Question

Calculate the solubility of cobalt (II) hydroxide, Co(OH)2 at pH 9. Solubility product of Co(OH)2 is 5.9*10^-15.

Answer 1

Ksp Co(OH)2 = [Co2+][OH-]^2

pH = 9

pOh = 14 - 9 = 5

pOH = -log[OH-]

[OH-] = 1 x 10^-5 M

Solubility of Co(OH)2 = [Co2+] = Ksp/[OH-]^2

                                                  = 5.9 x 10^-15/(1 x 10^-5)

                                                  = 5.9 x 10^-5 M