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.

We will balance the redox equation using the half-reaction method. In the redox equation, \(\mathrm{Cr_2O_7^{2-}}\) and \(\mathrm{C_2H_5OH}\) are the oxidizing and reducing agents, respectively. We can write the half-reactions as follows: Oxidation Half-Reaction: \(\mathrm{C_2H_5OH \rightarrow CO_2}\) Reduction Half-Reaction: \(\mathrm{Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq)}\) Balance the half-reactions: For oxidation half-reaction: i) Balance the carbon atoms by adding 2 molecules of CO₂. \(\mathrm{C_2H_5OH \rightarrow 2CO_2}\) ii) Balance the hydrogen atoms by adding 6H₂O to the right side. \(\mathrm{C_2H_5OH \rightarrow 2CO_2 + 6H_2O}\) iii) Balance the oxygen atoms by adding 6H⁺ to the left side. \(\mathrm{C_2H_5OH + 6H^+ \rightarrow 2CO_2 + 6H_2O}\) Now for the reduction half-reaction: i) Balance the chromium atoms by adding 2Cr³⁺. \(\mathrm{Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq)}\) ii) Balance the oxygen atoms by adding 7 water molecules. \(\mathrm{Cr_2O_7^{2-}(aq) + 7H_2O \rightarrow 2Cr^{3+}(aq)}\) iii) Balance the hydrogen atoms and charges by adding 14H⁺. \(\mathrm{Cr_2O_7^{2-}(aq) + 14H^+ \rightarrow 2Cr^{3+}(aq) + 7H_2O}\) Now, we can combine the balanced half-reactions: \(\mathrm{C_2H_5OH(aq) + 6H^+(aq) + Cr_2O_7^{2-}(aq) + 14H^+(aq) \rightarrow 2CO_2(g) + 6H_2O(l) + 2Cr^{3+}(aq) + 7H_2O(l)}\) Simplify the equation: \(\mathrm{C_2H_5OH(aq) + Cr_2O_7^{2-}(aq) + 20H^+(aq) \rightarrow 2CO_2(g) + 2Cr^{3+}(aq) + 13H_2O(l)}\) This is our balanced redox equation.

We are given that 31.05 mL of 0.0600 M potassium dichromate solution is required for titration. First, we find the moles of dichromate ions, and then the moles of ethanol using the stoichiometry of the balanced equation. moles of \(\mathrm{Cr_2O_7^{2-}} = \mathrm{Volume} \times \mathrm{Molarity}\) moles of \(\mathrm{Cr_2O_7^{2-}} = (31.05 \times 10^{-3} \mathrm{L}) \times 0.0600 \mathrm{M} = 1.863 \times 10^{-3}\; \mathrm{mol}\) From the balanced equation, we see that 1 mol of \(\mathrm{Cr_2O_7^{2-}}\) reacts with 1 mol of \(\mathrm{C_2H_5OH}\). Thus, the moles of ethanol are the same as the moles of dichromate ions. moles of \(\mathrm{C_2H_5OH} = 1.863 \times 10^{-3}\; \mathrm{mol}\)

Next, we'll find the mass of ethanol present in the blood plasma and calculate the mass percent. To do this, we will use the molar mass of ethanol (C₂H₅OH), which is approximately \(46.07 \mathrm{g/mol}\). mass of \(\mathrm{C_2H_5OH} = \mathrm{moles} \times \mathrm{molar \, mass}\) mass of \(\mathrm{C_2H_5OH} = (1.863 \times 10^{-3}\; \mathrm{mol}) \times 46.07 \mathrm{g/mol} = 0.0858 \mathrm{g}\) Now, calculate the mass percent of ethanol in the blood plasma: mass percent \(= \frac{\mathrm{mass \, of \, ethanol}}{\mathrm{mass \, of \, blood \, plasma}} \times 100\%\) mass percent \(= \frac{0.0858\; \mathrm{g}}{30.0\; \mathrm{g}} \times 100\% = 0.286\%\) Therefore, the mass percent of alcohol in the blood is approximately 0.286%.