In the Classroom edited by
Teaching with Technology
James P. Birk Arizona State University Tempe, AZ 85287
Graphing Calculator Strategies for Solving Chemical Equilibrium Problems
W
Henry Donato Jr. Department of Chemistry and Biochemistry, College of Charleston, Charleston, SC 29424;
[email protected] The chemical equilibrium state is described using algebraic equations. The ability to analyze information given in an equilibrium problem and write an algebraic expression appropriate to that system is a skill we would like to impart to our students. However, for many chemical systems, rigorous treatment of the equilibrium state can lead to polynomial equations of order three or higher (1). Graphical and numerical methods for finding the solution to these equations are well known and have been discussed in this Journal (2, 3). These methods are difficult to implement in introductory chemistry courses because they require either lengthy and tedious calculations or the use of a personal computer running some computational software. Neither time nor technology is usually available in the introductory chemistry classroom, so the logistics of teaching and testing when using these methods are difficult. Also, many introductory chemistry students lack the background and experience necessary to make use of personal-computer-based computational software without significant additional instruction. For these reasons, complete and rigorous treatments of chemical equilibria in introductory chemistry courses are not given. Reactions leading to polynomial equations of order three or more are not discussed. Also, in systems involving multiple equilibria, only conditions in which all but one of the equilibria can be ignored are considered. For example, the equilibrium system H2(g) + I2(g)
2HI(g)
is often used in examples illustrating equilibrium principles, whereas the system 3H 2(g) + N2(g)
2NH3(g)
is almost never discussed; yet the latter has considerable commercial and historical significance. Also, the pH of solutions of acids and bases is calculated without considering the self-ionization of water, unless, of course, one needs to consider that process. Useful but elaborate guidelines have been published to guide teachers and students in the appropriate use of approximations when dealing with weak acid and weak base solutions (4 ). Recently, similar guidelines have been suggested for aiding students with approximations associated with precipitation equilibria (5). Unfortunately, students often use an approximate method without a clear understanding of what approximation has been made or, in many instances, without knowing they have used an approximate method. The ideal situation would be to present complete and rigorous treatments of all chemical equilibria and then use some simple, easily understood method to find the roots of the polynomial equations that arise from this analysis. 632
The graphing calculator is a powerful computing device, which many students already own and use in their mathematics courses. It is relatively inexpensive, so each student can be required to have one. Students can use them in class, on tests, in the lab, and to complete homework assignments. The purpose of this paper is to investigate practical graphing calculator strategies for solving algebraic expressions arising from the complete and general treatment of chemical equilibrium problems. It is hoped that the reader will find these strategies useful enough to incorporate into classroom instruction. What follows is a general discussion of graphing calculator strategies. A detailed supplement is available via JCE Online, with directions for implementing these strategies on TI-85, TI-86, and TI-92 calculators. W Standard Equilibrium Problem Consider a reaction usually not discussed in introductory chemistry courses because of the mathematical difficulties involved in solving the algebraic expression arising from analysis of the equilibrium state. Suppose that at a temperature of 400 °C, a mixture of 3 parts H2(g) and 1 part N2(g) is placed in a container with a total pressure of 10 bar, and the partial pressures of H2, N2, and NH3 at equilibrium are desired. The chemical reaction involved is N2(g) + 3H2(g)
2NH3(g); Kp = 1.64 × 10 {4
The initial partial pressures of N2 and H2 can be extracted from information given in the problem. If one lets x be the partial pressure of the nitrogen gas molecules that react in order to reach equilibrium, then the equilibrium partial pressures of the gases are given below.
P N = 10 – x ; 2 4
P H = 30 – 3x ; 2 4
P NH = 2x 3
The equilibrium constant expression then becomes
Kp =
2x 10 – x 4
2
30 – 3x 4
3
This is a quartic equation with 4 roots, one of which is the chemically relevant solution. Rearranging the equation, we obtain
Kp –
2x 10 – x 4
2
30 – 3x 4
Journal of Chemical Education • Vol. 76 No. 5 May 1999 • JChemEd.chem.wisc.edu
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=0
In the Classroom
Figure 1. TI-85 calculator screen of the function graph. The graph window is defined by xMin = 0, xMax = 2.5, yMin = {1.64 × 10{4, yMax = 1.64 × 10{4. The root command has found the x -value for which the graph crossed the x -axis and has marked the graph with a cross at that spot. The coordinates are also shown on the graph.
