HH eqn for buffers

[Zellmer, Robert]

I've been asked the following before by students about the Henderson-
Hasselbalch eqn.  I received the same question this week from someone.
We covered this in lecture last week and again this week when we
were discussing weak acid/strong base and weak base/strong acid
titrations.

**********************
I was reading the book and saw that for the Henderson-Hasselbalch
equation, the concentrations of the acid and base equal to the
concentrations from the equilibrium but not the original concentrations.
Is that true?

**********************

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.

                            [A-]eq
pH = pKa + log (-----------)
                           [HA]eq

You would use this equation if you're trying to find the ratio of
[A-]/[HA] when trying to determine what this ratio needs to be when
making a buffer solution with a particular pH.  See EOC exercises
17.25, 17.26, 17.29 and 17.89.

Now on to the other form of the HH eqn, the one we use most often.

If you set up the ICE table, you would find for the equil. conc. above

[HA]eq = [HA]o - x
[A-]eq = [A-]o + x

where [HA]o and [A-]o are the initial conc. of the buffer components
and "x" is [H+]eq.  The pH above is -log([H+]eq).  Substituting for
[HA]eq and [A-]eq the above we get for the HH eqn,

                            ( [A-]o + x)
pH = pKa + log (----------------)
                            ([HA]o - x)

                             [A-]o
pH = pKa + log (----------)    when the "x" can be ignored.
                            [HA]o

So, in the normal practice of determining the pH of a buffer solution
we usually ignore the "x", just as we would when using an ICE table,
& use the HH eqn. w. the original starting conc.  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 often do it this way, using
the original conc. of the acid and base components, 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, their ratio is between 10 and 0.1 and the Ka is
relatively small.  That's because only then can you "ignore the x" and this
eqn is then valid using it for determining the pH.

Sample exercise 17.4 shows an example when it's not okay to use the
HH eqn.  Notice in this example, while the ratio of the base/acid for the
conj. acid-base pair of the buffer is 0.1, the conc. of each component
is small, [A-] = 1.00 x 10^-4 M and [HA] = 1.00 x 10^-3 M, and close to the
magnitude of Ka (1.8 x 10^-5).  The Ka needs to be about 1000 times
smaller than the conc. of acid and base in order to ignore the "x" and
safely use the HH eqn.

Dr. Zellmer
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