HH eqn for buffers

[Robert Zellmer]

I've been asked the following before by other students:

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Dr. Zellmer,

I was reading the book and saw that for the Henderson Hasselbach
equation, the concentrations of the acid and base equal to the
concentrations from the equilibrium but not the original concentrations.
is that true?  That's actually not what is stated in the book.

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In reality there are two answers to this (which I addressed in
lecture):

Technically they are the equil. conc. When asked to calc. the ratio
of base/acid from the pH and pKa you are actually getting the ratio
of the equil. conc. of  base and acid.

However, in the normal practice of determining the pH of a buffer
solution we use the original starting conc. because we usually ignore
the "x", just as we would when using an ICE table.  If we didn't do
this there would be no point in using the eqn. and we would just use
an ICE table every time (which you can do).  We do it this way, using
the original conc. of the acid and base, when we calc. the pH of a soln.
from conc. and the pKa.  Just remember, you can only use this eqn.
in this way if the conc. of the base and acid (the conj. acid-base pair)
are relatively large and the Ka is relatively small. That's because only
then can you "ignore the x" and this eqn is valid using it for determining
the pH.

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Can the HH eqn always be used?  No.  If the conc. of the acid and/or
base portion of the buffer system are small and/or the Ka is relatively
large you won't be able to ignore the "x" in the ICE table and thus the
HH eqn doesn't hold.

Sample Exercise 17.4 shows this.  The conc of acid is 1.00 x 10^-3
and the conc. of its conj. base is 1.00 x 10^-4 and the Ka is 1.8 x 10^-5 .
Just taking the Ka and dividing by the 1.00 x 10^-4 gives a ratio for
K/[base] of 1.8 x 10^-1 which is bigger than 10^-3 , which is what it
should be to get a small enough error.  If you ignore the "x" by using
the HH eqn you'll get a value of "x" which gives an error compared
to 1.00 x 10^-4 of 180%.  Pretty lousy.  You actually have to use an ICE
table and solve a quadratic eqn.  Using the HH eqn they get a pH or
3.74 and using an ICE table and solving the quadratic eqn gives a
pH of 4.06.  The 3.74 gives an error of 7.9%.

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  Dr. Zellmer
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