1. E 12 Water and Soil Solve problems relating to removal of heavy –metal ions and phosphates by chemical precipitation http://www.chem.purdue.edu/gchelp/howtosolveit/equilibrium/solubility_products.htm

2. Chemical Precipitation Many metal ions form insoluble / sparingly soluble salts Chemical precipitation of heavy metals and phosphates

3. Chemical Precipitation -1 Use of Hydrogen sulfide Removes sulfides of Hg, Cd, Pb, Zn Hg 2+ (aq) + H 2 S(aq) HgS(s) Pb 2+ (aq) + H 2 S(aq) PbS(s)

4. Chemical Precipitation -2 Use of Hydroxide to remove copper Cobalt Iron Cu 2+ (aq) + 2 OH - (aq) Cu(OH) 2 (s)

5. Solubility Rules

8. Quantitative Aspect of Solubility Equilibria

9. Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s)Ba 2+ (aq) + SO 4 2− (aq)

10. Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2− ] = 1.0 x 10 -10 mol 2 dm -6 at 25 ○ C where the equilibrium constant, K sp, is called the solubility product constant

11. Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M). Temperature dependent

12. Solubility Product Constant Concentration of solid is not in the expression Solution is saturated Precipitation depends on K sp

13. Will a Precipitate Form? In a solution, –If Q = K sp, the system is at equilibrium and the solution is saturated. –If Q < K sp, more solid will dissolve until Q = K sp. –If Q > K sp, the salt will precipitate until Q = K sp.

14. Solubility Product

15. Example 1 K sp = [Ba 2+ ] [SO 4 2− ] = 1.0 x 10 -10 mol 2 dm -6 at 25 ○ C X = 1.0 x 10 -5 mol dm -3

16. Factors Affecting Solubility The Common-Ion Effect –If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. BaSO 4 (s)Ba 2+ (aq) + SO 4 2− (aq)

17. Example 2 Consider solubility of BaSO 4 in 0.10 M sodium sulfate, strong electrolyte Calculate the con. Of Ba 2+ ions BaSO 4 Ba 2+ SO 4 2- Solubility of Barium sulfate = Y mol / dm3

18. K sp = [Ba 2+ ] [SO 4 2− ] = 1.0 x 10 -10 mol 2 dm -6 at 25 ○ C = 0.10

19. Application of Precipitation in Soil and Water Chemistry PPT of calcium ions as CaSO 4 by addition of sodium sulfate K sp = [Ca 2+ ] [SO 4 2− ] = 3.0 x 10 -5 mol 2 dm -6 at 25 ○ C

20. Lead as Lead Chloride Pb 2+ (aq) + 2 Cl - (aq) Pb(Cl) 2 (s) K sp = [Pb 2+ ] [Cl − ] 2 = 1.73 x 10 -7 mol 3 dm -9 at 25 ○ C

22. Lead as Lead Sulfate Pb 2+ (aq) + SO 4 2- (aq) PbSO 4 (s) K sp = [Pb 2+ ] [SO 4 2- ] = 6.3 x 10 -7 mol 2 dm -6 at 25 ○ C

23. Calcium as calcium phosphate 3 Ca 2+ (aq) + 2 PO 4 3- (aq) Ca 3 (PO 4 ) 2 (s) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 = 6.3 x 10 -7 mol 2 dm -6 at 25 ○ C

24. Phosphate as Aluminum or Iron or Calcium Fe 3+ (aq) + PO 4 3- (aq) FePO 4 (s)

25. Arsonate, AsO 4 3- Removal by addition of : Aluminum nitrate Iron(III)chloride Iron(III)sulfate Al 3+ (aq) + AsO 4 3- (aq) AlAsO 4(s)

26. Examples 1 Write the solubility product constant expression for the following compounds and indicate K sp value in terms of X if X is its molar solubility

27. Example A) AgCl B) PbCl 2 C) As 2 S 3 D) Ca 3 (PO 4 ) 2 E) Fe(OH) 3

28. Silver chloride AgCl (s) Ag + (aq) + Cl - (aq) x x K sp = [ Ag + ][ Cl - ] = x 2

29. Calcium Phosphate Ca 3 (PO 4 ) 2 (s) 3 Ca 2+ (aq) + 2 PO 4 3- (aq) 3x 2x K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 (3x) 3 (2x) 2 = 108 x 5

30. Example 2 A) Calcium ions present in hard water are ppted by adding sulfate ions. Write the net ionic equation for the reaction

31. 2 Ca 2+ (aq) + SO 4 2- (aq) CaSO 4 (s) B) Given the Ksp, = 3.0 x 10 -5 mol 2 dm -6 at 25 deg. C; calculate its molar solubility in water at 25 deg. C

