1. Chemistry 123 – Dr. Woodward 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+ 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. Chemistry 123 – Dr. Woodward Solubility Rules Soluble vs. Insoluble is not very helpful. It can really only be used to separate Ag + and Pb 2+ from the other ions.

3. Chemistry 123 – Dr. Woodward Separation into Groups Ag +, Pb 2+, Bi 3+, Cu 2+, Sb 3+, Sb 5+, Sn 2+, Sn 4+, Fe 2+, Fe 3+, Al 3+, Cr 3+, Ni 2+, Co 2+, Zn 2+ cold dilute HCl AgCl, PbCl 2 Remaining Cations & Pb 2+ Group I – Insoluble chlorides H 2 S (+ HNO 3 ) PbS, Bi 2 S 3, CuS, Sb 2 S 5, SnS 2 Remaining Cations Group II – Acid insoluble sulfides 1.NH 4 Cl, NH 3 2.H 2 S NiS, CoS, ZnS, Fe(OH) 3, Al(OH) 3, Cr(OH) 3 Groups 4 & 5 Group III– Base insoluble sulfides & hydroxides

4. Chemistry 123 – Dr. Woodward Group I – Insoluble chlorides AgCl (s) ↔ Ag + (aq) + Cl − (aq) K sp = 1.8  10 −10 PbCl 2 (s) ↔ Pb 2+ (aq) + 2Cl − (aq) K sp = 1.7  10 −5 AgCl & PbCl 2 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, AgCl 2 − & PbCl 4 2−, causes the precipitates to dissolve. Because PbCl 2 is only moderately insoluble adding HCl does not completely remove Pb 2+ from the mixture. So we need to consider Pb 2+ once again with the group II cations.

5. Chemistry 123 – Dr. Woodward Separating and Confirming Pb 2+ Because the solubility of PbCl 2 increases considerably upon heating (6.73 g/L at 0 °C vs. 33.4 g/L at 100 °C) we can dissolve PbCl 2 from AgCl by heating. To confirm the presence of lead we add potassium dichromate to precipitate PbCrO 4 which is a distinctive orange-yellow solid Cr 2 O 7 2− (aq) + H 2 O (l) ↔ 2CrO 4 2− (aq) + 2H + (aq) orangeyellow orangeyellow Pb 2+ (aq) + CrO 4 2− (aq) ↔ PbCrO 4 (s) K sp = 1.8  10 −14 orange-yellow orange-yellow Favored for low pH (acidic)Favored for high pH (basic)

6. Chemistry 123 – Dr. Woodward Confirming Ag + After separating Pb 2+ we should have a white precipitate of AgCl. How can we make sure the precipitate is AgCl and not undissolved PbCl 2 ? Concentrated NH 3 should dissolve the precipitate due to complex ion formation AgCl (s) + 2NH 3 (aq) ↔ Ag(NH 3 ) + (aq) + Cl − (aq) If Pb 2+ 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(NH 3 ) + (aq) + Cl − (aq) + 2H + (aq) ↔ AgCl (s) + 2NH 4 + (aq)

7. Chemistry 123 – Dr. Woodward Separation into Groups Ag +, Pb 2+, Bi 3+, Cu 2+, Sb 3+, Sb 5+, Sn 2+, Sn 4+, Fe 2+, Fe 3+, Al 3+, Cr 3+, Ni 2+, Co 2+, Zn 2+ cold dilute HCl AgCl, PbCl 2 Remaining Cations & Pb 2+ Group I – Insoluble chlorides H 2 S (+ HNO 3 ) PbS, Bi 2 S 3, CuS, Sb 2 S 5, SnS 2 Remaining Cations Group II – Acid insoluble sulfides 1.NH 4 Cl, NH 3 2.H 2 S NiS, CoS, ZnS, Fe(OH) 3, Al(OH) 3, Cr(OH) 3 Groups 4 & 5 Group III– Base insoluble sulfides & hydroxides Acidic solution Basic (buffered) solution

8. Chemistry 123 – Dr. Woodward Aqueous Chemistry of S 2- H 2 S (aq) ↔ H + (aq) + HS − (aq) K a1 = 9.5  10 −8 HS − (aq) ↔ H + (aq) + S 2− (aq) K a2 = 1  10 −19 Because K a2 is so small there are almost no S 2− 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. H 2 S (aq) ↔ H + (aq) + HS − (aq) K a1 = 9.5  10 −8 Co 2+ (aq) + HS − (aq) ↔ H + (aq) + CoS (s) 1/K sp = 2  10 +21 Increasing [H + ] (lowering pH) drives the equilibrium to the left. Co 2+ (aq) + H 2 S (aq) ↔ 2H + (aq) + CoS (s) K = 1.9  10 +14

9. Chemistry 123 – Dr. Woodward Separation of Group II from Group III Pb 2+ (aq) + 2HS − (aq) ↔ PbS (s) + H + (aq) K sp = 3  10 −28 Cu 2+ (aq) + 2HS − (aq) ↔ CuS (s) + H + (aq) K sp = 6  10 −37 Group II Cations Ni 2+ (aq) + 2HS − (aq) ↔ NiS (s) + H + (aq) K sp = 3  10 −20 Co 2+ (aq) + 2HS − (aq) ↔ CoS (s) + H + (aq) K sp = 6  10 −22 Group III Cations 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. Chemistry 123 – Dr. Woodward Group II – Acid Insoluble Sulfides Pb 2+ (aq) + 2HS − (aq) ↔ PbS (s) + H + (aq) Black ppt 2Bi 3+ (aq) + 3HS − (aq) ↔ Bi 2 S 3 (s) + 3H + (aq) Dk. Brown ppt Cu 2+ (aq) + HS − (aq) ↔ CuS (s) + H + (aq) Black ppt SnCl 6 2− (aq) + 2HS − (aq) ↔ SnS 2 (s) + 4H + (aq) + 6Cl − (aq) Yellow ppt Yellow ppt 2SbCl 6 − (aq) + 5HS − (aq) ↔ Sb 2 S 5 (s) + 5H + (aq) + 12Cl − (aq) Orange ppt Orange ppt

11. 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