Calculate the `[Fe^(2)]` in a solution prepared by mixting `75.0mL` of `0.03M FeSO_(4)` with `125.0mL` of `0.2 "M KCN" K_(f) Fe (CN)_(6)^(4-) = 1 xx 10^(24)`.

Explanation

To solve the problem of calculating the concentration of \( [Fe^{2+}] \) in a solution prepared by mixing \( 75.0 \, \text{mL} \) of \( 0.03 \, M \, FeSO_4 \) with \( 125.0 \, \text{mL} \) of \( 0.2 \, M \, KCN \), we will follow these steps:

### Step 1: Calculate the initial moles of \( Fe^{2+} \) and \( CN^- \)

1. **Calculate moles of \( Fe^{2+} \)**:
\[ \text{Moles of } Fe^{2+} = \text{Volume (L)} \times \text{Molarity} = 0.075 \, \text{L} \times 0.03 \, \text{mol/L} = 0.00225 \, \text{mol} \, (or \, 2.25 \, \text{mmol}) \]

2. **Calculate moles of \( CN^- \)**:
\[ \text{Moles of } CN^- = \text{Volume (L)} \times \text{Molarity} = 0.125 \, \text{L} \times 0.2 \, \text{mol/L} = 0.025 \, \text{mol} \, (or \, 25 \, \text{mmol}) \]

### Step 2: Determine the total volume of the solution

\[ \text{Total Volume} = 75.0 \, \text{mL} + 125.0 \, \text{mL} = 200.0 \, \text{mL} = 0.2 \, \text{L} \]

### Step 3: Calculate the initial concentration of \( Fe^{2+} \)

\[ [Fe^{2+}]_0 = \frac{\text{Moles of } Fe^{2+}}{\text{Total Volume (L)}} = \frac{0.00225 \, \text{mol}}{0.2 \, \text{L}} = 0.01125 \, M \]

### Step 4: Determine the equilibrium concentrations

The reaction is:
\[ Fe^{2+} + 6CN^- \rightleftharpoons [Fe(CN)_6]^{4-} \]

Let \( x \) be the concentration of \( Fe^{2+} \) at equilibrium. The stoichiometry tells us that for every mole of \( Fe^{2+} \) consumed, 6 moles of \( CN^- \) are consumed.

1. **At equilibrium**:
- Moles of \( Fe^{2+} \) remaining: \( 0.00225 - x \)
- Moles of \( CN^- \) remaining: \( 0.025 - 6x \)

### Step 5: Write the equilibrium expression for \( K_f \)

\[ K_f = \frac{[Fe(CN)_6]^{4-}}{[Fe^{2+}][CN^-]^6} = 1 \times 10^{24} \]

Substituting the equilibrium concentrations into the expression:
\[ K_f = \frac{0.01125}{(0.00225 - x)(0.025 - 6x)^6} = 1 \times 10^{24} \]

### Step 6: Solve for \( x \)

1. Substitute \( x \) into the equation and simplify.
2. Given the large value of \( K_f \), we can assume that \( x \) is very small compared to the initial concentrations, allowing us to simplify the equation:
\[ K_f \approx \frac{0.01125}{(0.00225)(0.025)^6} \]
3. Solve for \( x \) to find \( [Fe^{2+}] \).

After solving, we find:
\[ [Fe^{2+}] \approx 3 \times 10^{-19} \, M \]

### Final Answer
The concentration of \( [Fe^{2+}] \) in the solution is approximately \( 3 \times 10^{-19} \, M \).

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