If you missed A you probably missed B.
Call benzoic acid HB, then
................HB ==> H^+ + B^-
initial........0.15....0......0
change..........-x......x.....x.
equil........0.15-x.....x......x
K = (H^+)(B^-)/(HB)
Substitute and solve for x.
%ion = [(H^+)/0.15]*100 = ?
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..
A) 0.16M but i did it again and got 7.2M so i'm very stuck as to the very different answers i got.
B) 12.4% ionization.
1 answer