Chemical Education Today
Letters The pH at the First Equivalence Point in the Titration of a Diprotic Acid I read with interest the paper entitled “Easy Derivation of pH ⬇ (pKa1 ⫹ pKa2)/2 Using Autoprotolysis of HA᎑: Doubtful Value of the Supposedly More Rigorous Equation” by Stephen J. Hawkes (1). I believe the validity of this simple result can be seen even more easily than this paper indicates. First, the expression in the title of the paper, which is also presented by Hawkes in the form of equation 4, indicates that the pH of the solution of a diprotic acid, H2A, at the first equivalence point is half-way between the first and second pKa values. That is, if pKa1 ⫽ 2 and pKa2 ⫽ 6, the pH of the solution at the first equivalence point should be 4. (4) pKa1 ⫹ pKa2 ⬇ 2pH We can see that this must be true by thinking about the situation in this way. Assume that we have used aqueous sodium hydroxide to titrate 1 mole of H2A to the first equivalence point. Realizing that we have at this point a solution that we could have prepared by dissolving 1 mole of NaHA in water, we know that the average charge on each sodium ion is +1 (because every sodium ion has a charge of exactly +1), and that the average charge on the anion of the acid must be ᎑1 because the solution is neutral. The only way the average charge on the anion of the acid can be ᎑1, using the pKa values in this example, is for the pH of the solution to be 4. This conclusion follows from a consideration of the relationship between the pH of a solution, the pH at which the members of a conjugate pair of acid and base are exactly equal, (which is always given the symbol pKa), and the ratio of the concentrations of the members of the conjugate pair: pH ⫺ pKa = log
[basic member] [acidic member]
Thus in this example when the pH is 4, pH ⫺ pKa1 will be 4 ⫺ 2 ⫽ 2, and the ratio of HA1᎑ to H2A will be 100, or, saying the same thing in the opposite way, the ratio of H2A to HA1᎑ will be 0.01. In the other comparison when the pH is 4, pH ⫺ pKa2 will be 4 ⫺ 6 ⫽ ᎑2, and the ratio of A2᎑ to HA1᎑ will also be 0.01. In general, it is only at a pH that is
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exactly half-way between the two pKa values that the concentrations of both the conjugate acid of a species and the conjugate base of that same species will be equal. In this particular example, it is only at a pH of 4 (half-way between pKa1 and pKa2) that the concentrations of the uncharged species, H2A, and the doubly charged species, A2᎑, will be equal. To summarize this example: when the solution has the halfway pH of 4, approximately 98% of the anionic species, A, will be present at any instant as HA᎑, where the charge is ᎑1, approximately 1% as A2᎑, where the charge is ᎑2, and to exactly the same extent, approximately 1%, as H2A, where the charge is 0. When this is the instantaneous distribution of the anionic species among these three possible forms the average charge on each anionic species will be ᎑1 because each anionic species will be present 98% of the time with a charge of ᎑1, 1% of the time with a charge of ᎑2, and an equal 1% of the time with a charge of 0. Again, this can be the case only when the pH of the solution is exactly half-way between pKa1 and pKa2, which, in this example, is a pH of 4. Some readers will note a similarity between this approach and the one I took in a paper entitled “Do pH in Your Head” (2). In an example in that article the isoelectric pH of glycine (the pH at which the average charge of a glycine molecule is zero), has the value of 6.0, which is exactly half-way between 2.4, the pKa of the carboxyl group of glycine, and 9.6, the pKa of the ammonium group of glycine. This is what one would expect when realizing that a solution of neutral glycine right out of the bottle is equivalent to glycine obtained by titration of the conjugate acid of glycine to the first equivalence point. Those who are interested might want to consider why the isoelectric pH of an “acidic” amino acid, such as alanine, is exactly half-way between the pKa values of the two carboxyl groups, and why the isoelectric pH of a “basic” amino acid such as lysine is exactly half-way between the pKa values of the two ammonium groups. Literature Cited 1. Hawkes, Stephen J. J. Chem. Educ. 2000, 77, 1183–1184. 2. Ault, Addison. J. Chem. Educ. 1999, 76, 936–937. Addison Ault Department of Chemistry, Cornell College Mt. Vernon, IA 52314
[email protected] Journal of Chemical Education • Vol. 80 No. 12 December 2003 • JChemEd.chem.wisc.edu
