Titration Curves Of Diprotic Acids
A diprotic acid is an acid with two ionizable hydrogens such that it yields two H+ ions for each acid molecule. Examples of diprotic acids are H2CO3 (carbonic acid) , H2C2O4 (oxalic acid), and H2SO3 (sulfurous acid). The ionization of a general diprotic weak acid ionizes in water in two steps:
H2A(aq) H+(aq) + HA-(aq) Ka1 = [H+][HA-]/[H2A]
HA-(aq) H+(aq) + A2-(aq) Ka2 = [H+][A2-]/[HA-]
Because H2A has two ionizable hydrogens, its titration curve has two equivalence points, as shown below.
The equations for the acid-base reaction occurring between H2A and NaOH are:
from the beginning to the first equivalence point:
H2A(aq) + NaOH(aq) NaHA(aq) + H2O(l)
from the first to the second equivalence point:
NaHA(aq) + NaOH(aq) Na2A(aq) + H2O(l)
and for the overall reaction:
H2A(aq) + 2NaOH(aq) Na2A(aq) + 2H2O(l)
At the first equivalence point, all the H+(aq) from the first ionization of H2A have reacted with the NaOH. At the second equivalence point, all the H+(aq) from the second ionization have reacted with the NaOH. The volume of NaOH added at the second equivalence point is twice that of the first equivalence point.
It is possible to determine the H2A acid dissociation constants, Ka1 and Ka2, using the graph below.
Ka1 = [H+][HA-]/[H2A] (1)
Ka2 = [H+][A2-]/[HA-] (2)
On the graph, the red vertical line at 5.12 mL is the first half-titration point which represents when one-half of the H+(aq) ions in the first ionization have been titrated with NaOH. The first half-titration point volume is one-half the volume of the first equivalence point. At the point, 5.12 mL, [H2A] = [HA-] which can be substituted in Equation (1) above to yield:
Ka1 = [H+]
[HA-]/ [HA-]= [H+]
Taking the log of both sides yields:
log Ka1 = log [H+]
-log Ka1 = -log [H+]
pKa1 = pH
This means that the pH at the first half-titration point, 5.12 mL, equals the pKa1 of H2A. After pKa1 is determined from the graph, the equation Ka1 = 10-pKa1 can be used to determine the first dissociation constant for the acid H2A.
Using the same steps as above, it can be shown that pKa2 = pH at 15.39 mL of NaOH.
The same analysis can be used to determine the Ka's of triprotic acids such as phosphoric acid, H3PO4.