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## Ionic Equilibria III: The Solubility Product Principle

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**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 • Dissolving Precipitates**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.**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.**Solubility Product Constants**• The equilibrium constant expression for this dissolution is called a solubility product constant. • Ksp = solubility product constant**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.**Solubility Product Constants**• The solubility product expression for Ag2S is:**Solubility Product Constants**• The dissolution of solid calcium phosphate in water is represented as: • The solubility product constant expression is: You do it!**Solubility Product Constants**• In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:**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.**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:**Determination of Solubility Product Constants**• The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are:**Determination of Solubility Product Constants**• Substitution of the molar concentrations into the solubility product expression gives:**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.**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!**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.**Uses of Solubility Product Constants**• Make the algebraic substitution of x’s into solubility product expression and solve for x, giving the ion concentrations.**Uses of Solubility Product Constants**• Finally, to calculate the mass of BaSO4 in 1.00 L of saturated solution, use the definition of molarity.**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!**Uses of Solubility Product Constants**• Be careful, do not forget the stoichiometric coefficient of 2!**Uses of Solubility Product Constants**• Substitute the algebraic expressions into the solubility product expression.**Uses of Solubility Product Constants**• Solve for the pOH and pH.**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?)**The Common Ion Effect in Solubility Calculations**• Write equations to represent the equilibria.**The Common Ion Effect in Solubility Calculations**• Substitute the algebraic representations of the concentrations into the Ksp expression and solve for x.**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.**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?**The Reaction Quotient in Precipitation Reactions**• Write out the solubility expressions.**The Reaction Quotient in Precipitation Reactions**• Calculate the Qsp for PbSO4. • Assume that the solution volumes are additive. • Concentrations of the important ions are:**The Reaction Quotient in Precipitation Reactions**• Finally, calculate Qsp for PbSO4 and compare it to the Ksp.**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.**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?**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.**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.**Fractional Precipitation**• Repeat the calculation for silver chloride.**Fractional Precipitation**• Finally, for copper (I) chloride to precipitate.**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.**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.**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**Fractional Precipitation**• A similar calculation for the concentration of Ag+ ions unprecipitated before CuCl begins to precipitate is:**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**Simultaneous Equilibria Involving Slightly Soluble Compounds**• Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.**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-].**Simultaneous Equilibria Involving Slightly Soluble Compounds****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.**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+.**Simultaneous Equilibria Involving Slightly Soluble Compounds**