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Buffered Solutions
How Buffers Work
As you have seen in calculating the pH of solutions, only a small
amount of a strong acid is
necessary to drastically alter the pH. For certain experiments, however,
it is desirable to keep a fairly
constant pH while acids or bases are added to the solution either
by reaction or by the
experimenter. Buffers are designed to fill that role.
Chemists use buffers routinely to
moderate the pH of a reaction. Biology finds manifold uses for buffers which
range from
controlling blood pH to
ensuring that urine does not reach painfully acidic levels.
A buffer is simply a mixture of a weak acid and its conjugate base
or a weak base and
its conjugate acid. Buffers work by reacting with any added acid or
base to control the pH. For
example, let's consider the action of a buffer composed of the weak base
ammonia,
NH3, and its conjugate acid, NH4+. When
HCl is added
to that buffer, the NH3 "soaks up" the acid's proton to become
NH4+. Because that proton is locked up in the
ammonium ion, it
proton does not serve to significantly increase the pH of the solution.
When NaOH is added to the
same buffer, the ammonium ion donates a proton to the base to become
ammonia and water.
Here the buffer also serves to neutralize the base.
As the above example shows, a buffer works by replacing a strong acid or base
with a weak one. The
strong acid's proton is replaced by ammonium ion, a weak acid. The
strong base OH- was replaced by the weak base ammonia. These replacements of strong
acids and bases for
weaker ones give buffers their extraordinary ability to moderate pH.
Calculating the pH of Buffered Solutions
Buffers must be chosen for the appropriate pH range that they are called on to
control. The pH range of a buffered solution is given by the Henderson-
Hasselbalch equation. For the
purpose of the derivation,
we will imagine a buffer composed of an acid, HA, and its conjugate base,
A-. We
know that the acid dissociation constant pKa of the
acid is
given by this expression:
![]()
The equation can be rearranged as follows:
![]()
Taking the -log of this expression and rearranging the terms to make each
one positive gives the
Henderson-Hasselbalch equation:
![]()
Figure 1.3: The Henderson-Hasselbalch Equation
Note that the sample species HA and A- in the above
Expression are generalized to the terms
acid and base, respectively. To use
the equation, place the concentration of the acidic buffer species where
the equation says "acid" and
place the concentration of the basic buffer species where the equation
calls for "base". It is essential
that you use the pKa of the acidic species and not the
pKb of the basic
species when working with basic buffers--many students forget this point
when doing buffer
problems.
A buffer problem can be fairly simple to solve, provided you don't get
confused by all the other
chemistry you know. For example, let's calculate the pH of a
solution that is 0.5 M acetic acid and 0.5 sodium acetate both before and after
enough SO3
gas is dissolved to
make the solution 0.1 M in sulfuric acid. Before the acid is added, we can
use the
Henderson-Hasselbalch equation to calculate the pH.
![]()
Figure 1.4: Note that the pKa of acetic acid is 4.75.
This part of the problem does not require us to do the sort of equilibrium
calculations that we must use for Non-Buffered
Solutions, but many students still
try to do it the hard way. The hard way is a correct way of doing the
problem, but it may cost you valuable time on a test.
To calculate the pH after the acid is added, we assume that the
acid reacts with the base in
solution and that the reaction has a 100% yield. Therefore, we say that
0.1 moles per liter of acetate
ion reacts with 0.1 moles per liter of sulfuric acid to give 0.1 moles per
liter of acetic acid and
hydrogen sulfate. Here, we ignore the second dissociation of sulfuric acid
because it is minor in
comparison to the first. So the final concentration of acetic acid
is 0.6 M and acetate is 0.4M.
Plugging those values into the Henderson-Hasselbalch equation gives a pH of
4.57. Note that a
0.1 M solution of strong acid would give to a pH of 1 but the buffer gives a
pH of 4.57 instead.
To probe the useful range of the buffer, let's calculate the pH of the
solution resulting from the same
situation above but with different concentrations of the buffer. If the
buffer is 1.0 M in both acetate
and acetic acid, then the pH of the resulting solution after the
introduction of acid is 4.66. However,
if we make the solution only 0.11 M in acetic acid and acetate, then we
calculate a pH of 3.45!
Therefore, if you want a more effective buffer, make sure that the
concentration of the buffering
agents is large in comparison to the added acid or base.
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