Question

The solubility of Mg(OH)2 in water is approximately 9 mg/L.

(a) Calculate the Ksp of magnesium hydroxide.

(b) Calculate the hydroxide concentration needed to precipitate Mg2+ ions such that no more than 5.0 μg/L Mg2+ remains in solution.

(c) Calculate the maximum concentration of Mg2+ (as M) that can exist in a solution of pH = 12.

Answer 1

Mg(OH)₂ molar mass is 58.319 g/mol

so 9 mg will be

9 x 10^-3/58.319 = 0.15 x 10^-3 moles/L

When Mg(OH)₂ dissolves in water it dissociates into

Mg(OH)₂ (aq) Mg²⁺(aq) + 2OH^- (aq)

Ksp = [Mg²⁺][OH^-]²

When 0.15 x 10^-3 moles/L of Mg(OH)₂ dissolves in water it will produce 0.15 x 10^-3 moles of Mg²⁺ and 0.3 x 10^-3 moles of OH^- per litre each.

Ksp = 0.15 x 10^-3 x (0.3 x 10^-3 )²

a) Ksp = 1.35 x 10^-11

b) To precipitate Mg²⁺ as Mg(OH)₂ you need two equivalent of OH^- for every Mg²⁺

The final concentration that should remain is 5g/L which is 5 x 10^-6/58.319 =8.57 x 10^-8moles/L

so we need to reduce Mg²⁺ concentration from 0.15 x 10^-3 moles/L to 8.57 x 10^-8moles/L

so Mg²⁺ reduction is 0.15 x 10^-3 moles/L - 8.57 x 10^-8moles/L = 1.499 x 10^-4moles

To accomplish this we will need 1.499 x 10^-4moles x 2 = 2.998 x 10^-4 moles of OH^-

b) the hydroxide concentration needed is 2.998 x 10^-4 moles of OH^-

c) At pH=12 the OH- concentration is

pOH = 14-pH

pOH = 14-12

pOH = 2

[OH^-] = 10^-pOH

[OH^-] = 10^-2

[OH^-] = 0.01 M

at this pH the concentration of OH^- is 1 x 10^-2 moles/L

Ksp = [Mg²⁺][OH^-]²

1.35 x 10^-11 = [Mg²⁺] x (1 x 10^-2)²

[Mg²⁺] = 1.35 x 10^-11 / 1 x 10^-4

[Mg²⁺] = 1.35 x 10^-7 mole/L

c) the maximum concentration of Mg2+ (as M) that can exist in a solution of pH = 12 = 1.35 x 10^-7 mole/L or 1.35 x 10^-7 M