Question

A groundwater containing 10-4 M of Mg2+ is at equilibrium. Use the solubility constant Ksp = 1.35 x 10^-11. Assuming no other ions are present, (a) what is the equilibrium pH,

(b) will Mg2+ precipitate if the pH is increased to 12 (show calculations)?

Answer 1

(a) what is the equilibrium pH,

Solution-

Given

[Mg2+] = 10^-4 M

Ksp = 1.35 x 10^-11.

Here equation for Mg2+ is as folows

Mg²⁺ + 2 OH¯--> Mg(OH)₂

Lets write the Ksp equation

K_sp = [Mg²⁺] [OH¯]²

1.35 x 10^-11.= (10^-4 M) [OH¯]²

[OH¯]² = 1.35 x 10^-7

[OH¯] = 3.67 x 10^-5 M

Lets calculate bthe POH

pOH = -log[OH]

        = - log (3.67 x 10^{-5 )}= 4.43

pOH = 4.43

We know the relation between PH & POH

PH + POH =14

pH = 14 – 4.43 = 9.57

Answer = PH at equilibrium = 9.57

(b) will Mg2+ precipitate if the pH is increased to 12 (show calculations)?

Solution-

Given

PH = 12

We know the euation

PH = -log[H⁺]

-log[H⁺] = 12

Taking antilog on both the side

[H⁺] = 1 X 10^-12 M

POH = 14-12 =2

PoH =-log[OH]²

-log[OH^-]² = 2

Taking antilog on both the side

[OH^-]² = 0.01 M

[OH^-] = 0.1M

Answer = [OH-] ion is more therefore Mg2+ precipitates as hydroxide.