Question

Part A - Calculate the molar solubility in water

Answer 1

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