To `500 mL` of `0.150 M AgNO_(3)` solution were added `500 mL` of `1.09 M Fe^(2+)` solution and the reaction is allowed to reach an equilibrium at `25^(@)C`
`Ag^(o+)(aq)+Fe^(2+)(aq) hArr Fe^(3+)(aq)+Ag(s)`
For `25` mL of the solution, `30 mL` of `0.0832 M KMnO_(4)` was required for oxidation. Calculate the equilibrium constant for the the reaction `25^(@)C`.

Explanation

To calculate the equilibrium constant \( K_c \) for the reaction:

\[ \text{Ag}^+ (aq) + \text{Fe}^{2+} (aq) \rightleftharpoons \text{Fe}^{3+} (aq) + \text{Ag} (s) \]

we will follow these steps:

### Step 1: Calculate initial moles of reactants
1. **For AgNO₃**:
- Molarity = 0.150 M
- Volume = 500 mL = 0.500 L
- Moles of Ag⁺ = Molarity × Volume = \( 0.150 \, \text{mol/L} \times 0.500 \, \text{L} = 0.075 \, \text{mol} \) or 75 mmol.

2. **For Fe²⁺**:
- Molarity = 1.09 M
- Volume = 500 mL = 0.500 L
- Moles of Fe²⁺ = Molarity × Volume = \( 1.09 \, \text{mol/L} \times 0.500 \, \text{L} = 0.545 \, \text{mol} \) or 545 mmol.

### Step 2: Set up the equilibrium expression
Let \( x \) be the amount of Ag⁺ that reacts at equilibrium. Then at equilibrium:
- Moles of Ag⁺ = \( 75 - x \)
- Moles of Fe²⁺ = \( 545 - x \)
- Moles of Fe³⁺ = \( x \)

### Step 3: Use the information from KMnO₄ titration
From the problem, we know that 30 mL of 0.0832 M KMnO₄ was required to oxidize 25 mL of the solution.

1. Calculate the moles of KMnO₄ used:
\[ \text{Moles of KMnO₄} = \text{Molarity} \times \text{Volume} = 0.0832 \, \text{mol/L} \times 0.030 \, \text{L} = 0.002496 \, \text{mol} \]

2. Since KMnO₄ reacts with Fe²⁺ to form Fe³⁺, and the n-factor for the reaction is 5 (as 1 mole of KMnO₄ oxidizes 5 moles of Fe²⁺), we can find the moles of Fe²⁺ oxidized:
\[ \text{Moles of Fe}^{2+} \text{ oxidized} = 5 \times \text{Moles of KMnO₄} = 5 \times 0.002496 = 0.01248 \, \text{mol} \]

### Step 4: Relate the moles of Fe²⁺ to the equilibrium concentration
At equilibrium, the moles of Fe²⁺ will be:
\[ 545 - x = 0.01248 \]
Thus,
\[ x = 545 - 0.01248 = 544.98752 \approx 545 \, \text{mmol} \]

### Step 5: Calculate the equilibrium concentrations
1. **Concentration of Fe³⁺**:
\[ [\text{Fe}^{3+}] = \frac{x}{\text{Total Volume}} = \frac{0.01248 \, \text{mol}}{1.0 \, \text{L}} = 0.01248 \, \text{M} \]

2. **Concentration of Fe²⁺**:
\[ [\text{Fe}^{2+}] = \frac{545 - x}{\text{Total Volume}} = \frac{0.545 \, \text{mol} - 0.01248 \, \text{mol}}{1.0 \, \text{L}} = 0.53252 \, \text{M} \]

3. **Concentration of Ag⁺**:
\[ [\text{Ag}^+] = \frac{75 - x}{\text{Total Volume}} = \frac{0.075 \, \text{mol} - 0.01248 \, \text{mol}}{1.0 \, \text{L}} = 0.06252 \, \text{M} \]

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}][\text{Ag}^+]} \]

### Step 7: Substitute the values into the equilibrium expression
\[ K_c = \frac{0.01248}{(0.53252)(0.06252)} \]

### Step 8: Calculate \( K_c \)
1. Calculate the denominator:
\[ (0.53252)(0.06252) \approx 0.03324 \]
2. Now calculate \( K_c \):
\[ K_c \approx \frac{0.01248}{0.03324} \approx 0.375 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction at \( 25^\circ C \) is approximately **0.375**.