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 5.61×10 −11 . 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?

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          

Answer is to avoid experimental error

The net force is 100