1. Please Pick Up Solubility Products Constants Data Sheet Heterogeneous Equilibria Problem Set Sample page from the Handbook of Chemistry and Physics

2. Heterogeneous Equilibria Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

3. 7/15/2015 Heterogeneous Equilibria  Reading assignment: Zumdahl: Chapters 4.7-4.8, 8.8-8.9  Sparingly soluble salts  Solubility product  Common ion effect  Molar solubility  Applications

4. 7/15/2015 Some Salts are Soluble  Soluble salts Group IA, NH 4 +, NO 3 –, ClO 3 –, ClO 4 – Most halides (Cl –, Br –, I – ) ·except Ag +, Cu +, Hg 2 2+, Pb 2+ Most sulfates ·except CaSO 4, BaSO 4, Ag 2 SO 4, HgSO 4, PbSO 4

5. 7/15/2015 Some Salts are Not Very Soluble  Sparingly soluble salts Most hydroxides ·except Group IA, Ba(OH) 2 Carbonates, phosphates, sulfides ·except Group IA, NH 4 +  Solubility depends on Temperature and solvent  Strong electrolyte

6. 7/15/2015 Sparingly Soluble Salts AgCl (s)Ag + (aq) + Cl – (aq) The solubility product of silver chloride is 1.8 x 10 -10, how can it be a strong electrolyte? AgCl (s)AgCl (aq) This indicates that AgCl isn’t very soluble. (sparingly soluble) Ag + (aq) + Cl – (aq) This indicates that the amount of AgCl which dissolves ionizes extensively. (strong electrolyte)

7. 7/15/2015 Solubility Product, K sp  Products over reactants with simplifications [Ag + ] [Cl – ] [AgCl(s)]  K c [Ag + ] [Cl – ] = K c · [AgCl(s)] [Ag + ] [Cl – ] = K sp

8. 7/15/2015 For Each of the Following Determine the Solubility Product Expression  lead(II) carbonate  calcium fluoride  iron(III) hydroxide  iron(II) sulfide  lead(II) iodide  silver sulfate

9. 7/15/2015 For Each of the Following Determine the Solubility Product Expression  [Pb 2+ ] [CO 3 2 – ]  [Ca 2+ ] [F – ] 2  [Fe 3+ ] [OH – ] 3  [Fe 2+ ] [S 2 – ]  [Pb 2+ ] [I – ] 2  [Ag + ] 2 [SO 4 2 – ]

10. 7/15/2015 How to Determine If a Compound is Soluble or Not?  Solubility Rules  Table of Solubility Product Constants  Handbook of Chemistry and Physics Physical Constants of Inorganic Compounds

11. 7/15/2015 Physical Constants of Inorganic Compounds  Name  Formula  Molecular Weight  Crystalline Form Color  Density  Melting Point  Boiling Point  Solubility Cold Water Hot Water Other Solvents

12. 7/15/2015 Copper(II) Formate  Formula and Molecular Weight Cu(HCOO) 2, 153.58 amu  Color blue, monoclinic crystals  Is copper(II) formate soluble in water? yes, 12.5 grams per 100 mL of water greater than 1 gram per 100 mL water

13. 7/15/2015 Group Activity  An aqueous solution of copper(II) nitrate is mixed with aqueous sodium iodate. Write an equation for this process and use appropriate tables to determine if a reaction occurs. What observations are expected to be made? Is Cu(NO 3 ) 2 soluble? Is NaIO 3 soluble?

14. 7/15/2015 Copper(II) Iodate Cu 2+ (aq) + 2 NO 3 – (aq) + 2 Na + (aq) + 2 IO 3 – (aq) Cu(IO 3 ) 2 (s) + 2 NO 3 – (aq) + 2 Na + (aq) Ionic Equation: Cu 2+ (aq) + 2 IO 3 – (aq)Cu(IO 3 ) 2 (s) Net Ionic Equation:

15. 7/15/2015 Copper(II) Iodate  Sparingly soluble less than 1 gram per 100 mL of water 0.136 grams per 100 mL of water at 15 °C  Green, monoclinic crystals Cu 2+ (aq) + 2 IO 3 – (aq)Cu(IO 3 ) 2 (s)

16. 7/15/2015 Pre-laboratory Assignment  Experiment P: Qualitative Reactions  Complete the Predicted Reactions grid on page P-4 in laboratory manual  Predict double displacement reactions, use ·Solubility Rules ·Table of Solubility Product Constants ·Handbook of Chemistry and Physics

17. 7/15/2015 Al 2 (SO 4 ) 3 ZnCl 2 Al 2 (SO 4 ) 3 + ZnCl 2 AlCl 3 + ZnSO 4 predicted to be soluble NR

18. 7/15/2015 Al 2 (SO 4 ) 3 ZnCl 2 NR BaCl 2 Al 2 (SO 4 ) 3 + BaCl 2 AlCl 3 + BaSO 4 predicted to be soluble predicted to be an insoluble, white ppt BaSO 4 white ppt

19. Molar Solubility

20. 7/15/2015 Two Beakers Each Contain 500 mL of Water 50 grams of PbF 2 is added to Beaker 1 400 grams of PbF 2 is added to Beaker 2 Some of the solid dissolves. In which beaker will more lead(II) fluoride dissolve?

