Question

A solution contains 0.24 M CaCl₂ and 0.14 M MgCl₂. What concentration of hydroxide ions would be required to separate the magnesium as Mg(OH)₂ without precipitating the calcium as Ca(OH)₂?
Mg(OH)₂ will precipitate and Ca(OH)₂ will not precipitate if _ M <[OH-]<_ M

Answer 1

Ksp of Mg(OH)2 = 1.8*10^-11
[Mg2+] = 0.14 M

Mg(OH)2   <---> Mg2+ +   2 OH-
Ksp = [Mg2+] [OH-]^2
1.8*10^-11 = 0.14 * [OH-]^2
[OH-] =1.13*10^-5 M

if [OH-] > 1.13*10^-5 M , Mg(OH)2 will precipitate

Ksp of Ca(OH)2 = 5.5*10^-6
[Ca2+] = 0.24 M

Ca(OH)2   <---> Ca2+ +   2 OH-
Ksp = [Ca2+] [OH-]^2
5.5*10^-6 = 0.24 * [OH-]^2
[OH-] =4.79*10^-3 M

if [OH-] > 4.79*10^-3 M , Ca(OH)2 will precipitate

Answer:
4.79*10^-3 > [OH-] > 1.13*10^-5