Question

Mg(OH)2 is a sparingly soluble salt with a solubility product constant, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. Calculate the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.150 mol L−1 NaOH solution.

Answer 1

Sol :-

Let Solubility of Mg(OH)₂ in pure water = S mol/L

Partial dissociation of Mg(OH)₂ is :

Mg(OH)₂ (s) <-----------------> Mg²⁺ (aq) + 2OH^- (aq)

..............................................S mol/L.........2S mol/L

Expression of K_sp is :

K_sp = [Mg²⁺].[OH^-]²

5.61×10⁻¹¹ = S.(2S)²

4S³ = 5.61×10⁻¹¹

S = (1.40 x 10^-11 )^1/3

S = 2.40 x 10^-4 mol/L

Let Molar solubility of Mg(OH)₂ in 0.150 M NaOH is = S' mol/L

DSo, Total concentration of common ion i.e. OH^- = (2S' + 0.150 ) M

Expression is :

K_sp = [Mg²⁺].[OH^-]²_Total

5.61×10⁻¹¹ = S'(2S' + 0.150)²

Let 2S' <<<0.150, so neglect 2S'

S' = 5.61×10⁻¹¹ / (0.150)²

S' = 2.49 x 10^-9 mol/L

So, S/S' = 2.40 x 10^-4 mol/L / 2.49 x 10^-9 mol/L = 9.64 x 10⁴