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Ionic Equilibria III: The Solubility Product Principle

20. Ionic Equilibria III: The Solubility Product Principle. Chapter Goals. Solubility Product Constants Determination of Solubility Product Constants Uses of Solubility Product Constants Fractional Precipitation Simultaneous Equilibria Involving Slightly Soluble Compounds

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Ionic Equilibria III: The Solubility Product Principle

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  1. 20 Ionic Equilibria III: The Solubility Product Principle

  2. Chapter Goals • Solubility Product Constants • Determination of Solubility Product Constants • Uses of Solubility Product Constants • Fractional Precipitation • Simultaneous Equilibria Involving Slightly Soluble Compounds • Dissolving Precipitates

  3. Solubility Product Constants • Silver chloride, AgCl,is rather insoluble in water. • Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

  4. Solubility Product Constants • Silver chloride, AgCl,is rather insoluble in water. • Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

  5. Solubility Product Constants • The equilibrium constant expression for this dissolution is called a solubility product constant. • Ksp = solubility product constant

  6. Solubility Product Constants • The solubility product constant, Ksp, for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. • Consider the dissolution of silver sulfide in water.

  7. Solubility Product Constants • The solubility product expression for Ag2S is:

  8. Solubility Product Constants • The dissolution of solid calcium phosphate in water is represented as: • The solubility product constant expression is: You do it!

  9. Solubility Product Constants • In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:

  10. Solubility Product Constants • The same rules apply for compounds that have more than two kinds of ions. • One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.

  11. Determination of Solubility Product Constants • Example 20-1: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl. • The molar solubility can be easily calculated from the data:

  12. Determination of Solubility Product Constants • The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are:

  13. Determination of Solubility Product Constants • Substitution of the molar concentrations into the solubility product expression gives:

  14. Determination of Solubility Product Constants • Example 20-2: One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2. • Calculate the molar solubility of CaF2.

  15. Determination of Solubility Product Constants • From the molar solubility, we can find the ion concentrations in saturated CaF2. Then use those values to calculate the Ksp. • Note: You are most likely to leave out the factor of 2 for the concentration of the fluoride ion!

  16. Uses of Solubility Product Constants • The solubility product constant can be used to calculate the solubility of a compound at 25oC. • Example 20-3: Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x 10-10.

  17. Uses of Solubility Product Constants

  18. Uses of Solubility Product Constants • Make the algebraic substitution of x’s into solubility product expression and solve for x, giving the ion concentrations.

  19. Uses of Solubility Product Constants • Finally, to calculate the mass of BaSO4 in 1.00 L of saturated solution, use the definition of molarity.

  20. Uses of Solubility Product Constants • Example 20-4: The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25oC. You do it!

  21. Uses of Solubility Product Constants • Be careful, do not forget the stoichiometric coefficient of 2!

  22. Uses of Solubility Product Constants • Substitute the algebraic expressions into the solubility product expression.

  23. Uses of Solubility Product Constants • Solve for the pOH and pH.

  24. The Common Ion Effect in Solubility Calculations • Example 20-5: Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?)

  25. The Common Ion Effect in Solubility Calculations • Write equations to represent the equilibria.

  26. The Common Ion Effect in Solubility Calculations • Substitute the algebraic representations of the concentrations into the Ksp expression and solve for x.

  27. The Common Ion Effect in Solubility Calculations • The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8M. • The molar solubility of BaSO4 in pure water is 1.0 x 10-5M. • BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate! • Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution! • If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

  28. The Reaction Quotient in Precipitation Reactions • The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form. • Example 20-6: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?

  29. The Reaction Quotient in Precipitation Reactions • Write out the solubility expressions.

  30. The Reaction Quotient in Precipitation Reactions • Calculate the Qsp for PbSO4. • Assume that the solution volumes are additive. • Concentrations of the important ions are:

  31. The Reaction Quotient in Precipitation Reactions • Finally, calculate Qsp for PbSO4 and compare it to the Ksp.

  32. The Reaction Quotient in Precipitation Reactions • Example 20-7: Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8M? For HgS, Ksp=3.0 x 10-53.

  33. The Reaction Quotient in Precipitation Reactions

  34. The Reaction Quotient in Precipitation Reactions • Example 20-8: Refer to example 20-7. What volume of the solution (1.0 x 10-8M Hg2+ ) contains 1.0 g of mercury?

  35. Fractional Precipitation • The method of precipitating some ions from solution while leaving others in solution is called fractional precipitation. • If a solution contains Cu+, Ag+, and Au+, each ion can be precipitated as chlorides.

  36. Fractional Precipitation • Example 20-9: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal chlorides.

  37. Fractional Precipitation

  38. Fractional Precipitation • Repeat the calculation for silver chloride.

  39. Fractional Precipitation • Finally, for copper (I) chloride to precipitate.

  40. Fractional Precipitation • These three calculations give the [Cl-] required to precipitate AuCl ([Cl-] >2.0 x 10-11 M), to precipitate AgCl ([Cl-] >1.8 x 10-8 M), and to precipitate CuCl ([Cl-] >1.9 x 10-5 M). • It is also possible to calculate the amount of Au+ precipitated before the Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before the Cu+ begins to precipitate.

  41. Fractional Precipitation • Example 20-10: Calculate the percentage of Au+ ions that precipitate before AgCl begins to precipitate. • Use the [Cl-] from Example 20-9 to determine the [Au+] remaining in solution just before AgCl begins to precipitate.

  42. The percent of Au+ ions unprecipitated just before AgCl precipitates is: Therefore, 99.9% of the Au+ ions precipitates before AgCl begins to precipitate. Fractional Precipitation

  43. Fractional Precipitation • A similar calculation for the concentration of Ag+ ions unprecipitated before CuCl begins to precipitate is:

  44. The percent of Ag+ ions unprecipitated just before AgCl precipitates is: Thus, 99.905% of the Ag+ ions precipitates before CuCl begins to precipitate. Fractional Precipitation

  45. Simultaneous Equilibria Involving Slightly Soluble Compounds • Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

  46. Simultaneous Equilibria Involving Slightly Soluble Compounds • Example 20-12: If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? • For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5. • Calculate Qsp for Mg(OH)2 and compare it to Ksp. • Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M. • Aqueous ammonia is a weak base that we can calculate [OH-].

  47. Simultaneous Equilibria Involving Slightly Soluble Compounds

  48. Simultaneous Equilibria Involving Slightly Soluble Compounds • Once the concentrations of both the magnesium and hydroxide ions are determined, the Qsp can be calculated and compared to the Ksp.

  49. Simultaneous Equilibria Involving Slightly Soluble Compounds • Example 20-13: How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? (Note the similarity between this problem and Example 20-12.) You do it! • Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+.

  50. Simultaneous Equilibria Involving Slightly Soluble Compounds

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