5 The amount of ascorbic acid, C6HSO6, in orange juice was determined by oxidizing the ascorbic acid to dehydroascorbic acid, C6H6O6, with a known excess of I3, and back titrating the excess I3 with NazS2O3. A 5.00-mL sample of filtered orange juice was treated with 50.00 mL of excess 0.01023 M I. After the oxidation was complete, 13.82 mL of 0.07203 M Na2S2O3 was needed to reach the starch indicator end point. Report the concentration of ascorbic acid in milligrams per 100 mL.

5 The amount of ascorbic acid, C6HSO6, in orange juice was determined by oxidizing the ascorbic acid to dehydroascorbic acid, C6H6O6, with a known excess of I3, and back titrating the excess I3 with NazS2O3. A 5.00-mL sample of filtered orange juice was treated with 50.00 mL of excess 0.01023 M I. After the oxidation was complete, 13.82 mL of 0.07203 M Na2S2O3 was needed to reach the starch indicator end point. Report the concentration of ascorbic acid in milligrams per 100 mL.

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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Solid potassium hydroxide is slowly added to 175 mL of a 0.0631 M magnesium acetate solution. The concentration of hydroxide ion required to just initiate precipitation is М.
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Excess solid Ni(OH)2 was added to water and the resulting solution was allowed to stand for several days. The solution was then filtered to remove excess solid Ni(OH)2. A 0.250 L aliquot of the solution was then titrated with 10.54 mL of a standardized 0.000500 M HCl solution. What is the Ksp of Ni(OH)2?
13. The reaction of HCl with NaOH is represented by the equation: HCl(aq) + NaOH(aq) → NaCl(ag) + Hz0(1) What volume of 0.252 MHC1 is required to titrate 54.1 mL of 0.424 M NAOH? [ Select ]
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Mark and John, two vinegar enthusiasts, are each tasked to determine the acetic acid content of their respective vinegar concoctions by titration. First, a 1 M-labeled KOH solution was standardized against the KHP (MW = 204.22 g/mol) standard that is 99.4% pure. In the process, 0.540 g KHP was found to require 2.80 mL of the KOH solution to completely react up to the phenolphthalein endpoint. Then, Mark and John both prepared their samples by taking 10.0-mL aliquots of each vinegar and diluting them to 25.0 mL. Using the same titrant and indicator, Mark’s vinegar required 18.60 mL of the standardized titrant to reach the endpoint, while John’s vinegar required 16.50 mL of the same titrant to reach the same endpoint. Question: What is the acetic acid concentration of Mark’s and John’s vinegar in molarity given that the Concentration of KOH is 9386 M?
You are running an acid-base titration with hydrochloric acid and calcium hydroxide. The calcium hydroxide is your titrant; it has a known concentration of 0.0891 M. You have added 25.0 mL of the hydrochloric acid solution to an Erlenmeyer flask along with 2 drops of phenolphthalein (fee’-nol-thay-leen) indicator. To begin your first trial, you fill a clean 50 mL buret with the calcium hydroxide solution. Your starting volume is 0.25 mL. After the addition of solution from the buret, you see a color change in the flask, and the end point of the reaction is reached. You note the final volume in the buret to be 18.75 mL. You begin again for your second titration trial. You add 24.8 mL of the hydrochloric acid solution to a clean flask along with 2 drops of indicator. You refill the buret to an initial volume of 1.42 mL. When the reaction is complete, you note the final volume in the buret at 16.74 mL. Then you complete a third trial. You add 25.6 mL of hydrochloric acid solution…
mare 550 ml of 0,420 M Mg. COH)2 SOlunon?
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