1 / 78

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

colleen-lopez
Télécharger la présentation

Ionic Equilibria III: The Solubility Product Principle

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related