Figure 2. TI-85 calculator screen of the function graph. The graph window is defined by xMin = 0, xMax = 7.2 × 10{4, yMin = {1.64 × 10{-5, yMax = 1.64 × 10{5. The root command has found the x -value for which the graph crossed the x -axis and has marked the graph with a cross at that spot. The coordinates are also shown on the graph.
from which it is seen that a value of x that causes the left-hand side to vanish is a root. It seems obvious that a way to find the roots of the equation above is to graph the left-hand side at selected x values and determine graphically the x values where the function crosses the x-axis. Graphing calculators perform just this operation. TI graphing calculators have a “y =” window where expressions such as the left-hand side above may be entered. Then, rather than calculate the function at all values of x and display the graph, the operator may define a graph window so that only a portion of the function graph is calculated and displayed. Appropriate selection of the graph window, in addition to considerably reducing the difficulty of finding the chemically relevant root, can be used pedagogically by instructors of introductory chemistry students. We hope we can get students to recognize that x will not be less than zero or greater than 2.5. Also, y will not be larger than 1.6 × 10{4, the value of Kp. For this problem a convenient graphing window is:
Therefore
xMin = 0 xMax = 2.5 yMin = {1.64 × 10{4 yMax = 1.64 × 10{4
Looking at the graph in a larger window may be confusing. Note that as x values increase toward 2.5, y values tend to { ∞. So, for example, if a student makes the y window larger to keep the graph on scale as it plummets to { ∞, then the graph will not have enough resolution for one to see the chemically relevant root. But assuming one has selected an appropriate graph window, most calculators have a root function that quickly locates the value of x for which y = 0, which in this case is 0.179 bar. Figure 1 shows the TI-85 calculator screen of the function graph in the chemically relevant graph window. The root function of the calculator has located the root and displayed it on the screen. This method does not involve algebraic manipulation but does emphasize construction of the proper algebraic description of the equilibrium state and encourages the student to deduce the range of x values that contains the answer.
3
s=
K sp
= 0.016 M
4
Ksp = (s)(2s + 0.15)2
(ii)
Obviously, this equation leads to a cubic polynomial. In certain circumstances (but not this one) the assumption can be made that 2s is small compared to 0.15 and so can be neglected in that term, which makes the mathematics of solving the problem much easier. With our ability to find the chemically relevant roots of polynomial equations using the graphing calculator, we can solve the problem first approximately and then exactly so that the results can be compared. Suppose we ignore 2s in the second parentheses above. Then {5
s=
1.6 × 10 4
{4
= 7.11 × 10
Now, using this value as a guide to setting up our graphing window, we may graph Ksp – [(s)(2s + 0.15)2 ] = 0 in the graph window 0 < x < (7.2 × 10{4) and ({1.6 × 10{5) < y < (1.6 × 10 {5). Figure 2 shows this graph as it appears on the TI-85 calculator. Note that the root found by the graphical method is 6.98 × 10{4, which is different from the result obtained with the approximate method. (iii)
Ksp = (s + 0.08)(2s)2
If s can be ignored in the first parentheses, then s = 0.0071 M. However, graphical solution of the equation above (see Fig. 3), Ksp = (s + 0.08)(2s)2, yields s = 6.79 × 10{3. Students can demonstrate easily the limitations of approximations and hence obtain a better understanding of the ways chemists describe these slightly soluble salts.
Solubility Product Equilibria
Acid–Base Equilibria
Other problems presenting mathematical difficulties are those illustrating the common ion effect. For example, calculate the molar solubility of PbCl2 (Ksp = 1.6 × 10{5) in (i) pure water, (ii) in 0.15 M KCl solution, and (iii) 0.080 M Pb(NO3)2.
