heating to give ~0.10 M H2S(aq)

PbS (black), Bi2S3 (dark brown), CuS (black), Sb2S5 (orange), SnS2 (yellow)

NaOH

PbS, Bi2S3, CuS SbS43−, SbO4

3−, SnS43−,

SnO43−

SnS2 & Sb2S5 are amphoteric

Antimony subgroupCopper subgroup

Chemistry 123 – Dr. Woodward

Sulfide precipitation, pitfallsSulfide 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 SubgroupCopper SubgroupPbS, Bi2S3, CuS

HNO3 + heat

Pb2+, Bi3+, Cu2+

Sulfur

Pale yellow ppt - discard H2SO4 +

heat

Bi3+, Cu2+PbSO4(s)

White ppt

Heat to remove HNO3 and excess acid (sulfates

become more soluble in strong acid).

Dense white fumes of SO3 (a choking toxic gas that

forms from decomposition of SO4

2- ions) start to come off when you’ve heated

long enough

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 SubgroupCopper SubgroupPbS, Bi2S3, CuS

HNO3 + heat

Pb2+, Bi3+, Cu2+

Sulfur

Pale yellow ppt - discard

H2SO4 + heat

Bi3+, Cu2+PbSO4(s)

White pptConc.

NH3(aq)

Cu(NH3)42−Bi(OH)3(s)

Blue-violet solutionWhite ppt

The blue-green color of Cu2+ in

solution, and later the violet-blue color of Cu(NH3)4

2+ are a clear giveaway for

the presence of Cu2+

NH3 + H2O ↔ NH4+ +

OH−

Chemistry 123 – Dr. Woodward

Antimony SubgroupAntimony SubgroupSbS4

3−, SbO43−, SnS4

3−, SnO4

3−

Colorless Colorless solutionsolution

Split into two equal parts for confirming

tests

Tin testsAntimony tests

SbCl6−, SnCl62−

12M HCl & heatColorless Colorless solutionsolution

Neutralize with 3M HCl & react with

thioacetamideSb2S5(s), SnS2(s)SbSb22SS55 - -

orangeorangeSnSSnS22 - yellow - yellow

Chemistry 123 – Dr. Woodward

Confirmation of AntimonyConfirmation of Antimony

SbCl6−, SnCl62−

Add oxalic acid, H2C2O4, the oxalate ion C2O4

2− forms a stable complex with Sn4+, which sequesters it from further

reaction

SnCl62−(aq) + 3C2O4

2− ↔ 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 is gray-brown

Chemistry 123 – Dr. Woodward

Confirmation of TinConfirmation 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 Observation of black

precipitate confirms the precipitate confirms the presence of tin.presence of tin.

Chemistry 123 – Dr. Woodward

Example ProblemExample Problem1. The group II pretreatment followed by addition of

thioacetamide and heating forms a dark precipitate.

2. NaOH solution is added to the precipitate. The dark precipitate (A) is separated from a colorless solution (B).

3. Precipitate A is treated with 3 M HNO3 which caused it to dissolve to form a light blue solution.

4. H2SO4 was added and the mixture heated. There was no precipitate.

5. NH3 was added which led to the formation of a gelatinous precipitate, while the solution became deep blue in color.

6. When solution B (from step 2) was neutralized with 3 M HCl an orange-red precipitate was formed.

7. 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?