2. Monday, April 11 th : “A” Day Agenda  Homework questions/collect  Finish section 14.2: “Systems At Equilibrium”  Homework: Section 14.2 review, pg. 511: #1-9 Concept Review: Systems at Equilibrium, #15-20

3. The Solubility Product Constant, K sp  The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water.  Solubilities can be expressed in moles of solute per liter of solution (mol/L or M). –For example, the solubility of calcium fluoride in water is 3.4 × 10 −4 mol/L. –So, 0.00034 mol of CaF 2 will dissolve in 1 L of water to give a saturated solution. –If you try to dissolve 0.00100 mol of CaF 2 in 1 L of water, 0.00066 mol of CaF 2 will remain undissolved. (0.00100 – 0.00034 = 0.00066)

4. The Solubility Product Constant, K sp  Like most salts, calcium fluoride is an ionic compound that dissociates into ions when it dissolves in water  Calcium fluoride is one of a large class of salts that are said to be slightly soluble in water.  The ions in solution and any solid salt are at equilibrium.  Since solids are not part of the equilibrium constant expression, K eq = [Ca 2+ ] [F − ] 2, which is equal to a constant.

5. The Solubility Product Constant, K sp  Solubility product constants, K sp : the equilibrium constant for a solid that is in equilibrium with the solid’s dissolved ions. K sp = [Ca 2+ ][F − ] 2 = 1.6  10 −10  K sp values have NO units, just like K eq values.  This relationship is true whenever calcium ions and fluoride ions are in equilibrium with calcium fluoride, not just when the salt dissolves.

6. The Solubility Product Constant, K sp  For example, if you mix solutions of calcium nitrate and sodium fluoride, calcium fluoride precipitates.  The net ionic equation for this precipitation is the reverse of the dissolution.  This equation is the same equilibrium. So, the K sp for the dissolution of CaF 2 in this system is the same and is 1.6 × 10 −10

7. Solubility Product Constants at 25°C

8. Rules for Determining K sp 1. Write a balanced chemical equation. The solubility product is only for salts that have low solubility. Soluble salts, like NaCl, do not have K sp values. Make sure that the reaction is at equilibrium. Equations are always written so that the solid salt is the reactant and the ions are the products.

9. Rules for Determining K sp 2. Write a solubility product expression. Write the product of the ion concentrations. Concentrations of any solid or pure liquid are omitted. 3. Complete the solubility product expression. Raise each concentration to a power equal to the substance’s coefficient in the balanced chemical equation. (Remember: K sp values depend on temperature)

10. Sample Problem C, pg. 509 Calculating K sp from solubility Most parts of the oceans are nearly saturated with CaF 2. The mineral fluorite, CaF 2, may precipitate when ocean water evaporates. A saturated solution of CaF 2 at 25°C has a solubility of 3.4 X 10 −4 M. Calculate the solubility product constant for CaF 2. CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ] [F - ] 2

11. Sample Problem C, pg. 509 Calculating K sp from solubility CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ] [F - ] 2  The solubility of CaF 2 is 3.4 X 10 -4 M  That means [Ca 2+ ] = 3.4 X 10 -4  From the balanced equation [F - ] = 2[Ca 2+ ]  So [F - ] = 2 (3.4 X 10 -4 ) = 6.8 X 10 -4  K sp = (3.4 X 10 -4 ) (6.8 X 10 -4 ) 2 = 1.6 X 10 -10

12. Additional Practice Calculate the solubility product constant, K sp, of HgI 2 if the Hg 2+ concentration in a saturated solution is 1.9 X 10 -10 M. HgI 2 (s) Hg 2+ (aq) + 2 I - (aq) K sp = [Hg 2+ ] [ I - ] 2  [Hg 2+ ] = 1.9 X 10 -10  [ I - ] = 2 [Hg 2+ ] = 2 (1.9 X 10 -10 ) = 3.8 X 10 -10  K sp = (1.9 X 10 -10 ) (3.8 X 10 -10 ) = 2.7 X 10 -29

13. Sample Problem D, pg. 510 Calculating Ionic Concentrations Using K sp Copper (I) chloride has a solubility product constant of 1.2 × 10 −6 and dissolves according to the equation below. Calculate the solubility of this salt in ocean water in which the [Cl − ] = 0.55  K sp = 1.2 X 10 -6 = [Cu + ] [Cl - ]  1.2 X 10 -6 = [Cu + ] (0.55)  [Cu + ] = 2.2 X 10 -6  Solubility of CuCl = 2.2 X 10 -6 M

14. Additional Practice A chemist wishes to reduce the silver ion concentration in saturated AgCl solution to 2.0 X 10 -6 M. What concentration of Cl – would achieve this goal? AgCl (s) Ag + (aq) + Cl – (aq)  K sp = [Ag + ] [Cl – ]  [Ag + ] = 2.0 X 10 -6  From table 3 in book: K sp of AgCl = 1.8 X 10 -8  1.8 X 10 -8 = 2.0 X 10 -6 [Cl – ]  [Cl – ] = 9.0 X 10 -5

15. Using K sp to Make Magnesium  Though slightly soluble hydroxides are not salts, they have solubility product constants.  Magnesium hydroxide is an example: K sp =[Mg 2+ ][OH − ] 2 = 1.8 × 10 −11  This equilibrium is the basis for obtaining magnesium from seawater.

16. Using K sp to Make Magnesium  To get magnesium, calcium hydroxide is added to sea water, which raises the hydroxide ion concentration to a large value so that [Mg 2+ ][OH − ] 2 would be greater than 1.8 × 10 −11  As a result, magnesium hydroxide precipitates and can be collected.

17. Homework  Section 14.2 review, pg. 511: #1-9  Concept Review: Systems at Equilibrium, #15-20 Please use your class time wisely… Be ready for a quiz over this section next time..