1. Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Solubility Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 16.1

2. Chemistry 1011 Slot 52 16.1 Solubility Equilibrium YOU ARE EXPECTED TO BE ABLE TO: Write an expression for the solubility product constant, K sp, for a substance Calculate the concentration of ions at equilibrium, given K sp Calculate the solubility product constant for a substance given its solubility and formula Predict whether a combination of ions will form a precipitate, given K sp and ion concentrations Calculate the solubility of a substance in water, given K sp Use Le Chatelier’s Principle to determine the effect of adding a common ion to a solution. Calculate the solubility of a substance in the presence of a common ion

3. Chemistry 1011 Slot 53 Formation of Precipitates A precipitate is formed when –Two solutions are mixed, and –The cation from one solution combines with the anion from the other solution to form an insoluble solid NaCl (aq) + AgNO 3(aq) NaNO 3(aq) + AgCl (s) Na + (aq) + Cl  (aq) + Ag + (aq) + NO 3  (aq) Na + (aq) + NO 3  aq) + AgCl (s) Ag + (aq) + Cl  (aq) AgCl (s) An equilibrium is established between the solid and the corresponding ions in solution

4. Chemistry 1011 Slot 54 Solubility Equilibrium AgCl (s) Ag + (aq) + Cl  (aq) An expression can be written for the equilibrium constant: K sp = [ Ag + ]x[ Cl  ] [solid] does not appear in the equilibrium constant expression K sp is known as the solubility product constant Solubility product data are normally measured at 25 o C

5. Chemistry 1011 Slot 55 Determining Ion Concentrations Ag 3 PO 4(s) 3Ag + (aq) + PO 4 3  (aq) K sp = [Ag + ] 3 x[PO 4 3  ] = 1 x 10  16 PbCl 2(s) Pb 2+ (aq) + 2Cl  (aq) K sp = [Pb 2+ ]x[Cl  ] 2 = 1.7 x 10  5 Q: Calculate [Pb 2+ ] and [Cl  ] in a solution of PbCl 2 at 25 o C [Cl  ] = 2 x [Pb 2+ ] K sp = [Pb 2+ ] x [2Pb 2+ ] 2 = 1.7 x 10  5 K sp = 4 [Pb 2+ ] 3 = 1.7 x 10  5 [Pb 2+ ] = 1.6 x 10  2 mol/L [Cl  ] = 3.2 x 10  2 mol/L

6. Chemistry 1011 Slot 56 Calculating K sp The solubility of a salt can be determined by experiment K sp for the salt can be determined from these results Q: The solubility of magnesium hydroxide is found to be 8.4 x 10  4 g/100cm 3 at 25 o C. Find K sp Mg(OH) 2(s) Mg 2+ (aq) + 2OH  (aq) K sp = [Mg 2+ ]x[OH  ] 2 = ?? Solubility = 8.4 x 10  4 g/100cm 3 at 18 o C Solubility = (8.4 x 10  4 )g x 1000cm 3 /L = 1.44 x 10  4 mol/L 58.3 g/mol 100cm 3 [Mg 2+ ] = 1.44 x 10  4 mol/L; [OH  ] = 2.88 x 10  4 mol/L K sp = [Mg 2+ ]x[OH  ] 2 = 1.2 x 10  11

7. Chemistry 1011 Slot 57 Determining Precipitate Formation To determine whether a precipitate will form when two solutions are mixed: 1.Determine the concentrations of the reacting ions in the mixture 2.Calculate the ion product, P 3.Compare the ion product, P, with K sp 4.If P > K sp then precipitate will form 5.If P < K sp then no precipitate 6.If P = then no precipitate – solution is saturated

8. Chemistry 1011 Slot 58 Determining Precipitate Formation Q: Will a precipitate form when 5.0mL of 1.0 x 10  3 mol/L silver nitrate is added to 5.0mL of 1.0 x 10  5 mol/L potassium chromate? K sp Ag 2 CrO 4 = 1.0 x 10  12 2AgNO 3(aq) + K 2 CrO 4(aq) 2KNO 3(aq) + Ag 2 CrO 4(s) 2Ag + (aq) + CrO 4 2  (aq) Ag 2 CrO 4(s) K sp = [Ag + ] 2 x [CrO 4 2  ] = 1.0 x 10  12 [Ag + ] = 5.0 x 10  4 mol/L [CrO 4 2  ] = 5.0 x 10  6 mol/L Ion Product, P = [Ag + ] 2 x [CrO 4 2  ] = (5.0 x 10  4 ) 2 x (5.0 x 10  6 ) P = 1.25 x 10  12 P > K sp A precipitate will form

