Final answer:
The solubility of magnesium hydroxide in grams per liter is calculated as 0.0082 g/L using the Ksp value of 5.61 x 10^-12 and its molar mass of 58.3 g/mol.
Step-by-step explanation:
The solubility of magnesium hydroxide, Mg(OH)2, in grams per liter can be calculated using its solubility product constant (Ksp).
The balanced dissolution equation for Mg(OH)2 is:
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2 OH- (aq)
The Ksp expression for this reaction is:
Ksp = [Mg2+] [OH-]2
Given the Ksp value of 5.61 x 10-12, let 'x' be the molarity of Mg2+ at equilibrium. Then [Mg2+] = x and [OH-] = 2x. Substituting these into the Ksp expression gives:
5.61 x 10-12 = x * (2x)2 = 4x3
Solving for x yields approximately 1.4 x 10-4 mol/L. To find the solubility in grams per liter, we multiply the molarity of Mg(OH)2 by its molar mass, 58.3 g/mol:
Solubility = (1.4 x 10-4 mol/L) * (58.3 g/mol) ≈ 0.0082 g/L