1. K sp and Solubility Equilibria

2. Saturated solutions of salts are another type of chemical equilibrium. Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution.

3. When the solid is first added to water, no ions are initially present. As dissociation continues, the concentration of aqueous ions increases until equilibrium is reached.

4. This process can be represented by the solubility product constant or Ksp expression.

5. Even “insoluble” salts dissociate a little – their K sp values range from 10 -10 to 10 -50. A K sp value is unique to a given salt at a given temperature. Why would a change in temperature alter the value of K sp ?

6. Solubility mol/L g/L mg/L Solubility indicates the amount of salt that dissociates to form a saturated solution – think solubility curve! In essence, it indicates the equilibrium position for a given set of conditions. You can have different solubilities with the same K sp.

7. What you’ll need to be able to do Write K sp expresssions Calculate K sp given solubility Calculate solubility given K sp Compare solubilities of different salts

8. What you’ll need to be able to do Calculate the effect of a common ion or pH on solubility Determine if a precipitate will form given concentrations of ions and Ksp Determine the order of precipitation in a mixture of ions

9. Write K sp expresssions Write the dissocation equation first! Write the K sp expression – leaving out the solid Note: some of these have quite large exponents!

10. For a saturated solution of AgCl, the equation would be: AgCl (s)  Ag + (aq) + Cl - (aq) The solubility product expression would be: K sp = [Ag + ] [Cl - ]

11. For a saturated solution of Bi 2 S 3, the equation would be: Bi 2 S 3 (s)  2 Bi +3 (aq) + 3S -2 (aq) The solubility product expression would be: K sp = [Bi +3 ] 2 [S -2 ] 3

12. NiCO 3 NiCO 3 (s)  Ni +2 (aq) + CO 3 -2 (aq) K sp = [Ni +2 ] [CO 3 -2 ] Ag 2 SO 4 Ag 2 SO 4 (s)  2 Ag + (aq) + SO 4 -2 (aq) K sp = [Ag + ] 2 [SO 4 -2 ] Write Ksp expresssions

13. Calculate Ksp given solubility Example: Lead (II) chloride dissolves to a slight extent in water according to the equation: PbCl 2  Pb +2 + 2Cl - Calculate the K sp if the lead ion concentration has been found to be 1.62 x 10 -2 M.

14. Consider the equation, if lead’s concentration is “x”, then chloride’s concentration is “2x”. So.... K sp = [Pb +2 ] [Cl - ] 2 K sp = (1.62 x 10 -2 )(3.24 x 10 -2 ) 2 = 1.70 x 10 -5 PbCl 2  Pb +2 + 2Cl -

15. Example: When silver sulfide dissolves at 25 o C, the equilibrium concentration of silver ion is 5.8 x 10 -17 M. What is the K sp of silver sulfide? Ag 2 S  2 Ag + + S -2

16. Sulfide ion concentration is only ½ of silver’s So.... Ksp = [Ag + ] 2 [S -2 ] Ksp = (5.8 x 10 -17 ) 2 (2.9 x 10 -17 ) = 9.8 X 10 -50

17. Copper(I) bromide has a measured solubility of 2.0 X 10 -4 mol/L at 25°C. Calculate its K sp value. K sp = 4.0 X 10 -8

18. Calculate the K sp value for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0 X 10 -15 mol/L at 25°C. K sp = 1.1 X 10 -73

19. Write the dissociation equation Use the equation to consider the amount of ions given that x of the solid dissociates Write the Ksp expression and substitute your x values and solve Pay attention to freaky powers and roots! Calculate solubility given K sp

20. Copper(II) iodate has a Ksp of 1.4 X 10 -7 @ 20 o C. What is the molar solubility of the salt? Cu(IO 3 ) 2 (s)  Cu 2+ + 2 IO 3 - -x+x+2x K sp = [Cu +2 ] [IO 3 - ] 2 = (x) (2x) 2 = 4x 3 K sp = 1.4 X 10 -7 = 4x 3 x = 3.3 X 10 -3 M

21. In a saturated solution of silver carbonate, what is the molar solubility of the salt? Ksp = 8.1 X 10 -12 Ag 2 CO 3 (s)  2 Ag + + CO 3 -2 -x+2x+x K sp = [Ag + ] 2 [CO 3 -2 ] = (2x) 2 (x) = 4x 3 K sp = 8.1 X 10 -12 = 4x 3 x = 1.3 X 10 -4 M

22. The K sp for CaCO 3 is 3.8 x 10 -9 @ 25°C. Calculate the solubility of calcium carbonate in pure water in a) moles per liter b) grams per liter Calculate solubility given K sp

23. The relative solubilities can be deduced by comparing values of K sp. BUT, BE CAREFUL! These comparisons can only be made for salts having the same ION:ION ratio. Comparing Solubilities

24. Which salt is more soluble? Ag 2 SKsp = 1.0 X 10 -49 Ni(CN) 2 Ksp = 3.0 X 10 -23 Ag 2 S  2 Ag + + S -2 Ni(CN) 2  Ni +2 + 2 CN - Since both make 3 ions (4x 3 ) – the larger Ksp is the more soluble salt - Ni(CN) 2 !

