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Chapter 17 Solubility Equilibria

Chapter 17 Solubility Equilibria. Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1011.php. Solubility Equilibria. Dissolution M m X x (s)  m M n+ (aq) + x X y- (aq) Precipitation m M n+ (aq) + x X y- (aq)  M m X x (s)

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Chapter 17 Solubility Equilibria

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  1. Chapter 17Solubility Equilibria Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1011.php

  2. Solubility Equilibria • DissolutionMmXx (s) m Mn+ (aq) + x Xy- (aq) • Precipitation m Mn+ (aq) + x Xy- (aq)MmXx (s) • For a dissolution process, we give the equilibrium constant expression the name solubility product (constant) Ksp. For • MmXx (s) m Mn+ (aq) + x Xy- (aq) • Ksp= [Mn+]m [Xy-]x

  3. Problem • Write the expressions for the solubility product constantKsp of: • a) AgCl • b) PbI2 • c) Ca3(PO4)2 • d) Cr(OH)3

  4. Measuring Ksp and Calculating Molar Solubility from Ksp • For the dissolution of solid CaF2 in water, we might find that the concentrations of the ions Ca2+ and F- at equilibrium are • [Ca2+] = 3.3 x 10-4 mol/L • [F-] = 6.7 x 10-4 mol/L • Ksp= [Ca2+][F-]2 • = (3.3 x 10-4 mol/L)(6.7 x 10-4 mol/L)2 • = 1.5 x 10-10

  5. Measuring Ksp and Calculating Molar Solubility from Ksp • We could also find the Ksp of a solid by mixing solutions of known concentrations of the ions, leading to the precipitationof the solid. • When the system reaches equilibrium we can measure the ion concentrations to calculate Ksp.

  6. The advantage of using the dissolution of the solid method is that a clear relationship between the concentrations of the ions exists, based upon their relative stoichiometry in the solid.

  7. If we mix two separate solutions that are sources of calcium ions and fluoride ions, no such relationship exists, and we can have infinitely many mixtures that still obey the solubility product expression. • Since Ksp values are equilibrium constants, they will change with temperature!

  8. Molar solubility and saturation • If we know the Kspvalue for a solid, we can calculate its molar solubility, which is the number of moles of the solidthat can dissolve in the solvent before the solution becomessaturated (no more solid will dissolve). • Saturated means equilibrium!

  9. Problem • A saturated solution of Ca3(PO4)2 has • [Ca2+] =2.01 x 10-8 mol/L and • [PO43-] = 1.6 x 10-5 mol/L. • Calculate Kspfor Ca3(PO4)2

  10. Problem • If a saturated solution of BaSO4 is prepared by dissolving solid BaSO4 in water, and [Ba2+] = 1.05 x 10-5 mol/L, what is the Ksp for BaSO4?

  11. Problem • Which has the greater molar solubility? • AgCl with Ksp = 1.8 x 10-10 • or • Ag2CrO4 with Ksp = 1.1 x 10-12

  12. Factors that Affect Solubility • The Common-Ion Effect • If a solution already has a significant concentration of a common-ion the solution will dissolve LESS of the solid than the same volume of pure water can.

  13. The Common-Ion Effect • MmXx (s) m Mn+ (aq) + x Xy- (aq) • Ksp= [Mn+]m [Xy-]x • Imagine that instead of dissolving a solid in a solution of common-ion that we add a common ion to a solution created by dissolving the solid in pure water.

  14. The Common-Ion Effect • MmXx (s) m Mn+ (aq) + x Xy- (aq) • Ksp= [Mn+]m [Xy-]x • From Le Chatalier’s Principle, we know if we add one of the product ions (the common ion we added), the stress on the equilibrium is this added product. To relieve the stress the equilibrium will shift towards the reactantsmeaning more solid is created.

  15. Problem • Calculate the molar solubility of MgF2 • (Ksp = 7.4 x 10-11) • in pure water and • in 0.10 mol/L MgCl2.

  16. Precipitation of Ionic Compounds • Can we predict if a solid will precipitate if we mix two solutions of different ions? • YES! • Consider the mixing of a solution of Ca2+ ions and a solution of F- ions. A precipitation of solid is the dissolution reaction in reverse, so we can express the reaction as • CaF2 (s) Ca2+ (aq) + 2 F- (aq) Ksp = [Ca2+][F-]2

  17. Precipitation of Ionic Compounds • When we mix the solutions, the system is most likelynot at equilibrium. • For solid dissolution / precipitation reactions, we use a procedure similar to the reaction quotient Qc to define the ion product (IP) or Qsp. • Qsp = [Ca2+][F-]2

  18. If Qsp = Ksp, the solution is saturated, and the system is at equilibrium. If Qsp > Ksp, the solution is supersaturated, so the system is not at equilibrium. The concentration of the ionsis greater than it would be at equilibrium, and so the reaction proceeds from ions towards the solid. We expect precipitation to occur!

  19. If Qsp< Ksp, the solution is unsaturated, so the system is not at equilibrium. The concentration of the ionsis less than it would be at equilibrium, and so we expect more solid to be able to dissolve in this solution!

  20. Will a precipitate form when 0.150 L of 0.10 molL-1 Pb(NO3)2 and 0.100 L of 0.20 molL-1 NaClare mixed? • Ksp of PbCl2 is1.2 x 10-5 Be Careful!

  21. Problem • Will a precipitate form on mixing equal volumes of the following solutions? a) 3.0 x 10-3 mol/L BaCl2 and 2.0 x 10-3 mol/L Na2CO3 Be Careful! b) 1.0 x 10-5 mol/L Ba(NO3)2 and 4.0 x 10-5 mol/L Na2CO3 (Ksp of BaCO3 is2.6 x 10-9)

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