32. Ex B CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) Ksp = [ Ca 2+ ][ SO 4 2- ] x x = x 2 = 3.0 x 10 -5 mol 2 dm -6 x = 5.5 x 10 -3 mol dm -3

33. Ex C Determine if a ppt will form when its ion concentrations are [Ca 2+ ] = 1.0 x 10 -3 mol dm -3 [SO 4 2- ] = 1.0 x 10 -2 mol dm -3 B) Given the Ksp, = 3.0 x 10 -5 mol 2 dm -6 at 25 deg. C;

34. Ex 1 C K sp = [Ca 2+ ] [SO 4 2- ] (x) (x) = x 2 = ( 1.0 x 10 -3 mol dm -3 ) (1.0 x 10 -2 mol dm -3 ) = 1.0 x 10 -5 < Ksp Ppt will not form

35. Ex1D Calculate minimum ion of sulfate required to ppt. [Ca 2+ ] = 0.001 M [Sulfate] = x = ksp Calculate for X

36. Ex3

37. Ex4

38. Ex5

39. Applications of Ksp

40. 4.A 200.0 mL sample of a saturated solution of Mg(OH) 2 weighs 222.1210 g. When the beaker containing the solution is evaporated to dryness it weighs 22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp. Mass of Beaker + Mg(OH) 2 22.1213 g -Mass of Beaker-22.1200 g Mass of Mg(OH) 2 0.0013 gnote sig figs s=0.0013 gx1 mole 58.3 g 0.2000 L =1.1149 x 10 -4 M Mg(OH) 2 ⇌ Mg 2+ + 2OH - ss2s Ksp=[Mg 2+ ][OH - ] 2 = [s][2s] 2 = 4s 3 =4(1.1149 x 10 -4 ) 3 =5.5 x 10 -12

41. Ksp Solubility Product Special Keq Saturated solutions No Units Increasing Temperature increases the Ksp

42. 4.A 200.0 mL sample of a saturated solution of Mg(OH) 2 weighs 222.1210 g. When the beaker containing the solution is evaporated to dryness it weighs 22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp. Mass of Beaker + Mg(OH) 2 22.1213 g -Mass of Beaker-22.1200 g Mass of Mg(OH) 2 0.0013 gnote sig figs s=0.0013 gx1 mole 58.3 g 0.2000 L =1.1149 x 10 -4 M Mg(OH) 2 ⇌ Mg 2+ + 2OH - ss2s Ksp=[Mg 2+ ][OH - ] 2 = [s][2s] 2 = 4s 3 =4(1.1149 x 10 -4 ) 3 =5.5 x 10 -12

43. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1.The solubility (s) of BaCO 3 is 5.1 x 10 -5 M @ 25 0 C. Calculate the solubility product or Ksp. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ksp = [Ba 2+ ][CO 3 2- ] Ksp = [s][s] Ksp = s 2 Ksp = (5.1 x 10 -5 ) 2 Ba 2+ CO 3 2- BaCO 3(s)

44. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1.The solubility (s) of BaCO 3 is 5.1 x 10 -5 M @ 25 0 C. Calculate the solubility product or Ksp. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ksp = [Ba 2+ ][CO 3 2- ] Ksp = [s][s] Ksp = s 2 Ksp = (5.1 x 10 -5 ) 2 Ksp = 2.6 x 10 -9 Ba 2+ CO 3 2- BaCO 3(s)

45. 3.If 0.00243 g of Fe 2 (CO 3 ) 3 is required to saturate 100.0 mL of solution. What is the solubility product? Fe 2 (CO 3 ) 3 ⇌ 2Fe 3+ +3CO 3 2- s2s3s s=0.00243 g x 1 mole 291.6 g 0.100 L =8.333 x 10 -5 M Ksp = [Fe 3+ ] 2 [CO 3 2- ] 3 Ksp = [2s] 2 [3s] 3 Ksp = 108s 5 Ksp = 108(8.333 x 10 -5 ) 5 Ksp = 4.34 x 10 -19

46. Factors Affecting Solubility pH –If a substance has a basic anion, it will be more soluble in an acidic solution. –Substances with acidic cations are more soluble in basic solutions.

47. Factors Affecting Solubility Complex Ions –Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

48. Factors Affecting Solubility Complex Ions –The formation of these complex ions increases the solubility of these salts.

49. Factors Affecting Solubility Amphoterism –Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. –Examples of such cations are Al 3+, Zn 2+, and Sn 2+.

50. Amphoteric –Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. –Examples of such cations are Al 3+, Zn 2+, and Sn 2+.

51. Selective Precipitation of Ions

52. Precipitation One can use differences in solubility's of salts to separate ions in a mixture.