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

(b) Molarity of solution is 0.038 M

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 29.41 mL

Amount of pure ascorbic acid taken=0.197 g

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

Determine number of ascorbic acid

No of moles of ascorbic acid present

$\frac{0.197g}{176.124g/mol}=1.1185mmol$

Determine the molarity

Molarity of a substance is known as the number of moles of solute present per liter of solution

As from the reaction it can be stated that One mole of ascorbic acid reacts with 1 mole of ${I_3}^{-}$

Therefore, number of moles of ${I_3}^{-}$present = 1.1185

Molarity of ${I_3}^{-}$

$$\begin{gathered} =\frac{Noofmolsof{I_3}^{-}}{Volumeof{I_3}^{-}} \\ =\frac{1.1185mmol}{29.41mL} \\ =0.038M \end{gathered}$$