Question

The Ksp of Zn(OH)2 is 5.0*10^-17. What is the solubility of Zn(OH)2 in a buffer solution with a pH of 10.3 ?

Answer 1

we have below equation to be used:

pH = -log [H+]

10.3 = -log [H+]

log [H+] = -10.3

[H+] = 10^(-10.3)

[H+] = 5.012*10^-11 M

we have below equation to be used:

[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)/(5.012*10^-11)

[OH-] = 1.995*10^-4 M

At equilibrium:

Zn(OH)2 <----> Zn2+ + 2 OH-

   s 1.995*10^-4 + 2s

Ksp = [Zn2+][OH-]^2

5*10^-17=(s)*(1.995*10^-4+ 2s)^2

Since Ksp is small, s can be ignored as compared to 1.995*10^-4

Above expression thus becomes:

5*10^-17=(s)*(1.995*10^-4)^2

5*10^-17= (s) * 3.98*10^-8

s = 1.26*10^-9 M

Answer: 1.26*10^-9 M