|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
slight ionization of HA by shifting the following equilibrium to the
You should recognize this for the common ion effect which it is.
|HA <--------> H+ +
|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 [
pH = pKa - log [acid]
we end up with the following:
|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
|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
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.|