Vitamin C (ascorbic acid, C6H8O6, 176.12 g/mol) can be measured by redox titration with iodine solution. lodine is reduced to iodide ion, while ascorbic acid is oxidized to dehydroascorbic acid (C6H6O6). The iodine solution is usually made in the presence of iodide ion, forming the more stable triiodide ion, Is, giving the following overall reaction C6H8O6(aq) + 13(aq) + H2O(l) → C6H6O6(aq) + 31-(aq) + 2H+(aq) M 1 2 3 The titration is carried out in the presence of starch, which forms a dark blue complex with the excess iodine when the endpoint is reached. Ascorbic acid can be used to standardize the iodine solution. A 0.315 g sample of ascorbic acid was titrated with iodine solution, requiring 29.2 mL to reach the endpoint. Calculate [1], the molar concentration of Is ion in the solution.
The Correct Answer and Explanation is :
To calculate the molar concentration of iodide ion (I₃⁻) in the iodine solution, we need to use the stoichiometry of the redox reaction and the data from the titration.
Step 1: Write the balanced redox reaction
The balanced redox reaction between ascorbic acid (C₆H₈O₆) and iodine (I₂) is:
[
\text{C}_6\text{H}_8\text{O}_6 (aq) + \text{I}_3^- (aq) + \text{H}_2\text{O} (l) \rightarrow \text{C}_6\text{H}_6\text{O}_6 (aq) + \text{I}^- (aq) + 2 \text{H}^+ (aq)
]
This reaction shows that one mole of ascorbic acid reacts with one mole of iodine (I₃⁻) to form dehydroascorbic acid and iodide ions.
Step 2: Calculate the moles of ascorbic acid used in the titration
Given that the sample of ascorbic acid weighs 0.315 g, we can calculate the number of moles of ascorbic acid used:
[
\text{moles of ascorbic acid} = \frac{\text{mass of ascorbic acid}}{\text{molar mass of ascorbic acid}} = \frac{0.315 \, \text{g}}{176.12 \, \text{g/mol}} = 0.00179 \, \text{mol}
]
Step 3: Calculate the moles of iodine (I₃⁻) used in the titration
From the balanced equation, we see that 1 mole of ascorbic acid reacts with 1 mole of I₃⁻. Therefore, the moles of I₃⁻ used are the same as the moles of ascorbic acid:
[
\text{moles of I}_3^- = 0.00179 \, \text{mol}
]
Step 4: Calculate the molar concentration of I₃⁻
The volume of iodine solution used in the titration is 29.2 mL, which is 0.0292 L. To calculate the molar concentration of I₃⁻ (which is the same as the iodine concentration), we use the formula:
[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
]
[
\text{Molarity of I}_3^- = \frac{0.00179 \, \text{mol}}{0.0292 \, \text{L}} = 0.0613 \, \text{M}
]
Final Answer:
The molar concentration of the I₃⁻ ion in the iodine solution is 0.0613 M.
Explanation:
In this redox titration, ascorbic acid is oxidized to dehydroascorbic acid, and iodine is reduced to iodide. The titration measures the amount of iodine needed to oxidize the ascorbic acid, allowing us to determine the concentration of iodine in the solution. The calculation involved determining the moles of ascorbic acid and iodine and then using the volume of iodine solution to find its molarity.