Question

Stron U14 The Ksp of magnesium hydroxide, Mg(OH)2, is 5.61 x 10-12. Calculate the solubility of this compound in grams per liter.

Answer 1

Step 1: Explanation

The Ksp expression for a salt is the product of the concentrations of the ions, with each concentration raised to a power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium.

Step 2: Write the balanced decomposed reaction of Mg(OH)₂ at equilibrium

Mg(OH)₂(s) ⇌ Mg⁺² + 2 HO^-

Step 3: Calculate the solubility of compound

And now if we call the solubility of Mg(OH)₂ = S, then by definition, S = [Mg²⁺], and 2S = [⁻OH]

Mg(OH)₂(s) ⇌ Mg⁺² + 2 HO^-

Ksp = [Mg²⁺] × 2 [⁻OH]²

Ksp = S × ( 2S)² = 4S³

thus, Ksp = 4S³

if Ksp = 5.61 × 10^-12

then 5.61 × 10^-12 = 4S³

or, S = 3√ ( 5.61 × 10^-12 ) / 4

or, S = 1.12 × 10^-4 mol/L

to get solubility in mass multiply it by molar mass

S = 1.12 × 10^-4 mol/L × 58.32 g/mol = 6.53 × 10^-3 g/L