What is the gram solubility of Mg(OH)2 in a 0.55mol/L solution of KOH if the Ksp of Mg(OH)2 is 9.0x10^-12?

Answer 1

This is a 1:2 ionization in pure water => Solubility of any 1:2 or 2:1 salt ionization = Cube Root(Ksp/4).

Solubility (pure water) of Mg(OH)2 = Cube Root[(9x10^-12)/4] = 1.3x10^4 M.

1:1 ionization => S = Sqr Rt(Ksp)
1:2 or 2:1 => S = Cube Rt(Ksp/4)
1:3 or 3:1 => S = 4th Rt(Ksp/27)
1:4 or 4:1 => S = 5th Rt(Ksp/256)
2:3 or 3:2 => S = 5th Rt(Ksp/108)