1. Solubility Equilibria Chemistry 142 B Autumn Quarter, 2004 J. B. Callis, Instructor Lecture #25

2. Precipitation Precipitation - The formation of a solid from solution. The reverse of dissolution. Dissolution - The process by which a substance dissolves. The reverse of precipitation. Importance - (a) Selective precipitation is an important industrial purification process, especially when crystals are formed. (b) Scales that form on boilers and teeth are to be prevented, as are kidney stones. (c) Precipitation forms minerals - dissolution removes them.

3. Saturation Saturated Solution - One in which a dissolution - precipitation equilibrium exist between a solid and its dissolved form. Here the equilibrium is dynamic and the rate of dissolution is equal to the rate of precipitation. Unsaturated Solution - One in which the concentration of dissolved solid is not sufficient to cause precipitation. Obviously a quantitative description of this type of heterogeneous equilibrium is subject to the law of mass action, and equilibrium expressions can be written and deductions made concerning the concentration of various species at equilibrium.

4. Solubility Solubility - The greatest amount of a substance that will dissolve in equilibrium in a specified volume of solvent at a particular temperature. Example - The solubility of silver chloride in water at 25 o C is.0018 g/L or 1.3 x 10 -5 M. Most solubilities increase with temperature.

5. Classification of Ionic Materials by Solubility Soluble Ionic Materials - have solubilities in excess of 10 gL -1. Insoluble Ionic Materials - have solubilities less than 0.1 gL -1. Slightly Soluble Materials - have solubilities between 0.1 and 10 gL -1.

6. The Nature of Ionic Equilibria Most salts dissociate into ions when they dissolve. Equilibrium then exists between the solid salt and its aquated ions, and not between the solid salt and dissolved molecules of the salt. For example: PbSO 4 (s) = Pb 2+ (aq) + SO 4 2- (aq) This equilibrium system may be described by the mass-action expression: K sp = [Pb 2+ ][SO 4 2- ] Note that the pure solid does not enter into the equilibrium.

7. Solubility and the K sp One may provide solubility information as the solubility, S or as the solubility product, K sp. These two quantities are obviously related to each other.

8. Problem 25-1: Given Solubility, Calculate K sp Problem: What is the K sp of Pb(IO 3 ) 2 if.00896 g dissolves in 400 mL at 25 o C? Procedure: Ans:

9. Problem 25-2: Given K sp, Calculate Solubility Problem: Given the K sp at 20 o C for (NH 4 ) 2 (PtCl 6 ) calculate the solubility. (K sp = 5.6 x 10 -6 ) Procedure: Ans:

10. Problem 25-3:The Common Ion Effect What is the solubility of Ca(OH) 2 in 0.10 M Ca(NO 3 ) 2 ? Ca(OH) 2 (s) = Ca 2+ (aq) + 2 OH - (aq) ; K sp = 6.5 x 10 -6 Ans: Set up a reaction table, with S = [Ca 2+ ]from Ca(OH) 2 Conc. MCa(OH) 2 (s)=Ca 2+ (aq) + 2 OH - (aq) Initial Change Equil.

11. Problem 25-3 (cont.): The Common Ion Effect

12. Problem 25-4: Predicting the Effect of pH on Solubility Question: Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of (a) lead(II) bromide, (b) copper(II) hydroxide, and (c) iron(II) sulfide. Approach: Write the balanced dissolution equation and note the anion: weak-acid anions react with H 3 O + and shift the equilibrium position toward more dissolution. Strong acid anions do not react, so added H 3 O + has no effect. (a)PbBr 2 (s) = Pb 2+ (aq) + 2 Br - (aq)

13. Problem 25-4 (cont.): Predicting the Effect of pH on Solubility (b) Cu(OH) 2 (s) = Cu 2+ (aq) + 2 OH - (aq) (c) FeS(s) + H 2 O(l) = Fe 2+ (aq) + HS - (aq) +OH - (aq)

14. Problem 25-5: Predicting Whether a Precipitate Will Form Question: Does a precipitate form when 0.100 L of 0.30 M Ca(NO 3 ) 2 is mixed with 0.200 L of 0.06 M NaF? Approach: First decide whether any of the ions present will combine to form a slightly soluble salt. Then look up the K sp for the substance and from the initial concentrations after mixing, calculate Q sp and compare it to K sp.

