Let a be the dissociation of the weak acetic acid
                          Â
HA <---> H + + A-
initial
conc.          Â
c             Â
0Â Â Â Â Â Â Â Â 0
change          Â
  Â
-ca          Â
+ca     +ca
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 c = concentration = 0.1 M
      pH = 2.85
- log[H+] = 2.85
     [H+] =
10-2.85
           Â
= 1.41x10-3 M
But [H+] = ca
1.41x10-3 = 0.116 x a
          a =
0.0122
Ka = ca2 = 0.116x(0.0122)2
           Â
= 1.72x10-5
Therefore Ka = 1.72x10-5
