In the Classroom
Predicting Acid–Base Titration Curves without Calculations Dennis Barnum Department of Chemistry, Portland State University, P.O. Box 751, Portland, OR 97207-0751;
[email protected] A common approach to teaching acid–base chemistry is to have students calculate titration curves. Students often concentrate on setting up the equations or the spreadsheet, and they either lose sight of or fail to recognize the general principles that the calculations are intended to convey. Teaching students how to make a qualitative sketch of the expected titration curve helps to focus attention on the general principles. In this paper a simple qualitative system for sketching all manner of acid–base titration curves is described. Lengthy calculations, complicated algebraic equations, and computers are avoided altogether. All situations are handled with equal ease, including polyprotic acids, polyprotic bases, salts, and mixtures of salts with acids or bases. The only information required is approximate values of the acid or base ionization constants. No new principles are involved. Persons already familiar with acid– base titration curves will find the method obvious. It is presented as a teaching tool and a method to help students tie the several special cases together into a single comprehensive presentation. A qualitative approach takes advantage of the following well-known facts about acid base equilibria: 1. Acid–base reactions are rapid and reversible. 2. When titrating a weak acid with strong base the pH at the half-equivalence point is equal to the pK a of the acid: (1) pH1/2 = pK a; 2 < pK a < 12 3. When titrating a weak base with a strong acid the pOH at the half-equivalence point is equal to pK b of the base: pOH1/2 = pK b; 2 < pK b < 12 (2)
In either case, acid or base, the half-equivalence point is the middle of the buffer region. Consequently, the titration curve must be nearly flat at this point, sloping slightly upward if the titrant is a base or slightly downward if the titrant is an acid. By examining a few real or calculated titration curves one can get a good qualitative feeling for the slope. Alternatively, one can make use of Sturrock’s observation that for a monoprotic acid or base the pH changes by about one unit between 25% and 75% of the volume required to reach the equivalence point (1). 4. Beyond the equivalence point, in the region where excess titrant is being added, the pH must approach the pH of the titrant. 5. For monoprotic acids or bases, the pH at the equivalence point is approximately halfway between the pH at the half-equivalence point and either the upper or lower limit established by the pH of the titrant. The equivalence point is the steepest part of the titration curve.
Titration of a Weak Acid with a Strong Base Figure 1 illustrates the qualitative approach using the well-known titration of 0.100 M acetic acid with 0.100 M sodium hydroxide as an example. For convenience, assume that the initial volume of acid is 35.00 mL and that one is
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using a 50-mL buret. Step 1. Calculate the volume of titrant required to reach the equivalence point from the relationship Vacid Macid = Vbase Mbase In this case the equivalence point is calculated to be at 35.0 mL. The volume at the half-equivalence point is therefore 17.5 mL. Step 2. Since pKa of acetic acid is 4.76 the curve must pass through pH = 4.76 at 17.5 mL. This is the middle of the buffer region where the curve is nearly flat. Draw a short segment of the curve through point B, nearly horizontal, but sloping slightly upward as shown in Figure 1. For a rough sketch of the titration curve it is adequate to make the slope about the same as observed in other calculated or measured titration curves. For a somewhat better approximation to the slope the pH should increase by about one unit between the 25% point at 8.8 mL and the 75% point at 26.3 mL. Step 3. Beyond the equivalence point, in the region of excess titrant, the curve must approach the pH of the titrant. The pH of 0.100 M NaOH is 13.0, so one draws a short segment of the curve, well beyond the equivalence point, nearly flat, approaching pH = 13.0. This is segment D in Figure 1. Step 4. The pH at the equivalence point is roughly halfway between pH1/2 and the upper pH limit; in this case, about halfway between 4.76 and 13.0, which is 8.9. At this pH the curve has its steepest slope. Draw in a short segment, nearly vertical, through pH 8.9 and 35.0 mL (point C in Fig. 1). Step 5. The initial pH (point A in Fig. 1) can be estimated by observing that addition of 0.100 M HCl to the initial solution causes the pH to approach a lower limit equal to the pH of 0.100 M HCl. Thus, the initial pH is approximately halfway between 4.76 and pH 1.0; that is, at approximately 2.9. (This point is discussed further in a following section on reversibility.) Step 6. Complete the curve, as shown in Figure 1, by connecting the heavily lined known segments. Reversibility of Titration Curves Figure 1 also illustrates the reversible nature of titration curves. Note that the x-axis shows added 0.100 M NaOH going from left to right in the usual manner, but also shows added 0.100 M HCl going from right to left. For simplicity, the concentrations of the HCl and NaOH are taken to be equal. Because the acid–base reactions are reversible, one can move in either direction along the curve depending on whether NaOH or HCl is added. For example, in the titration of acetic acid with base one starts at point A. As 0.100 M NaOH is added the curve progresses from left to right through points B, C, and D. If the analyst stops adding base at any point and starts adding acid instead, the same curve is traced backward, from right
