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lect6_exp25b_notes - Separation of Group II Cations The...

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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+ Bi3+ Cu2+ Sb3+/Sb5+ Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq) (s) (aq) Sn2+/Sn4+ 2Bi3+(aq) + 3HS−(aq) ↔ Bi2S3(s) + 3H+(aq) Cu2+(aq) + HS−(aq) ↔ CuS(s) + H+(aq) (aq) (s) (aq) SnCl62−(aq) + 2HS−(aq) ↔ SnS2(s) + 4H+(aq) + 6Cl−(aq) (aq) 2SbCl6−(aq) + 5HS−(aq) ↔ Sb2S5(s) + 5H+(aq) + 12Cl−(aq) (aq) 2SbCl 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 SbS43−,CSbO43−,123 – 43−, SnO43− hemistry SnS 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 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 1 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) 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+ White ppt 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 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+ Pale yellow ppt discard 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 2 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 gray-brown SnS2 is yellow, SnS is123 – Dr. Woodward 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 3 ...
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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.

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