In aqueous solutions of weak acids and weak bases there are multiple equilibria, which must be satisfied simultaneously. Often the self-ionization of water can be ignored. For example, in calculating the pH of a 0.10 M acetic acid solution, students are generally instructed to solve the problem by completing the following steps. Students must recognize that acetic acid
(i)
Ksp = [Pb2+][Cl1{] 2 = (s)(2s)2 = 4s3
JChemEd.chem.wisc.edu • Vol. 76 No. 5 May 1999 • Journal of Chemical Education
633
In the Classroom
is a weak acid, which means that the following equilibrium exists in aqueous solutions of acetic acid: HAc
H+ + Ac{
Then they must express the equilibrium concentrations of the reactants and products in terms of a single unknown: Let x = the equilibrium concentration of H+. Then [H+] = [Ac{] = x [HAc] = 0.10 – x Next students must write down an algebraic equation involving the single unknown, x, which is obviously the equilibrium constant expression with K a = 1.8 × 10{5:
Ka =
x2 0.10 – x
To find the chemically relevant root, enter the left-hand side of the equation in the y = window of the TI-85. After selecting the appropriate graph window, graphing the equation, and determining the root, it is found that the root determined graphically is 0.0013 M. A more complete treatment of this problem is not given because it is not needed to get the correct answer and because it leads to a cubic polynomial equation. However, the method will solve a cubic equation as easily as a quadratic. Also, a more complete treatment is easier for students to understand and reproduce on their own. In the 0.10 M acetic acid solution, there are four chemical species: H+, OH {, HAc, and Ac{, whose concentrations need to be determined. One therefore needs four equations involving the concentrations of these species to be able to completely characterize the system. These equations are the following. 1. Mass balance: [HAc] + [Ac {] = CHAc, where C HA c is the initial concentration of acetic acid.
At this point the chemical analysis of the problem is over and all that is left is to find the chemically relevant root of this quadratic, which is [H+], and then calculate pH. With a graphing calculator, the roots of this polynomial equation can be found quickly. The equation above is rearranged to the following:
2. Charge balance: [Ac{] + [OH{] = [H+] 3. Ion product constant for water: K w = [H+][OH{] 4. Dissociation equilibrium for HAc: {
+
Ac
H
x2 Ka – =0 0.10 – x
Ka =
HAc
Now, using the dissociation equilibrium for HAc and substituting for the other variables using the eqs 1–3, one arrives at the following expression. +
H
+
H –
Kw +
H
Ka = C HAc – Figure 3. TI-85 calculator screen of the function graph. The graph window is defined by xMin = 0, xMax = 0.01, yMin = {1.64 × 10{5, yMax = 1.64 × 10{5. The root command has found the x- value for which the graph crossed the x -axis and has marked the graph with a cross at that spot. The coordinates are also shown on the graph.
+
H –
Kw +
H
This equation can be solved graphically as above by letting x = [H+], Kw = 1 × 10{14, Ka = 1.8 × 10{5, and CHAc = 0.10 M. Figure 4 shows the TI-85 screen for this problem. Again one finds that x = 0.0013. Certainly, there is more labor involved in setting up and solving this equation, but students can feel confident that they have completely specified the solution. Furthermore, identifying chemical species present and finding relationships among the concentrations of the species is an approach that can be used to solve all acid–base problems. Note
Figure 4. TI-85 calculator screen of the function graph. The graph window is defined by xMin = 0, xMax = 0.01, yMin = {1.8 × 10{5, yMax = 1.8 × 10{5. Note that xMax was decreased from 0.1 to better display the function graph. If xMax is set at 0.1, then the function graph and the y-axis are very close together. The calculator will find the root for either setting of the graph window. The root command has found the x -value for which the graph crossed the x-axis and has marked the graph with a cross at that spot. The coordinates are also shown on the graph.
634
W Supplementary materials including detailed directions for implementing these strategies on TI-85, TI-86, and TI-92 calculators are available on JCE Online at http://jchemed.chem.wisc.edu/Journal/issues/ 1999/May/abs632.html.
Literature Cited 1. 2. 3. 4. 5.
Gordus, A. A. J. Chem. Educ. 1991, 68, 215–217. Metz, C.; Donato, H. J. Chem. Educ. 1989, 67, A241. Gordus, A. A. J. Chem. Educ. 1991, 68, 291–293. Gordus, A. A. J. Chem. Educ. 1991, 68, 397–399. Cavaleiro, A. M. V. S. V. J. Chem. Educ. 1996, 73, 423–425.
Journal of Chemical Education • Vol. 76 No. 5 May 1999 • JChemEd.chem.wisc.edu
