Henry Donato, Jr.

Department of Chemistry and Biochemistry

College of Charleston

Charleston, SC 29424

The chemical equilibrium state is described using algebraic equations. The process of analyzing information given in an equilibrium problem and writing 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 higher¹. 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 either require 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 where one of the equilibria can be ignored, are considered. For example, the equilibrium system:

H_2(g) + I_2(g) Û 2HI_(g)

is often used in examples illustrating equilibrium principles while the system:

3H_2(g) + N_{2(g) Û} 2NH_3(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 solutions⁴. Recently, similar guidelines have been suggested for aiding students with approximations associated with precipitation equilibria⁵. 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 equation generated from this analysis.

The graphing calculator is a powerful computing devise 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. For purposes of illustration, a Texas Instruments-86 calculator (TI-86) is used. Screens captured from this calculator are incorporated into this document to help guide the reader through the manipulations

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 400⁰C, a mixture of 3 parts H_2(g) and 1 part N_2(g) is placed in a container with total pressure of 10 bar and the partial pressures of H₂, N₂, and NH₃ at equilibrium are desired. The chemical reaction involved is

N_2(g) + 3H_{2(g) Û} 2NH_3(g) K_p = 1.64 x 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:

The equilibrium constant expression then becomes:

One may then enter the difference between the left and right hand sides of the equation above into the y= screen of the graphing calculator which may be reached by pressing GRAPH followed by F1. When entering the formula, note that the calculator makes a distinction between the arithmetic operation of subtraction (the key marked with a dash) and a negative exponent (the key marked with a dash in parentheses). Also, the x used in the formula is entered with the x-VAR key rather than the alphanumeric x (ALPHA +).

Note that the entire equation is not displayed on the screen but one can scroll through the equation to check its accuracy. Since this is a quartic equation one expects four roots, but only one will be chemically reasonable. By examination of the equation one can set bounds on possible chemically reasonable values of x. For example, x will not be less than zero nor greater than 2.5. One can set the graphing window of the calculator so that only x values in that window will be graphed. The window command on the calculator (2^nd F2) allows one to set the graphing window.

The yMin and yMax values were chosen because it is apparent that y will not be larger than 1.64 x 10^-4. Then the graph of the function can be inspected by pressing F5.

The x value at which the function crosses the x axis is the chemically relevant root. That value can easily be found by the calculator using the root command under the math menu. To find the math menu, press MORE for a listing of additional commands under the graph menu. Then press F1 for the math menu and finally press F1 for the root command. The x and y coordinate of the current location of the cursor will appear on the screen. Use the arrow keys to move the cursor so that the x value is smaller than the root (to the left of the root), and press ENTER. Then move the cursor so that the x value is greater than the root and press ENTER. If one presses ENTER again, the calculator will search for the root and produce the result seen in the calculator screen to the left.

The root is seen to be 0.179 bar. Alternatively, one may keep expanding the graph window and read the value of the root directly from the graph. With a little practice, this procedure can be easily performed. It 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.

Other problems presenting mathematical difficulties are those illustrating the common ion effect. For example, calculate the molar solubility of PbCl₂ (Ksp = 1.6 x 10^-5 ) in (a) pure water, (b) in 0.15 M KCl solution and (c) 0.080 M Pb(NO₃)₂

1. Ksp = [Pb⁺²][Cl^-1]² = (s)(2s)² = 4s³

 

Therefore: = 0.016 M

 

2. Ksp = (s)(2s + 0.15)²

 

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

 



 

Now for the exact solution, first enter the equation below in the y = window (GRAPH, F1, and if a function is already there, press CLEAR to erase that function) of the graphing calculator using the x-VAR key to enter s.

 

1.6 x 10^-5 – [(s)(2s + 0.15)² ] = 0

Since 7.11 x 10^-4 represents an upper bound for possible s values, the range can be set as shown below. (Pressing 2^nd F2 allows one to adjust the window parameters.)

By pressing F5 one obtains a graph of the function. Then pressing MORE, F1, F1, setting a left and right bound and finally ENTER, one obtains the value of the chemically relevant root.



Note that the root found by the graphical method is 6.98 x 10^-4 which is different from the result obtained with the approximate method.

