Question

A. Calculate the solubility of Au(OH)3 in water (Ksp=5.5×10−46).

B. Calculate the solubility of Au(OH)3 in a solution maintained at a nitric acid concentration of 1.2 M .

Answer 1

A)

At equilibrium:

Au(OH)3 <----> Au3+ + 3 OH-

   s 3s

Ksp = [Au3+][OH-]^3

5.5*10^-46=(s)*(3s)^3

5.5*10^-46= 27(s)^4

s = 2.124*10^-12 M

Answer: 2.12*10^-12 M

B)

[H+] = [HNO3] = 1.2 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)/(1.2)

[OH-] = 8.33*10^-15 M

At equilibrium:

Au(OH)3 <----> Au3+ + 3 OH-

   s 8.33*10^-15

Ksp = [Au3+][OH-]^3

5.5*10^-46=(s)*(8.33*10^-15)^3

5.5*10^-46= (s) * 5.78*10^-43

s = 9.515*10^-4 M

Answer: 9.52*10^-4 M