Chemistry Solubility Math Question 8

Question

Compute the solubility of Mg(OH)₂ in a buffer solution at pH=10.93. The solubility product of magnesium hydroxide is 1.2E-11

Answer 1

Mg(OH)₂(s) <==> Mg²⁺(aq) + 2OH^-(aq)

Ksp = 1.2x10^-11 = [Mg²⁺][OH^-]²

From a pH = 10.93, we can find [OH-], and then solve for [Mg²⁺].

pH + pOH = 14

pOH = 14 - pH = 14 - 10.93

pOH = 3.07

[OH-] = 1x10^-3.07 = 8.5x10^-4 M

1.2x10^-11 = [Mg²⁺][8.5x10^-4]²

1.2x10^-11 = [Mg²⁺][7.2x10^-7]

[Mg²⁺] = 1.7x10^-5 M = molar solubility of Mg(OH)2 at pH 10.93