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30. A sample of vinegar weighing 10.52 g is titrated with NaOH. The end point is overstepped, and the solution is titrated back with HCl. From the following data, calculate the acidity of the vinegar in terms of percentage of acetic acid, CH;COOH (60.05 g/mol) Standardization Data: 1.050 mL HCI = 1.000 mL NaOH 1.000 mL NaOH = 0.06050 g benzoic acid, CHSCOOH (122.12 g/mol) Sample Analysis Data: Volume NaOH used = 19.03 ml Volume HCI used for back titration = 1.50 ml

30. A sample of vinegar weighing 10.52 g is titrated with NaOH. The end point is overstepped, and the solution is titrated back with HCl. From the following data, calculate the acidity of the vinegar in terms of percentage of acetic acid, CH;COOH (60.05 g/mol) Standardization Data: 1.050 mL HCI = 1.000 mL NaOH 1.000 mL NaOH = 0.06050 g benzoic acid, CHSCOOH (122.12 g/mol) Sample Analysis Data: Volume NaOH used = 19.03 ml Volume HCI used for back titration = 1.50 ml

Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Chapter18: Acids And Bases
Section: Chapter Questions
Problem 92A
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30. solution is titrated back with HC. From the following data, calculate the acidity of the vinegar in terms of percentage of acetic acid, CH;COOH (60.05 g/mol)- A sample of vinegar weighing 10.52 g is titrated with NaOH. The end point is overstepped, and the Standardization Data: 1.050 mL HCI = 1.000 mL N2OH 1.000 mL NAOH = 0.06050 g benzoic acid, CH;COOH (122.12 g/mol) Sample Analysis Data: Volume NaOH used = 19.03 mL Volume HCl used for back titration = 1.50 mL
Transcribed Image Text:30. solution is titrated back with HC. From the following data, calculate the acidity of the vinegar in terms of percentage of acetic acid, CH;COOH (60.05 g/mol)- A sample of vinegar weighing 10.52 g is titrated with NaOH. The end point is overstepped, and the Standardization Data: 1.050 mL HCI = 1.000 mL N2OH 1.000 mL NAOH = 0.06050 g benzoic acid, CH;COOH (122.12 g/mol) Sample Analysis Data: Volume NaOH used = 19.03 mL Volume HCl used for back titration = 1.50 mL
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