Question

A benzoic acid solution (C6H5CO2H) has a pH of 2.80. Given that Ka of benzoic acid is 6.4x10–5, calculate the concentration and percent dissociation of this acid. Answer is 4.1%

Answer 1

    Ka = 6.4*10^-5

    Pka   = -logKa

             = -log6.4*10^-5

             = 4.1938

PH   = 1/2Pka -1/2logc

2.8   = 1/2*4.1938 -1/2logc

2.8   = 2.0969-1/2logc

2.8-2.0969    = -1/2logc

-1/2logc         = 0.7031

   logc          = -0.7031*2

logc            = -1.4062

   C              = 10^-1.4062   = 0.03924M

PH   = 2.8

-log[H^+]   = 2.8

     [H^+]   = 10^-2.8 = 0.00158M

percent ionisation = Conc of H^+*100/inital con of acid

                                 = 0.00158*100/0.03924   = 4.1% >>>>answer