Chapter 4, Problem 118CP

The blood alcohol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ level can be determined bytitrating a sample of blood plasma with an acidic potassiumdichromate solution, resulting in the production of ${\text{Cr}}^{\text{3+}}\text{(}aq\text{)}$ and carbon dioxide. The reaction can be monitoredbecause the dichromate ion ${\text{(Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2–}}\text{)}$ is orange insolution, and the ${\text{Cr}}^{\text{3+}}$ ion is green. The unbalanced redoxequation is

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2–}}\text{(}aq{\text{)+C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH(}aq\text{)}\to {\text{Cr}}^{\text{3+}}\text{(}aq{\text{)+CO}}_{\text{2}}\text{(}g\text{)}$

If 31.05 mL of 0.0600 M potassium dichromate solutionis required to titrate 30.0 g of blood plasma, determinethe mass percent of alcohol in the blood.