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Topic: Henderson-Hasselbach Equation  (Read 13264 times)

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Offline lemonoman

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Henderson-Hasselbach Equation
« on: June 12, 2005, 09:33:58 PM »
Okay.  First, I'd like to remind the readers that I've finished second-year analytical chemistry in University.  And I have honestly worked at a simple acid-base titration problem (for a friend) for almost 5 hours   :'(.  And it's absolutely not making sense to me.

Here's the deal.  We're titrating 0.100 L of 0.1 M diethylamine (given Kb = 0.0855 with 1 M HCl.

I sincerely believe the reaction equation is:
Et2NH + HCL <--> Et2NH2+ + Cl-

and that the diethylamine acts as a base via:
Et2NH + H2O <--> Et2NH2+ + OH-

I've determined with a simple calculation that the pH of the initial solution (No HCl added) is 12.77

In my frustration, I've also figured out, that if the diethylamine completely dissociated the water, then the pH would reach its upper limit (for a 0.1 M solution) of 13.

Nonetheless, I'm looking to use the Henderson-Hasselbach equation...

pH = pKa + log ( [Base] / [BH+] ) according to MANY sources (internet and otherwise)

The pKa is for the conjugate acid, BH+ (in this case Et2NH2+), and so:
pKa = pKw / pKb = pKw / -log(Kb) = -log(10-14) / -log(0.0855) = 12.93

Now, if we add say, 1 mL of the HCl solution, we are 1/10 of the way to the equivalence point.  So base:conjugate acid is 9:1 and so [Base]/[BH+] = 9/1 = 9

So pH = pKa + log ([Base]/[BH+]) = 12.93 + log 9 = 12.93 + 0.95 = 13.88
which first of all, is an INCREASED pH, and second of all, violates the upper bound of pH on a solution of 0.1 M.

SOMETHING is wrong with this logic, and I'm almost embarrassed to ask.  But I have to.  Thanks for your help.  :)
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Offline AWK

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Re:Henderson-Hasselbach Equation
« Reply #1 on: June 13, 2005, 01:17:30 AM »
pKa for diethylamine is close to 3, not 0.95
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Offline lemonoman

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Re:Henderson-Hasselbach Equation
« Reply #2 on: June 13, 2005, 01:32:42 AM »
Okay...thanks for the help, but...

I do realize the given information about diethylamine is wrong.  This is how it was given in the question, and I too was like, "What?" - and also I think I had calculated the pKa to be 12.93, when it's really 10.98 or something comparable (3.02 is the pKb)

Thanks for the help, though...hopefully someone can find an answer...so far, I've had 7 university-student-calibre chemists give up because they can't figure it out...
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Offline Borek

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Re:Henderson-Hasselbach Equation
« Reply #3 on: June 13, 2005, 04:01:32 AM »
Quote from: lemonoman on June 12, 2005, 09:33:58 PM
pKa = pKw / pKb

pKw = pKa + pKb

Ka = Kw / Kb

You have mixed two equations.

Edit:
OK, I have checked that the calculation of pKa - althought you wrote wrong equation - were correct.

What I am writing here is valid for your calculation - so not for diethyleneamine, but for some amine with pKb = 1.

The problem is - you can't use HH equation in this case. HH equation works always when you want to calculate ratio of acid/base for known pH. In other direction it works too, but only if you use true values of acid and base concentrations.

Calculate dissociation fraction for pure amine. Now compare it with the dissociation fraction assumed by you for the 10% neutralization. You see where the problem comes from?

Check my BATE for titration curve calculation.
« Last Edit: June 13, 2005, 04:51:55 AM by Borek »
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Offline lemonoman

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Re:Henderson-Hasselbach Equation
« Reply #4 on: June 13, 2005, 07:46:47 AM »
I do see where the problem came from.

For some reason, I thought that the HH equation used the concentrations of each acid/base pair BEFORE dissociation occurred.  I'm going to runt hrough a few more examples to figure out when/why this works...perhaps the examples were for strong acids/bases, and so 100% dissociation occurred....

Thanks, Borek, for your help  :)
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