Question

The Ka of a monoprotic weak acid is 7.85 × 10-3. What is the percent ionization of a 0.136 M solution of this acid?

Answer 1

construct the ICE table

lets mono protic acid is HA

         HA + H2O <--------> H3O+ + A-

I             0.136                             0    0

C              -x                                 +x    +x

E                 0.136-x                          +x    +x

now write the dissociation expression

Ka = [H3O+][A-] / [HA]

7.85 × 10^-3 = [x][x] / [0.136-x]

x² + x7.85 *10^-3 - 1.0676 * 10^-3 =0

solve the quadratic equation

x = 0.029 M = [H3O+]

percentage of ionisation

= concentration of [H3O+] / HA x 100

= 0.029 / 0.136 x 100

= 21.32 %