# Chap17B - 1 Chapter 17 More Aqueous Equilibria-Part 2...

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Chapter 17: More Aqueous Equilibria-Part 2. Solubility of Slightly Soluble Salts Factors that Affect Solubility Separation of Ions and Precipitation IV. Solubility Equilibria for slightly soluble salts. A. When a slightly soluble substance is stirred in water only a very small amount dissolves but careful studies show that the amount that dissolves is completely dissociated. Sr 2 (CO 3 ) 3 (s) <---> 2 Sr 3+ (aq) + 3 CO 3 2- (aq) 1. The K c for this equilibrium is K c = [Sr 3+ ] 2 [CO 3 2- ] 3 but since this is the product of the ion concentrations for the solubility of a salt it is called the solubility product and the subscript “sp” is used rather than “c”. K sp = [Sr 3+ ] 2 [CO 3 2- ] 3 2. Write the K sp for the following: a. Calcium phoshpate b. Barium carbonate c. Magnesium hydroxide d. Lead (III) sulfide e. Silver chloride B. Experimental determination of K sp . 1. A 1.00 L solution saturated with Mg(OH) 2 is found to contain 0.0090 g of dissolved Mg(OH) 2 . Calculate the K sp for Mg(OH) 2 (molar mass = 58.3g/mol). a. First calculate the molar solubility of Mg(OH) 2 . DO ANS : 1.54 x 10 -4 . What is the solubility for each ion? Mg(OH) 2 <---> Mg 2+ (aq) + 2 OH - (aq) 1.54 x 10 -4 3.08 x 10 -4 [Mg 2+ ] = 1.54 x 10 -4 [OH - ] = 3.08 x 10 -4 b. Now write the K sp expression K sp = [Mg 2+ ] [OH - ] 2 = (1.54 x 10 -4 )(3.08 x 10 -4 ) 2 = 1.5 x 10 -11 . 2. K sp values are in Appendix D. These are experimental numbers and different values are often found in other tables. The K sp are constant at a specific T and will often increase as the T is raised. However, as we discussed earlier for some salts the solubility declines as T is raised. C. Calculation of solubility using the K sp . 1. Calculate the molar solubility of silver carbonate, the concentrations of each ion and the solubility of silver carbonate in units of grams/100mL if the K sp = 8.1 x 10 -12 . Ag 2 CO 3 (s) <---> 2 Ag + (aq) + CO 3 2- (aq) DO At equil. 2y y define y = [Ag 2 CO 3 ] that dissolves. 1

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K sp = [Ag + ] 2 [CO 3 2- ] = 8.1 x 10 -12 = (2y) 2 y 8.1 x 10 -12 = 4y 3 y = 1.3 x 10 -4 = [Ag 2 CO 3 ] (the molar solubility) [Ag + ] = 2y = 2.6 x 10 -4 [CO 3 2- ] = y = 1.3 x 10 -4 Solubility in g/100mL. {1.3 x 10 -4 moles Ag 2 CO 3 /1.00L} {276g Ag 2 CO 3 /1.00mole} {1.00L/1000mL} = 3.6 x 10 -5 g Ag 2 CO 3 /1.00mL {3.6 x 10 -5 g Ag 2 CO 3 /1.00mL }{100mL} = 3.6 x 10 -3 g Ag 2 CO 3 /100mL 2. Calculate the molar solubility and the concentration of each ion in a silver carbonate solution that is 0.0500M in Na 2 CO 3 . This is like a common ion problem. What will happen to the solubility of Ag 2 CO 3 if there is some Na 2 CO 3 present? Ag 2 CO 3 (s) <---> 2 Ag + (aq) + CO 3 2- (aq) DO At equil. 2y y + 0.0500 define y = [Ag 2 CO 3 ] that dissolves.We know that y is small compared to 0.05 (Why do we know this? Because in the previous problem y = 1.3 x 10 -4 and it will be smaller here) so we can assume that y + 0.0500 = 0.0500. K
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## This note was uploaded on 12/03/2009 for the course CHM CHM140 taught by Professor Krish during the Spring '09 term at University of Toronto.

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Chap17B - 1 Chapter 17 More Aqueous Equilibria-Part 2...

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