1. SOLUBILITY Saturated Solution BaSO 4(s)  Ba 2+ (aq) + SO 4 2- (aq) Equilibrium expresses the degree of solubility of solid in water. Equilibrium expresses the degree of solubility of solid in water. K sp = solubility product constant K sp = K eq [BaSO 4 ] (s) K sp = [Ba 2+ ] [SO 4 2- ] = 1.1 x 10 -10 K sp represents the amount of dissolution (how much solid dissolved into ions), the smaller the K sp value, the smaller the amount of ions in solution (more solid is present).

2. Table 1 Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, FormulaK sp Aluminum hydroxide, Al(OH) 3 Cobalt ( II ) carbonate, CoCO 3 Iron ( II ) hydroxide, Fe(OH) 2 Lead ( II ) fluoride, PbF 2 Lead ( II ) sulfate, PbSO 4 Silver sulfide, Ag 2 S Zinc iodate, Zn( I O 3 ) 2 3 x 10 -34 1.0 x 10 -10 4.1 x 10 -15 3.6 x 10 -8 1.6 x 10 -8 4.7 x 10 -29 8 x 10 -48 Mercury ( I ) iodide, Hg 2 I 2 3.9 x 10 -6

3. SOLUBILITY 1. Write the solubility product expression for each of the following: a) Ca 3 (PO 4 ) 2 b) Hg 2 Cl 2 c) HgCl 2. 2. In a particular sample, the concentration of silver ions was 1.2 x10 -6 M and the concentration of bromide was 1.7x10 -6 M. What is the value of K sp for AgBr?

4. 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 ) K sp : The equilibrium between the ionic solid and the saturated solution.

5. Interconverting solubility and Ksp SOLUBILITYOFCOMPOUND(g/L) MOLARSOLUBILITYOFCOMPOUND(mol/L) MOLARCONCENTRATIONOFIONS K sp

6. Sample Problem 1 Determining K sp from Solubility PROBLEM:(a) Lead ( II ) sulfate is a key component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25x10 -3 g/100mL solution. What is the K sp of PbSO 4 ? (b) When lead ( II ) fluoride (PbF 2 ) is shaken with pure water at 25 0 C, the solubility is found to be 0.64g/L. Calculate the K sp of PbF 2. PLAN:Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. K sp = [Pb 2+ ][SO 4 2- ] = 1.40x10 -4 M PbSO 4 K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10 -4 ) 2 = SOLUTION:PbSO 4 ( s ) Pb 2+ ( aq ) + SO 4 2- ( aq )(a) 1000mL L 4.25x10 -3 g 100mL soln 303.3g PbSO 4 mol PbSO 4 1.96x10 -8

7. Sample Problem 1 Determining K sp from Solubility continued (b) PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2 = 2.6x10 -3 M K sp = (2.6x10 -3 )(5.2x10 -3 ) 2 = 0.64g L soln245.2g PbF 2 mol PbF 2 7.0x10 -8

8. Sample Problem 2 Determining Solubility from K sp PROBLEM:Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water if the K sp is 6.5x10 -6. PLAN:Write out a dissociation equation and K sp expression; Find the molar solubility (S) using a table. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ][OH - ] 2 -Initial Change Equilibrium - - 00 +S+ 2S S2S K sp = (S)(2S) 2 S == 1.2x10x -2 M Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration (M)

9. Solubility vs. Solubility Product 1. 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? 2. Exactly 0.133 mg of AgBr will dissolve in 1.00 L of water. What is the value of Ksp for AgBr? 3. 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?

10. Relationship Between K sp and Solubility at 25 0 C No. of IonsFormulaCation:AnionK sp Solubility (M) 2MgCO 3 1:13.5 x 10 -8 1.9 x 10 -4 2PbSO 4 1:11.6 x 10 -8 1.3 x 10 -4 2BaCrO 4 1:12.1 x 10 -10 1.4 x 10 -5 3Ca(OH) 2 1:25.5 x 10 -6 1.2 x 10 -2 3BaF 2 1:21.5 x 10 -6 7.2 x 10 -3 3CaF 2 1:23.2 x 10 -11 2.0 x 10 -4 3Ag 2 CrO 4 2:12.6 x 10 -12 8.7 x 10 -5

11. Ion-Product Expression (Q sp ) & Solubility Product Constant (K sp ) At equilibrium Q sp = [M n+ ] p [X z- ] q = K sp For the hypothetical compound, M p X q

12. 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. 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.

