Question

Estimate solubility (in molarity units) of Ni(OH)2 at pH 8.5. KSP(Ni(OH)2) = 6×10-16

Answer 1

use:

pH = -log [H+]

8.5 = -log [H+]

[H+] = 3.162*10^-9 M

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(3.162*10^-9)

[OH-] = 3.162*10^-6 M

At equilibrium:

Ni(OH)2 <----> Ni2+ + 2 OH-

   s 3.162*10^-6

Ksp = [Ni2+][OH-]^2

6*10^-16=(s)*(3.162*10^-6)^2

6*10^-16= (s) * 9.998*10^-12

s = 6.00*10^-5 M

Answer: 6.00*10^-5 M