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K sp and Solubility Equilibria

K sp and Solubility Equilibria. Saturated solutions of salts are another type of chemical equilibrium. Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution. . When the solid is first added to water, no ions are initially present.

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K sp and Solubility Equilibria

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  1. Ksp and Solubility Equilibria

  2. Saturated solutions of salts are another type of chemical equilibrium. Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution.

  3. When the solid is first added to water, no ions are initially present. As dissociation continues, the concentration of aqueous ions increases until equilibrium is reached.

  4. This process can be represented by the solubility product constantor Ksp expression.

  5. Even “insoluble” salts dissociate a little – their Ksp values range from 10-10 to 10-50. A Ksp value is unique to a given salt at a given temperature. Why would a change in temperature alter the value of Ksp?

  6. Solubility mol/L g/L mg/L Solubility indicates the amount of salt that dissociates to form a saturatedsolution – think solubility curve! In essence, it indicates the equilibrium position for a given set of conditions.You can have different solubilities with the same Ksp.

  7. What you’ll need to be able to do • Write Ksp expresssions • Calculate Ksp given solubility • Calculate solubility given Ksp • Compare solubilities of different salts

  8. What you’ll need to be able to do • Calculate the effect of a common ion or pH on solubility • Determine if a precipitate will form given concentrations of ions and Ksp • Determine the order of precipitation in a mixture of ions

  9. Write Ksp expresssions • Write the dissocation equation first! • Write the Ksp expression – leaving out the solid Note: some of these have quite large exponents!

  10. For a saturated solution of AgCl, the equation would be: AgCl (s) Ag+ (aq) + Cl- (aq) • The solubility product expression • would be: • Ksp = [Ag+] [Cl-]

  11. For a saturated solution of Bi2S3, the equation would be: Bi2S3(s) 2 Bi+3 (aq) + 3S-2 (aq) • The solubility product expression • would be: • Ksp = [Bi+3]2 [S-2]3

  12. Write Ksp expresssions • NiCO3 NiCO3(s) Ni+2 (aq) + CO3-2 (aq) Ksp = [Ni+2] [CO3-2] • Ag2SO4 Ag2SO4(s) 2 Ag+ (aq) + SO4-2 (aq) Ksp = [Ag+]2 [SO4-2]

  13. Calculate Ksp given solubility • Example: • Lead (II) chloride dissolves to a slight extent in water according to the equation: • PbCl2 Pb+2 + 2Cl-Calculate the Ksp if the lead ion concentration has been found to be 1.62 x 10-2M.

  14. PbCl2 Pb+2 + 2Cl- Consider the equation, if lead’s concentration is “x” , then chloride’s concentration is “2x”. So. . . . Ksp = [Pb+2] [Cl-]2 Ksp = (1.62 x 10-2)(3.24 x 10-2)2 = 1.70 x 10-5

  15. Example: • When silver sulfide dissolves at 25oC, the equilibrium concentration of silver ion is 5.8 x 10-17M. What is the Ksp of silver sulfide? • Ag2S  2 Ag+ + S-2

  16. Ag2S  2 Ag+ + S-2 • Sulfide ion concentration is only ½ of silver’s • So. . . . Ksp = [Ag+]2 [S-2] • Ksp = (5.8 x 10-17)2 (2.9 x 10-17) • = 9.8 X 10-50

  17. Copper(I) bromide has a measured solubility of 2.0 X 10-4 mol/L at 25°C. Calculate its Ksp value. Ksp = 4.0 X 10-8

  18. Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 X 10-15 mol/L at 25°C. Ksp = 1.1 X 10-73

  19. Calculate solubility given Ksp Write the dissociation equation Use the equation to consider the amount of ions given that s of the solid dissociates Write the Ksp expression and substitute your s values and solve Pay attention to freaky powers and roots!

  20. Copper(II) iodate has a Ksp of 1.4 X 10-7 @ 20oC. What is the molar solubility of the salt? • Ksp = [Cu+2] [IO3-]2 = (s) (2s)2 = 4s3Ksp = 1.4 X 10-7 = 4s3 • s = 3.3 X 10-3M

  21. In a saturated solution of silver carbonate, what is the molar solubility of the salt? Ksp = 8.1 X 10-12 • Ksp = [Ag+]2 [CO3-2] = (2s)2 (s) = 4s3Ksp = 8.1 X 10-12 = 4s3 • s = 1.3 X 10-4 M

  22. Calculate solubility given Ksp The Ksp for CaCO3 is 3.8 x 10-9 @ 25°C. Calculate the solubility of calcium carbonate in pure water in a) moles per liter b) grams per liter

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