3. The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.0 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 М КОН: CH3COOC2H5 + OH° ®CH3COO¯ + C2H5OH After cooling, the excess OH was back-titrated with3.41 mL of 0.05042 M H2SO4. Calculate the number of grams of ethyl acetate (88.11 g/mol) per 100 mL of the original sample.

Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter16: Applications Of Neutralization Titrations
Section: Chapter Questions
Problem 16.20QAP
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3. The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.0
mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672
М КОН:
CH3COOC2H5 + OH° ®CH3COO¯ + C2H5OH
After cooling, the excess OH¯ was back-titrated with3.41 mL of 0.05042 M H2SO4. Calculate the number of grams of
ethyl acetate (88.11 g/mol) per 100 mL of the original sample.
Transcribed Image Text:3. The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.0 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 М КОН: CH3COOC2H5 + OH° ®CH3COO¯ + C2H5OH After cooling, the excess OH¯ was back-titrated with3.41 mL of 0.05042 M H2SO4. Calculate the number of grams of ethyl acetate (88.11 g/mol) per 100 mL of the original sample.
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