# POTENTIOMETRIC TITRATION OF A WEAK ACID

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2 A ph meter, or ph amplifier, such as the one used in this experiment, is essentially an electronic voltmeter. However, through calibration, its output reads directly in ph units instead of volts. The apparatus is sketched in Figure 1. One performs the titration by adding increments of titrant from the buret and reading the ph of the solution after each addition. Treatment of Potentiometric Data Figure 1 For each trial set of data, plot the entire titration curve as in Figure 2. The equivalence point is the point of maximum slope. You will notice, however, that the volume scale of such a graph is too compressed to read the equivalence point to ±0.02 ml as required in order to obtain part-per-thousand accuracy. One may prepare a second graph in which one plots the region from about 1-2 ml before the equivalence point to 1-2 ml beyond, as illustrated in Figure 3. The operator must judge the inflection point. The volume scale is now large enough that the volume at the inflection point can be read to the nearest ±0.02 ml. The main sources of error are the failure to draw a smooth curve through the data points and in judging the location of the inflection point. The latter error is somewhat analogous to a titration using a color indicator in which the main source of error is the operator's judgment as to the color at the end point. We will also use other mathematical methods to help determine the equivalence point. TABLE 1 Sample Titration Data Corrected Volume NaOH ph Corrected Volume NaOH ph Corrected Volume NaOH ph Potentiometric Titrations Revised Spring 2007 NF Page 2 of 14

3 Weak Acid Titration p H Estimation of pka at the Half-titration Point Volume of NaOH Figure 2 After you have located the equivalence point in the first titration curve, you can calculate the volume of sodium hydroxide at the equivalence point in the titration and therefore the volume at which exactly half of the acid will be neutralized. This is called the buffer point, half-titration point or halfequivalence point. This point is particularly significant because it provides an estimate of the acid dissociation constant of the acid being titrated. In general, for any weak acid we have the following equilibrium: HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) K a =!" H + # \$!" A - # \$ [ HA] Now, at the half-titration point, exactly half of the acid, HA, has been converted into its conjugate base. That is, the solution can be considered to be one containing equal concentrations of HA and A -. Thus, it should be clear that at this particular point in the titration [HA] = [A - ]. Therefore, K a = [H + ], or pk a = ph. Some typical data at the half-equivalence point are presented in Table 2. These data are for the same titration as the data in Figures 2 and 3. Thus, the equivalence point occurs at ml, the halfequivalence point at ml, and the pk a is approximately 4.90, as determined from Figure 4. Some disagreement in pk a from handbook is typical, and is a result of the use of molar concentrations in place of activities in the expression for the equilibrium constant. Nevertheless, it is valuable information when seeking to identify unknown acids. Table 2 Half-Equivalence Data Vol NaOH ml ph Figure 3 Potentiometric Titrations Revised Spring 2007 NF Page 3 of 14

4 First Derivative Plot We noted above that at the equivalence point, the slope of the titration curve is at a maximum. In other words, the rate of change of ph with addition of titrant is at its highest at the equivalence point. So, if we could plot the rate of change of ph with change in volume ( ph/ V) against volume, then a "spiked" curve should result and the peak of this spike should occur at the equivalence point. This is conveniently done by adding equal increments of titrant near the equivalence point. Consider the data collected during a titration, as shown in Table 3. We want to plot ph/ V against the volume to get the first derivative. Such a plot is shown in Figure 4. The volume used is the average of the two volumes used to calculate ph. So, the volume for ph/ V = 0.71 is ml, and so on. The equivalence point is the maximum of this plot, which, when extrapolated, occurs at ml. This extrapolation leads to an uncertainty that can be partially avoided by a second derivative plot (see below). Note that we have used equal volume increments here ( 0.20 ml), and so ph could have been plotted in place of ph/ V. These equal increments are not necessary but do shorten the calculations. Although the average volume may be calculated to ml for plotting, experimentally, we are not justified in reporting the equivalence point to more than 0.01 ml. Table 3 First Derivative Data Volume NaOH ph ph V Vol. Avg. ph/ V Figure 4 Potentiometric Titrations Revised Spring 2007 NF Page 4 of 14

5 Second Derivative Plot Mathematically, the second derivative of a titration curve should pass through zero at the equivalence point. The last four columns in Table 3 illustrate how such a plot can be accomplished. The second derivative is the rate of change of the first derivative with respect to the change in the average volume. The average of the two successive volumes used for the first derivative plot is also used for the second derivative plot (Figure 5). Again, there is some extrapolation, but it is less significant than in the first derivative plot. The equivalence point is taken as ml. As before, we are experimentally justified in reporting it to the nearest 0.01 ml. In using derivative methods, the volume increment should not be too large or there will not be sufficient points near the equivalence point. Caution should be used with derivative methods. Derivatives tend to emphasize noise or scatter in the data points, being worse for the second derivative. Therefore, if a particular titration is subject to noise, a direct plot (ph versus Volume) may be preferred. Table 4 Second Derivative Data Vol. Avg. ph/ V ( V) ( ph/ V) V avg ( ph/ V)/ V Figure 5 Potentiometric Titrations Revised Spring 2007 NF Page 5 of 14

