Calculate the solubilty and solubility product of `Co_(2) [Fe(CN)_(6)]` is water at `25^(@)C` from the following data: conductivity of a saturated solution of `Co_(2)[Fe(CN)_(6)]` is `2.06 xx 10^(-6) Omega^(-1) cm^(-1)` and that of water used `4.1 xx 10^(-7) Omega^(-1) cm^(-1)`. The ionic molar conductivites of `Co^(2+)` and `Fe(CN)_(6)^(4-)` are `86.0 Omega^(-1)cm^(-1) mol^(-1)` and `44.0 Omega^(-1) cm^(-1) mol^(-1)`.

Explanation

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

### Step 1: Calculate the conductivity of the saturated solution
Given:
- Conductivity of the saturated solution, \( \kappa_{solution} = 2.06 \times 10^{-6} \, \Omega^{-1} \, \text{cm}^{-1} \)
- Conductivity of water, \( \kappa_{water} = 4.1 \times 10^{-7} \, \Omega^{-1} \, \text{cm}^{-1} \)

The conductivity of the saturated solution can be calculated by subtracting the conductivity of water from the conductivity of the solution:

\[ \kappa = \kappa_{solution} - \kappa_{water} = 2.06 \times 10^{-6} - 4.1 \times 10^{-7} = 1.65 \times 10^{-6} \, \Omega^{-1} \, \text{cm}^{-1} \]

### Step 2: Calculate the molar conductivity of the complex compound
Using Kohlrausch's law, the molar conductivity \( \Lambda_m \) of the complex can be expressed as the sum of the molar conductivities of its ions:

\[ \Lambda_m = 2 \Lambda_{Co^{2+}} + \Lambda_{Fe(CN)_6^{4-}} \]

Where:
- \( \Lambda_{Co^{2+}} = 86.0 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)
- \( \Lambda_{Fe(CN)_6^{4-}} = 44.0 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)

Calculating:

\[ \Lambda_m = 2 \times 86.0 + 44.0 = 172.0 + 44.0 = 216.0 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

### Step 3: Relate molar conductivity to solubility
The relationship between molar conductivity, conductivity, and concentration is given by:

\[ \Lambda_m = \frac{\kappa \times 1000}{S} \]

Where \( S \) is the solubility in mol/L. Rearranging gives:

\[ S = \frac{\kappa \times 1000}{\Lambda_m} \]

Substituting the values:

\[ S = \frac{1.65 \times 10^{-6} \times 1000}{216.0} = \frac{1.65 \times 10^{-3}}{216.0} \approx 7.64 \times 10^{-6} \, \text{mol/L} \]

### Step 4: Calculate the solubility product \( K_{sp} \)
The solubility product for the dissociation of \( Co_2[Fe(CN)_6] \) is given by:

\[ K_{sp} = [Co^{2+}]^2 [Fe(CN)_6^{4-}] \]

Since the dissociation produces 2 moles of \( Co^{2+} \) and 1 mole of \( Fe(CN)_6^{4-} \):

\[ K_{sp} = (2S)^2 \times S = 4S^3 \]

Substituting the value of \( S \):

\[ K_{sp} = 4(7.64 \times 10^{-6})^3 \]

Calculating \( K_{sp} \):

\[ K_{sp} = 4 \times (4.46 \times 10^{-16}) \approx 1.78 \times 10^{-15} \]

### Final Results
- **Solubility \( S \)**: \( 7.64 \times 10^{-6} \, \text{mol/L} \)
- **Solubility Product \( K_{sp} \)**: \( 1.78 \times 10^{-15} \)