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 unbalanced redox equation is: Cr2O72-(aq) + C2H5OH(aq) --> Cr3+(aq) +CO2(g) If 17.11 mL of 0.0806 M potassium dichromate solution is required to titrate 15.9 g of blood plasma, determine the mass percent of alcohol in the blood.
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 unbalanced redox equation is: Cr2O72-(aq) + C2H5OH(aq) --> Cr3+(aq) +CO2(g) If 17.11 mL of 0.0806 M potassium dichromate solution is required to titrate 15.9 g of blood plasma, determine the mass percent of alcohol in the blood.
Chapter4: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 109AE: The blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with an...
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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 unbalanced redox equation is:
Cr2O72-(aq) + C2H5OH(aq) --> Cr3+(aq) +CO2(g)
If 17.11 mL of 0.0806 M potassium dichromate solution is required to titrate 15.9 g of blood plasma, determine the mass percent of alcohol in the blood.
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