[skibum143]

At 298 K, hypobromous acid (HBrO) dissociates in water with a Ka of 2.3e-09.

(a) Calculate capital deltaG° for the dissociation of HBrO. (Use Ka to calculate the answer.)
1Your answer is incorrect.
Your response is off by a multiple of ten. kJ/mol

(b) Calculate capital deltaG if [H3O+] = 0.00079 M, [BrO - ] = 0.25 M, and [HBrO] = 0.23 M.

For (a), I did deltaG0 = -RTln(K) = -(8.314)(298)(ln(2.3*10^-9))
= 49,279.79 J/mol

This says it's off by a multiple of 10: I think maybe I'm confused about why they wrote Ka the way they did, does 2.3e-09 not equal 2.3*10^-9?

For (b), I did deltaG = deltaG0 + RTln[(.00079)(.25)/(.23)]
= 31,797.89 J/mol, but this is wrong too.

Thanks for any help