A titration is performed as follows:

17.607 mL of Fe2+ solution of unknown concentration is charged into a 100 mL beaker

45 mL of deionized water is added

the solution is titrated with 0.655 M Ce4+ solution

the equivalence point is reached after 5.339 mL of Ce4+ solution is added

What is the concentration of the Fe2+ solution? Report your response to three digits after the decimal.

To determine the concentration of the \(\text{Fe}^{2+}\) solution, we can use the stoichiometry of the titration reaction and the concept of molarity, which is defined as:

\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
\]

### Step 1: Determine the moles of \(\text{Ce}^{4+}\) added at equivalence point
Given:
- Volume of \(\text{Ce}^{4+}\) added = \(5.339 \, \text{mL} = 0.005339 \, \text{L}\)
- Molarity of \(\text{Ce}^{4+}\) solution = \(0.655 \, \text{M}\)

The number of moles of \(\text{Ce}^{4+}\) added can be calculated as:

\[
\text{moles of } \text{Ce}^{4+} = \text{Molarity of } \text{Ce}^{4+} \times \text{Volume of } \text{Ce}^{4+}
\]
\[
= 0.655 \, \text{M} \times 0.005339 \, \text{L} = 0.003497 \, \text{mol}
\]

### Step 2: Stoichiometry of the reaction
In the redox titration between \(\text{Fe}^{2+}\) and \(\text{Ce}^{4+}\), the balanced chemical equation is:

\[
\text{Ce}^{4+} + \text{Fe}^{2+} \rightarrow \text{Ce}^{3+} + \text{Fe}^{3+}
\]

From the equation, we see that 1 mole of \(\text{Ce}^{4+}\) reacts with 1 mole of \(\text{Fe}^{2+}\). Thus, the moles of \(\text{Fe}^{2+}\) in the solution are equal to the moles of \(\text{Ce}^{4+}\) at the equivalence point.

\[
\text{moles of } \text{Fe}^{2+} = 0.003497 \, \text{mol}
\]

### Step 3: Calculate the concentration of \(\text{Fe}^{2+}\)
Now, we can calculate the concentration of the \(\text{Fe}^{2+}\) solution using the formula:

\[
\text{Concentration of } \text{Fe}^{2+} = \frac{\text{moles of } \text{Fe}^{2+}}{\text{Volume of } \text{Fe}^{2+} \text{ solution in liters}}
\]
\[
= \frac{0.003497 \, \text{mol}}{17.607 \, \text{mL} \times \frac{1}{1000} \, \frac{\text{L}}{\text{mL}}} = \frac{0.003497}{0.017607} \, \text{M} = 0.199 \, \text{M}
\]

### Final Answer:
The concentration of the \(\text{Fe}^{2+}\) solution is \(0.199 \, \text{M}\).

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. What is the significance of the equivalence point in a titration?
2. How does the volume of added water affect the titration result?
3. Can the molarity of \(\text{Fe}^{2+}\) be affected by impurities in the solution?
4. How would using a different oxidizing agent instead of \(\text{Ce}^{4+}\) affect the titration?
5. What are potential errors in titration experiments and how can they be minimized?

**Tip:** Always record the exact volume at the equivalence point in a titration to ensure accurate molarity calculations.

A titration is performed as follows: 17.607 mL of Fe2+ solution of unknown concentration is charged into a 100 mL beaker, 45 mL of deionized water is added, the solution is titrated with 0.655 M Ce4+ solution, the equivalence point is reached after 5.339 mL of Ce4+ solution is added. What is the concentration of the Fe2+ solution? Report your response to three digits after the decimal.

Learn how to calculate the concentration of Fe2+ in a titration with 0.655 M Ce4+ solution. This step-by-step solution covers key concepts like stoichiometry, molarity, and redox titration. Ideal for Grades 10-12 chemistry students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation