Question

What is the molar solubility (in mol/L) of Pb(OH)2 (Ksp = 5.06 ⋅ 10 − 13 ) in a solution buffered at 11.21.

Answer 1

Given:

pH = 11.21

use:

pH = -log [H+]

11.21 = -log [H+]

[H+] = 6.166*10^-12 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)/(6.166*10^-12)

[OH-] = 1.622*10^-3 M

At equilibrium:

Pb(OH)2 <----> Pb2+ + 2 OH-

   s 1.622*10^-3 + 2s

Ksp = [Pb2+][OH-]^2

5.06*10^-13=(s)*(1.622*10^-3+ 2s)^2

Since Ksp is small, s can be ignored as compared to 1.622*10^-3

Above expression thus becomes:

5.06*10^-13=(s)*(1.622*10^-3)^2

5.06*10^-13= (s) * 2.631*10^-6

s = 1.923*10^-7 M

Answer: 1.92*10^-7 M