Ka = [H+][C2H3O2-]/[HC2H3O2]
Since there is no other source of these ions (ignoring the water), let x be the hydrogen ion concentration. Then x is also the acetate ion concentration and .100-x is the concentration of HC2H3O2. However, HC2H3O2 is a weak acid, and .100-x is very nearly .100 . . .
Ka = (x)(x)/(.1-x) which is approximately x^2 /.100
1.8 x 10^-5 = x^2 /.100
x = 1.34 x 10^-3
Since this is less than 5 percent of .100, the approximation is valid.
Now take the log and change the sign.
pH = -log[H+] = -log (1.34 x 10^-3) = 2.87
take the -log(.100M) and you get 1. that is the Ph of the equation