Let α be the dissociation of the weak base,trimethylamine
                            BOH
<---> B + + OH-
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-α)      Â
cα     cα
Dissociation constant, Kb = (cα x cα) / (
c(1-α)              Â
              Â
                        Â
= c α2 / (1-α)
In the case of weak bases α is very small so 1-α is taken as
1
So Kb = cα2
==> α = √ ( Kb / c )
Given Kb = 6.46x10-5
         c =
concentration = 0.400 M
Plug the values we get α = 0.0127
So the fraction of dissociation is α = 0.0127
