If 21.35 mL of 0.005 M iodine solution is needed to titrate a solution of commercial vitamin C tablet (500 mg) prepared in a 100-ml volumetric flash, then what is the mass % of the vitamin C?

Question

If 21.35 mL of 0.005 M iodine solution is needed to titrate a solution of commercial vitamin C tablet (500 mg) prepared in a 100-ml volumetric flash, then what is the mass % of the vitamin C?

Answer 1

So the stoichiometry would look as follows:

ascorbic acid + I

Ascorbic acid + I2 → 2 I− + dehydroascorbic acid

moles of I2 used = 21.35 ml x 1L/2000 ml x 0.005 mol/L = 0.0001068 moles

Moles ascorbic acid = 0.0001068 mols

Mass ascorbic acid = 0.0001068 mol x176 g/ mol = 0.0188 g ascorbic acid

mass % = 0.0188 g/0.500 g (x100) = 3.76%