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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 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.

Answer :

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Answer:

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[tex]\frac{1 moleC_{2}H_{5}OH}{2moleCr_{2}O_{7}^{-2}}[/tex] = 9,780x10⁻⁴ moles of C₂H₅OH

These moles are:

9,780x10⁻⁴ moles of C₂H₅OH ₓ [tex]\frac{46,07g}{1mole}[/tex] = 0,04506 g of C₂H₅OH

The mass percent is:

[tex]\frac{0,04506gC_{2}H_{5}OH}{33,8 g of blood} *100 =[/tex] 0,1333% of alcohol in the blood

I hope it helps!

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