Learning Goal: To learn how to calculate the solubility from K spand vice versa.

Question

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 thesolubility, 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.

Part B

The value of Ksp for silver sulfate, Ag2SO4, is 1.20×105. Calculate the solubility of Ag2SO4 in grams per liter.

Express your answer numerically in grams per liter.

Explanation / Answer

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.

Solution :-

Lets write the dissociation equation of the PbF2

PbF2 ------ > Pb^2+ + 2F^-

Ksp equation is as follows

Ksp = [Pb^2+][F^-]2

Cocnnetration of the Pb^2+ = 2.08*10-3 M therefore concentration of the F^- is twice of the concentration of the Pb^2+

Therefore [F^-] = 2* 2.08*10-3 M = 4.16*10-3 M

Now lets use these concentrations in the ksp equation

Ksp = [2.08*10-3][ 4.16*10-3]2

Ksp = 3.60*10-8

Part B

The value of Ksp for silver sulfate, Ag2SO4, is 1.20×105. Calculate the solubility of Ag2SO4 in grams per liter.

Solution:-

Dissociation equation for the Ag2SO4 is as follows

Ag2SO4 ------ > 2Ag^+ + SO4^2-

                             2x                 x

Ksp equation is as follows

Ksp = [Ag^+]2[SO4^2-]

Ksp= [ 2x]2[x]

Lets put the ksp value in the formula and solve for the molar solubility

1.20*10-5 = [ 2x]2[x]

1.20*10-5 = 4x3

1.20*10-5 / 4 =x3

3*10-6 = x3

Taking cube root of both sides we get

0.01442 M =x

Now lets convert this molar solubility to g/L

Solubility in g/L= molar solubility * molar mass

                             = 0.01442 mol/L * 311.8 g per mol

                             = 4.50 g /L

Therefore the solubility of the Ag2SO4 = 4.50 g/L

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