Preparation: McMurry sections 20.3-20.5 (pp. 778-785). Review your general
chemistry text - acids & bases, buffers, and titration of a weak acid with a strong
base. A tutorial is available outlining
calculations for this type of titration which you studied in general chemistry.
The equilibrium constant for the reaction:
R-COOH + H2O ----> R-COO- + H3O+
could be expressed as K = [R-COO-][H3O+]/[R-COOH][H2O].
However, because the [H2O] is essentially constant it is convenient to multiply
the term with the formal equilibrium constant to form Ka , the acidity constant, a
property of all organic acids.
Hence, Ka = [H3O+][R-COO-]
/ [R-COOH]
If both sides of this equation are expressed in logarithmic form:
log Ka = log [H3O+] + log ( [R-COO- ]/[R-COOH] ) or making all terms negative:
-log Ka = -log [H3O+] - log ( [R-COO- ]/[R-COOH] ).
The -log Ka is known as the pKa , a property of a given acid related to its acidic strength. -log [H3O+] is the familiar pH. This gives
pKa = pH - log ( [R-COO-]/[R-COOH] ) which is the Henderson- Hasselbach equation.
This is a useful equation because it relates the pKa of an acid dissolved in any aqueous solution to the pH of the solution and the concentration ratio of the ionized conjugate base, R-COO-, to the unionized acidic form, R-COOH.
The following experiment will take advantage of this equation to determine the pKa and thus the Ka of an unknown organic acid in order to identify it. The strategy is to determine the pH of a solution of the acid when the concentrations of the ionized form and unionized form are equal. When this is the case,
[R-COO-]/[R-COOH] = 1; since log 1= 0, the equation then becomes pKa = pH.
In order to create the situation where [R-COO-] = [R-COOH] a titration is run and its progress is followed with a pH meter. The acid is dissolved in the solvent and the pH is measured; incremental amounts of base are added and the pH is measured. The pH measurements are plotted vs. the mL of base added and a smooth curve is drawn through these points as in the plot below.
The method begins by determining the endpoint, the point where the amount of base added equals the amount of acid initially present in the solution (in the example the endpoint occurs at roughly pH = 7.2 and (more accurately) at the 8.7 mL mark, Vep. One half of this amount is calculated (in the example, 4.35 mL); this is the point at which one half of the acid was converted to the ionized form by the base, the situation described above where [R-COO-] = [R-COOH]. At this point the pH of the solution equals the pKa of the acid examined. The pKa of the acid in the example is 4.80. The number whose log is -4.80 is 1.6x10-5; this is the Ka of the unknown acid. The aromatic acid that most closely corresponds to this example (Table 20.6 in McMurry) is ortho-aminobenzoic acid.
Procedure: Accurately weigh out approximately 20 mg of an unknown acid obtained from the stockroom. Record your unknown number! Dissolve the acid in 10 mL of ethanol in a 100 mL beaker. If the acid does not completely dissolve, continue to slowly add more ethanol in order to dissolve all the solid. Dilute the solution with 40 mL of water and add a magnetic stir bar. Calibrate a pH meter with "pH 7" and "pH 4" solutions and then carefully clamp the probe to measure the pH of the solution as a function of added strong base. Set up a clean burette filled with aqueous 0.01 N NaOH. Begin gentle stirring and record the initial pH and then after every 0.5 - 1.0 mL of base added. Near the endpoint record pH readings every 0.25 mL or less. (Why is this necessary?) Titrate at least 3-4 mL beyond the endpoint. Prepare a fresh sample of the acid and repeat the titration; make rough plots of the data and repeat the titration until the pKa values agree to within 0.2 units. Calculate the Ka and consult Table 20.6 in McMurry to identify your unknown acid.
Exercises: 1. Use your plot to determine pH when the acid was 75% neutralized. What was the value of the ratio [R- COO-]/[R-COOH] at this point? Take the negative log of this number and apply it to the Henderson-Hasselbalch equation above to determine the pKa of your acid. How well does this figure agree with the pKa you calculated initially?
2. The solvent used in your measurement contained ethanol. If ethanol has a lower ionizing power than water, consider the consequences of your pKa if you used only water as a solvent. Why was ethanol used?
3. Suppose 25mg instead of 20 mg of the acid are weighed out. How would this affect your result ? Why?
References:
"Mayo et al.": Mayo, D.W., Pike, R.M., Butcher, S.S. and Trumper, P.K. Microscale Techniques for the Organic Laboratory; Wiley: New York, 1991.
"M&B": Morrison, R.T., and Boyd, R.N. Organic Chemistry, 6th ed.,
Prentice Hall: Englewood Cliffs, NJ, 1992