Ascorbic acid (vitamin C) reacts with I3-according to the equation

Starch is used as an indicator in the reaction. The end point is marked by the appearance of a deep blue starch-iodine complex when the first fraction of a drop of unreacted I3-remains in the solution.

(a) Verify that the structures above have the chemical formulas written beneath them. You must be able to locate every atom in the formula. Use atomic masses from the periodic table on the inside cover of this book to find the formula mass of ascorbic acid.

(b) If 29.41 mL of I3- solution are required to react with 0.197 0 g of pure ascorbic acid, what is the molarity of the I3- solution?

(c) A vitamin C tablet containing ascorbic acid plus inert binder was ground to a powder, and 0.424 2 g was titrated by 31.63 mL of I3-. Find the weight percent of ascorbic acid in the tablet.

Short Answer

Expert verified

c) There is 49.95% of ascorbic acid in the tablet

Step by step solution

01

Given data

Ascorbic acid and I3-ion react and produce dehydroascorbic acid. Starch indicator results in formation of deep blue starch-iodine complex during the reaction.

Volume of I3-solution required 31.63 mL

Amount of tablet=0.424g

With reference to SID 135385-7-A E-a, the formula mass of ascorbic acid obtained 176.124 g/mol

With reference to SID 135385-7-A E-b, the molarity of I3-solution obtained 0.038 M. g/mol

02

Determine number  of moles

No of moles of I3- present

Molarity ×volume

=0.038M×31.63mL

=1.201 mmol

With 1mol of I3-, 1 mol of ascorbic acid reacted. Therefore, number of moles of ascorbic acid =1.201 mmol

03

Determine the amount of ascorbic acid

=Number of moles of ascorbic acid × Formula mass

=1.201mmol ×176.124 g/mol

=0.2119 g

Amount of tablet=0.4242 g

Therefore, weight percent of pure ascorbic acid in tablet

=0.21190.4242×100=49.95%

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