I disagree that Ba(NO3)2 isn't very soluble. The solubility is about 10g/100 mL H2O at 25 C. It doesn't have a Ksp in the sense of it being a slightly soluble salt.
dGrxn = (n*dGf products) - (n*dGf reactants)
Then dG = -RTlnK
Barium Nitrate is a not very soluble nitrate. Calculate the Ksp (at 25 degrees) from the following ΔG's:
Ba2+ = -561kj/mole
NO3 1- = - 109kj/mole
Ba(NO3)2 = - 797 kj/mole
2 answers
I am working on the same problem. I did the first step DrBob gave and I ended up with dGrxn = -127 kj/mol. When I plugged it into the second equation I got -127000/(-298)(8.314) = 51.2599. Then if I take the e of this I get 1.8x10^22. I feel like this is way off, since the options are:
a. 3.5 x 10^-3
b. 1.05 x 10^-5
c. 7.0 x 10^-4
d. 1.4 x 10^-6
a. 3.5 x 10^-3
b. 1.05 x 10^-5
c. 7.0 x 10^-4
d. 1.4 x 10^-6
1 answer