21. 7/15/2015 Solubility Product Write the chemical equation for the dissolution process. Let +x be the moles per liter of lead(II) ion that is produced. Write the mass-action expression. Substitute variables and solve the solubility product mass-action expression.solve What is the solubility of lead(II) fluoride in pure water?

22. 7/15/2015 Lead(II) Fluoride -x+x+2x PbF 2 (s)Pb 2+ (aq) + 2 F – (aq) K sp = [Pb 2+ ] [F – ] 2 = 3.7 x 10 –8 M 3 = (x) (2x) 2 = 3.7 x 10 –8 M 3 = 4x 3 = 3.7 x 10 –8 M 3 x = 2.1 x 10 –3 M

23. 7/15/2015 Lead(II) Fluoride -x+x+2x x = 2.1 x 10 –3 M  What does 2.1 x 10 –3 M represent? Molar solubility of PbF 2 in pure water Molarity of lead(II) ion in solution What is the molarity of the fluoride ion? PbF 2 (s)Pb 2+ (aq) + 2 F – (aq)

24. 7/15/2015 How many grams of lead(II) fluoride will dissolve in 500 mL of water?  Molar solubility of PbF 2 (FW 245.2 amu) in pure water is 2.1 x 10 –3 M 2.1  10 –3 moles PbF 2 liter        0.500 liter 245.2 grams PbF 2 moles PbF 2        0.26 g PbF 2

25. 7/15/2015 Each beaker contains 500 mL of water. In which beaker will more lead(II) fluoride dissolve? 50 grams of PbF 2 is added to Beaker 1 400 grams of PbF 2 is added to Beaker 2 If there is excess solid at equilibrium, the same amount dissolves!

26. 7/15/2015

27. Common Ion Effect

28. 7/15/2015 Two beakers each contain 500 mL of solution  50 grams of PbF 2 is added to pure water  50 grams of PbF 2 is added to 0.0200 M sodium fluoride Into which solution will more lead(II) fluoride dissolve? H2OH2O H2OH2O Na + F-F-

29. 7/15/2015 Common Ion Effect  Equilibrium concentrations of each ion depend on the number of those ions in solution, regardless of the source.  What is the molar solubility of lead(II) fluoride in a solution containing 0.0200 M sodium fluoride? Assume one liter of solution

30. 7/15/2015 Equation I D I R I  M C E Start with the initial conditions, assume one liter of solution 0.0200 0 0 lots 0 NaF Na + F¯ H2OH2OPbF 2 Pb 2+ Use moles or millimoles

31. 7/15/2015 Equation I D I R I  M C E Is NaF a strong electrolyte? Does it dissociate 100% Write the strong electrolyte, dissociation reaction here. 0.0200 0 0 lots 0 NaFNa + + F¯ -0.0200+0.0200 Is PbF 2 very soluble? 0 NaF Na + F¯ H2OH2OPbF 2 Pb 2+ 0 0 0.0200 lots 0

32. 7/15/2015 Cross-Reactions Which of these reactions are favorable?  Na + + Na +  Na + + F   Na + + PbF 2  Na + + H 2 O  F  + F   F  + PbF 2  F  + H 2 O  PbF 2 + PbF 2  PbF 2 + H 2 O  H 2 O + H 2 O

33. 7/15/2015 Na + + Na + Equilibrium Reaction Which of these reactions are favorable? K = 1K > 1K < 1 Na + + F  Na + + PbF 2 Na + + H 2 O F  + F  F  + PbF 2 F  + H 2 O PbF 2 + PbF 2 PbF 2 + H 2 O H 2 O + H 2 O

34. 7/15/2015 Equation I D I R I  M C E If there was a favorable reaction, its equation would be written here. 00000 0.0200 lots 0 Assume one liter of solution 0.0200 mol / 1.00 L 0.0200 lots 0 NaF Na + F¯ H2OH2OPbF 2 Pb 2+ 0.0200 0 0 lots -0.0200+0.0200 0 lots 0 NaFNa + + F¯ 0 0 0.0200 lots 0

35. 7/15/2015 What was the goal of the calculation? The molar solubility of lead(II) fluoride in a solution containing 0.0200 M sodium fluoride. That is, how many moles of lead(II) fluoride dissolve in a liter of solution containing 0.0200 M sodium fluoride?

