Solubility Equilibria

 Write a balanced chemical equation to represent equilibrium in a saturated solution.  Write a solubility product expression.  Answer questions about Ksp and various missing concentrations using I.C.E. tables.

C 6 H 12 O 6(s) C 6 H 12 O 6(aq)

There are 3 actions that affect solubility: 1. Nature of the solute and solvent “like dissolves like” Polar / ionic solute dissolve easier in polar solvent. Non-polar dissolve easier in non-polar. Even the most insoluble ionic solids are actually soluble in water to a limited extent

2. Temperature Solids in liquids: ↑ temperature - ↑ solubility. Gases in liquids: ↑ in temperature - ↓ solubility. 3. Pressure Changes in pressure do not affect the solubility of solids or liquids. Solubility of gases increases with an increase in pressure.

A a B b(s) aA + (aq) + bB¯ (aq) K sp, called the solubility product constant. K sp = [A + ] a [B - ] b Product of ion concentrations in a saturated solution. K c = [A + ] a [B - ] b [A a B b ]

Write the dissociation and the product constant equation for the solubility of calcium hydroxide. K sp = [Ca 2+ ][OH - ] 2 Ca(OH) 2 (s ) Pb 3 (PO 4 ) 2(s) 3 Pb 2+ (aq) + 2 PO 4 3- (aq) K sp = [Pb 2+ ] 3 [PO 4 3- ] 2 Write a solubility product expression for Pb 3 (PO 4 ) 2. Ca 2+ (aq) + OH - (aq) 2

At equilibrium, the [Ag + ] = 1.3 x M and the [Cl - ] = 1.3 x M, what is the K sp of silver chloride? K sp = [Ag + ][Cl - ] K sp =(1.3 x )(1.3 x ) K sp = 1.7 x AgCl (s ) Ag + (aq) + Cl - (aq) *NOTE: K sp has no units.

Solubility And I.C.E. Tables (Yeah!) Solubility - maximum amount of solute that can dissolve in a certain amount of solvent at a certain temperature.

Calculate K sp of lead (II) chloride if a 1.0 L saturated solution has of lead ions. I---00 C--- +x+2x E--- K sp = [Pb +2 ][Cl - ] 2 K sp = [1.62 x ][ x ] 2 K sp = 1.70 x PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) 1.62 x M 2(1.62 x ) 1.62 x M

The solubility of PbF 2 is. What is the value of the solubility product? PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) g 1 L g 1 mol = 1.90 x mol/L g/L Pb – (19) = 245g/mol *1L saturated solution – must add 1.90 x moles of PbF 2 *

K sp = [Pb 2+ ][F - ] 2 K sp = (1.90 x )(3.80 x ) 2 K sp = 2.74 x PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) [E][E] x M [I][I] 1.9 x mol/L 00 [C][C]- x+ x+ 2x 3.8 x M Saturated – all solid reactant dissociates.

Calculate K sp if 50.0 mL of a saturated solution was found to contain g of lead (II) chloride. I C-x +x +2x E M M K sp = [Pb 2+ ][Cl - ] g 278.1g 1 mol = M PbCl L PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) K sp = [0.0159][0.0318] 2 = 1.61 x 10 -5

K sp of magnesium hydroxide is 8.9 x What are the [equilibrium] of ions in saturated solution? I---00 C---+x+2x E--- x 2x Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) K sp = [Mg 2+ ][OH - ] x = [x][2x] x = [x] 4x x = 4x 3

[Mg 2+ ] = x = 1.3 x mol/L [OH - ] = 2x = 2.6 x mol/L 8.9 x = 4x x = x 3 3√3√ 3√3√ 1.3 x = x

Estimate the solubility in g/L of Ag 2 CrO 4 if the K sp is 1.1 x I---00 C---+2x+x E--- K sp = [Ag + ] 2 [CrO 4 2- ] 1.1 x = [2x] 2 [x] 1.1 x = 4x 3 x = 6.50 x M Ag 2 CrO 4(s)  2 Ag + (aq) + CrO 4 2- (aq) 1.30 x M 6.50 x M

[Ag 2 CrO 4 ] i = 6.50 x moles/1L Ag 2 CrO 4(s)  2 Ag + (aq) + CrO 4 2- (aq) 6.5 x mol 330 g 1 mol = g/L 1 L E x M 6.50 x M 1 1

Compare value of Q, with given K sp to determine if an aqueous solution is saturated or unsaturated. Q = K sp Saturated solution, no precipitate. Q >K sp Precipitate forms (“oversaturated”) Q < K sp Solution is unsaturated. Q sp = [A + ] a [B¯] b

PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) Pb – (19) = 245g/mol K sp of lead (II) fluoride is 1.6 x If 0.57 g are mixed with 1500 mL of water, is solution saturated? 0.57 g 1 L g 1 mol = 2.32 x mol/L K sp = [Pb 2+ ][F - ] 2 K sp = (2.32 x )(4.64 x ) 2 K sp = 5.0 x Q >K sp Precipitate forms

 Substances which are insoluble are actually slightly soluble.  The solubility product, K sp, describes the product of ion concentrations in saturated solutions.  Solubility can be determined from the solubility product.