Acid/Base Titrations
...cid: 100.0 mL Brand of vinegar used: Fred Meyer Percent weight of vinegar according to label: 5% Table1: Acetic acid and NaOH titration Run Acid buret readings (mL) Base buret readings (mL) Color at end point Vol. Acid used(mL) Vol. Base used(mL) Ratio acid/base 1 1.21 12.05 0.35 9.81 Light pink 10.84 9.46 0.8727* 2 12.05 22.45 9.81 19.00 Light pink 10.40 9.19 0.8837* 3 33.90 46.29 29.71 40.19 Light pink 12.39 10.48 0.8458 4 7.53 19.75 39.34 49.70 Light pink 12.22 10.36 0.8478 5 19.75 31.05 11.30 21.30 Light pink 11.30 10.00 0.8850 (*Note: the first two ratios were brought out to 4 significant figures so as not to prevent the proper amount of figures in answers to calculations. Runs 1 and 2 were originally omitted due to this problem but were included in report at professor’s guidance.) Observations: The endpoint of each titration was marked by the products turning a bright pink. Each run was balanced out with an end color of light pink that lasted for at least 20 seconds. A fading back to clear after 1 minute or more was desired. Both reactants were clear and the product was clear as well until equilibrium was reached and the phenolphthalein turned pink. Without the phenolphthalein there would be no indication of an end point. Calculations: 1. Acid/Base Ratios a. 10.78ml/10.95ml = 0.9845 b. 10.91ml/11.14ml = 0.9794 c. 10.68ml/10.92ml = 0.9780 d. 11.30ml/11.55ml = 0.9783 e. 10.79ml/10.14ml = 1.064 f. 11.05ml/12.34ml = 0.8955 2. Average A/B Ratio a. Sum(ratios)/ (# of ratios) = 0.9800 3. Molarity of NaOH a. 0.1025M x 0.9800 = 0.1004M 4. Base/Acid Ratios a. 9.46ml/10.84ml = 0.8767* b. 9.19ml/10.40ml = 0.8837* c. 10.48ml/12.39ml = 0.8458 d. 10.36ml/12.22ml = 0.8478 e. 10.00ml/11.30ml = 0.8850 5. Average B/A Ratio a. Sum(ratios)/ (# of ratios) = 0.8670 6. Molarity of Dilute Acetic Acid a. 0.1004M x 0.8670 = 0.08705M 7. Molarity of Vinegar a. 0.08705M(0.1000L)/0.01000L = 0.8705M 8. Grams of Acetic Acid per liter of Vinegar a. 0.8705mol/L(60.05g/1mol) = 52.27g/L 9. Percent by weight of Acetic Acid in the vinegar a. (52.27g/L / (1.011 g/ml x 1000ml)) x 100% = 5.170% Report Table: Fred Meyer Vinegar Molarity of Diluted Acetic Acid Determined % weight Acetic Acid in Vinegar % weight Acetic Acid in Vinegar from label % Error 0.08705M 5.170% 5% 3% High % Error = ((5%-5.170%)/5%) x 100% = 3% Discussion questions: 1. If the NaOH were to be obtained in a wet beaker there would not be an adverse affect on the over all calculations provided that the NaOH in the beaker was used for both sets of Titrations and the water in the beaker was distilled. Water in the beaker will change the molarity of the NaOH solution, but because we are doing the first titration to determine the molarity of the NaOH in the beaker it does not matter wh...