6 Fortunately for us, LoggerPro generates the first and second derivative data automatically, so it is simply a matter of plotting and analyzing the curves! Assembling the Data Once you have obtained the equivalence volumes and pk a s, you can use that information to determine the identity of your unknown acid. The molecular weight can be determined from the equivalence point data. The pk a values can be determined from the half-equivalence points. Your unknown acid can be either monoprotic or diprotic. Although this is often readily seen in the titration curve, some diprotic acids appear to have only one equivalence point, if the two pk a values are less than 2 ph units apart. However, the resulting calculated molecular weight will be the equivalent weight, and thus will be one-half the molecular weight. It is a match (or near-match) of both the pk a and molecular weight data that determines the identity of your unknown acid. Table 5 gives a list of possible organic acids for this experiment. Table 5 Selected Physical Data for Organic Acids Acid Molar mass(g/mol) pk a1 pk a2 Benzoic acid Chlorobenzoic acid Chlorobenzoic acid trans Crotonic acid Maleic acid Malonic acid Potassium hydrogen phthalate Succinic acid Sodium hydrogen sulfite d -Tartaric acid m Toluic acid Reference Table taken from Lange s Handbook of Chemistry, McGraw-Hill Book. Co Potentiometric Titrations Revised Spring 2007 NF Page 6 of 14

9 Processing the Data 1. Calculate the equivalent weight of your unknown using your equivalence point. 2. Find the pk a for your unknown acid from the half-equivalence point and identify the acid using the table attached. 3. Staple together the report forms, data tables, and the required graphs. Print your name and date on each graph. A portion of your grade will depend upon the presentation of your data and graphs. Graphs include: Complete titration curves based on the data for both titrations. The first derivative graph showing the equivalence point region on an expanded volume scale (see Figure 3). You should be able to read volumes to 0.01 ml. The second derivative graph showing the equivalence point region on an expanded volume scale (see Figure 4). You should be able to read volumes to 0.01 ml. Expanded half equivalence pt region for each titration. NOTE: If you have a diprotic acid, you will analyze derivative curves only for the second equivalence point (and half-equiv point): However, you will need to treat each equivalence point and halfequivalence point separately for calculations. Potentiometric Titrations Revised Spring 2007 NF Page 9 of 14

10 SAMPLE CALCULATIONS Calculations in this experiment are the same as those in the Standardization of NaOH experiment please review them! A. Weak Acid Strong Base Titrations The equivalence point of the titration occurs when the milliequivalents (meq) of base added exactly equals the milliequivalents of acid in the flask or when millimoles of base units added exactly equals the millimoles of acid units in the flask. meq acid = (ml acid)(n acid) meq base = (ml base)(n base) The usefulness of normality in volumetric analysis is demonstrated with following: at the equivalence point N Acid V Acid = N Base V Base To calculate the number of equivalents contained in a known acid sample, one needs the sample mass, its purity, and its equivalent weight. equiv. acid = or, in a more familiar form equiv. acid = g sample x sample weight (in g) x purity factor equiv. weight (g/eq) g acid 100 g sample x equiv acid g acid Example 1: A gram sample of a pure organic acid required ml (corrected) of N NaOH for to reach the equivalence point. Calculate the equivalent weight of the acid, and report with a relative error of 1 part per Assume a purity of 100% or 1. Equiv Wt = g acid x 1 equiv NaOH 1 equiv acid Equiv Wt = g/equiv (ppth precision) x 1 L equiv NaOH x L Potentiometric Titrations Revised Spring 2007 NF Page 10 of 14

11 Chem 135 Potentiometric Titration of a Weak Acid DATA REPORT SHEET Name Unknown Identification Normality of NaOH Approximate Equivalent Weight Data (Indicator Titration) Sample Mass = V NaOH = Equiv. Weight = Potentiometric Titration Data. Enter your data in to the appropriate spreadsheet in the lab. Show Sample Calculations on the back of this page! Sample Mass (g) Trial 1 Trial 2 Average V NaOH, 1 st Equiv. pt. V NaOH, 2 nd Equiv pt. (If applicable) Equiv. Wt. Molar Mass pk a1 pk a2 Identity of Unknown acid Show sample calculations for Equiv. Wt and Molar Mass on the back. Potentiometric Titrations Revised Spring 2007 NF Page 11 of 14

12 Sample calculations for Equiv. Wt. and Molar Mass Potentiometric Titrations Revised Spring 2007 NF Page 12 of 14

13 Chem 135 Potentiometric Titration of a Weak Acid PRELAB Name You may use this experiment handout, your textbook or any other resources you can find to answer these questions. 1. Distinguish between the terms endpoint and equivalence point. 2. Draw the correct Lewis structures for phosphoric acid and phosphorus acid and thoroughly explain why one of these is a triprotic acid and the other is diprotic. 3. Write a chemical equation which represents the acid dissociation constant for benzoic acid. 4. For a weak acid titrated with a strong base, where does the ph = pk a? a) At the equivalence point. b) At the half-equivalence point. c) At the beginning of the titration. - over - Potentiometric Titrations Revised Spring 2007 NF Page 13 of 14

14 5. A g sample of an unknown monoprotic acid is titrated with N NaOH ml of NaOH is required to reach the equivalence point. What is the molar mass of this acid? Molar Mass = 6. Look up the structures of the following acids and provide structural formulas for them. trans-crotonic acid Maleic Acid Malonic Acid Succinic Acid d-tartaric Acid m-toluic Acid Potentiometric Titrations Revised Spring 2007 NF Page 14 of 14

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