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),
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 SolubilitySource(s): https://shrinke.im/a0NWQ