3. Friday, March 21 st : “A” Day Monday, March 24 th : “B” Day Agenda  Homework questions/collect  Finish section 14.2: “Systems At Equilibrium”  Homework: Practice pg. 509: #1-2 Practice pg. 510: #1-2 Sec. 14.2 review, pg. 511: #1-9  Concept Review: “Systems at Equilibrium” #15-20

4. Dust off your Brains! At 450°C, K eq is 6.59 X 10 -3. If [NH 3 ] = 1.23 X 10 -4 and [H 2 ] = 2.75 X 10 -3 at equilibrium, what is [N 2 ]? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K eq = [NH 3 ] 2 [N 2 ] [H 2 ] 3 6.59 X 10 -3 = (1.23 X 10 -4 ) 2 1.51 X 10 -8 [N 2 ] (2.75 X 10 -3 ) 3 [N 2 ] 2.08 X 10 -8 [N 2 ] 1.37 X 10 -10 = 1.51 X 10 -8 [N 2 ] = 110.

5. 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 CaF 2 in water is 0.00034 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)

6. The Solubility Product Constant, K sp  Like most salts, calcium fluoride is an ionic compound that dissociates into ions when it dissolves in water  Like many salts, calcium fluoride is slightly soluble in water.  The ions in solution and any solid salt are at equilibrium as shown below:  Since solids are not included in the equilibrium constant expression, K eq = [Ca 2+ ] [F − ] 2, which is equal to a constant.

7. The Solubility Product Constant, K sp  Solubility product constant, 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.

8. The Solubility Product Constant, K sp  For example, if you mix solutions of calcium nitrate and sodium fluoride, calcium fluoride precipitates. Ca(NO 3 ) 2 (aq) + NaF (aq) CaF 2 (s) + NaNO 3 (aq)  The net ionic equation for this precipitation is the reverse of the dissolution of the salt:  This equation is the same equilibrium, so K sp is 1.6 × 10 −10

9. Solubility Product Constants at 25°C

10. Rules for Determining K sp 1. Write a balanced chemical equation. K sp 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.

11. 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)

12. Sample Problem C, pg. 509 Calculating K sp from Solubility Most parts of the oceans are nearly saturated with the mineral fluorite, CaF 2, which 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, K sp, for CaF 2. CaF 2 (s) Ca 2+ (aq) + 2 F - (aq)  Write solubility expression: K sp = [Ca 2+ ] [F - ] 2

13. Sample Problem C, pg. 509 Calculating K sp from Solubility, cont. CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) 3.4 X 10 -4 M K sp = [Ca 2+ ] [F - ] 2  Due to stoichiometry, [CaF 2 ] = [Ca 2+ ] = 3.4 X 10 -4  Due to stoichiometry, 2 [Ca 2+ ] = [F - ]  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

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

15. 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 Write solubility expression: K sp = [Cu + ] [Cl - ]  Plug in values: 1.2 X 10 -6 = [Cu + ] (0.55)  [Cu + ] = 2.2 X 10 -6 Since 1 mol of CuCl produces 1 mol of Cu +, the solubility of CuCl in ocean water is also 2.2 X 10 -6 M

16. 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) 2.0 X 10 -6 Write solubility expression: K sp = [Ag + ] [Cl – ]  From table 3 in book: K sp of AgCl = 1.8 X 10 -10 1.8 X 10 -10 = 2.0 X 10 -6 [Cl – ] [Cl – ] = 9.0 X 10 -5

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

18. Homework Practice pg. 509: #1-2 Practice pg. 510: #1-2 Concept Review: Systems at Equilibrium, #15-20 Please use your class time wisely…