6.03 Calorimetry Lab
Figure 1: Titration curve of 0.160 grams of an unknown diprotic acid that was dissociated in distilled water. Shown is the pH versus the volume in milliliters of 0.1 M NaOH, a strong base, added to the solution.
The initial pH reading of the solution was a pH of 2.60. Although the pH of the ½ equivalence point was unknown, it could be estimated by halving the volume of NaOH used at the first equivalence point. At the first equivalence point, 13.63 milliliters of NaOH had been added to the unknown acid solution. Once divided by two, that value ended up being 6.815 milliliters of NaOH added at the ½ equivalence point pH. In order to estimate the pH at that volume of NaOH being added, two data points around the volume of 6.815 milliliters were
used. At NaOH added to the unknown acid solution of 5.59 milliliters, the pH was 2.79 and at NaOH added of 8.41 milliliters, the pH was 2.99. By averaging the two pH values, an estimated pH of 2.89 was determined at 6.815 milliliters of NaOH added. At the first equivalence point, a pH of 4.62 was found from the graph. The pH of the second ½ equivalence point was calculated similarly to how the first ½ equivalence point was found. By using the pH and milliliters of two points around the 21.3725 milliliters (½ of the NaOH used at the second equivalence point), a rough pH estimate could be determined. At 20.92 milliliters of NaOH used, the pH was 6.60 and at 22.72 milliliters of NaOH used, the pH was 6.81. Once those two values were averaged, a pH of 6.71 was determined at 21.3725 milliliters of NaOH added. The final relevant pH was the second equivalence point, and it was found to be 9.82.
The initial pH reading of the solution was a pH of 2.60. Although the pH of the ½ equivalence point was unknown, it could be estimated by halving the volume of NaOH used at the first equivalence point. At the first equivalence point, 13.63 milliliters of NaOH had been added to the unknown acid solution. Once divided by two, that value ended up being 6.815 milliliters of NaOH added at the ½ equivalence point pH. In order to estimate the pH at that volume of NaOH being added, two data points around the volume of 6.815 milliliters were
used. At NaOH added to the unknown acid solution of 5.59 milliliters, the pH was 2.79 and at NaOH added of 8.41 milliliters, the pH was 2.99. By averaging the two pH values, an estimated pH of 2.89 was determined at 6.815 milliliters of NaOH added. At the first equivalence point, a pH of 4.62 was found from the graph. The pH of the second ½ equivalence point was calculated similarly to how the first ½ equivalence point was found. By using the pH and milliliters of two points around the 21.3725 milliliters (½ of the NaOH used at the second equivalence point), a rough pH estimate could be determined. At 20.92 milliliters of NaOH used, the pH was 6.60 and at 22.72 milliliters of NaOH used, the pH was 6.81. Once those two values were averaged, a pH of 6.71 was determined at 21.3725 milliliters of NaOH added. The final relevant pH was the second equivalence point, and it was found to be 9.82.