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Titration

We investigated titrations with sodium hydroxide (NaOH), potassium hydrogen phthalate (KHC8H7O4, abbreviated KHP), and unknown acid #1. We titrated a known amount of KHP with NaOH to find [NaOH]. We then used the NaOH to investigate pH during titration of an unknown weak acid and find its pKa. Our unknown acid #1 had a pKa of about 4.40. Our NaOH solution had a molarity of 9.1882 10-2. With the unknown acid's pKa of 4.74, we have a percent error of -7.17%.

Introduction

The measure of acidity is pH or the negative logarithm of the concentration of protons. To measure the pH of some solution, one can use titrations coupled with an indicator, a chemical pH indicator, or an electronic pH meter. Titrations rely on the equivalence point, or the point where all the molecules of the base have reacted with an acid or vice versa, to determine the pH. The equivalence point must be indicated by some sort of indicator or meter.

Weak bases and acids, unlike strong bases and acids, do not dissociate completely. Because they do not release all their protons, their pH's are a bit more difficult to calculate than stronger bases and acids. They have an equilibrium dissociation constant Kb and Ka respectively for bases and acids for their dissociation equations BOH B+ + OH-. It is possible to use the equilibrium solving methods to find the final concentrations of each of the reactants, but it is often convenient to use the Henderson-Hasselbach approximation: pH pKb + log, where pKb is the negative logarithm of Kb. With [B+] = [BOH], pH pKb. To find the point where the conjugate acid concentration equals the base concentration, we must define the half-equivalence point. Recall that the equivalence point is where all the base has reacted with the acid. If we had just titrated with half of the volume of acid required to react completely with the base, half the amount of acid reacts with half the amount of base. Thus the concentrations of conjugate acid and base are equal, and the pH of that solution is approximately equal to the pKb.

Weak bases and acids are useful to buffering because they exist both dissociated and undissociated in solution. They freely change forms allowing the absorption of additional protons and neutralizing stray hydroxide ions. Because the weak bases and acids sometimes prefer to join with H+ or OH-, their pH's at equivalence are not pH 7.

The graph of the pH versus the volume titrated roughly looks S-shaped. With weak bases and acids, there should be a flatter area where there is buffering. The equivalence point will be the steepest section of this graph. Near the equivalence point, any small change in the amount of base or acid would cause the dominant ion to change from H+ to OH-.

Our experiment would investigate the pH curve in titrations along with the equivalence and half-equivalence points.

Materials and Methods

The first part of the lab involved titrating unknown concentrations of NaOH. With potassium hydrogen phthalate (KHC8H7O4, abbreviated KHP). Starting out with a given mass of KHP, we neglected safety procedures and added the water to the acid. The solution, whose concentration would not need to be known, was then titrated with the unknown concentration NaOH with phenolphthalein, an indicator that turns pink when the solution turns slightly basic. The titration was repeated three times to ensure an accurate value.

The second part of the lab sought to find the pKa of an unknown acid which had a strong odor not unlike vinegar (acetic acid, HC2H3O2). Using the NaOH solution with the known concentration from the first part, we titrated the unknown acid. Every 0.3 mL or so, we would stop and measure the pH using an electronic pH meter which had been calibrated at the beginning of class. We would use the data collected to graph the pH during titration. The equivalence point would be indicated by a couple drops of phenolphthalein. This time, we would continue past the equivalence point so that our graph may show that the pH will level off. The procedure was repeated because the first attempt did not show a leveling off of the pH curve.

The lab dealt with acids and bases. Part one involved handling 6M NaOH which has a very high and dangerous pH. Approximate calculations were done to avoid solutions that would have dangerous pH's. Goggles and aprons, along with shoes and pants, were required. Neutralizing solutions were available and all chemicals were disposed of properly.

Results

Please see attached pages for data and graphs for part two. Please see lab book for calculations. All buret measurements are 0.025 mL. All pH measurements are 0.06 pH. All masses are 0.00005 g. All part two analysis uses the second attempt data.


Sample              1        2        3                     

Initial (mL)        24.50    24.00    24.20 ; refilled to   
                                      28.90                 

Final (mL)          46.05    48.9     50.00; after refill   
                                      29.45                 

Volume added (mL)   21.55    24.90    26.35                 

Mass of KHP (g)     0.4199   0.4717   0.4977                

[NaOH] 10-2 mol/L   9.40     9.14     9.02                  



Table 1: Part 1 Titrations

Our NaOH solutions had an average molarity of 9.1882 10-2.

The first attempt at the part two section of the lab showed some interesting data and clearly illustrates some problems with data noise. The noise in the data created the dual maxima on the pseudo-derivative plot. Because of this noise, it is not clear if the slope of the pH graph increases, stays the same, or decreases. Given this noise, the data should be scrutinized carefully and may not be completely reliable. Thus we repeated our experiment in the hope that cleaner data would result.

In order to calculated the steepest portion of the pH titration curve, a pseudo-derivative was used. The slope between the two data points would tell us how steep the curve is. By finding the highest value for the pseudo-derivative, we could find the steepest section of the pH titration curve. For the calculus students, we have just found the inflection point. The inflection point is the equivalence point. For our experiment, the point was at 11.375 mL added (pH 8.075).

Using the equivalence point, we could find the half equivalence point, 5.688 mL added, which would give us a pKa of approximately 4.40. With the unknown acid's pKa of 4.74, we have a percent error of -7.17%. Some rounding was done to make it convenient to look up the values from the measurements taken. The equivalence point was assumed to be maximal at the midpoint between the two adjacent readings.

The analysis of attempt two of the lab part two excludes the final point, 40.1 mL added, 12.40 pH, because the reading was taken under non-standard conditions and the volume added is uncertain because the buret had completely drained.

Discussion

Our unknown acid #1 had a pKa of about 4.40. Our NaOH solution had a molarity of 9.1882 10-2. Our measurements and calculations mirrored the results of the class. Also the graph of our titration pH curve looks similar to those shown as examples in the chemistry textbook. Our experiment also shows a buffered region roughly around pH 3.9 to 5.3. The range agrees with the established principle that a buffer, composed of a weak acid and its conjugate base, will work optimally when the pH pKa.

With the unknown acid's pKa of 4.74, we have a percent error of -7.17%. Errors could have resulted from improperly stirred solutions along with potentially slow responses from the pH meter. The consistency of the pre-droplet solution at the tip of the buret can also affect the readings. Other less significant errors could include pH meter drift where the batteries lower in voltage as a result from continuous use which could affect the electrodes and the measurement circuitry.

  • 1. Please see attached graphs.
  • 2. For our experiment, the point was at 11.375 mL added (pH 8.075). Using the equivalence point, we could find the half equivalence point, 5.688 mL added, which would give us a pKa of approximately 4.40.
  • 3. 11.375 mL 0.091882 = 0.1045 M
    The calculation above uses the equivalence point readings to determine that all of the acid has reacted with the base. Using the concentration found from part one of the lab, the number of moles of the acid can be found. Finally dividing by the initial volume results in the initial molarity of the unknown acid #1.