25. Comparing Solubilities Which salt is more soluble? PbCl 2 Ksp = 1.6 X 10 -5 PbBr 2 Ksp = 4.6 X 10 -6 PbI 2 Ksp = 1.4 X 10 -8

26. Comparing Solubilities Rank the following in order of increasing solubility: BaF 2 Ksp = 1.7 X 10 -6 BaCO 3 Ksp = 8.1 X 10 -9 Ag 2 CO 3 Ksp = 8.1 X 10 -12

27. Comparing Solubilities Rank the following in order of increasing solubility: BaF 2 Ksp = 1.7 X 10 -6 s = BaCO 3 Ksp = 8.1 X 10 -9 s = Ag 2 CO 3 Ksp = 8.1 X 10 -12 s= BaCO 3 Ag 2 CO 3 BaF 2

28. With some knowledge of the reaction quotient (Q), we can decide 1) whether a ppt will form, AND Determine if a precipitate will form

29. With some knowledge of the reaction quotient (Q), we can decide 1) whether a ppte will form, AND 2) what concentrations of ions are required to begin the ppte. of an insoluble salt. Determine if a precipitate will form

30. 1. Q = K sp, the system is at equil. (saturated) 2. Q < K sp, the system is not at equil. (unsaturated – shift right) 3. Q > K sp, the system is not at equil. (supersaturated – shift left) Determine if a precipitate will form

31. Precipitates form when the solution is supersaturated! Determine if a precipitate will form

32. Substitute molarities into Q Compare Q to Ksp If precipitation occurs, consider the stoichiometry and limiting reactant Readjust to equilibrium Look at sample 15.16 on P. 766 Determine if a precipitate will form

33. Determining Precipitation Conditions A solution is prepared by adding 750.0 mL of 4.00 X 10 -3 M Ce(NO 3 ) 3 to 300.0 mL of 2.00 X 10 -2 M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 X 10 -10 ) precipitate from this solution? Yes!

34. A solution is prepared by mixing 150.0 mL of 1.00 X 10 -2 M Mg(NO 3 ) 2 and 250.0 mL of 1.00 X 10 -1 M NaF. Calculate the concentrations of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp = 6.4 X 10 -9 ).

35. Solution [Mg 2+ ] = 2.1 X 10 -6 M [F - ] = 5.50 X 10 -2 M

36. Why Would I Ever Care About K sp ??? Keep reading to find out ! Actually, very useful stuff!

37. Solubility, Ion Separations, and Qualitative Analysis …introduce you to some basic chemistry of various ions. …illustrate how the principles of chemical equilibria can be applied.

38. Selective Precipitation Selective Precipitation A solution of 0.10 M potassium carbonate was added to a mixture of barium and silver ions. Which precipitate will form first? BaCO 3 K sp = 8.1 X 10 -9 Ag 2 CO 3 K sp = 8.1 X 10 -12

40. Separate the following metal ions: silver, lead, cadmium and nickel

41. From solubility rules, lead and silver chloride will ppt, so add dilute HCl. Nickel and cadmium will stay in solution.

42. Separate by filtration: Lead chloride will dissolve in HOT water… filter while HOT and those two will be separate.

43. Cadmium and nickel are more subtle. Use their K sp ’s with sulfide ion. Who ppt’s first???

45. Precipitation of Insoluble Salts Metal-bearing ores often contain the metal in the form of an insoluble salt, and, to complicate matters, the ores often contain several such metal salts.

46. Dissolve the metal salts to obtain the metal ion, concentrate in some manner, and ppt. selectively only one type of metal ion as an insoluble salt. Precipitation of Insoluble Salts

47. Selective Precipitation Selective Precipitation A solution contains 1.0 X 10 -4 M Cu + and 2.0 X 10 -3 M Pb 2+. If a source of I - is added gradually to this solution, will PbI 2 (K sp = 1.4 X 10 -8 ) or CuI (K sp = 5.3 X 10 -12 ) precipitate first?

48. Specify the concentration of I - necessary to begin precipitation of each salt.

49. Solution CuI will precipitate first. Concentration in excess of 5.3 X 10 -8 M required.