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John Alexander University of Cincinnati Cincinnati, OH 45221
A Puzzle Concerning Solution Equilibria Peter Gans School of Chemistry, The University of Leeds, Leeds LS2 9JT, England;
[email protected] Puzzle In an aqueous acid–base system how can three related species (analogous to an acid and its conjugate base) coexist in equilibrium with equal concentrations? Suggest one or more common substances that can exhibit this behavior and define the conditions under which the three species have equal concentrations. Answer The key to this puzzle is the recognition that for any equilibrium involving deprotonation of an acid, choice of pH fixes the ratio of the concentrations for the acid and its conjugate base. Therefore the third species must be involved in an equilibrium that does not involve proton transfer. A dimerization equilibrium (for example) will satisfy this requirement. To illustrate this point, consider the simplest case: what happens when a weak acid HL is dissolved in water and dissociates according to the equilibrium H + L HL (ionic charges are omitted from generic formulas). A protonation constant, β, can be defined in terms of concentrations (in this treatment it is assumed that activity quotients are constants that can be ignored): [HL] = β[H][L]. If the concentrations of the acid HL and its conjugate base L are equal then [HL] = [L] and consequently eq 1 holds. [H ]= 1/β ;
pH = log10 β
(1)
This is commonly expressed by saying that half-neutralization occurs when the pH of the solution is numerically equal to pKa (Ka, the acid dissociation constant, is the reciprocal of β, the protonation constant of the conjugate base: pKa = log β). In general, the pH for equal concentrations of two partners in any simple protonation reaction is fixed by an expression such as eq 1. Therefore, for any third species to have the same concentration as the other two it must be in an equilibrium that does not, by itself, depend on pH.
Example 1 For the equilibrium HL + L HL2 the concentrations are given by [HL2] = K2[HL][L]. Therefore when [HL2] = [HL], clearly [L] = 1/K2 is a possible condition, since it is independent
of pH. Therefore L could be the fluoride ion. Using the values log β1 = 3.17 for HF and log K2 = 0.58 for HF2 (1), [F] = 0.2630 mol dm3. In order to find TF, the amount of substance required to meet these conditions, the principle of mass balance may be applied: TF = [F] + [HF] + 2[HF2]. Thus the requisite conditions are then given by TF = 4[F] = 1.052 mol dm3, pH = 3.17. A solution of NaF made up to 1.052 mol dm3 and pH 3.17 would contain equal concentrations of F, HF, and HF2. To have equal concentrations of fluoride in each species the concentration of HF2 would have to be half that of HF. In that case, [HF] = 1/2K2 and TF = 3/2K2; that is, 0.3945 mol dm3. The following speciation diagram, Figure 1, produced directly by the program HySS (2), shows the three species each having 33% of the total fluoride.
Example 2 If the acid can dimerize (i.e., 2HL H2L2), a dimerization constant, KD, can be defined: [H2L2] = KD[HL]2. The concentrations of monomer and dimer are the same when [H2L2] = [HL], and so [HL] = 1/KD follows. This equality is also not dependant on pH, so the three species can have the same concentrations, namely, [L] = [HL] = [H2L2] = 1/KD. The amount of substance required is given by TL = [L] + [HL] + 2[H2L2]. Thus the requisite conditions in this example are then given by TL = 4/KD and pH = log β1. Recently published values for the protonation of the chromate ion are as follows (3): H+ + CrO42 2HCrO4
HCrO4
log β1 = 6.09
2
Cr2O7 + H2O
log KD = 1.77
100
% of total fluoride concentration
It is well known that half-neutralization of a weak acid occurs when the pH of a solution of the acid is adjusted to be numerically equal to pK a (K a is the dissociation constant of the acid). Under those conditions the concentrations of the two species, the acid and its conjugate base, are equal. The following puzzle is designed to reinforce understanding of simple acid–base equilibria and to give students an introduction to more complicated equilibria.
90 80 F
70
_
HF
60 50 40 30
HF 2
20
_
10 0 2.0
2.5
3.0
3.5
4.0
4.5
5.0
pH Figure 1. Hyss speciation diagram for a solution containing 0.3945 mol dm3 of NaF, showing three species with equal fluoride concentrations at pH 3.17.
JChemEd.chem.wisc.edu • Vol. 77 No. 4 April 2000 • Journal of Chemical Education
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In the Classroom
Thus, suitable conditions for [CrO42] = [HCrO4] = [Cr2O72] are obtained by dissolving Na2CrO4 to a concentration of 0.06793 mol dm3 and pH = 6.09. The HySS speciation diagram, Figure 2, shows the three species having equal chromium content at pH = 6.09. Since the dimer contains two chromium atoms, its concentration is half that of the monomer: [HL] = 1/2KD and TCr = 3/2KD, in this case.
% of total Cr concentration
100 90 CrO 4 2
80
_
70 60
Cr 2 O 7 2
_
50 40 30
Literature Cited
20 10
HCrO 4
_
0 4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
pH Figure 2. Hyss speciation diagram for a solution containing 0.02547 mol dm3 Na2CrO4, showing three species having equal chromium content at pH 6.09.
490
1. Broene, H. H.; DeVries, T. J. Am. Chem. Soc. 1947, 69, 1644– 1645. 2. Alderighi, L.; Gans, P.; Ienco, A.; Peters, D.; Sabatini, A.; Vacca, A. Coord. Chem. Rev. 1999, 184, 311–318; http://www.chim1. unifi.it/group/vacsab/hyss.htm (accessed Dec 1999). 3. Cruywagen, J. J.; Heyns, J. B. B.; Rohwer, E. A. Polyhedron 1998, 17, 1741–1746.
Journal of Chemical Education • Vol. 77 No. 4 April 2000 • JChemEd.chem.wisc.edu
