Problem 2 - Chemical Principles in the Laboratory

Question

A solution of \(\mathrm{I}_{2}\) was standardized with ascorbic acid. Using a 0.1000-g sample of pure ascorbic acid, \(25.32 \mathrm{mL}\) of \(\mathrm{I}_{2}\) were required to reach the starch end point. a. What is the molarity of the iodine solution? ___________\(\text{M}\) b. What is the titer of the iodine solution? ____________\(\mathrm{mg}\) asc/mL \(\mathrm{I}_{2}\)

Step-by-Step Solution

Answer

a. The molarity of the iodine solution is approximately 0.01120 M. b. The titer of the iodine solution is approximately \(0.008836 \, mg\, asc \, /mL\, I2\).

Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction between ascorbic acid (C6H8O6) and iodine (I2) to form dehydroascorbic acid (C6H6O6) and iodide ions (I-) is: C6H8O6 + 2I2 -> C6H6O6 + 4HI The stoichiometry for the titration is 1:2, meaning one mole of ascorbic acid reacts with two moles of iodine.

Step 2: Calculate moles of ascorbic acid

Calculate the moles of ascorbic acid, using the given mass (0.1000 g) and the molar mass of ascorbic acid (about \(176.12 g/mol\)): Moles of ascorbic acid = \( \frac{0.1000}{176.12} = 5.676 \times 10^{-4} \, mol \)

Step 3: Calculate molarity of iodine solution

Use the moles of ascorbic acid, the stoichiometry of the reaction, and the volume of iodine solution (25.32 mL) to calculate the molarity of the iodine solution: Molarity of iodine solution = \( \frac{moles\, of\, I2}{volume\, of\, I2} = \frac{0.5\, \times \, 5.676 \times 10^{-4}\, mol}{25.32 \times 10^{-3} L} = 0.01120\, Mol/L\) The molarity of the iodine solution is approximately 0.01120 M.

Step 4: Calculate the titer of the iodine solution

Titer is a measure of the concentration of a solution expressed in terms of the mass of solute per unit volume of solution. It refers to the mass in milligrams of a substance equivalent to the substance reacted with 1 mL of the solution. For this problem, it refers to the mass of ascorbic acid equivalent to 1 mL of iodine solution. Use the molarity and the molar mass of ascorbic acid to calculate the titer of the iodine solution: Titer = Molarity of iodine solution × \( \frac{2 × (mass\,of\,ascorbic\, acid)}{volume\, of\, I2}\) = 0.01120 × \( \frac{2\, \times \, 0.1000}{25.32}\) = 0.008836 mg/mL The titer of the iodine solution is approximately \(0.008836 \, mg\, asc \, /mL\, I2\).

Key Concepts

Titration

Titration is a vital technique in chemistry used to determine the concentration of a specific substance in a solution. In this process, a solution of known concentration, called a titrant, is gradually added to a solution of unknown concentration until the reaction between the two is complete.
A visible change, such as a color change, often indicates the end of a titration, known as the endpoint. This exercise involved the titration of iodine solution with ascorbic acid.

• Ascorbic acid served as the analyte, which is the substance whose concentration we aimed to find.
• Iodine was the titrant, the substance of known concentration added step by step to reach the endpoint.

This particular titration uses a starch indicator, which changes color when all ascorbic acid has reacted, marking the endpoint.

Stoichiometry

Stoichiometry is the mathematical bridge that allows us to calculate quantities in chemical reactions. It ensures precision in chemistry by using balanced chemical equations to relate proportions of reactants and products.
In the iodine and ascorbic acid reaction, the balanced equation is:\[ C_6H_8O_6 + 2I_2 \rightarrow C_6H_6O_6 + 4HI \]

• One mole of ascorbic acid reacts with two moles of iodine.
• This stoichiometric relationship shows a 1:2 ratio of ascorbic acid to iodine.

By determining the moles of ascorbic acid that react, we can accurately find out how much iodine is needed and calculate the molarity of the iodine solution accordingly.

Molarity

Molarity is a way of expressing the concentration of a solution, showing how many moles of a solute are present per liter of solution. The formula used is:\[ M = \frac{n}{V} \]where:

• \( M \) is the molarity.
• \( n \) is the number of moles of solute.
• \( V \) is the volume of the solution in liters.

In this exercise, the molarity of the iodine solution was calculated using the moles derived from the stoichiometry and the volume of iodine solution used. Calculating the molarity allows us to determine the concentration, which is crucial for understanding how much of the titrant is needed to achieve a complete reaction.

Chemical Reactions

Chemical reactions illustrate the transformation of substances into new products. Each reaction is characterized by a balanced equation that explains the conservation of mass and atoms. In the exercise, the chemical reaction involved ascorbic acid and iodine.
Atoms are neither created nor destroyed, but rearranged, a principle reflected in the balanced equation:
\[ C_6H_8O_6 + 2I_2 \rightarrow C_6H_6O_6 + 4HI \]This shows that:

• Ascorbic acid transforms into dehydroascorbic acid.
• Iodine converts to iodide ions.

Through this reaction, titration and stoichiometry work together to analyze how much of each starting material is used to form the products. This understanding is pivotal in chemical lab work, ensuring precise and predictable outcomes.