# Calculate the solubility of Au(OH)3 in 1.0 M nitric acid solution.?

Calculate the solubility of Au(OH)3 in water Ksp=5.5*10^-46.

Please show wrk for both problems if you can

• The equilibrium constant for the dissolution reaction:

Au(OH)3(s) -> Au+++(aq) + 3OH-(aq),

is:

K = ([Au+++]*[OH-]^3)/[Au(OH)3]

By convention, the concentration of a pure solid (Au(OH)3 in this case) is taken as 1, so

Ksp = ([Au+++]*[OH-]^3)

In an aqueous solution, the concentrations of H+ and OH- are related by the dissociation constant of water:

Kw = [OH-]*[H+]

and in pure water [OH-] = [H+].

At 25 degrees C, Kw ~= 10^-14, so [OH-] = [H+] = 10^-7

Assuming that the temperature is ~25C, we can write for the first part of this question:

Ksp = [Au+++] * (10^-7)^3 = 5.5*10^-46

[Au+++] = (5.5*10^-46)/(10^-21)

[Au+++] = 5.5*10^-25 M

Because each mole of Au(OH)3 that dissolves forms 1 mole of Au+++ ions, the solubility of Au(OH)3 in pure water at 25 C would be 5.5*10^-25 M. You can convert this to a concentration in terms of grams per liter by multiplying by the molecular mass of Au(OH)3.

For the next part of the question, we know that the concentration of H+ ions is 1M due to the nitric acid. We can ignore the small number of H+ ions contributed by the dissociation of water because that contribution is only 1 ten-millionth (10^-7 M versus 1M) of the amount due to the nitric acid.

Again using the dissociation constant of water, we have:

10^-14 = [H+][OH-] = 1*[OH-]

[OH-] = 10^-14

Now plug this into the expression for the solubility product and calculate [Au+++]:

5.5*10^-46 = [Au+++] * (10^-14)^3

(5.5*10^-46)/(10^-42) = [Au+++]

[Au+++] = 5.5*10^-4 M

• Nitric Acid Solubility