Question

A 0.1219 g sample of a Vitamin C (ascorbic acid) tablet was dissolved in acid. A 25.00 mL aliquot of 0.01739 M KIO3 was added along with excess KI. The resulting solution was titrated with 0.07211 M thiosulfate, requiring 21.44 mL to reach the endpoint. Compute the weight percent of ascorbic acid (FW = 176.12) in the tablet.

Answer 1

Step 1 Calculate the no. of mole of KIO₃

The no. of mole of KIO₃ = C x V

= 0.01739 M x 25/1000L

= 0.000435 moles = 4.35 x 10^-4

Step 2 Calculate the no. of mole of I₂

KIO₃ (aq) + 5KI(aq) + 3H₂SO₄(aq) → 3I₂(aq) + 3H₂O(l) + 3K₂SO₄(aq)

According to the equation,

no. of mole of I₂ generated = 3(4.35 x 10-4) = 1.305 x 10^-3 mol

Step 3 Calculate the no. of mole of Na₂S₂O₃ required in back titration = 0.07211 x 21.44/1000 mL

         = 0.00155 moles 1.55 x 10^-3

Step 4 Calculate the no. of mole of unreacted I₂

2 Na₂S₂O₃ (aq) + I₂(aq) → Na₂S₄O₆(aq) + 2NaI (aq)

Mole of unreacted I₂ = 1.55 x 10^-3 /2 = 7.75 x 10^-4 moles

Step 5 Calculate the no. of mole of I₂ reacted with ascorbic acid

= (1.305 x 10^-3 - 7.75 x 10^-4) = 5.3 x 10^-4 moles

Step 6 Calculate the no. of mole of ascorbic acid

Ascorbic acid and I₂ reacted in 1:1 molar ratio

So, moles of ascorbic acid = 5.3 x 10^-4 moles

Step 7 Calculate the mass of ascorbic acid

Molar mass of ascorbic acid = 176 .12 g/mol

Mass of ascorbic acid = 176 .12(5.3 x10^-4) = 0.0933 g

Step 8 Calculate the weight percent of ascorbic acid in the tablet

Weight percent of ascorbic acid in the tablet = (Mass of ascorbic acid/ Mass of tablet) x 100

                                                                                = (0.0933 g/0.1219 g) x 100

                                                                                 = 76.5%