The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by tirrating 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 $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If 31.05 \(\mathrm{mL}\) of 0.0600\(M\) potassium dichromate solution is required to titrate 30.0 \(\mathrm{g}\) of blood plasma, determine the mass percent of alcohol in the blood.

Step by step solution

01

Balancing the redox equation

First, we need to balance the given redox equation. We need to balance the atoms and charges on both sides. Our unbalanced equation is: $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ The balanced redox equation is: $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+16\mathrm{H}^{+}+3\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow 2\mathrm{Cr}^{3+}(a q)+11\mathrm{H}_{2}\mathrm{O}(l)+3\mathrm{CO}_{2}(g)$$

02

Calculate moles of potassium dichromate

We are given the volume (31.05 mL) and concentration (0.0600 M) of potassium dichromate solution. Convert the volume to liters and calculate the moles of potassium dichromate: $$\text{Moles of}\; \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} = \text{Volume} \times \text{Concentration}$$ $$= 0.03105\; \text{L} \times 0.0600\; \text{M}$$ $$= 0.001863\; \text{moles}$$

03

Calculate moles of alcohol

From the balanced redox equation, we can see that 1 mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) reacts with 3 moles of \(\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}\). Using the moles of potassium dichromate, calculate the moles of alcohol: $$\text{Moles of}\; \mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH} = \frac{3}{1} \times \text{Moles of}\; \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$$ $$= 3 \times 0.001863\; \text{moles}$$ $$= 0.005589\; \text{moles}$$

04

Calculate mass of alcohol and mass percent

Now, we will convert the moles of alcohol to mass using the molar mass of alcohol (46.07 g/mol): $$\text{Mass of alcohol} = \text{Moles of alcohol} \times \text{Molar mass of alcohol}$$ $$= 0.005589\; \text{moles} \times 46.07\; \text{g/mol}$$ $$= 0.257\; \text{g}$$ We are given that the blood sample weighs 30.0 g. Calculate the mass percent of alcohol in the blood sample: $$\text{Mass Percent} = \frac{\text{Mass of alcohol}}{\text{Total mass of blood sample}} \times 100$$ $$= \frac{0.257 \; \text{g}}{30.0 \; \text{g}} \times 100$$ $$= 0.857 \%$$ The mass percent of alcohol in the blood plasma is 0.857%.

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