Calculate the solubility product of `Co_(2)[Fe(CN)_(6)]` in water at `25^(@)C`.
Given, conductivity of saturated solutions of `Co_(2)[Fe(CN)_(6)]` is `2.06 xx 10^(-6) Omega^(-1)cm^(-1)` and that of water used is `4.1 xx 10^(-7) Omega^(-1)cm^(-1)`. The ionic molar conductivities of `Co^(2+)` and `[Fe(CN)_(6)]^(4)` are `86.0 Omega cm^(2)mol^(-1)` and `444.0 Omega^(-1) cm^(2)mol^(-1)`, respectivly.

Options

A. `7.87xx10^(-17)`
B. `7.87xx10^(-16)`
C. `7.87xx10^(-18)`
D. `7.87xx10^(-19)`

Explanation

To calculate the solubility product (Ksp) of the complex \( Co_2[Fe(CN)_6] \) in water at \( 25^\circ C \), we will follow these steps:

### Step 1: Determine the Specific Conductance of the Saturated Solution
The specific conductance (\( \kappa \)) of the saturated solution of \( Co_2[Fe(CN)_6] \) is given as:
\[ \kappa_{\text{saturated}} = 2.06 \times 10^{-6} \, \Omega^{-1} \text{cm}^{-1} \]
The conductivity of water is given as:
\[ \kappa_{\text{water}} = 4.1 \times 10^{-7} \, \Omega^{-1} \text{cm}^{-1} \]
To find the specific conductance of the solution, we subtract the conductivity of water from the conductivity of the saturated solution:
\[ \kappa_{\text{solution}} = \kappa_{\text{saturated}} - \kappa_{\text{water}} = 2.06 \times 10^{-6} - 4.1 \times 10^{-7} = 1.65 \times 10^{-6} \, \Omega^{-1} \text{cm}^{-1} \]

### Step 2: Calculate the Molar Conductivity of the Complex
The molar conductivity (\( \Lambda_m \)) of the complex can be calculated using the ionic molar conductivities of the ions:
- Molar conductivity of \( Co^{2+} \) is \( 86.0 \, \Omega \, \text{cm}^2 \text{mol}^{-1} \)
- Molar conductivity of \( [Fe(CN)_6]^{4-} \) is \( 444.0 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)

The molar conductivity of the complex is given by:
\[ \Lambda_m = 2 \cdot \Lambda_{Co^{2+}} + 1 \cdot \Lambda_{[Fe(CN)_6]^{4-}} = 2 \cdot 86.0 + 444.0 = 616.0 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Step 3: Calculate the Concentration of the Ions
Using the relationship between specific conductance and molar conductivity, we can find the concentration (\( C \)):
\[ C = \frac{1000 \cdot \kappa_{\text{solution}}}{\Lambda_m} \]
Substituting the values:
\[ C = \frac{1000 \cdot 1.65 \times 10^{-6}}{616.0} \approx 2.68 \times 10^{-3} \, \text{mol/L} \]

### Step 4: Calculate the Solubility Product (Ksp)
The dissociation of \( Co_2[Fe(CN)_6] \) in water can be represented as:
\[ Co_2[Fe(CN)_6] \rightleftharpoons 2 Co^{2+} + [Fe(CN)_6]^{4-} \]
Let the solubility of \( Co_2[Fe(CN)_6] \) be \( S \). Then:
- The concentration of \( Co^{2+} \) will be \( 2S \)
- The concentration of \( [Fe(CN)_6]^{4-} \) will be \( S \)

Thus, the solubility product \( Ksp \) can be expressed as:
\[ Ksp = [Co^{2+}]^2 \cdot [Fe(CN)_6]^{4-} = (2C)^2 \cdot C = 4C^3 \]
Substituting the concentration \( C \):
\[ Ksp = 4 \cdot (2.68 \times 10^{-3})^3 \]
Calculating this gives:
\[ Ksp \approx 4 \cdot (2.68 \times 10^{-3})^3 \approx 4 \cdot 2.43 \times 10^{-8} \approx 9.72 \times 10^{-8} \]

### Final Answer
The solubility product \( Ksp \) of \( Co_2[Fe(CN)_6] \) at \( 25^\circ C \) is approximately:
\[ \boxed{9.72 \times 10^{-8}} \]