Page 193 - Modern Analytical Chemistry

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176 Modern Analytical Chemistry

In a redox reaction, one of the reactants is oxidized while an- swer is needed, a ladder diagram may help you decide the equilib-
other reactant is reduced. Equilibrium constants are rarely used rium system’s composition.
when characterizing redox reactions. Instead, we use the electro- Solutions containing a weak acid and its conjugate base show
chemical potential, positive values of which indicate a favorable only a small change in pH upon the addition of small amounts of
reaction. The Nernst equation relates this potential to the concen- strong acid or strong base. Such solutions are called buffers.
trations of reactants and products. Buffers can also be formed using a metal and its metal–ligand
Le Châtelier’s principle provides a means for predicting how complex, or an oxidizing agent and its conjugate reducing agent.
systems at equilibrium respond to a change in conditions. When a Both the systematic approach to solving equilibrium problems
stress is applied to an equilibrium by adding a reactant or product, and ladder diagrams can be used to characterize a buffer.
by adding a reagent that reacts with one of the reactants or prod- A quantitative solution to an equilibrium problem may give an
ucts, or by changing the volume, the system responds by moving answer that does not agree with the value measured experimen-
in the direction that relieves the stress. tally. This result occurs when the equilibrium constant based on
You should be able to describe a system at equilibrium both concentrations is matrix-dependent. The true, thermodynamic
qualitatively and quantitatively. Rigorous solutions to equilibrium equilibrium constant is based on the activities, a, of the reactants
problems can be developed by combining equilibrium constant and products. A species’ activity is related to its molar concentra-
expressions with appropriate mass balance and charge balance tion by an activity coefficient, g, where a i = g i [] i . Activity coeffi-
equations. Using this systematic approach, you can solve some cients often can be calculated, making possible a more rigorous
quite complicated equilibrium problems. When a less rigorous an- treatment of equilibria.

6M Suggested EXPERIMENTS
Experiments “The Effect of Ionic Strength on an Equilibrium Constant (A “The Solubility of Silver Acetate.” In J. A. Bell, ed. Chemical
The following experiments involve the experimental determination of equilibrium constants and, in some cases,
demonstrate the importance of activity effects.

Class Study).” In J. A. Bell, ed. Chemical Principles in Practice.
Principles in Practice. Addison-Wesley: Reading, MA, 1967.
Addison-Wesley: Reading, MA, 1967.
In this experiment the importance of the soluble silver
–
acetate complexes AgCH 3COO(aq) and Ag(CH 3 COO) 2 (aq)
In this experiment the equilibrium constant for the
dissociation of bromocresol green is measured at several
in describing the solubility of AgCH 3COO(s) is investigated.
ionic strengths. Results are extrapolated to zero ionic strength
to find the thermodynamic equilibrium constant.
of the Thermodynamic Solubility Product, K sp , of PbI 2
Assuming Nonideal Behavior,” J. Chem. Educ. 1996, 73,
“Equilibrium Constants for Calcium Iodate Solubility and Green, D. B.; Rechtsteiner, G.; Honodel, A. “Determination
Iodic Acid Dissociation.” In J. A. Bell, ed. Chemical Principles 789–792.
in Practice. Addison-Wesley: Reading, MA, 1967. The thermodynamic solubility product for PbI 2 is
The effect of pH on the solubility of Ca(IO 3) 2 is studied in determined in this experiment by measuring its solubility at
this experiment. several ionic strengths.

6N PROBLEMS

1. Write equilibrium constant expressions for the following 2. Using a ladder diagram, explain why the following reaction
reactions. Determine the value for the equilibrium constant – –
H 3PO 4(aq)+F (aq) t HF(aq)+H 2PO 4 (aq)
for each reaction using appropriate equilibrium constants
from Appendix 3. is favorable, whereas
+
–
a. NH 3(aq) + HCl(aq) t NH 4 (aq)+Cl (aq) H 3PO 4 (aq)+2F (aq) t 2HF(aq)+H 2PO 4 (aq)
–
2–
2–
–
b. PbI 2(s)+S (aq) t PbS(s)+2I (aq)
4–
2–
–
2–
c. CdY (aq) + 4CN (aq) t Cd(CN) 4 (aq)+Y (aq) is unfavorable. Determine the equilibrium constant for these
4–
[Y is EDTA] reactions, and verify that they are consistent with your ladder
–
+
d. AgCl(s) + 2NH 3(aq) t Ag(NH 3) 2 (aq)+Cl (aq) diagram.
+
2+
e. BaCO 3 (s)+2H 3 O (aq) t Ba (aq)+H 2 CO 3 (aq)+2H 2 O(l)