acid base 2 postlab

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Dec 6, 2023

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Tiffany Lee CHEM 1034 Jim Erdei Section 003 Acid-Base Titration Curves II Post Lab 1. Using your data for the value for VEP1, calculate the molarity of H3PO4 at the 1st equivalence point, EP1. Show work clearly, with units and correct significant figures. Volume NaOH used to reach first equivalence point: 4.79 mL [H3PO4] = (Avg NaOH molarity) * (vol NaOH / total vol NaOH) * (½) [H3PO4] = (0.1491 M) * (4.79 mL / 10 mL) * (½) [H3PO4] = 0.0357 M 2. Using your data for the value for VEP2, calculate the molarity of H3PO4 at the 2nd equivalence point, EP2. Show work clearly, with units and correct significant figures. Volume NaOH used to reach second equivalence point: 7.39 mL [H3PO4] = (Avg NaOH molarity) * (vol NaOH / total vol NaOH) * (½) [H3PO4] = (0.1491 M) * (7.39 mL / 10 mL) * (½) [H3PO4] = 0.0551 M 3. Calculate the percent error between these two molarity values. Percent error = | First value - Second value | / Average value * 100 Percent error = | 0.0357 M - 0.0551 M | / 0.0454 M * 100 Percent error = 42.7% 4. Tabulate your average values of pKa1 and pKa2 and the accepted values. Calculate the percent error of pKa1 and pKa2. Calculated pKa1 = 1.21 Calculate pKa2 = 1.52 Accepted pKa1 = 2.12 Accepted pKa2 = 7.21

pKa1: Percent error = | Observed value - Expected value | / Expected value * 100 Percent error = | 1.21 - 2.12 | / 2.12 * 100 = 42.9% pKa2: Percent error = | 1.52 - 7.21 | / 7.21 * 100 = 78.9% 5. Calculate the molar mass of your unknown acid two different ways: Use the average mass from trials 1 and 2, and the average volume of the first equivalence point and use the average mass from trials 1 and 2, and the average volume of the second equivalence point. Show work clearly, with units and correct significant figures. First equivalence point Mass trial 1 = 0.1986 grams Mass trial 2 = 0.1919 grams Avg mass = 0.1953 grams Vol NaOH used trial 1 = 7.60 mL Vol NaOH used trial 2 = 5.44 mL Avg vol = 6.52 mL Molar mass = (2) * (avg mass) / (NaOH M * avg vol) Molar mass = (2) * (0.1953 g) / (0.1491 mol/L * 0.00652 L) Molar mass = 401.80 g/mol Second equivalence point Mass trail 1 = 0.1986 grams Mass trial 2 = 0.1919 grams Avg mass = 0.1953 grams Vol NaOH used trial 1 = 15.19 mL Vol NaOH used trial 2 = 10.88 Avg vol = 13.04 mL Molar mass = (2) * (avg mass) / (NaOH M * avg vol) Molar mass = (2) * (0.1953 g) / (0.1491 mol/L * 0.01304 L) Molar mass = 200.90 g/mol 6. Using the molar mass from ii, and your average values of pKa1 and pKa2, identify your unknown acid. Explain your answer. I believe the unknown acid we were given is maleic acid because the pKa1 value is 1.92, and our calculated pKa1 values were close to that number. I cannot really use the calculated molar mass as an indicator as to whether or not we had maleic acid because the accepted value for maleic acid’s molar mass is 116.1 g/mol, and our calculated molar masses were very far from that.

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