Part 1)
when the balanced equation for this reaction is:
and by using ICE table
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
initial 0 0
change +X +2X
Equ X 2X
When KSp expression = [Mg2+][OH-]
when we have KSp = 5.61 x 10^-11
and when we assumed [Mg2+] = X
and [ OH-] = (2X)^2
when we assume X is the value of molar solubility of Mg(OH)2
so, by substitution:
5.61 x 10^-11 = 4X^3
∴ X = 2.4 x 10^-4 M
∴ molar solubility of Mg(OH)2 = X = 2.4 x 10^-4 M
Part 2)
the molar solubility of Mg(OH)2 in 0.16 m NaOH we assumed it = X
by using the ICE table:
Mg(OH)2(s) → Mg2+(aq) + 2OH-
initial 0 0.16m
change +X +2X
equ X (0.16+2x)
when Ksp = [mg2+][OH-]^2
5.61 x 10^-11 = X * (0.16+2X)^2 by solving for X
∴ X = 1.3 x 10^-5 M
∴ the molar solubility = X = 1.3 x 10^-5 M