 

 

3. Ksp = (s + 0.08)(2s)²

If s can be ignored in the first parentheses then s = .010 M providing one with an approximate value of s which must be larger than the exact value of s. To find the exact value, enter the equation below into the y = window (GRAPH, F1, and if a function is present, CLEAR) of the graphing calculator using the x-VAR key to enter s.

1.6 x 10^-5 –[(s + 0.08)(2s)² ] = 0

One may use the approximate value of s to construct the chemically relevant graph window for the problem. (Press 2^nd, F2 to enter the graphing window.)

From the range window one selects F5, MORE, F1, F1, sets the left and right bounds and finally ENTER to obtain the chemically relevant root of 6.79 x 10^-3.

Hence students can demonstrate easily, the limitations of approximations and hence obtain a better understanding of the ways chemists describe these slightly soluble salts.

In aqueous solutions of weak acids and weak bases, there may be multiple equilibria which must be satisfied simultaneously. Often, the self-ionization of water can be ignored. For example, calculate the pH of a 0.10 M Acetic Acid solution. Students are generally instructed to solve this problem by completing the following steps. Students must recognize that acetic acid is a weak acid, which means that the following equilibrium exists in aqueous solutions of acetic acid:

HAc Û 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 Ka = 1.8 x 10^-5:

Ka =

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 quickly found. The equation above is rearranged to the following:

To find the chemically relevant root, enter the left hand side of the equation in the y = window of the TI-86 (GRAPH, F1, CLEAR) using Ka = 1.8 x 10^-5.

Note that x must be above zero and less that 0.10. Also, the y value can not be any larger than 1.8 x 10^-5. This allows us to define the graph window on the calculator. (2^nd, F2)

Once our window is set, the graph can be viewed: (F5)

One can barely see the function crossing the x axis at a very small value of x. Reset the graph window so that xMax is 0.01. To select the range window one only needs to press F2.

Now view the graph in the expanded viewing window by again pressing F5

Select the root option under the math menu and press enter. (MORE, F1, F1, set left and right bounds, ENTER)

Note that the root determined graphically is 0.0013 M.

The above procedure is perfectly general, is easy to understand, and can be performed in class and on tests by students. The emphasis is on writing the algebraic formula appropriate for the chemical situation, and defining the chemically relevant graphing window. In this case the value of x was small compared to the initial concentration of weak acid, so the student was required to expand the graphing window in order the view clearly the function which was graphed. Note that it is not necessary to expand the viewing window to use the root option in the math menu. It will find the root anyway.

The treatment described above is the standard treatment of a weak acid at the introductory chemistry course level. A more complete treatment is not given because it is not needed to get the correct answer to this problem, and because it leads to a cubic polynomial equation which must be solved. However, the method described above will solve a cubic equation as easily as a quadratic. Also a more complete treatment is easier for students to understand. In the 0.10 M acetic acid solution, there are four chemical species: H⁺, OH^-, HAc, and Ac^-, whose concentration needs 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^-] = C_HAc where C_HAc is the initial concentration of acetic acid.

2. Charge Balance [Ac^-] + [OH^-] = [H⁺]

3. Ion Product Constant for Water Kw = [H⁺][OH^-]

4. Dissociation Equilibrium for HAc

Now using the dissociation equilibrium for HAc and substituting for the other variables using the equations 1-3, one arrives at the following expression.

This equation can be solved graphically as above by letting x = [H⁺], Kw = 1 x 10^-14, Ka = 1.8 x 10^-5, and C_HAc = 0.10 M. Proceeding as before, enter the difference of the left and right hand sides of the equation above into the y = window of the graphing calculator (GRAPH, F1, CLEAR).

The same graph window can be used.

Finally, one can graph and determine the root of the equation. (F5, MORE, F1, F1, set left and right bounds, and ENTER)

Certainly, there is more labor involved in setting up and solving this equation, but the student can feel confident that he/she has completely specified the solution. Furthermore, the approach of identifying chemical species present and finding relationships among the concentrations of the species, is an approach which can be used to solve all acid base problems.

References

1. Gordus, A. A. J. Chem. Educ. (1991), 68, 215-217.

 

2. Metz. C. and Donato, H. J. Chem. Educ. (1989), 67, A241.

3. Gordus, A. A. J. Chem. Educ. (1991), 68, 291-293.

4. Gordus, A. A. J. Chem. Educ. (1991), 68, 397-399.

5. Cavaleiro, A. M. V. S. V. J. Chem. Educ. (1996), 73, 423-425.