9. Chemistry 1011 Slot 59 Determining Precipitate Formation Q: 1.0 mL of 1.0 mol/L barium chloride is added to 10.0 mL of a solution containing a small amount of magnesium sulfate. Determine the minimum concentration of magnesium sulfate that will cause a precipitate to form. K sp BaSO4 = 1.1 x 10  10

10. Chemistry 1011 Slot 510 Determining Precipitate Formation 1.0 mL of 1.0 mol/L barium chloride is added to 10.0 mL of a solution containing a small amount of magnesium sulfate. Determine the minimum concentration of magnesium sulfate that will cause a precipitate to form. K sp BaSO4 = 1.1 x 10  10 BaCl 2(aq) + MgSO 4(aq) MgCl 2(aq) + BaSO 4(s) Ba 2+ (aq) + SO 4 2  (aq) BaSO 4(s) K sp = [Ba 2+ ]x[SO 4 2  ] = 1.1 x 10  10 [Ba 2+ ] = 1.0 x 10  3 L x 1.0 mol/L  1.00 x 10  2 L = 1.0 x 10  1 mol/L [SO 4 2  ] = x mol/L K sp = [Ba 2+ ]x[SO 4 2  ] = (1.0 x 10  1 ) x (x) = 1.1 x 10  10 x = [SO 4 2  ] = minimum [MgSO 4 ] =1.1 x 10  10 mol/L

11. Chemistry 1011 Slot 511 Selective Precipitation Suppose that a solution contains two different cations, for example Ba 2+ and Ca 2+ Each forms an insoluble sulfate, BaSO 4 and CaSO 4 K sp BaSO 4 = 1.1 x 10  10 K sp CaSO 4 = 7.1 x 10  5 If sulfate ions are added to a solution containing equal amounts of Ba 2+ and Ca 2+, then the BaSO 4 will precipitate first Only when the Ba 2+ ion concentration becomes very small will the SO 4 2  ion concentration rise to the point that CaSO 4 will be precipitated

12. Chemistry 1011 Slot 512 Determining Solubility The solubility, s, of a salt can be determined from K sp data Q: Determine the solubility of lead chloride in water at 25 o C. K sp = 1.7 x 10  5 Let solubility of lead chloride = s mol/L For every mole of PbCl 2 that dissolves, 1 mole of Pb 2+ (aq) and 2 moles of Cl  (aq) are formed PbCl 2(s) Pb 2+ (aq) + 2Cl  (aq) [Pb 2+ ] = s mol/L [Cl  ] = 2 x [Pb 2+ ] = 2s mol/L K sp = [Pb 2+ ]x[Cl  ] 2 = (s) x(2s) 2 = 1.7 x 10  5 4s 3 = 1.7 x 10  5 s = 1.6 x 10  2 mol/L (Can also be expressed in grams/Litre)

13. Chemistry 1011 Slot 513 The Common Ion Effect The presence of a common ion will reduce the solubility of an ionic salt (Le Chatelier) If a common ion is added to a saturated solution of a salt, then the salt will be precipitated (Le Chatelier) For example, CaCO 3 is less soluble in a solution containing CO 3 2  ions than in pure water CaCO 3(s) Ca 2+ (aq) + CO 3 2  (aq) K sp = [Ca 2+ ] x [CO 3 2  ] = 4.9 x 10  9

14. Chemistry 1011 Slot 514 The Common Ion Effect CaCO 3(s) Ca 2+ (aq) + CO 3 2  (aq) K sp = [Ca 2+ ] x [CO 3 2  ] = 4.9 x 10  9 Solubility of CaCO 3 = [Ca 2+ ] In pure water [Ca 2+ ] = [CO 3 2  ] =  (4.9 x 10  9 ) = 7.0 x 10  5 mol/L Solubility of CaCO 3 in 1.0 x 10  1 sodium carbonate solution = ??? [CO 3 2  ] = 1.0 x 10  1 mol/L (ignore CO 3 2  from CaCO 3 ) [Ca 2+ ] = K sp = 4.9 x 10  9 = 4.9 x 10  8 mol/L [CO 3 2  ] 1.0 x 10  1

15. Chemistry 1011 Slot 515 One More Example K sp for manganese II hydroxide is 1.2 x 10  11 Solid sodium hydroxide is added slowly to a 0.10 mol/L solution of manganese II chloride. What will be the pH when a precipitate forms? Mn(OH) 2(s) Mn 2+ (aq) + 2OH  (aq) K sp = [Mn 2+ ] x [OH  ] 2 = 1.2 x 10  11 [Mn 2+ ] = 0.10 mol/L [OH  ] =  K sp  [Mn 2+ ] =  1.2 x 10  11  0.10 = 1.1 x 10  5 mol/L pH = 9.0