Question

Calculate the solubility of Cobalt (II) Hydroxide in a solution which has been buffered to have a pH of 10.00. Cobalt (II) Hydroxide has Ksp = 5.29 x 10^-15

Answer 1

use:
pH = -log [H3O+]
10 = -log [H3O+]
[H3O+] = 1*10^-10 M

use:
[OH-] = (1.0*10^-14)/[H3O+]
[OH-] = (1.0*10^-14)/(1*10^-10)
[OH-] = 1*10^-4 M

At equilibrium:
Co(OH)2   <---->     Co2+     +         2 OH-   
                     s                  1*10^-4 + 2s
Ksp = [Co2+][OH-]^2
5.29*10^-15=(s)*(1*10^-4+ 2s)^2
Since Ksp is small, s can be ignored as compared to 1*10^-4
Above expression thus becomes:
5.29*10^-15=(s)*(1*10^-4)^2
5.29*10^-15= (s) * 10^-8
s = 5.29*10^-7 M

Answer: 5.29*10^-7 M