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A 15.0 mL sample of whiskey was diluted to 500.0 mL. A 3.00 mL aliquot of the diluted sample was...

Question:

A 15.0 mL sample of whiskey was diluted to 500.0 mL. A 3.00 mL aliquot of the diluted sample was removed and the ethanol, C2H5OH, was distilled into 50.00 mL of 0.02500 M K2Cr2O7 and oxidized to acetic acid.

3C_2H_5OH+2Cr_2O_7^2-+16H^+ ---> 4Cr^3++3CH_3COOH+11H_2O

The excess Cr_2O_7^2- was then back titrated with 37.5 ml of 0.1020M Fe^2+, producing Cr^3+ and Fe^3+. Calculate the weight per volume percent, %w/v, of ethanol in the original whiskey sample

Back-Titration:

A back-titration is a type of indirect titration reaction wherein the concentration of an unknown analyte is determined by letting the unknown substance react with an excess of standard. The amount of standard left is titrated to determine the amount unreacted. From the amounts reacted and unreacted, we can determine the concentration of analyte.

Answer and Explanation: 1

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The balanced chemical equations for the reactions are given below.

{eq}\rm 3C_2H_5OH + 2Cr_2O_7^{2-} + 16H^+ \to 4Cr^{3+} + 3CH_3COOH+11H_2O...

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Titration of a Strong Acid or a Strong Base

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Chapter 11 / Lesson 9
23K

Learn about strong acid - strong base titration. Understand strong acid - strong base reactions and how to find an unknown substance concentration, and see the curve.


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