construct the ICE table
lets mono protic acid is HA
        HA + H2O
<--------> H3O+ + A-
IÂ Â Â Â Â Â Â Â Â Â Â Â
0.136
                          Â
0 Â Â 0
CÂ Â Â Â Â Â Â Â Â Â Â Â Â
-x                               Â
+x   +x
EÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0.136-x                        Â
+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 %
