Question

The Ksp of colbalt (II) hydroxide, Co(OH)2, is 5.92*10^-15 M^3. Calculate the solubility of this compound in grams/Liter.

Answer 1

Co(OH)2   ----------------------> Co+2   + 2 OH-

                                          S             2S

Ksp = [Co+2][OH-]^2

5.92 x 10^-15 = S x (2S)^2

5.92*10^-15 = 4S^3

S = 1.14 x 10^-5 M

solubility = 1.14 x 10^-5 M

              = 1.14 x 10^-5 x 92.95

              = 1.06 x 10^-3 g/L

solubility = 1.06 x 10^-3 g/L