Chap 20 - 20 Ionic Equilibria III: The Solubility Product...

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1 20 Ionic Equilibria III: The Solubility Product Principle
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2 Solubility Product Constants Silver chloride, AgCl,is rather insoluble in water. Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water. AgCl (s) Ag + (aq) + Cl - (aq)
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3 Solubility Product Constants Silver chloride, AgCl,is rather insoluble in water. Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.
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4 Solubility Product Constants The equilibrium constant expression for this dissolution is called a solubility product constant, K sp . -10 - sp 10 1.8 ] ][Cl [Ag K × = = +
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5 Solubility Product Constants The solubility product constant, K sp , for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. Consider the dissolution of silver sulfide in water. Ag 2 S (s) 2 Ag + (aq) + S 2-
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6 Solubility Product Constants The solubility product expression for Ag 2 S is: [ ] [ ] 49 10 0 1 + × = = . S Ag K 2 2 sp
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7 Solubility Product Constants The dissolution of solid calcium phosphate in water is represented as: [ ] [ ] 25 10 0 1 + × = = . PO Ca K 2 3 4 3 2 sp The solubility product constant expression is: Ca 3 (PO 4 ) 2(s) 3 Ca 2+ (aq) + 2 PO 4 3- (aq)
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8 Solubility Product Constants The same rules apply for compounds that have more than two kinds of ions. One example of a compound that has more than two kinds of ions is calcium ammonium phosphate. [ ][ ][ ] + + = 3 4 4 2 sp PO NH Ca K CaN H 4 PO 4(s) 2+ (aq) + NH 4 + (aq) + PO 4 3- (aq)
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9 Determination of Solubility Product Constants Example 20-1: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25 o C. Calculate the molar solubility of, and K sp for, AgCl. M L mol ?? L AgCl g . = = 00192 0
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10 Determination of Solubility Product Constants AgCl . AgCl g 143 AgCl mol L AgCl g . M 5 10 34 1 1 00192 0 × = ×
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11 Determination of Solubility Product Constants The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are: M M -5 -5 10 1.34 10 1.34 × × AgCl (s) Ag + (aq) + Cl - (aq) [ ][ ] + = Cl K sp
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12 Determination of Solubility Product Constants [ ][ ] 10 5 5 10 8 1 10 34 1 10 34 1 × = × × = . K . . K sp
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13 Determination of Solubility Product Constants Example 20-2: One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF 2 at 25 o C. Calculate the molar solubility of, and K sp for, CaF 2 . L 1.0 CaF g . L CaF mol ? 2 2 0167 0 = L CaF . g 78.1 1 L CaF g . 2 2 4 10 14 2 0167 0 × = ×
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14 Determination of Solubility Product Constants From the molar solubility, we can find the ion concentrations in saturated CaF 2 . Then use those values to calculate the K sp . ) M . 2( M . M . 4 4 4 10 14 2 10 14 2 10 14 2 × × × CaF 2(s) Ca 2+ (aq) + 2 F -
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15 Determination of Solubility Product Constants [ ][ ] 2 2 sp F Ca K + = ( )( ) 2 4 4 10 28 4 10 14 2 × × = . . K 11 10 92 3 × = . K
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16 Uses of Solubility Product Constants The solubility product constant can be used to calculate the solubility of a compound at 25 o C.
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This note was uploaded on 09/18/2010 for the course CHEM 1212 taught by Professor Suggs during the Fall '08 term at University of Georgia Athens.

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Chap 20 - 20 Ionic Equilibria III: The Solubility Product...

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