In the Classroom
Solving Chemical Equilibria David Uribe I.E.S. Antonio Gala, Avda. de la Cañada 44, 28820 Coslada (Madrid), Spain
The methods to solve chemical equilibrium problems are explained, in some detail, in most textbooks of general chemistry. The main difficulty for students appears when we have a set of simultaneous equilibria, in which case the technique employed is usually based on the application of the charge and mass balances and on the use of the different equilibrium constant expressions (1–5). A system of equations is obtained, where the unknowns are the equilibrium concentrations of the substances involved in the chemical mixture, and the number of equations obtained, equal to the number of unknowns, is in general greater than the number of equilibrium equations that form the system under study. However, since the various equilibrium reactions are governed by their equilibrium constants, it is logical to suppose that a number of algebraic equations, equal to the number of equilibrium reactions that form the chemical system, should be sufficient to describe the system. In fact, with the method employed in this paper it is not necessary to consider the charge and mass balances, but only the corresponding equilibrium constant expression for each chemical equilibrium. Besides, approximations that allow an easier solution of the problem can frequently be found. The treatment is presented in a simple form, which is intended for teachers as well as for students.
Description of the Method The technique is based on considering, exclusively, the variations in the concentrations of the species participating in the equilibria, so that a number of algebraic equations (and unknowns) equal to the number of equilibrium equations is obtained. An important quantity in stoichiometry is the extent of reaction, ξ, which, for a given chemical equation, is defined by ∆ξ = ∆ni / νi (6 ), where ∆ni denotes the change in the molar amount of the i th substance and νi the corresponding stoichiometric number (the stoichiometric coefficient, negative for reactants and positive for products). If we work at constant volume, which is the most usual condition, we have ∆ci = ∆ni /V, where ∆ci is the change in concentration. Calling x = ∆ξ /V, we finally get ∆ci = νi x
(1)
For instance, for the equilibrium aA + bB cC + dD, the changes in concentration of A, B, C and D will be ax, bx, cx, and dx, respectively. Let us now suppose that we have the following three simultaneous equilibria Continued on page 1178
JChemEd.chem.wisc.edu • Vol. 75 No. 9 September 1998 • Journal of Chemical Education
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In the Classroom
2A + B
3C
A + 2C
2D
A+D
B
We will assume that the initial concentrations of A, B, C and D are a, b, c, and d, and that x, y, and z denote the quotient ∆ξ /V corresponding to the first, second, and third equilibrium, respectively, so that the changes in concentration, according to eq 1, will be for A: 2x, y, and z; for B: x and +z; for C: +3x and 2y; and for D: +2y and z. Thus we have the following equilibrium concentrations: 2A + B a – 2x – y – z b–x+z A + 2C a – 2x – y – z c + 3x – 2y A + D d + 2y – z a – 2x – y – z
2D d + 2y – z B b–x+z
Calculate the concentrations of H+, acetic acid (HA), and acetate ion (A) in a 107 M acetic acid solution. Ka = 1.74 × 105 mol/L.
This example has been taken from a well-known chemistry problems book (5), where the conventional technique is used, and the following four equations are considered: [A][H+]/[HA]
equilibrium constant for acid dissociation: Ka = equilibrium constant for water dissociation: Kw = [H+][OH] charge balance: [H+] = [A] + [OH] mass balance: c = [HA] + [A]
where c (= 107 mol/L) is the analytical acetic acid concentration and [H+], [OH], [HA], and [A] are the four unknowns. This set of equations leads to the following cubic equation: [H+]3 + Ka [H+]2 – (Kw + Ka c) [H+] – Ka Kw = 0 The H+ concentration produced by dissociation of pure water is, at room temperature, 107 mol/L, which is equal to the analytical concentration of acetic acid. Thus, water dissociation must be taken into account when solving this problem. Only when the acid concentration is much greater than 107 mol/L may the H+ ions be regarded as yielded exclusively from the acid, and water dissociation may be ignored. Now, we are going to solve the problem by means of the method proposed in this paper, which leads to a simple set of two equations in two unknowns. To simplify the symbolism, we will work with the numerical values of concentrations, which will be denoted by greek letters. Acetic acid and water dissociate, according to H+ + OH β
β
where the unknown α denotes the numerical value of the H+ concentration obtained from the acetic acid dissociation, and the unknown β the one obtained from the water dissociation. The total numerical value of the H+ concentration will
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H2 O
H+ + OH
α+β
β
Making δ = α + β and taking into account the equilibrium constants, the following two equations can be obtained: 5
1.74 × 10 =
αδ 7 10 – α
(2)
10 14 = δ( δ – α)
(3)
From eq 2 we get 12
α=
Example 1
H2 O
HA A + H+ 107 – α α α + β
3C c + 3x – 2y
Then, using the equilibrium constant expressions, a system of three equations in three unknowns (x, y, z) is obtained. We can observe that, as it is logical, the equilibrium concentrations of A, B, C and D are the same in all the equilibria. The following two examples have been developed in enough detail to illustrate the method.
HA A + H+ 107 – α α α
then be α + β, and
1.74 × 10
(4) 5 1.74 × 10 + δ We know that the H+ concentration in pure water, at room temperature, is 107 mol/L. In the case of a complete dissociation of the acetic acid, it will contribute to the H+ concentration with 107 mol/L. It is therefore evident that δ must be in the range between 107 (no dissociation of the acetic acid at all) and 2 × 107 (complete dissociation of the acetic acid); hence δ
