The sp cond of a saturated solution of `AgCI` at `25^(@)C` after substractin the sp conductances of conductivity of water is `2.28 xx 10^(-6) mho cm^(-1)`. Find the solubility product of `AgCI` at `25^(@)C(lambda_(AgCI)6(oo) = 138.3 mho cm^(2))`

Explanation

To find the solubility product (Ksp) of AgCl at 25°C, we can follow these steps:

### Step 1: Understand the Dissociation of AgCl
AgCl is a sparingly soluble salt that dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

Let the solubility of AgCl be \( S \) moles per liter. Therefore, at equilibrium:
- The concentration of \( \text{Ag}^+ \) ions = \( S \)
- The concentration of \( \text{Cl}^- \) ions = \( S \)

### Step 2: Write the Expression for Ksp
The solubility product (Ksp) can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S \cdot S = S^2 \]

### Step 3: Use the Given Specific Conductance
The specific conductance of the saturated solution of AgCl after subtracting the conductivity of water is given as:
\[ \kappa = 2.28 \times 10^{-6} \, \text{mho cm}^{-1} \]

### Step 4: Relate Conductivity to Solubility
The equivalent conductivity (\( \Lambda \)) at infinite dilution is given as:
\[ \Lambda_{AgCl} = 138.3 \, \text{mho cm}^2 \]

The relationship between conductivity (\( \kappa \)), equivalent conductivity (\( \Lambda \)), and solubility (\( S \)) is:
\[ \kappa = \Lambda \cdot \frac{1000}{S} \]

### Step 5: Substitute Values and Solve for S
Substituting the known values into the equation:
\[ 2.28 \times 10^{-6} = 138.3 \cdot \frac{1000}{S} \]

Rearranging to find \( S \):
\[ S = 138.3 \cdot \frac{1000}{2.28 \times 10^{-6}} \]

Calculating \( S \):
\[ S = \frac{138300}{2.28 \times 10^{-6}} \]
\[ S \approx 6.06 \times 10^{10} \]

### Step 6: Calculate Ksp
Now, substituting \( S \) back into the Ksp expression:
\[ K_{sp} = S^2 \]
\[ K_{sp} = (1.648 \times 10^{-5})^2 \]
\[ K_{sp} = 2.72 \times 10^{-10} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of AgCl at 25°C is:
\[ K_{sp} \approx 2.72 \times 10^{-10} \, \text{mol}^2/\text{L}^2 \]

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