Repeat steops 1-16 for Vinegar columns 2 and 3 on Data Table 1. 17 17 Calculate the concentration of acetic acid in your vinegar sample based on your experimental data. Calculate the concentration of acetic acid in your vinegar sample based on your experimental data. 18 18 Determination of Acetic Acid Concentration in Vinegar Using Titration 2 ©2014, Carolina Biological Supply Company
Determination of Acetic Acid Concentration in Vinegar Using Titration Laboratory Questions 1. Based on the experimental data, what is the molarity of the vinegar tested? How does this value compare with those determined in the pre-lab questions? Molarity of the tested vinegar was 0.93. It was slightly lower than the one in the pre-lab questions. It is lower because the experiment consists molarity of 1M instead of 0.5 M. 2. List at least two possible sources of error from the experiment. Would these errors result in an overestimation or an underestimation of the concentration of acetic acid in the vinegar. An overestimation of concentration would occur due to adding the NaOH solution way too quickly, skewing the endpoint. Also, the amount of acetic acid may not have been accurate. 3 ©2014, Carolina Biological Supply Company
Determination of Acetic Acid Concentration in Vinegar Using Titration Discussion Questions 1. Describe on a molecular level the difference between two titrations—one, of a strong acid with a strong base and the other, of a weak acid with a strong base. Draw a titration curve for each situation to further illustrate the differences between strong and weak acids. 2. Given the following experimental data, calculate the molarity and the p K a of the acid analyte (HA) in the titration, if the balance neutralization reaction is HA + NaOH NaA + H 2 O. A pH reading of 3.65 was obtained when 13.0 mL of 0.413 M NaOH was added to the 25.0 mL analyte. The endpoint was achieved when 31.25 mL of NaOH had been added. Molarity of Acid (31.25 x 0.413)/(25) = 0.516 m Molarity of HA = 0.516 m Moles of acid = 25 mL x 0.516 m = 1209 mmol Moles of NaOH = 13 x 0.415= 5.4 mmol PH = Pka + log [NaA]/[HA] 3.65 = Pka + log [5.4/ 7.5] = Pka – 0.143 Pka = 3.79 Ka = 10^-3.79 Vinegar Sample 1 Vinegar Sample 2 Vinegar Sample 3 4 ©2014, Carolina Biological Supply Company
Determination of Acetic Acid Concentration in Vinegar Using Titration Mass of Vinegar 5.06 g 5.03 g 5.04 g Volume of Vinegar Density= 1.0005 g/mL 5.00 5.01 5.03 Initial NaOH volume in syringe 10 mL 10 mL 10 ml Final NaOH volume in syringe 4.6 mL 5.2 mL 4.7mL Volume of NaOH Delivered 4.4 mL 4.6 mL 4.2 mL Moles of NaOH Delivered 0.0044 0.0046 0.0042 Moles of Acetic Acid 0.0044 0.0046 0.0042 Mass of Acetic Acid 0.31 g 0.31 g
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