The concentration of a NaOH solution was determined in a single experiment by titrating a weighed sample of pure, dried KHP (FW=204.23 g/mol) with the NaOH. The data are as follows: Weighing by difference: Weight KHP + container (initial) = 11.6723 ±0.0001 g Weight KHP + container (final) = 10.8364±0.0001 g Titration: Volume NaOH (initial) = 1.00±0.02 mL Volume NaOH (final) = 29.50±0.02 mL This NaOH solution was used to analyze tartaric acid in wine. A 50.00±0.02 mL sample of a white dinner wine required 15.30±0.02 mL of the NaOH using phenolphthalein indicator. Calculate the %w/v tartaric acid (and uncertainty) (FW=150.09 g/mol) {pka1=3.17, pKa2=4.91} of the wine.
The concentration of a NaOH solution was determined in a single experiment by titrating a weighed sample of pure, dried KHP (FW=204.23 g/mol) with the NaOH. The data are as follows: Weighing by difference: Weight KHP + container (initial) = 11.6723 ±0.0001 g Weight KHP + container (final) = 10.8364±0.0001 g Titration: Volume NaOH (initial) = 1.00±0.02 mL Volume NaOH (final) = 29.50±0.02 mL This NaOH solution was used to analyze tartaric acid in wine. A 50.00±0.02 mL sample of a white dinner wine required 15.30±0.02 mL of the NaOH using phenolphthalein indicator. Calculate the %w/v tartaric acid (and uncertainty) (FW=150.09 g/mol) {pka1=3.17, pKa2=4.91} of the wine.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The concentration of a NaOH solution was determined in a single experiment by titrating a weighed sample of pure, dried KHP (FW=204.23 g/mol) with the NaOH. The data are as follows:
Weighing by difference:
Weight KHP + container (initial) = 11.6723 ±0.0001 g
Weight KHP + container (final) = 10.8364±0.0001 g
Titration:
Volume NaOH (initial) = 1.00±0.02 mL
Volume NaOH (final) = 29.50±0.02 mL
This NaOH solution was used to analyze tartaric acid in wine. A 50.00±0.02 mL sample of a white dinner wine required 15.30±0.02 mL of the NaOH using phenolphthalein indicator. Calculate the %w/v tartaric acid (and uncertainty) (FW=150.09 g/mol) {pka1=3.17, pKa2=4.91} of the wine.
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