25 mL of a solution of `Fe^(2+)` ions was titrated with a solution of the oxidizing agent `Cr_(2)O_(7)^(2-)` . 50 mL of 0.01 M `K_(2)Cr_(2)O_(7)` solution was required. What is the molarity of the `Fe^(2+)` solution ?

Answer

To find the molarity of the `Fe^(2+)` solution, we can follow these steps:

### Step 1: Write the balanced redox reaction
The redox reaction between `Fe^(2+)` and `Cr_(2)O_(7)^(2-)` in an acidic medium can be written as:
\[ \text{6Fe}^{2+} + \text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} \rightarrow 6\text{Fe}^{3+} + 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O} \]

### Step 2: Determine the n-factor for `Fe^(2+)` and `Cr_(2)O_(7)^(2-)`
- For `Fe^(2+)` to `Fe^(3+)`, the change in oxidation state is from +2 to +3, which means the n-factor is **1** (since 1 electron is lost).
- For `Cr_(2)O_(7)^(2-)` to `Cr^(3+)`, the change in oxidation state is from +6 to +3. Since there are 2 chromium atoms, the total change is \(2 \times (6 - 3) = 6\). Thus, the n-factor is **6**.

### Step 3: Use the equivalence concept
According to the law of equivalence:
\[ \text{Equivalence of } Fe^{2+} = \text{Equivalence of } Cr_{2}O_{7}^{2-} \]

### Step 4: Calculate the equivalence of `Cr_(2)O_(7)^(2-)`
Given:
- Volume of `K_(2)Cr_(2)O_(7)` solution = 50 mL = 0.050 L
- Molarity of `K_(2)Cr_(2)O_(7)` solution = 0.01 M

The number of equivalents of `Cr_(2)O_(7)^(2-)` can be calculated as:
\[ \text{Equivalence of } Cr_{2}O_{7}^{2-} = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor} \]
\[ = 0.01 \, \text{mol/L} \times 0.050 \, \text{L} \times 6 = 0.003 \, \text{equivalents} \]

### Step 5: Calculate the equivalence of `Fe^(2+)`
Let the molarity of `Fe^(2+)` solution be \( M \).
The volume of `Fe^(2+)` solution is 25 mL = 0.025 L. The equivalence of `Fe^(2+)` can be expressed as:
\[ \text{Equivalence of } Fe^{2+} = M \times \text{Volume (L)} \times \text{n-factor} \]
\[ = M \times 0.025 \, \text{L} \times 1 \]

### Step 6: Set the equivalences equal
\[ M \times 0.025 = 0.003 \]

### Step 7: Solve for \( M \)
\[ M = \frac{0.003}{0.025} = 0.12 \, \text{M} \]

### Final Answer
The molarity of the `Fe^(2+)` solution is **0.12 M**.

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