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 the following. cr2o72−(aq) + c2h5oh(aq) → cr3+(aq) + co2(g) if 31.91 ml of 0.0613 m potassium dichromate solution is required to titrate 33.8 g of blood plasma, determine the mass percent of alcohol in the blood.

0,1333% of alcohol in the blood

Explanation:

The equation of titration is:

Cr₂O₇²⁻(aq) + C₂H₅OH(aq) → Cr³⁺(aq) + CO₂(g)

The balance of this reaction is:

16H⁺ + 2Cr₂O₇²⁻(aq) + C₂H₅OH(aq) → 4Cr³⁺(aq) + 2CO₂(g) + 11H₂O

31,91 mL of 0,0613 M K₂Cr₂O₇ are:

0,03191 L ₓ 0,0613 M K₂Cr₂O₇ = 1,956x10⁻³ moles of Cr₂O₇²⁻

As 2 moles of Cr₂O₇²⁻ reacts with 1 mole of C₂H₅OH the moles of C₂H₅OH in blood plasma are:

1,956x10⁻³ moles of Cr₂O₇²⁻ ₓ = 9,780x10⁻⁴ moles of C₂H₅OH

These moles are:

9,780x10⁻⁴ moles of C₂H₅OH ₓ = 0,04506 g of C₂H₅OH

The mass percent is:

0,1333% of alcohol in the blood

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