Learning Goal To learn how to calculate the solubility from Ksp and vice versa.

Question

Learning Goal

To learn how to calculate the solubility from Ksp and vice versa. Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution:

CaF2 (s) --> Ca + 2F

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

Ksp = [Ca][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 perunit volume of saturated solution.

PartA

A saturated Solution of lead (II) fluoride, PbF2 in water. The concentration of Pb ion in the solution was found to be 2.08*10^-3 M

Calculate Ksp for PbF2

Part B

The Value of Ksp for silver carbonate, Ag2CO3, is 8.10*10^-12

Calculate the solubility of Ag2CO3 in grams per liter.

Explanation / Answer

PbF2 ----> Pb2+ + 2F-

x 2x

and given x = 2.08*10^-3 M

ksp = x*(2x)^2 = 4x^3

ksp = 3.6*10^-8

let solubility be x

Ag2CO3 ----> 2Ag+ + CO32-

-x 2x x

ksp = x*(2x)^2 = 4x^3 = 8.1*10^-12

x = 1.27*10^-4

solubility = 1.27*10^-4 M

solubility in g/L = 275.74*1.27*10^-4 = 0.035gm/L

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