± Introduction to Solubility and the Solubility Product Constant Learning Goal:

Question

± Introduction to Solubility and the Solubility Product Constant

Learning Goal:

To learn how to calculate the solubility from Kspand vice versa.

CaF2(s)Ca2+(aq)+2F(aq)

Ksp=[Ca2+][F]2

Part A

A saturated solution of lead(II) fluoride, PbF2, was prepared by dissolving solid PbF2 in water. The concentration of Pb2+ ion in the solution was found to be 2.08×103M . Calculate Ksp for PbF2.

Express your answer numerically.

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Part B

The value of Ksp for silver carbonate, Ag2CO3, is 8.10×1012. Calculate the solubility of Ag2CO3 in grams per liter.

Express your answer numerically in grams per liter.

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± Introduction to Solubility and the Solubility Product Constant

Learning Goal:

To learn how to calculate the solubility from Kspand vice versa.

Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution:

CaF2(s)Ca2+(aq)+2F(aq)

At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid CaF2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is

Ksp=[Ca2+][F]2

Ksp is called the solubility product and can be determined experimentally by measuring the solubility, which is the amount of compound that dissolves per unit volume of saturated solution.

Part A

A saturated solution of lead(II) fluoride, PbF2, was prepared by dissolving solid PbF2 in water. The concentration of Pb2+ ion in the solution was found to be 2.08×103M . Calculate Ksp for PbF2.

Express your answer numerically.

Ksp =

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Part B

The value of Ksp for silver carbonate, Ag2CO3, is 8.10×1012. Calculate the solubility of Ag2CO3 in grams per liter.

Express your answer numerically in grams per liter.

solubility =   g/L  

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Explanation / Answer

part A

ksp = [Pb2+][F-]^2

    = (2.08*10^-3)(2*2.08*10^-3)^2

    = 3.6*10^-8

part B

ksp of Ag2co3 = (2S)^2(S)

8.1*10^-12 = 4S^3

S = Solubility = 1.265*10^-4 M

solubility in g/L = (1.265*10^-4)*275.7453 = 0.035 g/L

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