15. Problem 25-5(cont.): Predicting Whether a Precipitate Will Form

16. Problem 25-6: Calculating the Concentrations of Complex Ions Problem: How much Zn(NH 3 ) 4 2+ is made by mixing 50.0 L of.0020 M Zn(H 2 O) 4 2+ and 25.0 L of 0.15 M NH 3. The K f of Zn(NH 3 ) 4 2+ = 7.8 x 10 8. Approach: We write the equation for formation and the K f expression. Then we set up a reaction table.

17. Problem 25-6(cont.(i)): Calculating the Concentrations of Complex Ions The formation reaction: The formation constant:

18. Problem 25-6(cont.(ii)): Calculating the Concentrations of Complex Ions The reaction table: Assume that nearly all the Zn(H 2 O) 4 2+ is converted to Zn(NH 3 ) 4 2+. Set up a table with x = [Zn(H 2 O) 4 2+ ] at equilibrium. Since 4 mol NH 3 are needed per mole of Zn(H 2 O) 4 2+, the change in [NH 3 ] is: [NH 3 ] reacted = And, [Zn(NH 3 ) 4 2+ ] =

19. Problem 25-6(cont.(iii)): Calculating the Concentrations of Complex Ions Conc., MZn(H 2 O) 4 2+ + 4 NH 3 =Zn(NH 3 ) 4 2+ Initial Change Equilibrium

20. Problem 25-6(cont.(iv)): Calculating the Concentrations of Complex Ions

21. Problem 25-7: Calculating the Effect of Complex-Ion Formation on Solubility Problem: A critical step in black and white film developing is the removal of excess AgBr from the film negative by ‘hypo’, an aqueous solution of sodium thiosulfate (Na 2 S 2 O 3 ), through formation of the complex ion Ag(S 2 O 3 ) 2 3-. Calculate the solubility of AgBr in (a) H 2 O: (b) 1.0 M hypo. K f of Ag(S 2 O 3 ) 2 3- = 4.7 x 10 13 and K sp of AgBr = 5.0 x 10 -13. Approach: (b) Find the overall equation for dissolution of AgBr in hypo by combining the equations of dissolution of AgBr in water and of formation of the complex of silver thiosulfate.

22. Problem 25-7(cont.(i)): Calculating the Effect of Complex-Ion Formation on Solubility Ans (a): K sp = [Ag + ][Br - ] S = [AgBr] dissolved = K sp = Ans (b): Combine the following reactions AgBr(s) = Ag + (aq) + Br - (aq) Ag + (aq) + 2 S 2 O 3 2- (aq) = Ag(S 2 O 3 ) 2 3- (aq) Sum: K overall =

23. Problem 25-7(cont.(ii)): Calculating the Effect of Complex-Ion Formation on Solubility Ans (b) (cont.): Let x = [AgBr] dissolved Conc., M2 S 2 O 3 2- =Ag(S 2 O 3 ) 2 3- Br - Initial Change Equilibrium

24. Problem 25-7(cont.(iii)): Calculating the Effect of Complex-Ion Formation on Solubility Ans (b) (cont.):

25. Problem 25-8: Separating Ions by Selective Precipitation Problem: A solution consists of 0.20 M MgCl 2 and 0.10 M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = 6.3 x 10 -10 ; K sp of Cu(OH) 2 = 2.2 x 10 -20 Approach: Since the two hydroxides have the same formula type (1:2), we can directly compare their K sp values and see that Mg(OH) 2 is about 10 10 times more soluble than Cu(OH) 2. Thus Cu(OH) 2 precipitates first. We solve for the the [OH - ] that will just give a saturated solution of Mg(OH) 2 because this amount of OH - will precipitate the greatest amount of Cu 2+ ion.

26. Problem 25-8 (cont.): Separating Ions by Selective Precipitation Ans: First write the equilibria and ion products Mg(OH) 2 (s) = Mg 2+ (aq) + 2 OH - (aq) K sp =[Mg 2+ ][OH - ] 2 Cu(OH) 2 (s) = Cu 2+ (aq) + 2 OH - (aq) K sp =[Cu 2+ ][OH - ] 2

27. Answers to Problems 1.2.60 x 10 -13 2.0.011 mol L -1 = 5.0 g L -1 3.S = 4.0 x 10 -3 M 4.(a) no effect, (b) increases solubility, (c) increases solubility. 6.4.1 x 10 -7 7.(a) 7.1 x 10 -7 M; (b) 0.45 M 8.[OH - ] = 5.6 x 10 -5 M 5.