One way of determining blood alcohol levels is by titrating a sample of blood according to the following net ionic equation.

C₂H₅OH(aq) + 2 Cr₂O₇^2-(aq) + 16 H⁺(aq) $\rightarrow$ 2 CO₂(g) + 4 Cr³⁺(aq) + 11 H₂O(l)
If 8.76 mL of 0.04988 M Cr₂O₇^2- is required for the titration of a 10.00 mL sample of blood, what was the molarity of the alcohol present in the blood sample?

Question

One way of determining blood alcohol levels is by titrating a sample of blood according to the following net ionic equation.

C₂H₅OH(aq) + 2 Cr₂O₇^2-(aq) + 16 H⁺(aq)

\rightarrow

2 CO₂(g) + 4 Cr³⁺(aq) + 11 H₂O(l)
If 8.76 mL of 0.04988 M Cr₂O₇^2- is required for the titration of a 10.00 mL sample of blood, what was the molarity of the alcohol present in the blood sample?

Quizplus · Accepted Answer

The molarity of the alcohol in the blood sample can be calculated using the stoichiometry of the reaction and the volume and molarity of the titrant. The reaction shows that 1 mole of alcohol reacts with 2 moles of Cr2O7^2-. First, calculate the moles of Cr2O7^2- used: 8.76 mL * 0.04988 M = 0.000437 moles. Since the ratio of alcohol to Cr2O7^2- is 1:2, the moles of alcohol are half that of Cr2O7^2-, so 0.000437 / 2 = 0.0002185 moles. The molarity of alcohol in the blood sample is then 0.0002185 moles / 0.01 L = 0.02185 M, which falls between 0.01 and 0.03 M.

One Way of Determining Blood Alcohol Levels Is by Titrating →\rightarrow→

One Way of Determining Blood Alcohol Levels Is by Titrating $\rightarrow$