Question

The Ka of a monoprotic weak acid is 5.13 × 10-3. What is the percent ionization of a 0.179 M solution of this acid?

Answer 1

                     HA ------> H+ + A-

initial              C            0       0

Change     -C        C     C

at equilibrium C(1-)    C    C

       Ka = [H+][A-]/[HA]

      Ka    = C*C/C(1-)

                                   1>>>>

       Ka    = C²

       ² = Ka/C

             = 5.13*10^-3/0.179

            = 2.865*10^-2

          =1.692*10^-1

%     = 16.92%