The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{~mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{~g}\) blood plasma, determine the mass percent of alcohol in the blood.

To balance the given redox equation, first, we should identify the oxidation and reduction half-reactions. The ethanol \(\mathrm{C_{2}H_{5}OH}\) molecule loses electrons (it is oxidized) to form \(\mathrm{CO_{2}}\), while the \(\mathrm{Cr_{2}O_{7}^{2-}}\) ion gains electrons (it is reduced) to form \(\mathrm{Cr^{3+}}\). The balanced half-reactions are: Oxidation half-reaction: \(\mathrm{C_{2}H_{5}OH} \rightarrow \mathrm{CO_{2}} + 12\mathrm{H^{+}} + 12\mathrm{e^-}\) Reduction half-reaction: \(\mathrm{Cr_{2}O_{7}^{2-}} + 14\mathrm{H^{+}} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_{2}O}\) To combine these half-reactions, multiply the oxidation half-reaction by 6 and the reduction half-reaction by 2, then add them together to form the balanced redox equation: \[\begin{aligned} &6\left(\mathrm{C_{2}H_{5}OH} \rightarrow \mathrm{CO_{2}} + 12\mathrm{H^{+}} + 12\mathrm{e^-}\right) \\ + &2\left(\mathrm{Cr_{2}O_{7}^{2-}} + 14\mathrm{H^{+}} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_{2}O}\right) \\ \rightarrow &6\mathrm{C_{2}H_{5}OH} + 2\mathrm{Cr_{2}O_{7}^{2-}} + 52\mathrm{H^{+}} \rightarrow 6\mathrm{CO_{2}} + 4\mathrm{Cr^{3+}} + 14\mathrm{H_{2}O} \end{aligned}\]

We are given the volume \((31.05\ \mathrm{mL})\) and the concentration \((0.0600\ \mathrm{M})\) of potassium dichromate solution. To determine the moles, use the formula: Moles = Volume × Concentration Moles of potassium dichromate = \(31.05 \times 0.0600\) = \(1.86 \times 10^{-3}\ \mathrm{mol}\)

According to the balanced redox equation, 2 moles of \(\mathrm{Cr_{2}O_{7}^{2-}}\) react with 6 moles of \(\mathrm{C_{2}H_{5}OH}\). Therefore, the ratio between the moles of potassium dichromate and alcohol is 1:3. To find the moles of alcohol, we can use the following equation: \(\mathrm{moles\ of\ C_{2}H_{5}OH} = 3 \times \mathrm{moles\ of\ Cr_{2}O_{7}^{2-}}\) Moles of alcohol = \(3 \times 1.86 \times 10^{-3}\) = \(5.58 \times 10^{-3}\ \mathrm{mol}\)

To find the mass of alcohol, multiply the moles of alcohol by its molar mass: Mass of alcohol = moles of alcohol × molar mass of alcohol Molar mass of alcohol = \(2\mathrm{(C)} + 6\mathrm{(H)} + 1\mathrm{(O)} = 46.07\ \mathrm{g/mol}\) Mass of alcohol = \(5.58 \times 10^{-3} \times 46.07\) = \(0.257\ \mathrm{g}\)

The mass percent of alcohol in the blood can be calculated using the following formula: Mass percent of alcohol = \(\frac{\text{mass of alcohol}}{\text{mass of blood plasma}} \times 100\) Mass percent of alcohol = \(\frac{0.257}{30.0} \times 100\) = \(0.857\ \%\) The mass percent of alcohol in the blood is approximately \(0.857\ \%\).