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Qualitative Analysis of Metallic Elements

Qualitative Analysis of Metallic Elements. Ag + , Pb 2+ , Bi 3+ Cu 2+ , Al 3+ , Cr 3+ Ni 2+ , Co 2+ , Zn 2+ Sb 3+ /Sb 5+ Sn 2+ /Sn 4+ , Fe 2+ /Fe 3+.

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Qualitative Analysis of Metallic Elements

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  1. Qualitative Analysis of Metallic Elements Ag+, Pb2+, Bi3+ Cu2+, Al3+, Cr3+ Ni2+, Co2+, Zn2+ Sb3+/Sb5+ Sn2+/Sn4+, Fe2+/Fe3+ Given a solution that contains one or more of these cations how can we use chemistry to identify which ions are present and which are not? • Differences in solubility • Soluble vs. insoluble • Solubility with temperature • Solubility and pH • Complex ion formation • Amphoteric behavior • Changes in solubility with change in oxidation state

  2. Solubility Rules Soluble vs. Insoluble is not very helpful. It can really only be used to separate Ag+ and Pb2+ from the other ions.

  3. Separation into Groups cold dilute HCl AgCl, PbCl2 Remaining Cations & Pb2+ Group I – Insoluble chlorides H2S (+ HNO3) PbS, Bi2S3, CuS, Sb2S5, SnS2 Remaining Cations • NH4Cl, NH3 • H2S Group II – Acid insoluble sulfides Group III– Base insoluble sulfides & hydroxides Groups 4 & 5 NiS, CoS, ZnS, Fe(OH)3, Al(OH)3, Cr(OH)3 Ag+, Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+,Fe2+, Fe3+, Al3+, Cr3+, Ni2+, Co2+, Zn2+

  4. Group I – Insoluble chlorides AgCl(s)↔Ag+(aq) + Cl−(aq) Ksp = 1.8  10−10 PbCl2(s)↔Pb2+(aq) + 2Cl−(aq) Ksp = 1.7  10−5 • AgCl & PbCl2 are both white solids. • Dilute HCl is used rather than a chloride salt (i.e. NaCl) to avoid precipitating SbOCl and BiOCl (which are soluble in acidic solutions). • If the chloride concentration is too high (conc. HCl) complex ion formation, AgCl2− & PbCl42−, causes the precipitates to dissolve. • Because PbCl2 is only moderately insoluble adding HCl does not completely remove Pb2+ from the mixture. So we need to consider Pb2+ once again with the group II cations.

  5. Separating and Confirming Pb2+ • Because the solubility of PbCl2 increases considerably upon heating (6.73 g/L at 0 °C vs. 33.4 g/L at 100 °C) we can dissolve PbCl2 from AgCl by heating. • To confirm the presence of lead we add potassium dichromate to precipitate PbCrO4 which is a distinctive orange-yellow solid Favored for low pH (acidic) Favored for high pH (basic) Cr2O72−(aq) + H2O(l) ↔ 2CrO42−(aq) + 2H+(aq) orange yellow Pb2+(aq) + CrO42−(aq) ↔ PbCrO4(s) Ksp = 1.8  10−14 orange-yellow

  6. Confirming Ag+ After separating Pb2+ we should have a white precipitate of AgCl. How can we make sure the precipitate is AgCl and not undissolved PbCl2? Concentrated NH3 should dissolve the precipitate due to complex ion formation AgCl(s) + 2NH3(aq) ↔ Ag(NH3)+(aq) + Cl−(aq) If Pb2+ remains in solution the white precipitate Pb(OH)2 will form on making the solution basic. To confirm Ag+ we reverse this by making the sol’n acidic Ag(NH3)+(aq) + Cl−(aq) + 2H+(aq)↔ AgCl(s) + 2NH4+(aq)

  7. Separation into Groups cold dilute HCl AgCl, PbCl2 Remaining Cations & Pb2+ Group I – Insoluble chlorides H2S (+ HNO3) PbS, Bi2S3, CuS, Sb2S5, SnS2 Remaining Cations • NH4Cl, NH3 • H2S Group II – Acid insoluble sulfides Group III– Base insoluble sulfides & hydroxides Groups 4 & 5 NiS, CoS, ZnS, Fe(OH)3, Al(OH)3, Cr(OH)3 Ag+, Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+,Fe2+, Fe3+, Al3+, Cr3+, Ni2+, Co2+, Zn2+ Acidic solution Basic (buffered) solution

  8. Aqueous Chemistry of S2- Co2+(aq) + H2S(aq)↔ 2H+(aq) + CoS(s) K = 1.9  10+14 H2S(aq)↔ H+(aq) + HS−(aq) Ka1 = 9.5  10−8 HS−(aq)↔ H+(aq) + S2−(aq) Ka2 = 1  10−19 Because Ka2 is so small there are almost no S2− ions in solution. We can neglect the second reaction. Formation of metal sulfides occurs through reaction of HS− with metal cations, and the concentration of HS− is dependent on the pH. Consider the precipitation of CoS. H2S(aq)↔ H+(aq) + HS−(aq) Ka1 = 9.5  10−8 Co2+(aq) + HS−(aq)↔ H+(aq) + CoS(s) 1/Ksp = 2  10+21 Increasing [H+] (lowering pH) drives the equilibrium to the left.

  9. Separation of Group II from Group III Group II Cations Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq) Ksp = 3  10−28 Cu2+(aq) + 2HS−(aq) ↔ CuS(s) + H+(aq) Ksp = 6  10−37 Group III Cations Ni2+(aq) + 2HS−(aq) ↔ NiS(s) + H+(aq) Ksp = 3  10−20 Co2+(aq) + 2HS−(aq) ↔ CoS(s) + H+(aq) Ksp = 6  10−22 By making the solution acidic we shift the equilibrium to the left (favors reactants) which makes the Group III sulfides dissolve, but not the Group II sulfides.

  10. Group II – Acid Insoluble Sulfides Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq) Black ppt 2Bi3+(aq) + 3HS−(aq) ↔ Bi2S3(s) + 3H+(aq) Dk. Brown ppt Cu2+(aq) + HS−(aq) ↔ CuS(s) + H+(aq) Black ppt SnCl62−(aq) + 2HS−(aq) ↔ SnS2(s) + 4H+(aq) + 6Cl−(aq) Yellow ppt 2SbCl6−(aq) + 5HS−(aq) ↔ Sb2S5(s) + 5H+(aq) + 12Cl−(aq) Orange ppt

  11. HNO3 acts as an oxidizing agent Cl− acts as a complexing agent HNO3 + HCl (Aqua regia) Pb2+, Bi3+, Cu2+, SbCl61−, SnCl62− Removes excess acid, be careful not to overdo it. Evaporate to a paste CH3CSNH2 (thioacetamide) decomposes on heating to give ~0.10 M H2S(aq) HNO3 CH3CSNH2, heat PbS (black), Bi2S3 (dark brown), CuS (black), Sb2S5 (orange), SnS2 (yellow) SnS2 & Sb2S5 are amphoteric NaOH Copper subgroup Antimony subgroup PbS, Bi2S3, CuS SbS43−, SbO43−,SnS43−, SnO43− Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+

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