chem 12
posted by Sarah .
a) Calculate [H3O^+] in a 0.15M solution of benzonic acid, HC7H5O2(aq) having K of 6.4x10^5.
b) Calculate the per cent ionization of the HC7H5O2.
I got..
Kb = [HC7H5O3][OH]/HC7H5O2
6.3x10^5 x Kb =1.0x10^14
Kb = 1.0x10^14/6.3x10^4 = 1.5x10^10
1.5x10^10 = (x)(x)/0.15
i'm not sure where i got from there.

Where i go from there **

You start over. Benzoic acid is an acid, not a base. Let's call benzoic acid HB.
..........HB ==> H^+ + B^
init.....0.15.....0.....0
change....x......x.....x
equil....0.15x...x.....x
Ka + (H^+)(B^)/(HB)
Substitute the equilibrium line above into the Ka expression and solve for x = (H^+) = (H3O^+).
b. %ion = [(H^+)/0.15]*100 = ? 
Okay so i got
Ka = (x)(x)/0.15 = 6.5x10^5 
OK. Ka = 6.5E5 = (x*x)/0.15; now solve for x which is (H^+) (H3O^+ to be more exact) which is what the question asked. Then use the x value to solve for percent ionization as I showed you in part b.