1. Solubility Equilibria Solubility Product Constant, K sp Precipitation Complex Ions

2. Solubility Rules Consider an ionic solid that completely dissociates into hydrated cations and anions: NaF (s)  Na + (aq) + F - (aq) Recall the “Solubility Rules” for substances that are 100% ionized.

3. Solubility Product Constant  Salts that are not listed on your solubility rules chart, are sparingly soluble to the degree indicated by the solubility product constant, or K sp.  With more concentration of ions and more collisions, the ions recombine to form the salt again: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ][ F - ] 2 [CaF 2 ](s) Since the solid is not aqueous, the concentration designation is meaningless, and it is not included in the equilibrium expression. Therefore, for K sp problems, the equilibrium expression is simply one in this form: K sp = [Ca 2+ ][ F - ] 2 K sp is >> 1 for all the substances that are 100% ionized, or listed on the solubility rules.

4. Solubility of Solids  Increasing surface area (crushing, etc.) will increase the rate that equilibrium occurs, but not the equilibrium position: More surface area increases rate of dissolving, but also rate of re- formation.  Solubility product = a constant at a given temperature  Solubility = an equilibrium position  At the saturation point, which is controlled by the K sp expression, it does not matter how much solid you add; no more can dissolve.  If the K sp expression is not exceeded, no precipitation will occur.

5. Concentration Affects Recombination CaF 2 (s) Ca 2+ (aq) + 2F - (aq) If concentration of either ion is increased, the equilibrium shifts back toward the reactants, causing more solid, or precipitant to be formed. O H H - + + - + + - + + - + + - + + - + + Cu F - ++ - + + F - ++ - + + - + + - + + - + + - + + - ++ - + + F - ++ - + + - + + - + +

6. K sp Example  One liter of saturated barium sulfate solution contains 0.0025 g of the solute. What is the molar solubility of BaSO 4 ? Find the solubility product constant for BaSO 4. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq)K sp = [Ba 2+ ][SO 4 2- ] Molar solubility = mol :0.0025 g BaSO 4 |1.0 mol BaSO 4 = 1.1 x 10 -5 mol BaSO 4 L1.0 L|233 g BaSO 4 L One mole of each ion is formed by one mole of BaSO 4 : 1.1 x 10 -5 M BaSO 4, forms 1.1 x 10 -5 mol/L Ba 2+ and 1.1 x 10 -5 mol/L SO 4 2- K sp = [Ba 2+ ][SO 4 2- ] = (1.1 x 10 -5 ) (1.1 x 10 -5 ) = 1.2 x 10 -10 If the coefficient of either of the products is 2, [ion - ] is squared.

7. In Real Life… Calcium Carbonate CaCO3 (s) Ca 2+ (aq) + CO 3 2- (aq)K sp = [Ca 2+ ][CO 3 2- ] = 5.0 x 10 -10 ΔH is negative Sea shells are mainly made of calcium carbonate (reacts with acid but does not dissolve appreciably in water). When the sea creatures die and the remains sink to great depth in the oceans, the shells dissolve because of changes in the position of two equilibria: 1. At great depths the pressure is much higher and the temperature lower Both factors favor more CO 2 dissolving The forward reaction is promoted: CaCO 3 (aq) + CO 2 (aq) + H 2 O (l) Ca 2+ (aq) + 2HCO 3 - (aq) 2.Lower temperature favors a higher K sp value because the dissolving process is slightly exothermic, so more of the calcium carbonate will dissolve. There are no shells lying at the bottom of the oceans! The limestone cliffs you see in the landscape were formed in warm shallow tropical seas. http://www.docbrown.info/page07/equilibria4a.htm