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 2.08×10⁻³M . Calculate Ksp for PbF2.

Express your answer numerically.

Part B

The value of Ksp for silver sulfate, Ag2SO4, is 1.20×10−5. Calculate the solubility of Ag2SO4 in grams per liter.

Express your answer numerically in grams per liter.

Answer 1

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 2.08×10⁻³M . 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 1.20×10−5. 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