1. SOLUBILITY I. Saturated Solution BaSO 4(s)  Ba 2+ (aq) + SO 4 2- (aq) Equilibrium expresses the degree of solubility of solid in water. Ksp = solubility product constant Ksp = [Ba 2+ ] [SO 4 2- ] = 1.1 x 10 -10 Represents the amount of dissolution, the smaller the Ksp value, the less amount of solid dissolved. Q1. Write the solubility product expression for Hg 2 Cl 2 & HgCl 2 Q2. Exactly 0.133 mg of kgBr will dissolve in 1.00 L of water. What is the value of Ksp for AgBr?

2. Interconverting solubility and Ksp SOLUBILITY OF COMPOUND (g/L) MOLAR SOLUBILITY OF COMPOUND (mol/L) MOLAR CONCENTRATION OF IONS Ksp

3. II. Solubility vs Solubility Product Solubility: The quantity of solute that dissolves to form a saturated solution. (g/L) Molar Solubility: (n solute/L saturated solution) Ksp: The equilibrium between the ionic solid and the saturated solution. Q1. A student finds that the solubility of BaF 2 is 1.1 g in l.00 L of water. What is the value of Ksp? Q2. Calomel (Hg 2 Cl 2 ) was once used in medicine. It has a Ksp = 1.3 x 10 -18. What is the solubility of Hg 2 Cl 2 in g/L?

4. III. Solubility and Common Ion effect CaF 2(s)  Ca 2+ (aq) + 2F - (aq) The addition of Ca 2+ or F - shifts the equilibrium. According to Le Chatelier’s Principle, more solid will form thus reducing the solubility of the solid. Solubility of a salt decreases when the solute of a common ion is added. Q1. What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 x 10 -2 mol of HCl?

5. IV. CRITERIA FOR PRECIPITATION OF DISSOLUTION BaSO 4(s)  Ba 2+ (aq) + SO 4 2- (aq) Equilibrium can be established from either direction. Q1. (Ion Product) is used to determine whether or not precipitation will occur. Q < K  solid dissolves Q = K equilibrium (saturated solution) Q > K  ppt Q2. Calcium phosphate has a Ksp of 1.0 x 10 -26, if a sample contains 1.0 x 10 -3 M Ca 2+ and 1.0 x 10 -8 M PO 4 3- Ions, calculate Q and predict whether Ca 3 (PO 4 ) 2 will precipitate?

6. Q3. Exactly 0.400 L of 0.50 M Pb 2+ and 1.60 L of 2.5 x 10 -8 M Cl - are mixed together to form 2.00L. Calculate Q and predict if a ppt will occur. Ksp = 1.6 x 10 -5

7. 1. When 50.0 mL of 0.100 M AgNO 3 and 30 mL of 0.060 M Na 2 CrO 4 are mixed, a precipitate of silver chromate is formed. The solubility product is 1.9 x 10 -12. Calculate the [Ag + ] and [CrO 4 2- ] remaining in solution at equilibrium. 2. Suppose 300 mL of 8 x 10 -6 M solution of KCl is added to 800 mL of 0.004 M solution of AgNO 3. Calculate [Ag + ] and [Cl - ] remaining in solution at equilibrium.

8. COMPLEX ION EQUILIBRIA Transition metal Ions form coordinate covalent bonds with molecules or anions having a lone pair of e-. AgCl (s)  Ag + + Cl - Ksp 1.82 x 10 -10 Ag + + 2NH 3  Ag(NH 3 ) 2 + Kf 1.7 x 10 7 AgCl + 2NH 3  Ag(NH 3 ) 2 + + Cl - Complex Ion H 3 N:AG:NH 3 metal = Lewis is acid ligand = Lewis base Kf = [Ag(NH 3 ) 2 + ]formation constant [Ag + ][NH 3 ] 2

9. Kf = Formation Constant M + + L -  ML Kd = Dissociation constant ML  M + + L - Kd = 1 Kf

10. Q 1. Calculate [Ag + ] present in a solution at equilibrium when concentrated NH 3 is added to a 0.010 M solution of AgNO 3 to give an equilibrium concentration of [NH 3 ] = 0.20M. Q2. Silver chloride usually does not ppt in solution of 1.0 M NH 3. However AgBr has a smaller Ksp. Will AgBr ppt form a solution containing 0.010 M AgNO 3, 0.010 M NaBr and 1.0 M NH 3 ? Ksp = 5.0 x 10 -13 Q3.Calculate the molar solubility of AgBr in 1.0M NH 3 ?

11. EFFECT OF pH ON SOLUBILITY CaF 2  Ca 2+ + 2F - 2F - + 2H +  2HF CaF 2 + 2H +  Ca 2+ + 2HF Salts of weak acids are more soluble in acidic solutions. Salts with anions of strong acids are largely unaffected by pH. Q1. Consider the two slightly soluble salts BaF 2 and AgBr. Which of these two would have its solubility more affected by the addition of a strong acid? Would the solubility of that salt increase or decrease.

12. 3 STEPS TO DETERMINING THE ION CONCENTRATION AT EQUILIBRIUM I.Calculate the [Ion]i that occurs after dilution but before the reaction starts. II.Calculate the [Ion] when the maximum amount of solid is formed. - we will determine the limiting reagent and assume all of that ion is used up to make the solid. - The [ ] of the other ion will be the stoicometric equivalent. III. Calculate the [Ion] at equilibrium. Since we assume the reaction went to completion but by definition a slightly soluble can’t. We must account for some of that solid re-dissolving back into solution.