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

Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2(OH)¯(aq)

Mg(OH)2(s) <==> Mg2+ + 2OH- .......... Ksp = 5.61x10^-11
...........................0 ....... 0.100M ........ initial
.......................... +x ..... +2x ............. change
...........................x ....... 0.150+2x ..... equilibrium

Ksp = [Mg2+] [OH-]²
5.61x10^-11 = x(0.150+2x)²
x = 2.49x10^-9

Mg(OH)2 solubility = 2.49x10^-9M in NaOH

In plain water.....
Mg(OH)2(s) <==> Mg2+ + 2OH- .......... Ksp = 5.61x10^-11
...........................0 .......... 0 ........ initial
.......................... +x ..... +2x ....... change
...........................x ......... 2x ....... equilibrium

Ksp = [Mg2+] [OH-]²
5.61x10^-11 = 4x³
x = 2.31x10^-4

The solubility of Mg(OH)2 in plaine water is 2.31x10^-4M

The ratio ...... 2.31x10^-4M / 2.49x10^-9M = 92,771 ..... Mg(OH)2 is over 92 thousand times more soluble in water than in 0.15M NaOH.