Journal of Chemical Education • Vol. 76 No. 7 July 1999 • JChemEd.chem.wisc.edu
In the Classroom
to left. There is, of course, some dilution due to added titrant but the dilution effect is small and can be ignored when one is interested only in the qualitative features of the curve. One might also start at point A and add 0.100 M HCl instead of base. In this case the analyst is just adding HCl to a solution of acetic acid and the titration curve simply starts at point A and approaches pH = 1.0, which is the lowest the pH can possibly go owing to addition of 0.100 M HCl. Because the curve is reversible, Figure 1 also contains the titration curve for an acetate salt titrated with an acid. The solution at point C is exactly the same as a solution prepared by dissolving solid sodium acetate in water. Thus, titration of an acetate salt with an acid starts at point C and progresses from right to left giving an equivalence point at point A. The half-equivalence point at B occurs at pOH1/2 = pKb for the acetate ion, which, of course, is the same as pH1/2 = pKa for the conjugate acid of acetate ion. Titration of Polyprotic Acids Titrating a polyprotic acid is the same as titrating an equimolar mixture of two or more monoprotic acids having different pKa values. The strongest acid is titrated first, followed by the second strongest acid, then the third strongest, etc. To illustrate, the curve for 15.0 mL of 0.100 M phosphoric acid titrated with 0.100 M NaOH is developed in Figure 2. For phosphoric acid we have pKa = 2.17, pKa = 7.21, and pKa = 12.36. One can think of phosphoric acid as a mixture of the three monoprotic acids H3PO4, H2PO4{, and HPO42{. One would then expect equivalence points at 15.0, 30.0, and 45.0 mL. From zero to 15.0 mL, H 3PO4 is being converted to H2PO4{ and the half-equivalence point occurs at 7.50 mL. This is point B in Figure 2. The pH at this point is approxi1
2
3
mately equal to pKa = 2.2. The solution at this point is a buffer of H 3PO 4 and H2PO 4{, so one draws a nearly flat segment of the curve at this point. From 15.0 to 30.0 mL the second strongest acid, H2PO4{, is being converted to HPO42{. A second half-equivalence point therefore occurs at 22.5 mL and pH = 7.2, so one draws a nearly flat segment of the curve through point D. Finally, from 30.0 to 45.0 mL the third acid, HPO42{, is being titrated. For this acid pKa = 12.36. But, recall that the relation between pH1/2 and pKa is only valid in the range approximately 2 to 12, so the curve passes a little lower than 12.4 at the third half-equivalence point at 37.5 mL. Draw a short, nearly flat segment through point F. Also, draw a segment well beyond the third equivalence point showing the curve approaching the pH of the titrant. After drawing segments through the half-equivalence points, one can go back and locate the equivalence points. Point C is approximately halfway on the vertical scale between B (pH = 2.2) and D (pH = 7.2), that is, at about 4.7. The second equivalence point, E, is approximately halfway between D and F. At the third equivalence point the curve is essentially flat. That is, the third proton of phosphoric acid is so weak there is no discernible step at the third equivalence point.1 Because the titration curve is reversible, Figure 2 also contains the curves for phosphate salts titrated with a strong acid. Point G in Figure 2 represents a solution of Na3PO4. Consequently, the titration of trisodium phosphate starts at point G and proceeds from right to left through points F, E, D, C, B, and A. Similarly, the titration of Na2HPO4 starts at point E and gives an equivalence point at C. Because the three acid ionization constants of phosphoric acid are fairly far apart, two relatively steep equivalence points are observed. We now ask, what happens when the acid ionization constants are close together? Tartaric acid is a good example, having pKa = 3.04 and pKa = 4.37. In Figure 3 the curve for titration of 20.0 mL of 0.100 M tartaric acid with 0.100 M NaOH is sketched. One expects equivalence points at 20.0 mL and 40.0 mL. The first half-equivalence point is at 10.0 mL and the second at 30.0 mL. Consequently, we draw one short, nearly flat segment through the point 10.0 mL and pH 3.0 (point B) and a second through the point 30.0 mL and pH 4.4 (point D). The pH at the first equivalence point is approximately halfway between 3.0 and 4.4; that is, about 3.8. Clearly, there is no room for a large “step” in pH at the first equivalence point. The pH at the second equivalence point is approximately halfway between 4.4 and the upper limit of 13.0. Draw in a short steep segment at 40.0 mL and pH 8.8 and a segment approaching the pH of the titrant. Connect the known segments. 1
1
1
Figure 1. Sketching a titration curve qualitatively for 35.0 mL of 0.100 M acetic acid titrated with 0.100 M sodium hydroxide. Addition of 0.100 M NaOH progresses from left to right; addition of 0.100 M HCl progresses from right to left. A is a solution of acetic acid only, HOAc; B is the middle of the buffer region where [HOAc] = [OAc {] and pH1/2 = pKa of acetic acid; C is a solution of sodium acetate, NaOAc; D is the region of sodium acetate and excess sodium hydroxide; and E represents a region of acetic acid and excess HCl.
