Buffers?

[Robert Zellmer]

I've been getting a lot of questions about buffers and what happens when
strong acid or base is added.  Here's a little bit of what was discussed.
Hopefully this helps and doesn't cause confusion.

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 a weak base and add some strong
acid to convert some of the weak base to its conj. acid, leaving some
of the weak base behind (such as starting with NH3 and adding some HCl
to react with the NH3 to produce NH4+ but leave some NH3).

A strong acid with a strong base (HCl/NaOH) is NOT a buffer system.

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 this 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 has to be added.

The above eqn could be written as:

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

where [base]_eq = [base]_orig + x and [acid]_eq = [acid]_orig - x

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 we ignore the "x"
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.

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 this 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 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 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 since they're
in the same volume of soln and the ratio of moles is the same as conc.,

    [A-]     mol A-
   ------ = --------
    [HA]     mol HA

That's about it.  See you in a short while.

Dr. Zellmer
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