36. 7/15/2015 Equation I D I R I  M C E NaF Na + F¯ H2OH2OPbF 2 Pb 2+ Write the equilibrium reaction here. PbF 2 Pb 2+ +2 F¯ Let -x equal the moles of PbF 2 which dissolve per liter of solution. -x+x+2x 0.0200.0200+2x lots-x x lots 00000 0.0200 lots 0 0.0200 lots 0 Assume one liter of solution 0.0200 0 0 lots -0.0200+0.0200 0 lots 0 NaFNa + + F¯ 0 0 0.0200 lots 0

37. 7/15/2015 The Molar Solubility of Lead(II) Fluoride in 0.0200 M Sodium Fluoride PbF 2 (s)Pb 2+ (aq) + 2 F – (aq) 0.0200 -x+x+2x K sp = [Pb 2+ ] [F – ] 2 = 3.7 x 10 –8 M 3 K sp = (x) (0.0200 + 2x) 2 = 3.7 x 10 –8 M 3 x = 9.2 x 10 –5 M x = 0.023 g in one liter

38. 7/15/2015 Group Activity Why is the molar solubility of lead(II) fluoride different in a solution containing sodium fluoride?

39. 7/15/2015 Group Activity  Why is the molar solubility of lead(II) fluoride different in a solution containing sodium fluoride? Explain in terms of Le Châtelier’s Principle PbF 2 (s)Pb 2+ (aq) + 2 F – (aq)

40. Applications

41. 7/15/2015 Selective Precipitation A solution contains 0.100 M cadmium ion 0.050 M iron(III) ion 0.200 M nickel(II) ion If sodium hydroxide is slowly added to the solution What precipitates will form? In what order will the precipitates form?

42. 7/15/2015 Selective Precipitation Strategy  Determine the hydroxide ion concentration needed to precipitate each metal.  Determine the order in which those hydroxide ion concentrations will be reached. 0.100 M cadmium ion 0.050 M iron(III) ion 0.200 M nickel(II) ion

43. 7/15/2015 Hydroxide Ion Concentration Needed to Precipitate 0.100 M Cd 2+ Cd(OH) 2 (s)Cd 2+ (aq) + 2 OH – (aq) [Cd 2+ ] [OH – ] 2 = K sp [0.100] [OH – ] 2 = 2.0 x 10 –14 M 3 [OH – ] = 2.0 x 10 –14 M 3 [0.100]      1/2 [OH – ] = 4.5 x 10 –7 M

44. 7/15/2015 Hydroxide Ion Concentration Needed to Precipitate  0.100 M Cd 2+ [OH – ] = 4.5 x 10 –7 M  0.050 M Fe 3+ [OH – ] = 1.1 x 10 –12 M  0.200 M Ni 2+ [OH – ] = 2.8 x 10 –8 M As the solution is slowly made basic, which precipitates first? 10 –5 10 –6 10 –7 10 –8 10 –9 10 –10 10 –11 10 –12 10 –13 10 –14 [OH – ]  Cd(OH) 2  Fe(OH) 3  Ni(OH) 2 Basic

45. 7/15/2015 Selective Precipitation At what pH will the last metal ion begin to precipitate? 9 8 7 6 5 4 3 2 1 0 pH  7.65 When the cadmium ion begins to precipitate how much of the iron(III) ion is still in solution? 10 –5 10 –6 10 –7 10 –8 10 –9 10 –10 10 –11 10 –12 10 –13 10 –14 [OH – ]  Cd(OH) 2  Ni(OH) 2  Fe(OH) 3

46. 7/15/2015 Selective Precipitation  0.100 M cadmium ion begins to precipitate when the hydroxide ion concentration is 4.5 x 10 –7 M.  What is the maximum iron(III) ion concentration under these conditions?

47. 7/15/2015 Selective Precipitation Fe(OH) 3 (s)Fe 3+ (aq) + 3 OH – (aq) [Fe 3+ ] [OH – ] 3 = K sp [Fe 3+ ] (4.5 x 10 –7 M) 3 = 6.0 x 10 –38 M 4 [Fe 3+ ] = 6.0 x 10 –38 M 4 (4.5 x 10 –7 M ) 3 [Fe 3+ ] = 6.6 x 10 –19 M

48. 7/15/2015 Determination of Molar Solubility and K sp  The Handbook of Chemistry and Physics indicates that the solubility of zinc fluoride (FW 103.37) is 1.62 grams per 100 mL of water. What is the molar solubility of zinc fluoride? What is the solubility product of zinc fluoride?

49. 7/15/2015 Zinc Fluoride Molar Solubility 1.62 g /100 mL water 16.2 g /1000 mL water   16.2 g /1000 mL solution 16.2 g /103.37 g·mol –1 = 0.157 mol ZnF 2 The molar solubility of zinc fluoride is 0.157 M.

50. 7/15/2015 Zinc Fluoride Solubility Product Zn(F) 2 (s)Zn 2+ (aq) + 2 F – (aq) K sp = [Zn 2+ ] [F – ] 2 K sp = (0.157 M) (2 x 0.157 M) 2 Two moles of fluoride ion form for every zinc ion. K sp = (0.157 M) (0.314 M) 2 K sp = 1.54 x 10 –2 M 3

51. 7/15/2015

52. or     

53. 7/15/2015