Buffer questions

[Zellmer, Robert]

I got some questions about buffers and what happens when strong acid or
base is added.


A buffer is a special type of common-ion problem.  It consists of the
conjugate acid-base pair of a weak acid and its conj. base (or weak base
and its conj. acid).  Examples are:

    HF/F-,    CH3CO2H/CH3CO2-,   NH3/NH4+

You could get a buffer by simply adding both members of the weak conj.
acid-base pair.  You could also get it by starting with a weak acid
and adding some strong base to convert some of the weak acid to its
conj. base but leaving some of the weak acid behind (such as starting
with HF and adding some NaOH to react with the HF to produce F- but
leave some HF).  You could start with NH3 and add some strong acid to
convert some of the weak NH3 to its conj. acid, NH4+, leaving some of
the NH3 and so you wind up with both NH3 and NH4+ in solution (like
the example I did in lecture).

A strong acid with a strong base (HCl/NaOH) is NOT a buffer system.
A strong acid with it's conjugate base is not a buffer.  A strong base
with its conjugate acid won't be a buffer (like NaOH and Na+).

For buffers you want to remember, the Henderson-Hasselbalch eqn.  It
can be used in two ways.  It's technically,

                  [base]eq
pH = pKa  +  log -----------
                  [acid]eq

where the ratio gives the equilibrium conc. of the two components of
the buffer (the weak conj. acid-base pair).

You can use the above eqn if you're given the pH and pKa and asked for
the ratio of the conc. of the base and acid in the solution or you're
trying to make a buffer at a specific pH and you need to figure out
how much of one of the components must be added.


The above eqn can also be written as:

                  [base]orig + x
pH = pKa  +  log ----------------
                  [acid]orig - x

where [base]eq = [base]orig + x and [acid]eq = [acid]orig - x .  This comes
from the ICE table you could use instead of this eqn.

If the original conc. of acid and base are relatively large, their ratio
is between 0.1 and 10 and the Ka is relatively small we ignore the "x"
in both the numerator and denominator to get the eqn below using just
the original conc. given in the problem.

                  [base]orig
pH = pKa  +  log -----------
                  [acid]orig


If the above conditions aren't true, you wouldn't be able to ignore the
"x" and you would have to set up an ICE table and solve a quadratic.
See Sample Exercise 17.4 in the textbook for an example of this.

Finding the pH of a buffer soln given conc. for the base and acid is just
a matter of plugging in numbers in this eqn.

The problem comes in when adding a strong acid or strong base to the buffer
system.  The discussion below is for a buffer of a weak acid HA and its
conj. base, A-.  The equilibrium for the ICE table would be

        HA  <==>   H+  +  A-

Again, you could likely use the HH eqn and not have to set up the ICE table.

You need to understand what happens when you add strong acid or base to the
system.  This seems to be the main question I'm getting.  Here's what happens
in a nutshell.

Strong Acid - reacts with the A- (base) part of the buffer system to create
              additional HA and leaves some A-.  You'll still have a buffer
              as long as the amount of strong acid added was small (in terms
              of the moles added).  You need to do a stoichiometry table for
              the neutralization reaction between the strong acid and A-.
              This is a BCA table (the ICC tables I used in lecture), in MOLES.
              Using H+ to stand for the strong acid this is the net ionic
        eqn needed,

                H+  +  A-  ==>  HA      (in moles)


Strong Base - reacts with the HA (acid) part of the buffer system to create
              additional A- and leaves some HA.  You'll still have a buffer
              as long as the amount of strong base added was small (in terms
              of the moles added).  You need to do a stoichiometry table for
              the neutralization reaction between the strong base and HA.
              This is a BCA table (the ICC tables I used in lecture), in MOLES.
              Using OH- to stand for the strong base this is the net ionic
        eqn needed,

                HA  +  OH-  <==>  A-  +  H2O   (in moles)

Once you do the proper neutralization reaction you can plug the new conc.
of A- and HA back into the HH eqn and calculate the new pH.  Technically,
if you remember from class, you can use the moles of A- and HA you get from
doing the above neutralization reactions since they're in the same volume of
soln and the ratio of moles is the same as the ratio of conc.,

   [A-]     (mol A-/V_total)     mol A-
  ------ = ------------------ = --------
   [HA]     (mol HA/V_total)     mol HA

You see buffers in titrations as well.

That's about it.  I hope this helps.

Dr. Zellmer
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