1. Chemistry 123 – Dr. Woodward Separation of Group II Cations Pb 2+ Bi 3+ Cu 2+ Sb 3+ /Sb 5+ Sn 2+ /Sn 4+ The group II cations are separated from all other cations by forming acid insoluble sulfide salts. Pb 2+ (aq) + 2HS − (aq) ↔ PbS (s) + H + (aq) 2Bi 3+ (aq) + 3HS − (aq) ↔ Bi 2 S 3 (s) + 3H + (aq) Cu 2+ (aq) + HS − (aq) ↔ CuS (s) + H + (aq) SnCl 6 2− (aq) + 2HS − (aq) ↔ SnS 2 (s) + 4H + (aq) + 6Cl − (aq) 2SbCl 6 − (aq) + 5HS − (aq) ↔ Sb 2 S 5 (s) + 5H + (aq) + 12Cl − (aq)

2. Chemistry 123 – Dr. Woodward Pb 2+, Bi 3+, Cu 2+, Sb 3+, Sb 5+, Sn 2+, Sn 4+ HNO 3 + HCl (Aqua regia) Pb 2+, Bi 3+, Cu 2+, SbCl 6 1−, SnCl 6 2− HNO 3 acts as an oxidizing agent Cl − acts as a complexing agent Removes excess acid, be careful not to overdo it. Evaporate to a paste HNO 3 CH 3 CSNH 2, heat CH 3 CSNH 2 (thioacetamide) decomposes on heating to give ~0.10 M H 2 S(aq) PbS (black), Bi 2 S 3 (dark brown), CuS (black), Sb 2 S 5 (orange), SnS 2 (yellow) NaOH PbS, Bi 2 S 3, CuSSbS 4 3−, SbO 4 3−, SnS 4 3−, SnO 4 3− SnS 2 & Sb 2 S 5 are amphoteric Antimony subgroupCopper subgroup

3. Chemistry 123 – Dr. 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 H 2 S from thioacetamide by heating CH 3 CSNH 2 + 2H 2 O  CH 3 COO − + NH 4 + + H 2 S If you heat to rapidly or much above 80 °C H 2 S will bubble out of solution and there won’t be enough to fully precipitate the cations. 3. The nitrate ion (NO 3 − ) 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. 3H 2 S (aq) + 2NO 3 − (aq) + 2H + (aq)  3S (s) + 2NO (g) + 4H 2 O (l)

4. Chemistry 123 – Dr. Woodward Copper Subgroup PbS, Bi 2 S 3, CuS HNO 3 + heat Pb 2+, Bi 3+, Cu 2+ Sulfur Pale yellow ppt - discard H 2 SO 4 + heat Bi 3+, Cu 2+ PbSO 4 (s) White pptHeat to remove HNO 3 and excess acid (sulfates become more soluble in strong acid). Dense white fumes of SO 3 (a choking toxic gas that forms from decomposition of SO 4 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.

5. Chemistry 123 – Dr. Woodward Copper Subgroup PbS, Bi 2 S 3, CuS HNO 3 + heat Pb 2+, Bi 3+, Cu 2+ Sulfur Pale yellow ppt - discard H 2 SO 4 + heat Bi 3+, Cu 2+ PbSO 4 (s) White ppt Conc. NH 3 (aq) Cu(NH 3 ) 4 2− Bi(OH) 3 (s) Blue-violet solutionWhite ppt The blue-green color of Cu 2+ in solution, and later the violet- blue color of Cu(NH 3 ) 4 2+ are a clear giveaway for the presence of Cu 2+ NH 3 + H 2 O ↔ NH 4 + + OH −

6. Chemistry 123 – Dr. Woodward Antimony Subgroup SbS 4 3−, SbO 4 3−, SnS 4 3−, SnO 4 3− Colorless solution Split into two equal parts for confirming tests Tin testsAntimony tests SbCl 6 −, SnCl 6 2− 12M HCl & heat Colorless solution Neutralize with 3M HCl & react with thioacetamide Sb 2 S 5 (s), SnS 2 (s) Sb 2 S 5 - orange SnS 2 - yellow

7. Chemistry 123 – Dr. Woodward Confirmation of Antimony SbCl 6 −, SnCl 6 2− Add oxalic acid, H 2 C 2 O 4, the oxalate ion C 2 O 4 2− forms a stable complex with Sn 4+, which sequesters it from further reaction SnCl 6 2− (aq) + 3C 2 O 4 2− ↔ Sn(C 2 O 4 ) 3 2− (aq) + 6Cl − Next add thioacetamide CH 3 CSNH 2 and heat to reacts with Sb 5+ Sb 2 S 5 (orange ppt) What other precipitates could form? SnS 2 is yellow, SnS is gray-brown

8. Chemistry 123 – Dr. Woodward Confirmation of Tin SbCl 6 −, SnCl 6 2− 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. + SnCl 6 2− (aq) + Fe(s) + 5OH − ↔ Sn(OH) 3 − (aq) + Fe(OH)(s) + 6Cl − (aq) + 2SbCl 6 − (aq) + 5Fe(s) ↔ 2Sb(s) + 5Fe 2+ (aq) + 12Cl − (aq) Centrifuge to remove the solids, and mix the decantate with Bi(OH) 3, which triggers a redox reaction between Bi 3+ and Sn 2+ 2Bi(OH) 3 + 3Sn(OH) 3 − + 3OH − ↔ 2Bi(s) + 3Sn(OH) 6 2− (s) Observation of black precipitate confirms the presence of tin.

9. Chemistry 123 – Dr. Woodward Example Problem 1.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 HNO 3 which caused it to dissolve to form a light blue solution. 4.H 2 SO 4 was added and the mixture heated. There was no precipitate. 5.NH 3 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?