Chemical Education T oday
Letters Graphing Calculator Strategies for Solving Chemical Equilibrium Problems Several articles on the use of graphing calculators to solve chemical problems appeared in the May 1999 issue of this Journal, including Henry Donato’s paper on solving chemical equilibrium problems (1). Although I was intrigued to learn that it is so easy to obtain chemically reasonable roots to complex polynomial equations using modern graphing calculators, I found myself asking “So what?” Is this really useful for our students? Will they learn important chemistry by using this technique? Donato teaches his introductory chemistry students to use their graphing calculators to solve for exact roots of polynomial equations that stem from equilibrium constant expressions. He gives three types of examples: gas phase, solubility, and acid–base equilibria. With Keq’s of 105 and 104, standard approximation methods lead to errors of ca. 3 and 16%, respectively. Donato believes that solving for exact polynomial roots allows students to appreciate “the limitations of approximations and hence obtain a better understanding of the ways chemists describe [equilibria]”. And further, that “students can feel confident that they have completely specified the solution to the problem. I would disagree on two counts. First, by solving for the exact root of a polynomial equation, one has in no way “completely specified” what actually happens in an equilibrium chemical system. Since 1966, authors have repeatedly reported that, owing to the effects of pressure, temperature, and ionic strength, one can obtain only the vaguest idea of a salt’s actual solubility from the value of its Ksp (2–5). This point was confirmed by a laboratory project in the very same issue of this Journal as Donato’s article, which showed that Kf for the FeSCN2+ formation reaction decreased by over 50% as ionic strength increased from 0.05 to 0.3 M (6 ). Owing to the effects of both non-ideality and competing side reactions, calculations based on known equilibrium constants may give answers that are off by an order of magnitude or more compared with experimental results. It therefore does not seem particularly useful to ask students to solve a polynomial in order to get an “exact” solution that may still be wrong by an order of magnitude. Second, even if one ignores the above problem and pretends that Keq’s accurately describe what happens in real reactions, I’m still not convinced of the usefulness of this technique in the introductory chemistry classroom. As Stephen J. Hawkes has stressed repeatedly (3, 7 ), when deciding what to teach, we should always ask ourselves “Why should they know that?” If the standard approximation algorithm yields a solution that is correct within 4%, is it really worth the extra effort to teach students to use a graphing calculator to solve a polynomial in order to get the “exact” solution? Especially when the “exact” solution is probably not really accurate anyway? I believe that the time in class would be better spent explaining the chemical intuition that lies behind the approximation method, and in engaging in a more rigorous qualitative discussion of equilibrium systems (3, 5), leaving the solving for exact polynomial roots to math courses. 1120
Literature Cited 1. Donato, H. J. Chem. Educ. 1999, 76, 632–634. 2. Meites, L.; Pode, J. S. F.; Thomas, H. C. J. Chem. Educ. 1966, 43, 667–672. 3. Hawkes, S. J. J. Chem. Educ. 1998, 75, 1179–1181. 4. Clark, R. W.; Bonicamp, J. M. J. Chem. Educ. 1998, 75, 1182–1185. 5. Hawkes, S. J. J. Chem. Educ. 1996, 73, 421–423. 6. Stoltzberg, R. J. J. Chem. Educ. 1999, 76, 640–641. 7. Hawkes, S. J. J. Chem. Educ. 1994, 69, 178–181. Todd P. Silverstein Department of Chemistry Willamette University Salem, OR 93701
[email protected] The author replies: I would like to thank Todd Silverstein for his careful reading of my paper and for contemplating using graphing calculator strategies in his classroom. I am disappointed he finds little or no value in the technique but welcome the opportunity to address his specific objections as well as to comment in general on use of these strategies in the introductory chemistry classroom. Silverstein suggests that solving polynomial equations arising from the analysis of chemical equilibria “exactly” has little value because the calculated “exact” results can disagree substantially with experiment owing to both non-ideality and competing side reactions. The extent of the disagreement may depend on the pressure, temperature, and the ionic strength. Furthermore, even if agreement with experiment were not so poor and since the standard approximation algorithm yields a solution that is correct within 4%, “is it really worth the extra effort to teach students to use a graphing calculator to solve a polynomial in order to get the ‘exact’ solution?” In response, I would like to offer the following comments. 1. The “extra effort to teach students to use a graphing calculator” is not that much extra. I find that the vast majority of introductory chemistry students already own a graphing calculator and have received years of instruction in its use in math classes up through calculus. Most students readily see the logic behind the method and already know graphing calculator tricks for implementing the method. Try inviting your students to show you how to use the graphing calculator as a way to assess their capabilities. 