lect5_exp25a

lect5_exp25a - Qualitative Analysis of Metallic Elements...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Qualitative Analysis of Metallic Elements Ag+, Pb2+, Bi3+ Cu2+, Al3+, Cr3+ Ni2+, Co2+, Zn2+ 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 Differences • Soluble vs. insoluble Sb3+/Sb5+ • Solubility with temperature Sn2+/Sn4+, • Solubility and pH Fe2+/Fe3+ • Complex ion formation • Amphoteric behavior • Changes in solubility with change in oxidation state Chemistry 123 – Dr. Woodward Solubility Rules Soluble vs. Insoluble is not very helpful. It can really only be used to separate Ag+ and Pb2+ from the other ions. Chemistry 123 – Dr. Woodward 1 Ag+, Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+, Fe2+, Fe3+, Al3+, Cr3+, Ni2+, Co2+, Zn2+ Separation into Groups cold dilute HCl AgCl, PbCl2 Remaining Cations & Pb2+ Group I – Insoluble chlorides H2S (+ HNO3) PbS, Bi2S3, CuS, Sb2S5, SnS2 Group II – Acid insoluble sulfides Group III– Base insoluble sulfides & hydroxides Remaining Cations 1. NH4Cl, NH3 2. H2S NiS, CoS, ZnS, Fe(OH)3, Al(OH)3, Cr(OH)3 Groups 4 & 5 Chemistry 123 – Dr. Woodward Group I – Insoluble chlorides AgCl(s) ↔ Ag+(aq) + Cl−(aq) (s) (aq) (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 Dr. Woodward Chemistry 123 – cations. 2 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 Ksp = 1.8 × 10−14 Pb2+(aq) + CrO42−(aq) ↔ PbCrO4(s) orange-yellow orange- Chemistry 123 – Dr. Woodward 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) (s) (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) (aq) (s) Chemistry 123 – Dr. Woodward 3 Separation into Groups Ag+, Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+, Fe2+, Fe3+, Al3+, Cr3+, Ni2+, Co2+, Zn2+ cold dilute HCl AgCl, PbCl2 Group I – Insoluble chlorides Acidic solution Remaining Cations & Pb2+ H2S (+ HNO3) Basic (buffered) solution PbS, Bi2S3, CuS, Sb2S5, SnS2 Group II – Acid insoluble sulfides Group III– Base insoluble sulfides & hydroxides Remaining Cations 1. NH4Cl, NH3 2. H2S NiS, CoS, ZnS, Fe(OH)3, Al(OH)3, Cr(OH)3 Groups 4 & 5 Chemistry 123 – Dr. Woodward Aqueous Chemistry of S2H2S(aq) ↔ H+(aq) + HS−(aq) (aq) (aq) Ka1 = 9.5 × 10−8 HS−(aq) ↔ H+(aq) + S2−(aq) (aq) (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) (aq) (aq) Ka1 = 9.5 × 10−8 Co2+(aq) + HS−(aq) ↔ H+(aq) + CoS(s) (aq) (aq) (s) 1/Ksp = 2 × 10+21 Co2+(aq) + H2S(aq) ↔ 2H+(aq) + CoS(s) K = 1.9 × 10+14 (s) Chemistry to the left. Increasing [H+] (lowering pH) drives the equilibrium123 – Dr. Woodward 4 Separation of Group II from Group III Group II Cations Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq) (s) (aq) Ksp = 3 × 10−28 Cu2+(aq) + 2HS−(aq) ↔ CuS(s) + H+(aq) (s) (aq) Ksp = 6 × 10−37 Group III Cations Ni2+(aq) + 2HS−(aq) ↔ NiS(s) + H+(aq) (s) (aq) Ksp = 3 × 10−20 Co2+(aq) + 2HS−(aq) ↔ CoS(s) + H+(aq) (s) (aq) Ksp = 6 × 10−22 By making the solution acidic we shift the equilibrium to the left (favors reactants) which makes the Group III Chemistry 123 – Dr. Woodward sulfides dissolve, but not the Group II sulfides. Group II – Acid Insoluble Sulfides Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq) (s) (aq) Black ppt 2Bi3+(aq) + 3HS−(aq) ↔ Bi2S3(s) + 3H+(aq) Dk. Brown ppt Cu2+(aq) + HS−(aq) ↔ CuS(s) + H+(aq) (aq) (s) (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 Orange Chemistry 123 – Dr. Woodward 5 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 SbS4 3−, 3− , SbO4 SnS Dr. Woodward Chemistry 123 – 4 , SnO4 3− 3− 6 ...
View Full Document

Ask a homework question - tutors are online