2
Amino Acids Amino acids are a particularly interesting example of polyprotic acids. We will take the titration of glutamic acid as an example: O HO
C
O CH
CH2 CH2 C
OH
+
NH3
Cl-
Glutamic acid hydrochloride, H3glu+ Cl { ; p Ka 1 = 2.19
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In the Classroom
O HO
C
O CH
CH2 CH2 C
O-
+ NH3
Glutamic acid, H2glu; pKa 2 = 4.25 O -O
C
O CH
CH2 CH2 C
+ NH3
O-
Na+
Glutamic acid monosodium salt, Hglu{ Na+; pK a 3 = 9.67
Acid ionization constants are counted from the most highly protonated species; that is, from the hydrochloride, in this case. We have: pKa = 2.19 for ionization of the first –COOH group pKa = 4.25 for ionization of the second –COOH group pKa = 9.67 for ionization of the –NH3+ group The titration of 15.0 mL of 0.100 M glutamic acid hydrochloride with 0.100 M NaOH is shown in Figure 4. Because there are three titratable protons we expect equivalence points at 15.0, 30.0, and 45.0 mL. The curve must be relatively flat as it passes through the middle of buffer regions with pH = 2.2 at 7.5 mL, pH = 4.3 at 22.5 mL, and pH = 9.7 at 37.5 mL. The pH values at the three equivalence points are approximately 3.3, 7.0, and 11.3. Although Figure 4 is the titration curve for the hydrochloride salt of glutamic acid, it also contains the curve for other forms. For example, if one were to start with glutamic acid (rather than the hydrochloride) the curve would start at point C and progress from left to right through D, E, F, and G. Similarly, if one were to titrate the disodium salt with strong acid the curve would start at point G and progress from right to left.2 1
2
3
Figure 2. Titration of 15.0 mL of 0.100 M phosphoric acid with 0.100 M NaOH. To cover the tip of the pH electrode, 15.0 mL of water was added at the start.
Titration of Salts and Weak Bases Textbooks often treat the titration of salts such as sodium acetate, trisodium phosphate, or potassium carbonate separately from neutral bases. This implies that the ions in salts are different from neutral weak bases such as ammonia or ethylamine. However, if one compares ammonia with the acetate ion it is seen that there is no difference:
Figure 3. Titration of 20.0 mL of 0.100 M tartaric acid with 0.100 M NaOH. pKa 1 = 3.04; pKa 2 = 4.37.
NH3(aq) + H2O NH4+(aq) + OH{(aq) CH3COO{(aq) + H2O CH 3COOH(aq) + OH{(aq) Both equilibria are hydrolysis reactions and both represent base ionization. In fact, most anions are bases and many are strong enough that they can be analyzed by titration with standard acid. 3 We have already seen that Figures 1 and 2 contain the titration curves for acetate and phosphate salts. Values of base ionization constants of anions are not commonly tabulated in textbooks.4 However, to sketch the titration curve for an anion it is usually more convenient to take the pH at the half-equivalence point to be equal to the pKa of the conjugate acid of that base. An Application: The Effect of Dilution on the Titration of Weak Acids—an Apparent Paradox As analyte and titrant are made more and more dilute, titrations become less feasible because the change in pH at the equivalence point becomes smaller and less sharp. However, 940
Figure 4. Titration of 15.0 mL of 0.100 M glutamic acid hydrochloride with 0.100 M NaOH. pKa 1 = 2.19; pKa 2 = 4.25; pKa 3 = 9.67.
