Solubility Product (Ksp)

The solubility product constant (Ksp) is a measure of the extent to which a solid salt dissolves in water to produce its constituent ions. It is the equilibrium constant for the dissociation reaction of the salt in water.

Ni(OH)2 (s) ⇌ Ni2+ (aq) + 2OH- (aq)

The expression for the solubility product constant (Ksp) is given by:

Ksp = [Ni2+][OH-]^2

Given that the Ksp of Ni(OH)2 is 2 x 10^-15, we can use this information to determine the molar solubility of Ni(OH)2 in 0.01M NaOH.

Using the Ksp to Determine Molar Solubility

To find the molar solubility of Ni(OH)2 in 0.01M NaOH, we need to compare the concentrations of Ni2+ and OH- ions in the solution with the Ksp value.

Since NaOH is a strong base, it completely dissociates in water to produce Na+ and OH- ions:

NaOH (aq) ⇌ Na+ (aq) + OH- (aq)

Therefore, the concentration of OH- ions in the solution is 0.01M.

Calculating Molar Solubility

Let's assume the molar solubility of Ni(OH)2 is x.

From the dissociation reaction of Ni(OH)2, we know that one mole of Ni(OH)2 produces one mole of Ni2+ and two moles of OH-.

Therefore, the concentration of Ni2+ ions in the solution is also x.

Using the expression for Ksp, we can write:

Ksp = [Ni2+][OH-]^2

Substituting the known values:

2 x 10^-15 = x * (0.01)^2

Simplifying the equation:

2 x 10^-15 = x * 0.0001

x = (2 x 10^-15) / 0.0001

x = 2 x 10^-11 M

Therefore, the molar solubility of Ni(OH)2 in 0.01M NaOH is 2 x 10^-11 M.