Question

Nickel (II) Hydroxide in aqueous solution has a solubility product of 7.9 * 10⁻¹⁶ at 25°C. Determine the solution pH at equilibrium.

Answer 1

At equilibrium:
Ni(OH)2   <---->     Ni2+     +         2 OH-   

                     s                  2s      

Ksp = [Ni2+][OH-]^2
7.9*10^-16=(s)*(2s)^2
7.9*10^-16= 4(s)^3
s = 5.824*10^-6 M

So,
[OH-] = 2s
= 2*5.824*10^-6 M
= 1.165*10^-5 M

use:
pOH = -log [OH-]
= -log (1.165*10^-5)
= 4.9337


use:
PH = 14 - pOH
= 14 - 4.9337
= 9.0663
Answer: 9.07