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Solubility Equilibria

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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.

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

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

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2. Temperature Solids in liquids: temperature - solubility. Gases in liquids: in temperature - solubility. 3. Pressure Does not affect the solubility of (s)/(l). (g): pressure solubility.

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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 ]

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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

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At equilibrium, the [Ag + ] = 1.3 x 10 -5 M and the [Cl - ] = 1.3 x 10 -5 M, what is the K sp of silver chloride? K sp = [Ag + ][Cl - ] K sp =(1.3 x 10 -5 )(1.3 x 10 -5 ) K sp = 1.7 x 10 -10 AgCl (s ) Ag + (aq) + Cl - (aq) *NOTE: K sp has no units.

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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.

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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 10 -2 ][ 3. 24 x 10 -2 ] 2 K sp = 1.70 x 10 -5 PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) 1.62 x 10 -2 M 2(1.62 x 10 -2 ) 1.62 x 10 -2 M

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The solubility of PbF 2 is. What is the value of the solubility product constant? PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) 0.466 g 1 L 245.2 g 1 mol = 1.90 x 10 -3 M PbF 2 0.466 g/L Pb – 207 + 2 (19) = 245g/mol K sp = [Pb 2+ ][F - ] 2

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K sp = (1.90 x 10 -3 )(3.80 x 10 -3 ) 2 K sp = 2.74 x 10 -8 PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) [E][E] 0 1.9 x 10 -3 M [I][I] 1.9 x 10 -3 mol/L 00 [C][C]- x+ x+ 2x 3.8 x 10 -3 M Saturated – all solid reactant dissociates.

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Calculate K sp if 50.0 mL of a saturated solution was found to contain 0.2207 g of lead (II) chloride. I0.0159 0 0 C-x +x +2x E0 0.0159 M0.0318 M K sp = [Pb 2+ ][Cl - ] 2 0.2207 g 278.1g 1 mol = 0.0159 M PbCl 2 0.05 L PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) K sp = [0.0159][0.0318] 2 = 1.61 x 10 -5

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K sp of magnesium hydroxide is 8.9 x 10 -12. 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 - ] 2 8.9 x 10 -12 = [x][2x] 2 8.9 x 10 -12 = [x] 4x 2 8.9 x 10 -12 = 4x 3

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[Mg 2+ ] = x = 1.3 x 10 -4 mol/L [OH - ] = 2x = 2.6 x 10 -4 mol/L 8.9 x 10 -12 = 4x 3 44 2.23 x 10 -12 = x 3 3 3 1.3 x 10 -4 = x

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Estimate the solubility in g/L of Ag 2 CrO 4 if the K sp is 1.1 x 10 -12. I---00 C---+2x+x E--- K sp = [Ag + ] 2 [CrO 4 2- ] 1.1 x 10 -12 = [2x] 2 [x] 1.1 x 10 -12 = 4x 3 x = 6.50 x 10 -5 M Ag 2 CrO 4(s) 2 Ag + (aq) + CrO 4 2- (aq) 1.30 x 10 -4 M 6.50 x 10 -5 M

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[Ag 2 CrO 4 ] i = 6.50 x 10 -5 moles/1L Ag 2 CrO 4(s) 2 Ag + (aq) + CrO 4 2- (aq) 6.5 x 10 -5 mol 330 g 1 mol = 0.022 g/L 1 L E---1.30 x 10 -4 M 6.50 x 10 -5 M 1 1

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Precipitation

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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

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PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) Pb – 207 + 2 (19) = 245g/mol K sp of lead (II) fluoride is 1.6 x 10 -5. If 0.57 g are mixed with 1500 mL of water, is solution saturated? 0.57 g 1 L 245.2 g 1 mol = 2.32 x 10 -3 mol/L Q sp = (2.32 x 10 -3 )(4.64 x 10 -3 ) 2 Q sp = 5.0 x 10 -8 Q >K sp Precipitate forms K sp = [Pb 2+ ][F - ] 2

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0.01 M NaCl0.02 M Pb(NO 3 ) 2 Predict if there is a precipitate of PbCl 2 if 100 mL each of of and are added together. K sp of PbCl 2 = 1.7 x 10 -5 PbF 2(s) Pb 2+ (aq) + 2 Cl¯ (aq) Q sp = [Pb 2+ ][Cl - ] 2 0.01 M NaCl0.02 M Pb(NO 3 ) 2

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C 1 V 1 = C 2 V 2 (0.01 M)(0.1 L) = (C 2 )(0.2 L) = 0.005 M Cl - (0.02 M)(0.1 L) = (C 2 )(0.2 L) = 0.01 M Pb 2+ Q sp = [0.01][0.005] 2 = 2.5 x 10 -7 Q sp < K sp A precipitate will not form. Because you are adding volumes, you must account for the new diluted concentrations.

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If 20.0 mL of 0.0010 M silver nitrate is mixed with 20.0 mL of 3.0 x 10 -5 M potassium bromide, does silver bromide (K sp = 5.0 x 10 -13 ) precipitate? Assume the volumes are additive. AgBr (s) Ag + (aq) + Br¯ (aq) K sp = [Ag + ][Br - ] (0.001 M)(0.02 L) = (C 2 )(0.04 L) = 5.0 x 10 -4 M Ag + (2e -5 M)(0.02 L) = (C 2 )(0.04 L) = 1.5 x 10 -5 M Br - Q sp = [5.0 x 10 -4 ][1.5 x 10 -5 ] = 7.5 x 10 -9 Q sp > K sp A precipitate will form.

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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.

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