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Solubility Equilibrium in Chemistry

Learn how to calculate the solubility product constant, predict precipitate formation, and determine ion concentrations in solubility equilibrium.

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Solubility Equilibrium in Chemistry

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  1. Chemistry 1011 TOPIC Solubility Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 16.1 Chemistry 1011 Slot 5

  2. 16.1 Solubility Equilibrium YOU ARE EXPECTED TO BE ABLE TO: • Write an expression for the solubility product constant, Ksp, for a substance • Calculate the concentration of ions at equilibrium, given Ksp • Calculate the solubility product constant for a substance given its solubility and formula • Predict whether a combination of ions will form a precipitate, given Ksp and ion concentrations • Calculate the solubility of a substance in water, given Ksp • 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 5

  3. 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) + AgNO3(aq) NaNO3(aq) + AgCl(s) Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) Na+(aq) + NO3-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 5

  4. Solubility Equilibrium AgCl(s) Ag+(aq) + Cl-(aq) • An expression can be written for the equilibrium constant: Ksp = [Ag+]x[Cl-] • [solid] does not appear in the equilibrium constant expression • Ksp is known as the solubility product constant • Solubility product data are normally measured at 25oC Chemistry 1011 Slot 5

  5. Determining Ion Concentrations Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = [Ag+]3x[PO43-] = 1 x 10-16 PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+]x[Cl-]2 = 1.7 x 10-5 Q: Calculate [Pb2+] and [Cl-] in a solution of PbCl2 at 25oC [Cl-] = 2 x [Pb2+] Ksp = [Pb2+] x [2Pb2+]2 = 1.7 x 10-5 Ksp = 4 [Pb2+]3 = 1.7 x 10-5 [Pb2+] = 1.6 x 10-2 mol/L [Cl-] = 3.2 x 10-2 mol/L Chemistry 1011 Slot 5

  6. Calculating Ksp • The solubility of a salt can be determined by experiment • Ksp 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/100cm3 at 25oC. Find Ksp Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+]x[OH-]2 = ?? Solubility = 8.4 x 10-4 g/100cm3 at 18oC Solubility = (8.4 x 10-4)g x 1000cm3/L = 1.44 x 10-4 mol/L 58.3 g/mol 100cm3 [Mg2+] = 1.44 x 10-4 mol/L; [OH-] = 2.88 x 10-4 mol/L Ksp = [Mg2+]x[OH-]2 = 1.2 x 10-11 Chemistry 1011 Slot 5

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

  8. 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? Ksp Ag2CrO4 = 1.0 x 10-12 2AgNO3(aq) + K2CrO4(aq) 2KNO3(aq) + Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq) Ag2CrO4(s) Ksp = [Ag+]2 x [CrO42-] = 1.0 x 10-12 [Ag+] = 5.0 x 10-4 mol/L [CrO42-] = 5.0 x 10-6 mol/L Ion Product, P = [Ag+]2 x [CrO42-] = (5.0 x 10-4 )2 x (5.0 x 10-6 ) P = 1.25 x 10-12 P > Ksp A precipitate will form Chemistry 1011 Slot 5

  9. 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. Ksp BaSO4 = 1.1 x 10-10 Chemistry 1011 Slot 5

  10. 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. Ksp BaSO4 = 1.1 x 10-10 BaCl2(aq) + MgSO4(aq) MgCl2(aq) + BaSO4(s) Ba2+(aq) + SO42-(aq) BaSO4(s) Ksp = [Ba2+]x[SO42-] = 1.1 x 10-10 [Ba2+] = 1.0 x 10-3L x 1.0 mol/L  1.00 x 10-2L = 1.0 x 10-1 mol/L [SO42-] = x mol/L Ksp = [Ba2+]x[SO42-] = (1.0 x 10-1 ) x (x) = 1.1 x 10-10 x = [SO42-] = minimum [MgSO4] =1.1 x 10-10 mol/L Chemistry 1011 Slot 5

  11. Selective Precipitation • Suppose that a solution contains two different cations, for example Ba2+ and Ca2+ • Each forms an insoluble sulfate, BaSO4 and CaSO4 Ksp BaSO4 = 1.1 x 10-10 Ksp CaSO4 = 7.1 x 10-5 • If sulfate ions are added to a solution containing equal amounts of Ba2+ and Ca2+, then the BaSO4 will precipitate first • Only when the Ba2+ ion concentration becomes very small will the SO42- ion concentration rise to the point that CaSO4 will be precipitated Chemistry 1011 Slot 5

  12. Determining Solubility • The solubility, s, of a salt can be determined from Ksp data Q: Determine the solubility of lead chloride in water at 25oC. Ksp = 1.7 x 10-5 Let solubility of lead chloride = s mol/L • For every mole of PbCl2 that dissolves, 1 mole of Pb2+(aq) and 2 moles of Cl-(aq) are formed PbCl2(s) Pb2+(aq) + 2Cl-(aq) [Pb2+] = s mol/L [Cl-] = 2 x [Pb2+] = 2s mol/L Ksp = [Pb2+]x[Cl-]2 = (s) x(2s)2 = 1.7 x 10-5 4s3 = 1.7 x 10-5 s = 1.6 x 10-2 mol/L (Can also be expressed in grams/Litre) Chemistry 1011 Slot 5

  13. 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, CaCO3 is less soluble in a solution containing CO32- ions than in pure water CaCO3(s) Ca2+(aq) + CO32-(aq) Ksp = [Ca2+] x [CO32-] = 4.9 x 10-9 Chemistry 1011 Slot 5

  14. The Common Ion Effect CaCO3(s) Ca2+(aq) + CO32-(aq) Ksp = [Ca2+] x [CO32-] = 4.9 x 10-9 • Solubility of CaCO3 = [Ca2+] • In pure water [Ca2+] = [CO32-] = (4.9 x 10-9) = 7.0 x 10-5 mol/L Solubility of CaCO3 in 1.0 x 10-1 sodium carbonate solution = ??? • [CO32-] = 1.0 x 10-1 mol/L (ignore CO32- from CaCO3 ) • [Ca2+] = Ksp = 4.9 x 10-9 = 4.9 x 10-8 mol/L [CO32-] 1.0 x 10-1 Chemistry 1011 Slot 5

  15. One More Example • Ksp 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) Mn2+(aq) + 2OH-(aq) Ksp = [Mn2+] x [OH-]2 = 1.2 x 10-11 • [Mn2+] = 0.10 mol/L • [OH-] =  Ksp  [Mn2+] =  1.2 x 10-11 0.10 = 1.1 x 10-5 mol/L • pH = 9.0 Chemistry 1011 Slot 5

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