Let volume of sodium benzoate =x
     volume of benzoic
acid     = 100-x
     no of moles of sodium benzoate =
molarity * volume
                                                   Â
= 0.22*x
    no of moles of bezoic
acid          Â
= molarity * volume
                                                     Â
= 0.1*(100-x)
      PH = PKa +log[sodium
benzoate]/[benzoic acid]
      4    Â
= 4.2 + log0.22*x/0.1*(100-x)
      4-4.2  =
log0.22*x/0.1*(100-x)
     Â
-0.2Â Â Â Â = log0.22*x/0.1*(100-x)
log0.22*x/0.1*(100-x) = -0.2
0.22*x/0.1*(100-x)Â Â Â Â Â Â =
10-0.2 = 0.63
0.22*x/0.1*(100-x)Â Â = 0.63
0.22*x
            Â
= 0.63*0.1(100-x)
                           Â
= 0.063(100-x)
0.22x
                 Â
=6.3-0.063x
0.22x-6.3+0.063x =0
0.263x-6.3 =0
  0.263x = 6.3
        x  =
6.3/0.263Â Â = 23.95
volume of benzoic acid  = 23.95ml
volume of benzoate      = 100-x =
100-23.95 = 76.05ml
  Â
     Â