Question

A 0.250 m ascorbic acid solution is 1.79% ionized at 25 degrees. what is the pka for the ascorbic acid at this temp??

Answer 1

Let ascorbic acid be represented by HA . Acid dissociation is given by

1.79 % of 0.25 = (1.7/100) x 0.25 = 0.004475

                            HA (aq) <------------>   H+ (aq)   + A-(aq)

Initially                0.25                                 0              0

equilibrium          0.25 - 0.004475              0.004475     0.004475

Ka = [H+] [A-] /[HA]

   = ( 0.004475)( 0.004475) / ( 0.25-0.004475)

        = 8.16 x 10^ - 5

pka = -log Ka = -log ( 8.16 x 10^-5) = 4.09