A) Dissociation constant of benzoic acid , Ka = 6.3 x
10-5
C)
               Â
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
                                                 Â
= [-log ( 0.16 x 10-5) - log(0.01)]/2
                                                 Â
= 3.9
   Therefore,
        pH = 14 -pOH =
14 - 3.1 = 8.1
      Therefore,
               Â
pH of of 0.01 M Sodium benzoate = 8.1
B) hydrolysis constant of sodium benzoate
       Kh = Kw/ kb
           Â
= Kw/ Kb = 1.0 x 10-14/0.16 x 10-5
               Â
= 6.3 x 10-5