Asolution of benzoic acid (c6h5cooh) that is 0.20 m has a ph of 2.45. what is ka for benzoic acid?

We first calculate for the H+ ion concentration from the pH of the solution of benzoic acid. We do as follows:

pH = -log[H+]
2.45 = -log [H+]
[H+] = 3.55x10^-3

We then use the initial concentration and the h+ ion concentration as follows for Ka calculation:
   C6H5COOH = C6H5COO-   +   H+
I       0.20                      0                    0                
C        -x                        x                    x
E      0.20-x                   x                   x

x = [H+] = 3.55x10^-3
[C6H5COO-] = 3.55x10^-3
[C6H5COOH] = 0.1965

Ka = 3.55x10^-3 (3.55x10^-3) / 0.1965 = 6.42x10^-5

Any photo? or something