Question

According to the lab manual, ascorbic acid (vitamin C) is a mild reducing agent that reacts rapidly with triiodide. In this experiment, we will generate a known excess of I3- by the reaction of iodate with iodide, allow the reaction with ascorbic acid to proceed, and then back titrate the excess I3- with thiosulfate. If you take a tablet (containing approximately 500 mg of ascorbic acid), dissolve it in 39 mL of 0.3 M H2SO4 (using a glass rod to help break the solid), add 2.8 g of solid KI and 50.00 mL of 0.064 KIO3 and then titrate it with 41.7 mL of 0.24 M Na2S2O3... the measured amount (expressed in mg/tablet) would be:

Answer 1

When iodate ions (IO3 ⁻) are added to an acidic solution containing iodide ions (I⁻), an oxidation-reduction reaction occurs:

Moles of IO₃^- = 0.061 x (50/1000) = 0.00305

Moles of H⁺ = 2 x 0.4 x (31/1000) = 0.0248

Moles of I^- = 2.2 / 166 = 0.012 (Limiting Reactant)

ascorbic acid + I₂ → 2 I⁻ + dehydroascorbic acid

Moles of I₂ formed = (3/5) x 0.012 = 0.0072 (limiting reactant)

Moles of S₂O₃^2- = 0.25 x (36/1000) = 0.009

Moles of Dehydroascorbic acid = 0.0072

Mass of dehydroascorbic acid = 0.0072 x 176.12 = 1.268 g