Acid Base problems?

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Offline Bill.D.Katt.

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Acid Base problems?
« on: 15/01/2011 03:43:20 »
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?

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Offline lightarrow

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Acid Base problems?
« Reply #1 on: 15/01/2011 22:33:21 »
I have not used your formula but solved another way.
I write the equilibrium reaction:

EtNH2 + H2O ↔ EtNH3+ + OH-

Using 7/90 as the initial concentration of EtNH2 and 1/90 as initial concentration of EtNH3+.
So I can say that, at the equilibrium:

[EtNH2] = 7/90 - x
[EtNH3+] = 1/90 + x

where x = [OH-]

So: [EtNH3+]*[OH-]/[EtNH2] = Kb

(1/90 + x)*x/(7/90 - x) = 6.41*10-4

Solving the second grade equation in x gives: x = 3.31*10-3

--> pH = 11.52