The blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of Cr3+ (aq) and carbon dioxide. The reaction can be monitored because the dichromate ion (Cr2O72-) is orange in solution, and the Cr3+ ion is green. The balanced equation is 16H+(aq) + 2Cr2O72-(aq) + C2H5OH(aq) -> 4Cr3+(aq) + 2CO2(g) + 11H20. If the blood alcohol limit is 0.08%, what concentration in M of Cr2O72-(aq) would you use , such that titrant volume will be 2.00 mL for a 5.00 mL sample that would be over the limit ? Specify what change you will monitor at equivalence point.
The blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of Cr3+ (aq) and carbon dioxide. The reaction can be monitored because the dichromate ion (Cr2O72-) is orange in solution, and the Cr3+ ion is green. The balanced equation is 16H+(aq) + 2Cr2O72-(aq) + C2H5OH(aq) -> 4Cr3+(aq) + 2CO2(g) + 11H20. If the blood alcohol limit is 0.08%, what concentration in M of Cr2O72-(aq) would you use , such that titrant volume will be 2.00 mL for a 5.00 mL sample that would be over the limit ? Specify what change you will monitor at equivalence point.
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter6: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 111AE
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The blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of Cr3+ (aq) and carbon dioxide. The reaction can be monitored because the dichromate ion (Cr2O72-) is orange in solution, and the Cr3+ ion is green. The balanced equation is
16H+(aq) + 2Cr2O72-(aq) + C2H5OH(aq) -> 4Cr3+(aq) + 2CO2(g) + 11H20.
If the blood alcohol limit is 0.08%, what concentration in M of Cr2O72-(aq) would you use , such that titrant volume will be 2.00 mL for a 5.00 mL sample that would be over the limit ?
Specify what change you will monitor at equivalence point.
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