Obviously I am embarrassed in asking this because I always try to solve chem problems on my own, especially if they are related to school [:-[]. This wasn't assigned, but I have had a horrific time trying to align my answers with the books answer. The way I solved is below.

The Question: The base ionization constant of ethylamine (C2H5NH2) in aqueous solution is Kb=6.41x10^-4 at 25 C. Calculate the pH for the titration of 40 ml of a .1 M solution of ethylamine with 5ml .1 M HCl.

The equation needed is pH= pKa+log(Ao/HAo). pKa=-log((1x10^-14)/(6.41x10^-4))

5ml=.005 L, .005L x .1M= .0005mol HCl, 40ml x .1M=.004mol

the .0005mol HCl reacts with .004mol base to yield a (7/90)M [C2H5NH2] and a (1/90)M [C2H5NH3+], so the equation comes out to be pH=10.81+log((1/90)/(7/90))=9.965. Which is much different than the books 11.52.

I tried switching the Ao and HAo but the answer still was .3 off. Thoughts?