Calculate the `[Fe^(2+)]` in a solution containing `0.2M [Fe(CN)_(6)]^(4-)` and `0.1 M CN^(Theta). K_(f)Fe (CN)_(6)^(4-) = 1 xx 10^(24)`.

Explanation

To calculate the concentration of \([Fe^{2+}]\) in a solution containing \(0.2M [Fe(CN)_{6}]^{4-}\) and \(0.1M CN^{-}\), we will use the formation constant \(K_f\) for the complex ion formation.

### Step-by-Step Solution:

1. **Write the Formation Reaction**:
The formation of the complex ion can be represented as:
\[ Fe^{2+} + 6CN^{-} \rightleftharpoons [Fe(CN)_{6}]^{4-} \]

2. **Write the Expression for the Formation Constant**:
The formation constant \(K_f\) is given by:
\[ K_f = \frac{[Fe(CN)_{6}]^{4-}}{[Fe^{2+}][CN^{-}]^6} \]
Given \(K_f = 1 \times 10^{24}\).

3. **Substitute Known Concentrations**:
We know the concentrations:
- \([Fe(CN)_{6}]^{4-} = 0.2M\)
- \([CN^{-}] = 0.1M\)
Let the concentration of \(Fe^{2+}\) be \(x\).

4. **Set Up the Equation**:
Substitute the known values into the \(K_f\) expression:
\[ 1 \times 10^{24} = \frac{0.2}{x \cdot (0.1)^6} \]

5. **Simplify the Equation**:
Rearranging the equation gives:
\[ x = \frac{0.2}{1 \times 10^{24} \cdot (0.1)^6} \]

6. **Calculate \((0.1)^6\)**:
\[ (0.1)^6 = 1 \times 10^{-6} \]

7. **Substitute Back to Find \(x\)**:
Now substituting this back into the equation:
\[ x = \frac{0.2}{1 \times 10^{24} \cdot 1 \times 10^{-6}} = \frac{0.2}{1 \times 10^{18}} = 2 \times 10^{-19} \, M \]

8. **Final Answer**:
Thus, the concentration of \([Fe^{2+}]\) in the solution is:
\[ [Fe^{2+}] = 2 \times 10^{-19} \, M \]