Question

A 0.25 M solution of benzoic acid HC7H5O2 at 35 degree celcius has a pH of 2.19. If [H+]=[C7H5O-] calculate the Ka of benzoic acid.

Answer 1

PH = 2.19

-log[H^+]   = 2.19

[H^+]     = 10^-2.19   = 0.006456M

[H+]=[C7H5O-]   = 0.006456M

HC7H5O2 (aq) ------------------> C7H5O2^- (aq) + H^+ (aq)

                  Ka   = [C7H5O2^-][H^+]/[HC7H5O2]

                          = 0.006456*0.006456/0.25   = 0.000167

                          = 1.67*10^-4