Solubility Product, Ksp Write the dissociation equation and the solubility product (Ksp) expression for the dissociation of each of the following substances. a) PbSO4 (s) d) AlCl3 (s) 2) The solubility of calcium hydroxide (Ca(OH)2) is 0.02 mol/L.  Calculate the solubility product constant for this substance. 3) Calculate the solubility (in mol/L) for each of the following compounds: a)FeS Ksp = 3.7 x 10-19 b)Mg(OH)2 KSp = 9.0 x 10-12

a)

Ksp = s²

d)

Ksp = 27s³

2)

Ksp = 3.2 * 10^-5

3)

a)

solubility of FeS = 6.08 * 10^-10 mole / L

b)

solubility of Mg(OH)2 = 1.31 * 10^-4 mole / L

Step-by-step explanation

 a)

s = molar solubility.

PbSO4 (s).............> Pb^2+(aq) + SO42- (aq)

................................................s (M).........s (M)

Ksp = [Pb^2+][SO42-]

or

Ksp = s * s

or

Ksp = s²

d) AlCl3 (s) ............> Al^3+ (aq) + 3 Cl- (aq)

............................................s (M).................3s (M)

Ksp = [Al^3+][Cl-]^3

or

Ksp = s * (3s)^3

or

Ksp = 27s³

2) The solubility of calcium hydroxide (Ca(OH)2) is 0.02 mol/L. Calculate the solubility product constant for this substance.

Ca(OH)2.............> Ca^2+ + 2 OH-

........................................s (M).......2s (M)

Ksp = [Ca^2+][OH-]^2

or

Ksp = s * (2s)^2

or

Ksp = 4s^3 = 4 * (0.02)^3 = 3.2 * 10^-5

3) Calculate the solubility (in mol/L) for each of the following compounds:

a)FeS Ksp = 3.7 x 10-19

Ksp = [Fe^2+][S^2-]

Ksp = s * s

3.7 x 10-19 = s^2

s = 6.08 * 10^-10 M

solubility of FeS = 6.08 * 10^-10 M

b)Mg(OH)2 KSp = 9.0 x 10-12

Ksp = [Mg^2+][OH-]^2

or

Ksp = s * (2s)^2

or

Ksp = 4s^3

or

9.0 * 10^-12 = 4s^3

or

s = 1.31 * 10^-4 M

thus

solubility = 1.31 * 10^-4 mole / L