Question

An aqueous ethylene glycol (HOCH2CH2OH, FW = 62.07 g/mol) solution with a mass of 207.9 mg is titrated with 48.9 mL of 0.0829 M Ce4 in 4 M HClO4. The solution is held at 60°C for 15 minutes to oxidize the ethylene glycol to formic acid (HCO2H) and carbon dioxide. The excess Ce4 is titrated with 11.07 mL of 0.0409 M Fe2 to a ferroin end point. What is the mass percent of ethylene glycol in the unknown solution?

Answer 1

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Alcohols oxidize in the presence of an acid solution of Ce(IV) following the reaction:

This reaction can be used as an analytical method to determine the quantity of ethylene glycol in the aqueous ethylene glycol solution. Adding an excess of Ce(IV) ensures that the reaction is complete and that all of the ethylene glycol has reacted. The excess Ce(IV) is determined by a method called cerimetry. Cerimetry is a method of volumetric chemical analysis that uses a redox titration and a solution of Fe(II). The end point of the titration is obtained using 1,10-phenanthroline complex (ferroin) for the color change. The reactions stoichiometry is:

As it can be seen in the reaction we use one mol of iron to titrate 1 mol of cerium. We determine the number of mol from the solution of iron as:

n = M x Vol = 0.0409 M x 11.07 mL = 0.4528mmol

This will also be the number of mol of cerium in excess from the first reaction. Now, to find the number of mol of cerium that were used by the ethylene glycol during the oxidation we find the total number of mol and subtracts the excess. Like this:

n[Ce(IV) total] = M x Vol = 0.0829 M x 48.9 mL = 4.0538 mmol.

n[Ce(IV) oxidation] = n[Ce(IV) total] – 0.4528 mmol = 4.0538 mmol – 0.4528 mmol = 3.601 mmol

Determination of the number of mol of ethylene glycol:

From the first reaction we can see that each mol of ethylene glycol uses two mol of Ce(IV). This means that we have to divide the number of mol of Ce(IV) by two:

n(mol of ethylene glycol) = 3.601 mmol / 2 = 1.80 mmol

Weight of ethylene glycol = n(mol of ethylene glycol) x MW(ethylene glycol) =

Weight of ethylene glycol = 1.80 mmol x 62.07 mg/mmol = 111.75 mg

Mass % = Weight of ethylene glycol x 100/ Weight of solution = 111.75 mg x 100 / 207.9 mg = 53.75%

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