100 ml of a saturated solution of Mg(OH)2 contains 8.2  10-4 g solute. Mol. wt. of Mg(OH)2 is 58.3 g mole-1; calculate its Ksp.

Sure, here is the step by step solution:

Step 1: Calculate the molarity of Mg(OH)2 Molarity = mass of solute (g) / molar mass (g/mol) / volume of solution (L) Molarity = 8.2 x 10^-4 g / 58.3 g/mol / 0.1 L = 1.41 x 10^-5 mol/L

Step 2: Write the solubility product expression for Mg(OH)2 Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq) Ksp = [Mg2+][OH-]^2

Step 3: Substitute the molarity into the Ksp expression Since the stoichiometry of the reaction shows that for every 1 mole of Mg(OH)2 that dissolves, 1 mole of Mg2+ and 2 moles of OH- are produced, the molar concentration of Mg2+ is the same as the molarity of Mg(OH)2, and the molar concentration of OH- is twice the molarity of Mg(OH)2.

[Mg2+] = 1.41 x 10^-5 mol/L [OH-] = 2 x 1.41 x 10^-5 mol/L = 2.82 x 10^-5 mol/L

Ksp = (1.41 x 10^-5) x (2.82 x 10^-5)^2 = 1.12 x 10^-14

So, the Ksp of Mg(OH)2 is 1.12 x 10^-14.