Question

A solution consist of 0.20 M MgCl2 and 0.1 M CuCl2. A) What concentration of sodium hydroxide would be needed to separate the magnesium and copper (II) ions? B) For the ion that precipitates, how much will remain in the solution?

Answer 1

Sol :-

Solubility product (K_sp) is the product of the molar concentration of products raised to the power of stoichiometric coefficient at equilibrium stage of the reaction.

Solubility product (K_sp) of sparingly soluble salt that is Cu(OH)₂ = 1.6 x 10^-19

Cu(OH)₂ (s) <------------------> Cu²⁺ (aq) + 2e^- , K_sp = 1.6 x 10^-19

and

Solubility product (K_sp) of sparingly soluble salt that is Mg(OH)₂ = 5.6 x 10^-12

Mg(OH)₂ (s) <------------------> Mg²⁺ (aq) + 2e^- , K_sp = 5.6 x 10^-12

Because, K_sp of Mg(OH)₂ > K_sp of Cu(OH)₂, therefore Cu(OH)₂ will form precipitates firstly.

Now, Expression of K_sp of Mg(OH)₂ is :

K_sp = [Mg²⁺].[OH^-]²

5.6 x 10^-12 = (0.20 M) [OH^-]²

[OH^-] = (28.0 x 10^-12)^1/2

[OH^-] = 5.29 x 10^-6 M

Hence, [OH^-] = 5.29 x 10^-6 M