Question

Determine the pH at the equivalence point when 0.05 M KOH is titrated with 50.0 ml of 0.01 M benzoic acid HC₇H₅O₂. The K_a for benzoic acid is 6.3 x10^-5.

Answer 1

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