Part A - Calculate the molar solubility in water Mg(OH) 2 is a sparingly soluble

Question

Part A - Calculate the molar solubility in water

Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×1011. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.

Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in pure H2O?

Part B - Calculate the molar solubility in NaOH

Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.160 M NaOH?

Explanation / Answer

PART A:

The equilibrium is

Mg(OH)2 (s) <=> Mg2+ (aq) + 2 OH- (aq)

in pure water : [Mg2+] = x and [OH-]= 2x

Ksp = [Mg2+][OH-]2 = (x)(2x)2 = 4x3

5.61 x 10-11 = 4x3

x3 = 1.40 x 10-11

x = 2.41 x 10-4 M = molar solubility

PART B:

In 0.160 M NaOH;

[Mg2+]= x and [OH-] = 2x + 0.160

we can assume 2x + 0.160 = 0.160 (since x << 0.160)

5.61 x 10-11 = x ( 0.160)2

x = 2.19 x 10-9 M = molar solubility

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