Calculate [H3O^-] in a 0.15M solution of benzoic acid, HC7H5O2(aq) having a Ka of 6.4 x 10^-5

Question

I got.
6.4x10^-5=(x)(x)/0.15

6.4x10^-5= x^2/0.15

6.4x10^-5 x 0.15 = x^2

square root of 9.6x10^-6 = square root of x^2

x= 0.003098
The pH solution is -log [0.003098]
= 2.50

b.) Calculate the percent ionization of the HC7H5O.

%ion= (H^+)/0.15x100???? not sure of this?

Answer 1

a is ok.
b is ok if you used
%ion = [(H^+)/0.15]*100 = about 2%