Question

a) Calculate the solubility (in grams per 1.00×10^2mL of solution) of magnesium hydroxide in a solution buffered at pH = 12.

S1 = _________________ g/(1.00×10^2mL)

b) Calculate the solubility (in grams per 1.00×10^2mL of solution) of magnesium hydroxide in pure water.

S = __________________ g/(1.00×10^2mL)

c) How does the solubility of Mg(OH)2 in a buffered solution compare to the solubility of Mg(OH)2 in pure water?

S1/S = ________________

TIA.

Answer 1

Mg2+(aq) + OH(aq) → Mg(OH)2(s) Ksp = [Mg2+] · [OH-]

Hydroxide concentration is [OH-] = 10^(pH-14)

Hence the maximum magnesium ion concentration is
[Mg2+] = Ksp / [OH-]
= 1.2 * 10^-11 / 10^(12-14)
= 1.2 * 10^-9 M

The maximum soluble amount of magnesium hydroxide per liter is the same. Multiply by molar mass and you get the maximum solubility in g/L:
c(Mg(OH)2) = [Mg(OH)2] · M(Mg(OH)2)
= 1.2 * 10^-9 mol/L * (58.34 g/mol)
S1 = 7.0 * 10^-8 g/L

Mg2+(aq) + OH(aq) → Mg(OH)2(s) Ksp = [Mg2+] · [OH-]

Hydroxide concentration is [OH-] = 10^(pH-14)

Hence the maximum magnesium ion concentration is
[Mg2+] = Ksp / [OH-]
= 1.2 * 10^-11 / 10^(7-14)
= 1.2 * 10^-4 M

The maximum soluble amount of magnesium hydroxide per liter is the same. Multiply by molar mass and you get the maximum solubility in g/L:
c(Mg(OH)2) = [Mg(OH)2] · M(Mg(OH)2)
= 1.2 * 10^-4 mol/L * (58.34 g/mol)
S = 7.0 * 10^-3 g/L

S1/S = (7.0 * 10^-8) / (7.0 * 10^-4)

S1/S = 1 * 10^-4