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

Chemistry 1011 Slot 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

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

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

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

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

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

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

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

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

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

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)

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

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

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