Ethanol-dichromate titration

Acidified potassium dichromate can be used to oxidise ethanol to ethanoic acid. These days instruments are used to quickly determine the concentration of ethanol in blood, but at one time blood alcohol levels (for drink-driving convictions) were determined by titration. The ethanol was separated from the blood sample and reacted with the dichromate solution. Calculate the concentration of ethanol contained in a 10.0 mL sample of blood that reacted with 13.7 mL of 0.0209 mol L–1 solution of potassium dichromate, and hence determine whether the blood exceeded the legal limit of 80 mg of ethanol per 100 mL of blood.

Step 1: Determine the amount of the known reagent (the dichromate) V(Cr2O72–) = 13.7 mL c(Cr2O72–) = 0.0209 mol L–1 = 13.7  10–3 L 0.0209 mol L–1 13.7  10–3 L n(Cr2O72–) = cV =  = 2.863  10–4 mol

3CH3CH2OH + 2Cr2O72– + 16H+ → 3CH3COOH + 4Cr3+ + 11H2O 3 2 Step 2: Use the equation for the chemical reaction to determine the amount of ethanol present 3CH3CH2OH + 2Cr2O72– + 16H+ → 3CH3COOH + 4Cr3+ + 11H2O 3 2 Unknown on top n(CH3CH2OH) Rearrange to get the unknown on its own. = Known on the bottom n(Cr2O72–) n(CH3CH2OH) 3 = n(Cr2O72–) 2 3  2.863  10–4 mol = 2 = 4.295  10–4 mol

Step 3: Calculate the concentration of ethanol in the sample n(CH3CH2OH) = 4.295  10–4 mol V(CH3CH2OH) = 10.0 mL = 10.0  10–3 L 4.295  10–4 mol Don’t forget to convert mL to L. 10.0  10–3 L n c(CH3CH3OH) = V = = 4.295  10–2 mol L–1

Step 4: Convert the concentration of ethanol in the blood sample into mg per 100 mL of blood c(CH3CH2OH) = 4.295  10–2 mol L–1 M(CH3CH2OH) = 46.0 g mol–1 4.295  10–2 mol L–1 46.0 g mol–1 To convert concentration in mol L–1 into g L–1, multiply by the molar mass (in g mol–1). c(CH3CH2OH) =  = 1.976 g L–1 = 0.198 g per 100 mL = 198 mg per 100 mL