2. Silverstein asks, why not use an approximate method to calculate results? After all, if one uses an exact method, then the answer may still deviate significantly from experimental results. That is, the error introduced by standard approximations is small compared to the error from other sources. There seems to be a lack of consistency between Silverstein’s two objections. If poor agreement implies that “exact” calculations have little value, then approximate calculations also have little value. Poor agreement between calculations based on some model and experimental results places scientists at
Journal of Chemical Education • Vol. 77 No. 9 September 2000 • JChemEd.chem.wisc.edu
Chemical Education T oday
a crossroad. Either the scientist rejects the model as a poor description of reality, or searches for modifications to the model that more closely describe the actual system. Of course the pedagogical problem is that introducing modifications into simple equilibrium descriptions of systems often leads to algebraic equations with no simple pencil and paper solutions. Pedagogical problems aside, scientists still evaluate models by comparing quantitative predictions of the model to experimental results. 3. While I discuss this only briefly in the last section of my paper, I contend that the computational power of the graphing calculator allows incorporation into the introductory chemistry classroom of modifications to the description of simple equilibrium systems so that the description more closely agrees with experimental data. Consider the solubility of PbCl 2(s). Many textbooks show this calculation as (Ksp/4)1/3 = 0.016 M, assuming the only equilibrium existing in solution is PbCl2(s)
Keq = 1.66 × 105
Pb2+(aq) + 2Cl (aq)
As pointed out by Hawkes (1), the solubility of PbCl2(s) reported in the Handbook of Chemistry and Physics is 0.036 M. This discrepancy is in part due to the existence of other equilibria occurring in solution: PbCl2(s)
PbCl+(aq) + Cl (aq)
PbCl2(s)
PbCl2(aq)
Keq = 6.6 × 10 4 Keq = 1.05 × 10 3
Having no fear that we will generate an algebraic equation that our students can not solve, we attempt to assess quantitatively the effect of these competing equilibria on the solubility of PbCl2(s). There are four chemical species whose concentration we want to determine: Pb2+, Cl , PbCl+, PbCl2(aq). Consequently four equations involving these chemical species are required. Three of those equations are the equilibrium constant expressions for the three equilibria present in solution. The fourth can be found from the charge balance condition for the solution. Expressing all the concentrations in terms of [Cl ] using the relevant equilibrium constant expressions and subtracting the right-hand side from the left, we have 5
1.66 × 10 Cl
2
4
+
6.6 × 10 Cl
Therefore, [Pb2+] = 0.011 M, [PbCl+] = 0.017 M, and [PbCl2(aq)] = 1.1 × 103 M. The total solubility is the sum of these three lead species, which is 0.029 M. This simple graphical analysis has quantitatively evaluated the effect of competing equilibria on the solubility of PbCl2(s) and reproduces the results given in Table 1 of ref 1. Since our result still does not agree with experiment, we have the opportunity to introduce the concepts of activity and activity coefficients. So how should we teach chemical equilibria? Should we leave out competing equilibria and deviations from ideal behavior so that we can describe systems with single equilibria, which can be analyzed to produce algebraic equations that students can work with even if only approximately? Or should we discuss all the processes that occur in solution and all the species that we suspect exist and only try to analyze the solution qualitatively, knowing that a quantitative analysis can lead to some messy algebra very quickly? I would like to suggest a third option. For at least some systems, it should be possible to do both—include all known chemical species and all equilibria and analyze the system quantitatively. This is the potential advantage of using the computing power of graphing calculators in introductory chemistry courses. Literature Cited 1. Hawkes, S. J. J. Chem. Educ. 1998, 75, 1179–1181. Henry Donato Jr.
2[Pb2+ ] + [PbCl+] = [Cl ]
2×
root is found graphically at 0.0389 M (see the calculator screen below).
Department of Chemistry College of Charleston Charleston, SC 29424
– Cl = 0
This cubic polynomial has a chemically relevant root, which is [Cl]. One enters the polynomial into the graphing calculator with [Cl] = x and the graph is investigated to find values of x that make the polynomial vanish. As always, it is convenient to define a range of chemically reasonable x values in which to search ( y values are much less critical). The chemically relevant root must be greater than zero. It also seems logical that the root would be less than the sum of [Cl] produced by each of the first two equilibria if they alone were present in solution; that is, 0.016 + 0.026 = 0.042. The
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