Solubility (Precipitation Equilibria: The solubiloity Product- Section 10.5 (p 326 -332)

1. Solubility (Precipitation Equilibria: The solubiloity Product- Section 10.5 (p 326 -332) Chapter 10

2. Precipitation (Insoluble Salts) • Many metal ions form compounds that are insoluble in water. • We call them insoluble or slightly soluble salts or precipitates. • Common precipitates are carbonates, hydroxides, sulfates, and sulfides. • An insoluble salt in contact with water maintains an equilibrium with the ions. In simple cases where there are no common ions or competing equilibria, the ion concentrations depend only on the equilibrium constant for the particular precipitate. When we talk about solubility equilibria we always write the equilibrium with the solid on the left. For example: Ba(IO3)2 (s) Ba2+(aq) + 2 IO3-(aq) • The equilibrium constant expression for an insoluble salt is written following the same rules as for any other equilibrium. • The equilibrium constant is called the solubility product, Ksp. The Ksp expression for the above equilibrium is: Ksp = [Ba2+][IO3-]2

3. Ksp Values for Some Precipitates

4. Solubility (Precipitation Equilibria) Solubility is defined as moles/L, g/L, or mg/L of the dissolving species in solution. Example : • How is the solubility of lead chloride, PbCl2, related to the solution concentrations of Pb2+ and Cl-? • Each formula unit of PbCl2 that dissolves produces one lead ion, Pb2+, and two chloride ions, Cl-. • The molar solubility of PbCl2 is equivalent to the concentration of Pb2+, SPbCl2 = [Pb2+]. • Since two Cl-s are produced per PbCl2 formula unit that dissolves, the molar solubility of PbCl2 equals one-half the solution concentration of Cl-, SPbCl2 = 0.5*[Cl-]. • SPbCl2 = [Pb2+] = 0.5*[Cl-].

5. Example 2 What is the solubility of barium iodate, Ba(IO3)2, in pure water at 25 oC? The solubility equilibrium is: Ba(IO3)2 (s) Ba2+(aq) + 2 IO3-(aq) x 2x Ksp = [Ba2+][IO3-]2 = 1.5x10-9 [Ba2+] = x [IO3-] = 2x Ksp = (x) (2x)2 = 1.5x10-9 4*[x]3 = 1.5x10-9 *[Ba2+]3 x3 = [Ba2+]3 = 3.75x10-10 M X = S (solubility) = [Ba2+] = 7.2x10-4 M Or in mass terms: S = (7.2x10-4 M)(487 g/mol) = 0.35 g/L

6. Why did we choose of Ba(IO3)2 as an example?? • Determining the solubility of Ba(IO3)2 is simple compared to determining the solubility of many insoluble salts. • Ba2+ is a spectator ion so we do not worry about. It does not react with water. • What about the iodate anion? It is a base and can react with water: IO3-(aq) + H2O HIO3(aq) + OH-(aq) • The extent to which this reaction proceeds is given by the value of Kb. Ka for iodic acid is 0.17 (it is a reasonably strong acid). Kb = Kw/Ka = 1x10-14/0.17 = 5.9x10-14 • This value of Kb is so small that we can conclude that the hydrolysis of iodate does not affect the solubility of Ba(IO3)2.

7. Common –ion effectExample 3 • What is the solubility of barium iodate, Ba(IO3)2, at 25 oC in a solution of 0.10 M barium nitrate, Ba(NO3)2? • The problem is set up the same as before: • Ba(IO3)2 (s) Ba2+(aq) + 2 IO3-(aq) Initial x 2x After addition 0.1 + x 2x • Ksp = [Ba2+][IO3-]2 = 1.5x10-9 • The difference is that there are two sources of Ba2+, the 0.10 M Ba(NO3)2 and the dissolution of Ba(IO3)2:

8. Common –ion effectExample 3 Ksp = [Ba2+][IO3-]2 = 1.5x10-9 • Ksp = (0.10 M +x)*[2x-]2 = 1.5x10-9 • We could solve a cubic equation or we could try neglecting the [Ba2+] that comes from dissolution of Ba(IO3)2 compared to 0.10 M barium nitrate. • Since 0.1>>> x then 0.1 + x  0.1 • Ksp = (0.1)(2x)2 = 1.5x10-9

9. Ksp = (0.10)[2x]2 = 1.5x10-9 (2x)2 = 1.5x10-8 (x)2 = 1.5x10-8 / 4 = 3.75x 10-9 X = S = [Ba2+] = = 6.1x10-5 M S = 0.5*[IO3-], so [IO3-] = 2x = 2 x = 6.1x10-5 = 1.2x10-4 M • Our assumption that S << 0.1 M is justified. • Compare this result to the solubility in pure water, S = 7.2x10-4 M. The solubility of barium iodate is much less when one of the ions involved in the equilibrium is present in solution. • This effect is called the common-ion effect, and is explained qualitatively by LeChatelier's principle.

10. Will addition of other ions such as Na+ and Cl- affect this equilibrium? To a first approximation no. However, adding spectator ions to a solution increases the ionic strength and affects the activity coefficients. • In which of the following examples should we worry about activity? • solubility of barium iodate in water • solubility of barium iodate in a solution of 0.10 M barium nitrate • solubility of barium iodate in a solution of 0.50 M NaCl

11. Example 4 • Will a precipitate form when 100.0 mL of 0.2 M Fe3+ in 0.5 M H2SO4 is mixed with 100.0 mL of 1.0 M NaOH? • Before thinking about equilibrium, look at this problem and decide if anything will happen when the two solutions are mixed. • A solution of a strong base is mixed with a solution containing strong acid. • The acid and base will react until one of both are consumed. • In this case an equal number of moles of H+ and OH- are mixed so the resulting solution will be at pH=7.

12. The solubility equilibrium is: Fe(OH)3 (s) Fe3+(aq) + 3OH-(aq) Ksp = 2x10-39 = [Fe3+] [OH-]3 • To determine if a precipitate will form: find Q and compare it to Ksp. • (this quantity is called the ion product, P, but it is the same expression as Q.) • If Q > Ksp the ion concentrations are greater than their equilibrium concentrations and a precipitate will form. • If Q < Ksp the ion concentrations are below their equilibrium concentrations and no precipitate forms. • Q = CFe3+COH-3 = (0.100 M)(1x10-7 M)3 = 1.00x10-22 • 1x10-22 > 2x10-39Q > KspSo a precipitate will form.

13. What will be the [Fe3+] of the resulting solution? 2x10-39 = [Fe3+](1x10-7 M)3[Fe3+] = 2x10-18 M • This is the predicted [Fe3+] in water. • In real water systems the total amount of iron in solution can be much higher since the iron can exist in a variety of forms such as iron complexes and colloids.