| The pH of a Buffered Solution |
|
| Because HA is a weak acid, very little of it is
dissociated at equilibrium.
Even if HA were the only solute. But the solution also contains A-
from the dissolved salt. The presence of A- suppresses the
already
slight ionization of HA by shifting the following equilibrium to the
left.
You should recognize this for the common ion effect which it is.
|
| HA <--------> H+ +
A-
|
| ie. The value of [HA] at equilibrium = [HA] initially
and the value of [A-] at equilibrium = [A-] initially. Therefore we can make 2 assumptions: [HA]equilibrium = [HA]from the initial [acid] = [acid] [A-]equilibrium = [A-]from the initial [salt] = [anion] Therefore we can do this: [H+] = Ka x [acid]
|
| If we take the -log of both sides:
-log[H+] = -log Ka + (-log [
acid]
)
pH = pKa - log [acid]
|
Since 
we end up with the following:
|
 (Henderson-Hasselbach
Equation)
|
| Two factors govern the pH in a buffered solution.
|
| (1) pKa of the weak acid;
|
| (2) ratio of the initial molar []'s of the acid and it's
salt.
|
| If we prepare a solution where [anion] = [acid] then the
log [anion] = log 1 = 0
|
| Therefore the pH of the solution will turn out equal to
the pKa
of the weak acid.
|
| What mostly determines where on a pH scale a buffer can work best is the pKa of the weak acid. Then by adjusting the ratio of [anion] to [acid], we can cause shifts so that the pH of the buffered solution comes out on one side or the other of this value of pH. |