Chemical Education Today
Letters Easy Derivation… The title of the recently published article by Stephen J. Hawkes, “Easy Derivation of pH ≈ (pK a1 + pK a2)/2 Using Autoprotolysis of HA᎑: Doubtful Value of the Supposedly More Rigorous Equation” (J. Chem. Educ. 2000, 77, 1183), contains two dubious premises. The “easy derivation” presented is not as simple as it could be using the material and charge balance equations and the more rigorous equation is essential for understanding when the simpler equation will give accurate results. Moreover, the rigorous equation can cope with a case, NaHSO4, where the simple equation completely fails. At best, the “easy derivation” of the title equation (eq 4 in the article) is a useful way to give some insight into the linked chemical equilibria. It is an example of the intuitive approach widely used in general chemistry textbooks, but limited in scope to problems in which there is a single dominant equilibrium. At worst, it could lead students down the wrong path, by discouraging them from learning to use material and charge balance equations and encouraging them to think that if they are clever enough, they can find a simple short-cut way to deal with complex problems. Let’s look at the chain of logic in the author’s derivation. The crucial step in the derivation presented is to show that [H2A] ≅ [A2᎑]. First he must think of all possible reactions of HA᎑ (reaction as an acid, as a base, and autoprotolysis). Then he must write the equilibrium constant expression for each possible reaction, show that the equilibrium constant is some combination of K a1, K a2, and Kw, look up the values of the relevant equilibrium constants, and make three separate calculations. The author then appeals to intuition to convince us that if one of these calculated equilibrium constants is much larger than the others, then we can safely believe that it is the dominant equilibrium. The question of how much larger it needs to be is not addressed. Next, if the autoprotolysis reaction appears to be the dominant equilibrium, it is reasonable to suppose that [H2A] ≅ [A2᎑]. All of this is plausible but the derivation contains hidden pitfalls. An easier and more rigorous way to arrive at the crucial result is to start with the mass and charge balance (MB and CB) equations for the system MHA–water: MB:
C = [H2A] + [HA᎑] + [A2᎑]
(1)
CB:
[M+] + [H+] = [HA᎑] + 2[A2᎑] + [OH᎑]
(2)
where C is the total (or stoichiometric) concentration of the diprotic acid. By definition of the first equivalence point, we may set [M+] = C (because it is the point where one mole of MOH has been added per mole of diprotic acid). Using this equality, we rearrange eq 2 to give [M+] = C = [HA᎑] + 2[A2᎑] + [OH᎑] – [H+]
(3)
Next we set the right-hand sides of eqs 1 and 3 equal to one another, simplify by removing canceling terms, and rearrange to obtain the final result: [H2A] + [H+] = [A2᎑] + [OH᎑]
(4)
(Substitution of the expressions for [H2A], [A ], and [OH᎑] 2᎑
from Ka1, Ka2, and Kw into eq 4 and multiplication by Ka1 × [H+] leads easily to the author’s eq 8, the rigorous equation.) Looking at eq 4 of this letter we discover one of the hidden pitfalls in the author’s derivation. The relation [H2A] ≅ [A2᎑] is justified, provided that [H+] and [OH᎑] are negligibly small. The accuracy of simple eq 4 begins to deteriorate exactly when [H+] is no longer negligible and [A2᎑] becomes greater than [H2A], or, looking at the denominator term of rigorous eq 8, when K a1 is not negligible compared to [HA᎑]. However, you wouldn’t be aware of this from the author’s derivation of eq 4. Using the material and charge balance equations we arrive at a rigorously exact result, using very simple algebra. There is no need to think about the possible reactions of HA᎑ or calculate values of their equilibrium constants. The systematic approach has the advantage of simplicity, economy, accuracy, and elegance. After learning how to write basic equilibrium constant expressions relevant to the system and mass and charge balance equations, it requires no special insight or intuition to extend the calculations to more complex systems. Rigorous eq 8, [H+]2 = {Ka1(Ka2[HA᎑] + Kw)}/{K a1 + [HA᎑]}, of the author’s article has not just one but two special cases. If [HA᎑] >> K a1, eq 8 reduces to the simple eq 4 {pH = (pK a1 + pKa2)/2}. If Ka1 >> [HA᎑], eq 8 reduces to [H+] ≅ {[Ka2[HA᎑]}1/2, equivalent to a pseudo-monoprotic acid with K a = K a2. This corresponds exactly to the NaHSO4 case, which was not considered by the author. Simple eq 4 completely fails for this case because K a1 is large and unknown, so you can’t make a calculation from eq 4. However, for 0.05 M NaHSO4 both eq 8 and its reduced form give the same value of pH = 1.69, when activity coefficients are included. It should be noted that using K a1 ≥ 1000 and [HA᎑] = C = 0.05 in eq 8, with activity coefficients for ionic strength 0.05 M included, gives a calculated pH of 1.56. When the correct value ([HA᎑] ≅ 0.026 M) is used in eq 8, with correction of the activity coefficients for increased ionic strength (I ≅ 0.098), the pH value is 1.69. The author makes two more statements that need correction. The first is that “It (eq 8) will always reduce to eq 4 at the intermediate endpoint of a titration because either of the conditions that would prevent this (low Ka2 or high Ka1) would make the endpoint impossibly diffuse.” Endpoint sharpness and the criteria for deciding when eq 8 reduces to the simpler eq 4 are separate issues, as described in the next paragraphs. The second is that “I can think of no example that is significant in any real world situation” (where eq 8 would not reduce to eq 4, except 0.001 M HSO3᎑ [H2SO3, pKa1 = 1.91; pKa2 = 7.18]). There is a simple criterion for eq 8 to reduce to eq 4 without significant error (> Kw.) However, a small pKa1 value, corresponding to a high (or large) Ka1 value, does not preclude a sharp endpoint at either the first or second endpoint. Oxalic acid (with pKa1 = 1.23; pKa2 = 4.19) gives a diffuse intermediate endpoint and a sharp final endpoint; the diprotic amino acid leucine hydrochloride (pKa1 = 2.33, pKa2 = 9.75) gives a sharp intermediate endpoint and a diffuse final endpoint and for a hypothetical
JChemEd.chem.wisc.edu • Vol. 79 No. 2 February 2002 • Journal of Chemical Education
161
Chemical Education Today
Letters acid with pKa1 = 2 and pKa2 = 7 both endpoints would be tolerably sharp. In the titration of a diprotic acid, the sharpness of the intermediate endpoint (first equivalence point) is determined by the ratio Ka1/Ka2. The larger this ratio, the sharper the intermediate endpoint. The sharpness of the final endpoint (second equivalence point) is determined by the value of K a2, the value of Ka1 being irrelevant. Diprotic acids with pKa2 < 7 give feasibly sharp endpoints at the final (second) equivalence point (for 0.05 M solutions). A casual scan of tables of acid ionization constants turned up a number of real-world examples of diprotic acids whose pKa1 values are such that the calculated pH has a significant error when eq 4 is used in place of eq 8. The list of acids (pKa1; pKa2 values in parenthesis) includes the aforementioned leucine hydrochloride (2.33; 9.75) and sulfurous (1.91; 7.18) acids, nicotinic acid (2.05; 4.81), maleic (1.83; 6.07), oxalic (1.23; 4.19), picolinic (1.01; 5.39), and, of course, sulfuric acid (pKa1
