Let a be the dissociation of the weak acid
                          Â
HA <---> H + + A-
initial
conc.          Â
c             Â
0Â Â Â Â Â Â Â Â 0
change             Â
-cα          Â
+cα     +cα
Equb. conc.       Â
c(1-a)Â Â Â Â Â Â Â Â
ca     ca
Dissociation constant , Ka = ca x ca / ( c(1-a)
                                       Â
= c a2 / (1-a)
In the case of weak acids α is very small so 1-a is taken as
1
So Ka = ca2
==> a = √ ( Ka / c )
Given Ka = 5.55x10-3
         c =
concentration = 0.182 M
Plug the values we get a = 0.175
∴ % ionization = 0.175x 100 = 17.5
Therefore the percent ionization is 17.5%
