Volume of KOH used to reach equivalence point = 0.01 M x 50
ml/0.05 M
                                                                              =
10 ml
At eqivalence point all of benzoic acid has been neutralized
with added KOH
[C6H5COO-] formed = 0.01 M x 50 ml/60 ml = 0.0083 M
C6H5COO- + H2O <==> C6H5COOH + OH-
let x amount of salt has hydrolyzed
Kb = 1 x 10^-14/6.3 x 10^-5 = x^2/0.0083
x = [OH-] = 1.15 x 10^-6 M
pOH = -log[OH-]
        = 5.94
pH = 14 - pOH
    = 8.06