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In the Classroom
the behavior of weak acids is different from that of strong acids, as one can see by comparing Figures 5 and 6. The curves behave the same in the region of excess strong base. However, in the region before the equivalence point the two figures differ because of the requirement that the titration curve pass through pH1/2 = pKa at the half-equivalence point. Here is an apparent paradox. Even though the analytical concentration of the weak acid is decreasing, the hydrogen ion concentration remains nearly constant! The same phenomenon is often seen in the general rule that the pH of a buffer is, roughly, independent of dilution. How can the hydrogen ion concentration (activity) remain constant when the concentration of the acid is decreased? The reason is that when a weak acid or base is diluted the percentage ionized increases. Of course, if the solution is made too dilute the relation pH1/2 = pKa fails because approximations made in its derivation are no longer valid. In Figure 6 one can see that for acetic acid the relation is valid in the range of 10{1 to about 10{4 M but fails at 10{5 M. An Application: Carbonate Mixtures Carbonate mixtures are particularly important because of the central role that carbon dioxide, bicarbonate ion, and carbonate minerals play in physiology and geochemistry. Information about the kind and quantity of dissolved carbonate species is obtained by titration. A common exercise given to students in analytical chemistry courses is to identify the component(s) present in solution and then calculate the concentration of each from titration data. Interpretation of the titration data is best related to the pH curve for carbonic acid shown in Figure 7. The curve is easily sketched by the method described above. It is also important to realize that the titration curve is just as valid going from right to left as in the conventional direction. Figure 7 is for a natural water with pH = 6.7. The sample is titrated with 0.002 M HCl to obtain “alkalinity” and a second sample titrated with 0.002 M NaOH to obtain “acidity.”
Figure 5. Effect of dilution on the titration of 30.0 mL of a strong acid (HCl) with a strong base (NaOH). The molarity is indicated on the curves.
Conclusion Students often learn to calculate titration curves but miss the broader generalizations and insights that the calculations are intended to illustrate. When doing quantitative calculations students are confronted with a variety of special cases; for example, strong acids and bases, weak acids and bases, before the equivalence point, beyond the equivalence point, monoprotic, diprotic, and what approximations might be valid. All these cases are accompanied by the problem of solving the resulting algebraic equations. This is not to suggest that quantitative calculations be neglected in teaching acid–base chemistry. Rather, it is suggested that the qualitative approach described here helps to bring the entire topic together into a cohesive overview. Once the qualitative approach is understood students can readily answer such questions as Given the pKa values of an acid, is it feasible to titrate it with strong base? Is the compound monoprotic, diprotic, etc.? What is the equivalent weight? Is it possible to titrate a given salt as an acid? As a base? Notes 1. The equivalent weight of phosphoric acid is seldom, if ever, one-third of the formula weight. If titration is stopped at the first equivalence point, then the equivalent weight is equal to the molecular weight because only one mole of titrant is required per mole of phosphoric acid. If the titration is carried to the second equivalence point, then two moles of titrant are titrated and the equivalent weight of H3PO4 is one-half of its molecular weight. As seen in Figure 2, there is no discernible step in pH after addition of 45 mL of titrant (the third equivalence point). Consequently, in a titration, the equivalent weight of phosphoric acid will never be one-third of the formula weight. 2. Some textbooks, mainly in biochemistry, have published acid–base titration curves for amino acids that, while correct in the midrange, show all titrations beginning abruptly at pH 1.0 and ending abruptly at pH 14.0. Perhaps the system presented here will lead to a correction. Mathews, C. K.; Van Holde, K. E. Biochemistry, 2nd ed.; Benjamin/Cummings: Menlo Park, CA, 1996; p 44. Zubay, G.; Parson, W. W.; Vance, D.E. Principles of Biochemistry; Wm. C. Brown: Dubuque, IA, 1994; pp 53–54. Garrett, R. H.; Grisham, C. M. Biochemistry, 1st ed.; Harcourt Brace Jovanovich: Orlando, FL, 1997; p 48. Lehninger, A. L.; Nelson, D. L.; Cox, M. M. Principles of
Figure 6. Effect of dilution on the titration of 30.0 mL of a weak acid (acetic) with a strong base (NaOH). The molarity is indicated on the curves.
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Figure 7. Titration curve of bicarbonate in a natural water showing the definition of alkalinity and acidity. Acid titrant is 0.002 M HCl; basic titrant is 0.002 M NaOH.
Biochemistry, 2nd ed.; Worth: New York, 1993; p 119. 3. Most anions are weak bases. The only common anions that are too weak to be considered bases in water are the conjugate bases of strong acids, namely, Cl {, Br{, I{, HSO4{, ClO 4{, and NO 3{, a rather short list compared with the large number of possible anions. 4. Of course, one can always find pKb from the relation pKa + pKb = pK w .
Literature Cited 1. Sturrock, P. E. J. Chem. Educ. 1968, 45, 258, 621.
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