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]                    [anion] If we take the -log of both sides: -log[H+] = -log Ka + (-log  [ acid] )                                           [anion] pH = pKa - log   [acid]                          [anion] 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       [acid] 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.