Conductivity of a saturated solution of a sparingly soluble salt AB at 298K is 1.85×`10^(−5) Sm^(−1)` . Solubility product of the salt AB at 298K is :
[Given `Λ^(o)m`(AB)=140×`10 ^(−4)Sm^2mol^(−1)`]

Options

A. 1.32×`10^(−12)`
B. 1.74×`10^(−12)`
C. 5.7×`10^(−12)`
D. 7.5×`10^(−12)`

Answer

To find the solubility product (Ksp) of the sparingly soluble salt AB, we will follow these steps:

### Step 1: Calculate the Molar Conductivity (Λm)
Given:
- Conductivity (K) = \(1.85 \times 10^{-5} \, \text{S m}^{-1}\)
- Molar conductivity (Λm) = \(140 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1}\)

Using the formula for molar conductivity:
\[ Λ_m = \frac{K \times 1000}{C} \]
Where C is the concentration in mol/m³.

### Step 2: Rearranging the Formula
Rearranging the formula to find C:
\[ C = \frac{K \times 1000}{Λ_m} \]

### Step 3: Substitute the Values
Substituting the values into the equation:
\[ C = \frac{1.85 \times 10^{-5} \times 1000}{140 \times 10^{-4}} \]

### Step 4: Perform the Calculation
Calculating the concentration:
\[ C = \frac{1.85 \times 10^{-2}}{1.4} \approx 1.32 \times 10^{-6} \, \text{mol/m}^3 \]

### Step 5: Write the Dissociation Equation
The dissociation of the salt AB can be represented as:
\[ AB \rightleftharpoons A^+ + B^- \]
At equilibrium, if the solubility of AB is C, then:
- \([A^+] = C\)
- \([B^-] = C\)

### Step 6: Write the Expression for Ksp
The solubility product (Ksp) is given by:
\[ K_{sp} = [A^+][B^-] = C \times C = C^2 \]

### Step 7: Substitute the Concentration
Substituting the value of C:
\[ K_{sp} = (1.32 \times 10^{-6})^2 \]

### Step 8: Perform the Calculation
Calculating Ksp:
\[ K_{sp} = 1.7424 \times 10^{-12} \approx 1.74 \times 10^{-12} \]

### Final Answer
The solubility product of the salt AB at 298K is:
\[ K_{sp} \approx 1.74 \times 10^{-12} \]

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