Question

Find the fraction of dissociation, α of an aqueous 0.400M triethylamine [(CH₂CH₃)N] solution.

Answer 1

Let α be the dissociation of the weak base,trimethylamine
                            BOH <---> B ⁺ + OH^-

initial conc.            c               0         0

change               -cα            +cα      +cα

Equb. conc.         c(1-α)        cα      cα

Dissociation constant, K_b = (cα x cα) / ( c(1-α)               

                                          = c α² / (1-α)

In the case of weak bases α is very small so 1-α is taken as 1

So K_b = cα²

==> α = √ ( K_b / c )

Given K_b = 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