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...

See similar textbooks

Similar questions

Please don't provide handwritten solution ......
6. Chloride amount was determined by Mohr method in the sewage water. Chloride in 250 mL of the sewage water was titrated with 3.2 mL of 0.01020 M AgNO; solution. a) Write all the reactions related with this titration and write their functions. b) Calculate the amount of C in terms of mg/L (ppm) in the sewage water (At. Mass of Cl=35.453 g/mol).
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.
5 b) Benzoic acid, HC7H5O2, is a weak monoprotic acid. A stock HC,H5O2 solution with unknown concentration was prepared. A 25.00 mL sample of the stock HCH5O2 solution was titrated with 0.1234 M sodium hydroxide to the equivalence point, using phenolphthalein as an indicator. The titration took 20.26 mL of titrant (sodium hydroxide). (b) Write a balanced complete ionic equation for the chemical reaction that occurs in this titration.
7B-3. A 15.00 mL sample of an unknown perchloric acid solution is titrated with a 0.687 M barium hydroxide solution. It takes 29.31 mL of the base solution to reach the endpoint. а. Write a balanced chemical equation for the reaction that occurs. b. What is the molarity of the unknown acid solution?
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 ]
A reg1.hu.edu.jo العربية ENGLISH Question 4/12 ميز هذا السؤال Titration of 0.2121 g of primary standard Na2C204 (133.999 g/mol) required 21.66 mL of KMNO4 (158.034 g/mol) to reach the end point. Calculate the molar concentration of KMNO4? 2 MnO4¯+ 5 C204²-+ 16 H+ → 2 Mn2+ 10 CO2 +8 H2O 1. 00.0146 2. 00.0292 3. 00.00731 4. 00.1462
17.69 mL of the I2 solution from above is required to titrate a sample containing As2O3. Calculate the mass of As2O3 (197.8 g/mol) in the sample. As2O3 + 5(H2O) + 2I2 → 2(H3AsO4) + 4HI
The protein in a 0.2976 -g sample of cheese is determined by the Kjeldahl method. The sample is digested with H2SO4, the resulting solution made basic with NaOH, and the ammonia distilled into 50.00 mL of 0.09756 M HCl. The excess HCl is then back titrated using 39.87 mL of 0.05345 M NaOH. Calculate the % N and % protein of the sample.
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…