Re: Question 3.7

Guiedo the Magnificant (ottende@PLU.edu)
Wed, 25 Sep 1996 21:52:39 -0800 (PST)

The first thing to do is to recognize that the anilinium ion is
the congigate acid of aniline. The congigate acid of methylamine is
methylaminium ion, with a pKa of 10.6. How do I know? Look up at the
first picture above the problem and that is the reaction of Methylamine
with water, which also shows the pKa for its conjugate acid (10.6). Now
compare numbers:

Anilinium ion: 4.6
Methylaminium ion: 10.6

Anilinium has a lower pKa which means (by the definition of pKa = -log
Ka) that Ka is a larger number. This means that it favors the reaction
of Anilinium Ion + OH- <--> Aniline + Water towards the products. This
means that less OH- is left in the water meaning it is a weaker base.
to look at it from the other point of view:
Methylamine ion has a higher pKa which means (by the definition of pKa =
-log Ka) that Ka is a smaller number. This means that it favors the
reaction of Methylaminium Ion + OH- <--> Methylamine + Water twoards the
reactants. This means that more OH- is left in the water meaning it is a
stronger base.