edited by:
J. ALEXANDER University 01 Cincinnati
JOHN
Cincinnati, Ohio 45221
A Problem for Buffer Buffs J.
K,
a conservation mass equation
Ram6n Leis and M. Elana Peiia
Dpto. QuimicaFisica
Colegio Universitario de Lug0 27001 Lugo. Spain C. Herrero-Latorre Dpto. QuimicaAnalitica. Nutricion y Bmmatolagia Colegio Universitario de Lugo 27001 Lugo, Spain This problem is particularly intended for undergraduates studying chemistry as a part of some other discipline such as biochemistry, biology,.or medicine. I t illustrates the perils of taking for granted the simplifying assumptions made by some textbooks for general chemistry when discussing chemical equilibrium in solution. The student (or teacher!) who unthinkingly applies simplified formulas to equilibrium problems is especially likely to slip up when calculating acid-base equilihria in buffer solutions; indiscriminate use of the Henderson-Hasselbalch equation, or consideration of chemical equilibria alone, without regard to other constraints on the solution, caflall too easily lead to errors of the kind discussed below. Problem A buffer solution is prepared by mixing equal volumes of 0.020 M solutions of an acid HA of DK. 12 and its sodium salt NaA. Its pH can a t first glance be calckated by either of the following two methods. Method 1. The equilibrium equation HA (0.010- x )
+ H20 e
A-
(0.010+ x)
+H,Ot r
together with the law of mass action implies
K, = 1.00 X lo-''
= [H30t][OH-]
+
-x )
= ~(0.010 x)/(O.OlO
C. + Cb = [A-I
+ [HA]
where C, and Cb are the nominal concentrations of HA and NaA, respectively,and a conservation of charge equation
[Nat]
+ [H30+] = [OH-] + [A-]
(4)
The last is used by neither Method 1nor Method 2, which are both solved for only three of the four species present (Method 1ignores OH- and Method 2 H30+). Equation 4 is in fact far from being satisfied by the solution given by 0.010 Method 1;0.010 M (Na+) 1.00 X 10-l2 M (&of) 0.010 M (OH-), since the concentration of the M (A-) species ignored, OH-, is not negligible in comparison with the others. Method 2 performs better; 0.010 M (Na+) 10-11.62M (H30+)= 4.14 10-3 M (OH-) 5.86 10-3 M (A-), because the concentration of the species ignored in this case, &0+, is negligible. A rigorous solution of eqs 1-4 leads to the cubic equation
+
+
+
+
+
whose solution other than by a computer program involves an amount of labor that one would not wish to have to perform every time the pH of a buffer is to be calculated. The "trick" is therefore to use Method 1or Method 2 but to check in either case that values of [H30+] and [OH-] obtained are consistent with the assumption that [OH-] is negligible (if Method 1is used); ifresults are not compatible with the assumptions of the method used, then the other method could be employed. Alternatively, sketching the graph of logarithmic concen,trations against pH makes it clear a t once whether i t is [H30C]or [OH-] that can he ignored; see the figure.
and hence that [H30+] = 1.00 X 10-12, i.e., the buffer is pH 12.00. Method 2. The conjugate base A- has pKb = pK, pK. = 2. The equation A+H,O ~1 HA +OH-
-
(0.010- x)
+
(0.010 r )
x
together with the law of mass action implies
K b= 1.00 X lo-'
+
= ~(0.010 x)/(O.OlO
- r)
and hence that [OH-] = 4.14 X 10-3, i.e., the buffer is of pH 11.62. Why do the methods yield different results? Acceptable Solution Apart from sodium ions and water molecules, the solution in question contains four species: HA, A-, H@+, and OH-. Calculation of the concentrations of all four requires the use of four equations: the law of mass action for K, (or K b )
K. = [H30t][A-]/[HA] or Kb = [HA][OH-]/[A-I (1) the relationship between [HzO+]and [OH-], i.e., the definition of the ion product of water, K, Volume 68 Number 2 February 1991
141
