Question

Ascorbic acid (vitamin C) reacts with ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}$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 ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}$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 ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}$ solution are required to react with 0.197 0 g of pure ascorbic acid, what is the molarity of the ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}$ 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 ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}$. Find the weight percent of ascorbic acid in the tablet.

Short answer

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

Step-by-step solution

Given data

Ascorbic acid and ${I_3}^{-}$ion react and produce dehydroascorbic acid. Starch indicator results in formation of deep blue starch-iodine complex during the reaction.

Volume of ${I_3}^{-}$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 ${I_3}^{-}$solution obtained 0.038 M. g/mol

Determine number  of moles

No of moles of ${I_3}^{-}$ present

Molarity ×volume

=0.038M×31.63mL

=1.201 mmol

With 1mol of ${I_3}^{-}$, 1 mol of ascorbic acid reacted. Therefore, number of moles of ascorbic acid =1.201 mmol

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

$$\begin{gathered} =\frac{0.2119}{0.4242}\times 100 \\ =49.95\% \end{gathered}$$