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

2 Answers

• 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