Question

The Ka of a monoprotic weak acid is 4.40 × 10-3. What is the percent ionization of a 0.185 M solution of this acid?

Answer 1

             HA -------> H+ + A-

I          C                 0        0

C    -C               C     C

E   C(1-)          C      C

       Ka   = [H+][A-]/[HA]

       Ka    = C * C/C(1-)

     Ka    = C²

²     = Ka/C

          = 4.4*10^-3 /0.185

         = 15.42 *10^-2

%      = 15.42*10^-2 *100

              = 15.42%