1. SOLUBILITY EQULIBRIUM 18.4

2. So what can the K expression be used for?

3. 18.4 The K expression is used for any reversible reaction:  K w is used for the ionization of water into H 3 O + and OH -  K a is used for the ionization of weak acids  K b is used for the ionization of weak bases  K sp is used for the dissolving of a solid

4. 18.4 In all cases:  K = products/reactants  Only ions in solution or gases are part of the equation  The larger the K value, the more products are produced and the more the forward reaction is favored.

5. 18.4 What is K sp ? AgCl(s)  Ag 1+ (aq) + Cl 1- (aq) So if the K sp is large, is it more or less soluble?

6. 18.4 K sp is the solubility product constant. It is used to quantitatively determine how much of a solid will dissolve. It also gives the concentrations of ions in a saturated solution. The larger the K value, the more ions form and the more it dissolves in water.

7. 18.4 K sp is the solubility product constant. p. 615 What ions are missing from this chart? Why? What connection is there between this chart and the pink sheet solubility rules? The bigger the K sp the more soluble or less soluble? If a solution’s ions are at K sp what word can be used to describe it?

8. 18.4 Example 1: Write K sp expressions for the solvation of the following salts. MgCO 3 Fe(OH) 3 Ag 2 CO 3

9. 18.4 Example 2: Which of the above is the most soluble? Least soluble?

10. 18.4 Example 3: It is determined that 0.0781 g/L of BaC 2 O 4 dissolves in water. What is the K sp ?

11. 18.4 Example 4: The solubility of Pb(IO 3 ) 2 is 4.0 x 10 -5 mol/L. What is the K sp of this salt?

12. 18.4 Example 5: What is the molarity of the ions of AgCl in water?

13. 18.4 Example 6: What is the molar solubility of Ag 2 CrO 4 in water? K sp of Ag 2 CrO 4 is 1.9 x 10 -12.

14. 18.4 Example 7: The solubility of Ag 2 SO 4 in water is 4.99 g/L. What is the K sp of this salt?

15. 18.4 Example 8: How many grams of CaSO 4 can dissolve in 600 mL of water?

16. 18.4 The real value of K sp comes in determining if there are enough ions present to form a precipitate.

17. 18.4 The real value of K sp comes in determining if there are enough ions present to form a precipitate. If a precipitate forms then the solution must be saturated.

18. 18.4 What is the ion product?

19. 18.4 The ion product is a measurement of the ions dissolved in solution.

20. 18.4 The ion product is a measurement of the ions dissolved in solution. If the ion product is larger than K sp then there are more ions than should be and a precipitate will form. If the ion product is smaller than K sp then there are not enough ions dissolved and no precipitate will form. If the ion product equals K sp then the solution is exactly saturated.

21. 18.4 Example 9: Would a precipitate form if 5.0 x 10 -2 mol of AgNO 3 and 1.0 x 10 -3 mol of NaC 2 H 3 O 2 are dissolved in 1.0 L of solution?

22. 18.4 Example 10: Would a precipitate form if a 0.010 M solution of Pb(NO 3 ) 2 is mixed with a 0.010 M HCl solution?

23. 18.4 Example 11: Will a precipitate form when 0.025 M CaCl 2 is mixed with 0.0050 M Na 2 CO 3 ?

24. 18.4 Example 12: Will a precipitate form when 100 mL of a 0.0030M Pb(NO 3 ) 2 solution is mixed with 400 mL of 0.040 M Na 2 SO 4 ?

25. CHAPTER 18 TEST NOTES 20 multiple choice (3 points each) 8 problems (5 points each) 1 extra credit (5 points)

26. CHAPTER 18 TEST NOTES Le Chatelier’s Principle Use it to predict which way the system will move to compensate for being stressed  Temperature  Concentration  pressure Figuring K values (K, K a, K b, K sp, or ion product)  gases and aqueous solutions only included  [ ] = M  bigger the K value, the more products are formed  coefficients become powers  Don’t forget to use ICE box for K a or K b

27. CHAPTER 18 TEST NOTES Writing equations acid/base conjugate pairs Buffers Use equations to explain how they work Comparing ion product to K sp