Question

Purifying Mg2+ from sea water. Aqueous sodium hydroxide is added to an aqueous solution that contains 0.0099 M magnesium chloride and 0.021 M calcium chloride. What will be the concentration of the metal ion that precipitates first (as M(OH)2) at the time the second metal ion begins to precipitate? Ksp (Mg(OH)2) = 6.3 x 10-10 and Ksp (Ca(OH)2) = 6.5 x 10-6

Hint: Use Ksp values to determine the solubility of each product to determine the order of precipitation.

Answer 1

Given : concentration of Mg²⁺ = [Mg²⁺] = 0.0099 M

concentration of Ca²⁺ = [Ca²⁺] = 0.021 M

(a.) The dissociation equation of Mg(OH)₂ is : Mg(OH)₂ (s) Mg²⁺ (aq) + 2 OH^- (aq)

Let molar solubility of Mg(OH)₂ = s

[Mg²⁺] = [Mg(OH)₂] + concentration of Mg²⁺

[Mg²⁺] = s + 0.0099 M

[OH^-] = 2 * [Mg(OH)₂]

[OH^-] = 2s

K_sp [Mg(OH)₂] = [Mg²⁺][OH^-]²

6.3 x 10^-10 = (s + 0.0099 M) * (2s)²

Assuming s << 0.0099 M

6.3 x 10^-10 = (0.0099 M) * (2s)²

6.3 x 10^-10 = (0.0099 M) * (4s²)

s² = (6.3 x 10^-10) / (0.0099 M)

s² = 6.36 x 10^-8

s = (6.36 x 10^-8)^1/2

s = 2.52 x 10^-4 M

[OH^-] = 2s

[OH^-] = 2 * (2.52 x 10^-4 M)

[OH^-] = 5.04 x 10^-4 M

(b.)

The dissociation equation of Ca(OH)₂ is : Ca(OH)₂ (s) Ca²⁺ (aq) + 2 OH^- (aq)

Let molar solubility of Ca(OH)₂ = s

[Ca²⁺] = [Ca(OH)₂] + concentration of Ca²⁺

[Ca²⁺] = s + 0.021 M

[OH^-] = 2 * [Ca(OH)₂]

[OH^-] = 2s

K_sp [Ca(OH)₂] = [Ca²⁺][OH^-]²

6.5 x 10^-6 = (s + 0.021 M) * (2s)²

Solving for x, s = 7.545 x 10^-3 M

[OH^-] = 2s

[OH^-] = 2 * (7.545 x 10^-3 M)

[OH^-] = 0.0151 M

Since less concentration of OH^- is required to precipitate Mg(OH)₂, therefore, Mg(OH)₂ will precipitate first and Ca(OH)₂ will precipitate second.

concentration of OH^- when Ca²⁺ begins to precipitate = 0.0151 M

K_sp Mg(OH)₂ = [Mg²⁺][OH^-]²

6.3 x 10^-10 = [Mg²⁺] * (0.0151 M)²

[Mg²⁺] = (6.3 x 10^-10) / (0.0151 M)²

[Mg²⁺] = 2.8 x 10^-6 M