Question

Given that the Ksp for Magnesium Hydroxide is 1.2 x 10-11 . calculate the solubility of this compound in grams per 100 mL of solution.

Answer 1

Consider reaction Mg(OH)_{2 (s)}   Mg ²⁺_{(aq )} + 2 OH ^-_(aq)  

For above reaction, K sp = [ Mg ²⁺] [ OH ^-] ²  

Assume S is the solubility of Mg(OH)₂ in mol / L then [ Mg ²⁺] = S mol / L and [ OH ^-]= 2 S mol / L .

Therefore, K sp = S ( 2 S) ² = 1.2 10 ^-11

4 S ³ = 1.2 10 ^-11

S ³ = 1.2 10 ^-11  /4

S ³ = 3 10 ^-12

S = 1.44 10 ^-04 mol / L

solubility of Mg(OH)₂ in g / L = solubility of Mg(OH)₂ in mol / L Molar mass

solubility of Mg(OH)₂ in g / L = 1.44 10 ^-04 mol / L 58.32 g / mol = 0.008398 g / L

solubility of Mg(OH)₂ in g / 100 ml = solubility of Mg(OH)₂ in g / L / 10

= 0.008398 g / L / 10

=8.40 10 ^-04 g / 100 ml

ANSWER : solubility of Mg(OH)₂ in g / 100 ml = 8.40 10 ^-04 g / 100 ml