8. A chemist carries out an iodimetric titration to determine the vitamin C cont

Question

8. A chemist carries out an iodimetric titration to determine the vitamin C content of some supplement tablets. A KIO, solution was prepared by dissolving 1.057 gof KIO, in a 500 mL volumetric flask. To 50.00 mL taken from this solution were added 2 g of KI and 10 mL of 0.5 M H2SO4, and this solution was used to standardize a NazS10, solution by titration. The titration required 3527 mL of thiosulfate. Finally, a sample of crushed supplement tablets weighing 1.216 g was dissolved in dilute H2SO4 and treated with 2 g of KI and 50.00 mLof the KIO, solution. The excess triiodide in the solution required 13.98 mL of thiosulfate solution to reach the titration end point. (a) This type of titration is called a back tiuration. What does this mean? (b) Comment on the use of exact (e.g. 00 mL) and approximate (e.g. 2 g) quantities in the description of this experimental procedure on are oxidizing agents, and which are reducing agents? (c) the (d) What was the concentration of the thiosulfate solution? is the mass percentage of vitamin C (Fw 176.13) in the supplement tablets?

Explanation / Answer

From the given data

(a) back titration is when we add a reagent in excess, some part of it reacts and what is left is titrated back with some other reagent after.

(b) We have used KIO3 as the limiting reagent and KI (2 g) excess reagent.

(c) C6H8O6 is reducing agent, I3- is oxidizing agent

molarity of KIO3 solution = 1.057/214 x 0.5 = 0.0099 M

moles of KIO3 in 50 ml solution = 0.0099 x 50 = 0.495 mmol

moles of KI = 2/166 = 0.012 mol

KIO3 is the limiting reagent

moles of I3- produced = 3 x 0.495 = 1.485 mmol

moles of Na2S2O3 required for this = 2 x 1.485 = 2.97 mmol

(d) molarity of Na2S2O3 solution = 2.97/35.27 = 0.084 M

Vitamine C tablet

excess I3- = 0.084 M x 13.98 ml = 1.174 mmol

I3- reacted = 1.485 - 1.174 = 0.311 mmol

moles of Vitamin C in tablet = 0.311 mmol

mass of Vitamin C in tablet = 0.311 x 176.13 = 0.055 g

(e) mass percent of vitamin C = 0.055 x 100/1.216 = 4.523%

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