Question

(a) Determine the equilibrium constant of Mg(OH)₂(s) dissolution at 25 ^oC.

(b) For a solution containing 10^-3 M dissolved Mg, at what pH will Mg(OH)₂(s) precipitate?

Answer 1

(a) Equilibrium constant

The dissolution of a slightly soluble solid is an equilibrium process given by the general chemical reaction:

The expression of the solubility product constant is given by the equation:

In this case, the dissolution of Mg(OH)_2(s)is under the equilibrium:

And the equilibrium constant is:

At 25ºC in pure water, solubility of Mg(OH)_2(s) is S = 12mg/L. We need to convert it into molar concentration units (mol/L) to be used in the equation of K_SP by using the molecular weight of Mg(OH)_2(s)

According to the equilibrium:

For every mol of Mg(OH)₂ dissolved we will have 1 mol dissolved of Mg⁺² and 2 moles dissolved of OH^-. Hence, for a saturated solution in the equilibrium we will have:

	Mg(OH)2		Mg+2	+	2OH-
Equilibrium solubility	S		S		2S

Substituting the equilibrium solubility in the equation (1)

Substituting known values (units are simplified):

THE EQUILIBRIUM CONSTANT OF Mg(OH)_2(s) DISSOLUTION AT 25ºC IS KSP = 3.54*10^-11

(B) pH of precipitation of a solution containing 10^-3M of Mg⁺²:

According to the equilibrium:

The precipitation will occur when solution is saturated, it is to say, when K_SP value is reached by equilibrium concentrations. We know the value of K_SP and the concentration of Mg⁺², we solve equation (1) to calculate [OH^-] value:

Substituting known values:

Calculating pOH by using the equation:

Applying the known relationship:

The pH of precipitation of the solution given is: