2. Solubility Equilibrium

3. Solubility Product Constant Ionic compounds (salts) differ in their solubilities Most “insoluble” salts will actually dissolve to some extent in water Better said to be slightly, or sparingly, soluble in water

4. The equilibrium between solids and ions is different from the equilibrium between gases The equilibrium between solids and ions is a “phase” equilibrium NaCl(s) Na + (aq) + Cl - (aq) Ksp deals with a phase equilibrium: (s)  (aq) Solubility Product

5. Consider: NaCl  Na + (aq) + Cl - (aq) NaCl(s) Na + (aq) + Cl - (aq) Ksp = [Na + (aq)] [Cl - (aq)] K = [Na + (aq)] [Cl - (aq)] [NaCl(s)] [NaCl(s)] doesn’t change (it’s a constant) K [NaCl(s)] =[Na + (aq)] [Cl - (aq)] Solubility Product

6. A solid always dissolves until no more can dissolve - called molar solubility (mol/L or M) An equilibrium is established when the amount dissolving equals the amount precipitating This can only be true if there is some solid (we can usually see if there is an equilibrium) A solution with solid remaining (i.e. in equilibrium) is called “saturated” Note: adding more solid will not affect equilibrium Solubility Product

8. Ksp = [7.1x10 –7 ][7.1x10 –7 ] = 5.0 x 10 –13 Ksp = [Ag + (aq)][Br – (aq)] 1.00 L of water dissolves 7.1x10 -7 mol of AgBr. What is the Ksp? AgBr(s)  Ag + (aq) + Br – (aq) Solubility Product

9. Ksp = [1.7x10 -3 ][0.1034] 2 = 1.7 x 10 –5 Ksp = [Pb 2+ (aq)][Cl – (aq)] 2 The molar solubility of PbCl 2 is 1.7x10 -3 M in a 0.10 M NaCl solution. What is Ksp? NaCl(s)  Na + (aq) + Cl – (aq) PbCl 2 (s)  Pb 2+ (aq) + 2Cl – (aq) (Cl - from NaCl must be added to Cl - from PbCl 2 ) Solubility Product

10. Find the concentration of ions present in calcium fluoride (in water) and the molar solubility. CaF 2 (s)  Ca +2 + 2 F - -x +x +2x K sp = [Ca +2 ] [F - ] 2 (from table) 2 x 10 -10 = [x] [2x] 2 3.68 x 10 -4 = x x = [Ca +2 ] = 3.68 x 10 -4 2x = [F - ] = 7.37 x 10 -4 Solubility of CaF 2 = 3.68 x 10 -4

11. Solubility Product Find the molar solubility of silver chloride (in water). AgCl (s)  Ag + + Cl - Ksp = 1.6 x 10 -10

12. Copper (II) azide has Ksp = 6.3 x 10 –10. Find the solubility of Cu(N 3 ) 2 in water, in g/L. Solubility Product

13. At any given time, the ion product Q = [ Ba 2+ ] [SO 4 2– ] If Q > Ksp... ppt If Q < Ksp...no ppt If Q = Ksp...equilibrium selective precipitation: using the different solubilities of ions to separate them Solubility Product

14. Will a precipitate form from mixing 50.0 mL of 8.0 x 10 –3 M Pb(NO 3 ) 2 (aq) and 50.0 mL of 5.0 x 10 –3 M Na 2 SO 4 (aq)? K sp = 1.6 x 10 -8 Solubility Product

15. The Ksp of CuS is 2.5 x 10 -41 at 18°C. If 2.0 x 10 -6 mole of Cu(NO 3 ) 2 and 2.0 x 10 -6 mole of Na 2 S are mixed in enough water to give a total volume of 10.0 liters, will a precipitate form? Solubility Product

16. Calculate the hydrogen ion concentration of a 0.25 M solution of hypochlorous acid, HOCl, for which K a = 3.5 x 10 -8. Solubility Product