13. The effect of a common ion on solubility PbCrO 4 ( s ) Pb 2+ ( aq ) + CrO 4 2- ( aq ) CrO 4 2- added

14. Sample Problem 3Calculating the Effect of a Common Ion on Solubility PROBLEM:In Sample Problem 19.6, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10M Ca(NO 3 ) 2 ? K sp of Ca(OH) 2 is 6.5x10 -6. PLAN:Set up a reaction equation and table for the dissolution of Ca(OH) 2. The Ca(NO 3 ) 2 will supply extra [Ca 2+ ] and will relate to the molar solubility of the ions involved. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration(M) Initial Change Equilibrium - - - 0.100 +S+2S 0.10 + S2S K sp = 6.5x10 -6 = (0.10 + S)(2S) 2 = (0.10)(2S) 2 S << 0.10 S = = 4.0x10 -3 Check the assumption: 4.0% 0.10M 4.0x10 -3 x 100 =

15. Solubility and Common Ion effect CaF 2(s)  Ca 2+ (aq) + 2F - (aq) 1. The K sp of the above equation is 3.2 x 10 -11. (a) Calculate the molar solubility in pure water. (b) Calculate the molar solubility in 3.5 x 10 -4 M Ca(NO 3 ) 2. 2. What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 x 10 -2 mol of NaCl?

16. CRITERIA FOR PRECIPITATION OF DISSOLUTION BaSO 4(s)  Ba 2+ (aq) + SO 4 2- (aq) Equilibrium can be established from either direction. Q (the Ion Product) is used to determine whether or not precipitation will occur. Q < K  solid dissolves Q = K equilibrium (saturated solution) Q > K  ppt

17. Sample Problem 3 Predicting Whether a Precipitate Will Form PROBLEM:A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO 3 ) 2 is mixed with 0.200L of 0.060M NaF? PLAN:Write out a reaction equation to see which salt would be formed. Look up the K sp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION:CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = 3.2x10 -11 mol Ca 2+ = 0.100L(0.30mol/L) = 0.030mol[Ca 2+ ] = 0.030mol/0.300L = 0.10M mol F - = 0.200L(0.060mol/L) = 0.012mol[F - ] = 0.012mol/0.300L = 0.040M Q = [Ca 2+ ][F - ] 2 =(0.10)(0.040) 2 = 1.6x10 -4 Q is >> K sp and the CaF 2 WILL precipitate.

18. CRITERIA FOR PRECIPITATION OF DISSOLUTION 1. Calcium phosphate has a K sp of 1 x 10 -26, if a sample contains 1.0 x 10 -3 M Ca 2+ & 1.0 x 10 -8 M PO 4 3- ions, calculate Q and predict whether Ca 3 (PO 4 ) 2 will precipitate? 2.Exactly 0.400 L of 0.50 M Pb 2+ & 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. K sp = 1.6 x 10 -5

19. 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. Thus shifting the solubility to the right. Salts with anions of strong acids are largely unaffected by pH.

20. Sample Problem 4 Predicting the Effect on Solubility of Adding Strong Acid PROBLEM:Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of these ionic compounds: (a) Lead ( II ) bromide(b) Copper ( II ) hydroxide(c) Iron ( II ) sulfide PLAN:Write dissolution equations and consider how strong acid would affect the anion component. Br - is the anion of a strong acid. No effect. SOLUTION:(a) PbBr 2 ( s ) Pb 2+ ( aq ) + 2Br - ( aq ) (b) Cu(OH) 2 ( s ) Cu 2+ ( aq ) + 2OH - ( aq ) OH - is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS( s ) Fe 2+ ( aq ) + S 2- ( aq )S 2- is the anion of a weak acid and will react with water to produce OH -. Both weak acids serve to increase the solubility of FeS. FeS( s ) + H 2 O( l ) Fe 2+ ( aq ) + HS - ( aq ) + OH - ( aq )

21. EFFECT OF pH ON SOLUBILITY 1. 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. 2. What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 x 10 -2 mol of HCl?

22. 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 stoichiometric equivalent. III. Calculate the [Ion] at equilibrium*. *Since we assume the reaction went to completion, yet by definition a slightly soluble can’t, we must account for some of the solid re-dissolving back into solution.

23. 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.