Problem 2
Question
A solution of \(\mathrm{I}_{2}\) was standardized with ascorbic acid (Asc). Using a 0.1032-g sample of pure ascorbic acid, \(25.15 \mathrm{mL}\) of \(\mathrm{I}_{2}\) titrant was required to reach the starch endpoint. a. What is the molarity of the iodine solution?____\(M\) b. What is the titer of the iodine solution?_____\((m g A s c) /\left(m L I_{2} \text { solution }\right)\)
Step-by-Step Solution
Verified Answer
a. 0.023283 M
b. 0.004102 mg Asc/mL I2 solution
1Step 1: Determine the Moles of Ascorbic Acid
First, calculate the moles of ascorbic acid (Asc) using its molar mass, which is 176.12 g/mol. The total mass provided is 0.1032 g. Use the formula:\[\text{moles of Asc} = \frac{\text{mass of Asc}}{\text{molar mass of Asc}} = \frac{0.1032\, \text{g}}{176.12\, \text{g/mol}}\].Calculating gives:\[\text{moles of Asc} \approx 0.0005856\, \text{mol}\].
2Step 2: Use the Stoichiometry of the Reaction
The reaction between ascorbic acid (Asc) and iodine (I\(_2\)) is a 1:1 stoichiometry. Thus, the moles of iodine required for the reaction are equal to the moles of ascorbic acid: \[\text{moles of I}_2 = 0.0005856\, \text{mol}\].
3Step 3: Calculate the Molarity of the Iodine Solution
Molarity is defined as the moles of solute per liter of solution. Convert the volume of iodine titrant from mL to L:\[25.15\, \text{mL} = 0.02515\, \text{L}\].Then, calculate the molarity (M) of the iodine solution:\[M = \frac{\text{moles of I}_2}{\text{volume in L}} = \frac{0.0005856\, \text{mol}}{0.02515\, \text{L}} \approx 0.023283\, \text{M}\].
4Step 4: Calculate the Titer of the Iodine Solution
The titer is calculated as the mass of ascorbic acid per mL of iodine solution:\[\text{Titer} = \frac{\text{mass of Asc}}{\text{volume of } I_2} = \frac{0.1032\, \text{g}}{25.15\, \text{mL}} \approx 0.004102\, \frac{\text{mg of Asc}}{\text{mL of } I_2}\].
Key Concepts
Molarity CalculationStoichiometry ReactionTitration Analysis
Molarity Calculation
Molarity allows us to understand concentration in terms of moles per liter of solution. It's a key concept in solution stoichiometry as it gives us an accurate measure of how much solute is dissolved. In the example exercise, to find the molarity of the iodine solution, we first calculate the moles of ascorbic acid using its molar mass. Remember
- Molarity (\(M\)) can be represented by the formula: \[M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]
- First, determine the moles of the ascorbic acid, knowing its mass and molar mass.
- The iodine solution's molarity then uses the same number of moles due to the 1:1 stoichiometry.
Stoichiometry Reaction
Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It's crucial for calculating exact amounts in chemical equations. In this exercise, ascorbic acid and iodine react at a 1:1 stoichiometry, meaning one mole of ascorbic acid reacts with one mole of iodine. Important points include:
- A balanced equation ensures that the atoms in reactants equal those in products.
- A 1:1 stoichiometry simplifies calculations since moles of one substance directly equal moles of another.
- Knowing stoichiometric ratios helps determine how much of each reactant is needed or produced.
Titration Analysis
Titration is a technique to determine the concentration of a solute in a solution by slowly adding a titrant until a reaction is complete. The endpoint indicates when the reaction has finished, often marked by a color change facilitated by an indicator like starch. When performing a titration:
- Measure the volume of titrant needed to reach the endpoint.
- Use the known concentration of the titrant to find the unknown concentration of the other reactant.
- The titer provides a way to express this value as mass per volume of titrant used: \[\text{Titer} = \frac{\text{mass of solute}}{\text{volume of titrant}}\]