The blood alcohol (C_2H_5 OH) level can be determined by titrating a sample of b

Question

The blood alcohol (C_2H_5 OH) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of Cr^3+ (aq) and carbon dioxide. The reaction can be monitored because the dichromate ion (Cr207 ~) is orange in solution, and the Cr3+ ion is green. The unbalanced redox equation is Cr_2O_7^2-(aq) + C_2H_5 OH (aq) rightarrow Cr^3+(aq) + CO_2(g) If 38.20 111L of 0.0300 M potassium dichromate solution is required to titrate 30.0 g blood plasma, determine the mass percent of alcohol in the blood.

Explanation / Answer

moles Cr2O7^2- = 0.03820 L x 0.0300 = 0.001146
Moles ethanol = 0.001146 / 2 =0.000573
Mass ethanol = 0.000573 mol x 46.06 g/mol = 0.02639 g
% = 0.02639 x 100 / 30.0g = 0.087

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