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<tt>I've been getting a lot of questions about buffers and what
happens when</tt><tt><br>
</tt><tt>strong acid or base is added. Here's a little bit of what
was discussed.<br>
Hopefully this helps and doesn't cause confusion.<br>
</tt><tt><br>
A buffer is a special type of common-ion problem. It consists of
the<br>
conjugate acid-base pair of a weak acid and its conj. base (or
weak base<br>
and its conj. acid). Examples are:<br>
<br>
HF/F-, CH3CO2H/CH3CO2-, NH3/NH4+<br>
</tt><tt><br>
</tt><tt>You could get a buffer by simply adding both members of the
weak conj.<br>
acid-base pair. You could also get it by starting with a weak
acid<br>
and adding some strong base to convert some of the weak acid to
its<br>
conj. base but leaving some of the weak acid behind (such as
starting<br>
with HF and adding some NaOH to react with the HF to produce F-
but<br>
leave some HF). You could start with a weak base and add some
strong<br>
acid to convert </tt><tt><tt>some of the weak base to its conj.
acid, leaving some<br>
of the weak base behind </tt></tt><tt><tt><tt>(such as starting
with NH3 and adding some HCl<br>
to react with the NH3 to produce NH4+ but leave some NH3)</tt>.<br>
<br>
A strong acid with a strong base (HCl/NaOH) is NOT a buffer
system.<br>
<br>
For buffers you want to r</tt>emember, the Henderson-Hasselbalch
eqn. It<br>
can be used in two ways.</tt><tt> </tt><tt>It's technically,</tt><tt><br>
</tt><tt><br>
</tt><tt> [base]<sub>eq</sub> </tt><tt><br>
</tt><tt>pH = pKa + log -----------<br>
</tt><tt><tt> [acid]<sub>eq</sub> <br>
<br>
where the ratio gives the equilibrium conc. of the two
components of<br>
the buffer (the weak conj. acid-base pair).<br>
<br>
You can use this eqn if you're given the pH and pKa and asked
for the<br>
ratio of the conc. of the base and acid in the solution or
you're trying<br>
to make a buffer at a specific pH and you need to figure out how
much of<br>
one of the components has to be added.<br>
<br>
The above eqn could be written as:<br>
</tt></tt><tt><tt><tt><tt><tt><br>
</tt><tt> [base]<sub>orig</sub> + x </tt><tt><br>
</tt><tt>pH = pKa + log ----------------<br>
</tt><tt><tt> [acid]<sub>orig</sub> - x </tt></tt></tt></tt></tt></tt><tt><br>
</tt><tt><br>
</tt><tt><tt><tt>where </tt></tt><tt><tt><tt>[base]<sub>eq</sub></tt>
= </tt></tt><tt><tt><tt><tt><tt>[base]<sub>orig</sub> + x
and </tt></tt></tt></tt></tt><tt><tt><tt><tt>[acid]<sub>eq</sub>
</tt></tt>= </tt></tt><tt><tt><tt><tt><tt><tt>[acid]<sub>orig</sub>
- x<br>
<br>
If the </tt></tt></tt></tt></tt></tt></tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt>original
conc. of acid and base are relatively
large, their ratio<br>
is between 0.1 and 10 and the Ka is
relatively small we </tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt>we
ignore the "x"<br>
to get the eqn below using
just the original conc. given
in the problem.<br>
<br>
</tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt></tt>
[base]<sub>orig</sub> </tt><tt><br>
</tt><tt>pH = pKa + log -----------<br>
</tt><tt><tt> [acid]<sub>orig</sub> <br>
<br>
</tt></tt><tt><tt><tt><tt><tt><tt><br>
If the above conditions aren't true you wouldn't be able
to ignore the "x"<br>
and you would have to set up an ICE table and solve a
quadratic.<br>
<br>
Finding the pH of a buffer soln given conc. for the base
and acid is just<br>
a matter of plugging in numbers in this eqn.<br>
<br>
The problem comes in when adding a strong acid or strong
base to the buffer<br>
system. The discussion below is for a buffer of a weak
acid HA and its<br>
conj. base, A-. The equilibrium for the ICE table would
be<br>
<br>
HA <==> H+ + A-<br>
<br>
Again, you could likely use the HH eqn and not have to
set up this ICE table.<br>
<br>
You need to understand what happens when you add strong
acid or base to the<br>
system. This seems to be the main question I'm
getting. Here's what happens<br>
in a nutshell.<br>
<br>
Strong Acid - reacts with the A- (base) part of the
buffer system to create<br>
additional HA and leaves some A-. You'll
still have a buffer<br>
as long as the amount of strong acid added
was small (in terms<br>
of the moles added). You need to do a
stoichiometry table for<br>
the neutralization reaction between the
strong acid and A-.<br>
This is a BCA table (the ICC tables I used
in lecture), in MOLES.<br>
Using H+ to stand for the strong acid this
is the eqn needed,<br>
<br>
H+ + A- <==> HA (in
moles)<br>
<br>
</tt></tt></tt></tt></tt></tt><br>
<tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt><tt>Strong Base - reacts
with the HA (acid) part of the buffer system
to create<br>
additional A- and leaves some
HA. You'll still have a buffer<br>
as long as the amount of
strong base added was small (in terms<br>
of the moles added). You need
to do a stoichiometry table for<br>
the neutralization reaction
between the strong base and HA.<br>
This is a BCA table (the ICC
tables I used in lecture), in MOLES.<br>
Using OH- to stand for the
strong base this is the eqn needed,<br>
<br>
HA + OH- <==> A-
+ H2O (in moles)<br>
</tt></tt></tt></tt></tt></tt><br>
Once you do the proper neutralization reaction you can
plug the new conc.<br>
of A- and HA back into the HH eqn and calculate the new
pH. Technically,<br>
if you remember from class, you can use the moles of A-
and HA since they're<br>
in the same volume of soln and the ratio of moles is the
same as conc.,<br>
<br>
[A-] mol A-<br>
------ = --------<br>
[HA] </tt></tt></tt></tt> mol HA<br>
<br>
That's about it. See you in a short while.<br>
<br>
Dr. Zellmer<br>
</tt></tt>
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