Question

Calculate the pH of a 0.01 M solution of sodium benzoate.

A: Dissociation constant of benzoic acid.

B: Calculate the hydrolysis constant of sodium benzoate.

C: Calculate the pH of the 0.01 M solution.

Answer 1

A) Dissociation constant of benzoic acid , Ka = 6.3 x 10^-5

C)

                 Kb of sodium benzoate = Kw/ Ka = 1.0 x 10^-14/6.3 x 10^-5 = 0.16 x 10^-5

   Hence,

   pOH of 0.01 M Sodium benzoate = [ -log Kb- logC]/2

                                                   = [-log ( 0.16 x 10^-5) - log(0.01)]/2

                                                   = 3.9

    Therefore,

         pH = 14 -pOH = 14 - 3.1 = 8.1

       Therefore,

                 pH of of 0.01 M Sodium benzoate = 8.1

B) hydrolysis constant of sodium benzoate

        Kh = Kw/ kb

             = Kw/ Kb = 1.0 x 10^-14/0.16 x 10^-5

                 = 6.3 x 10^-5