The specific conductance and equivalent conductance of a saturated solution of `BaSO_(4)` are `8xx10^(-5)ohm^(-1)cm^(-1)` and 8000 `ohm^(-1)cm^(2)"equi"^(-1)` respectively. Hence `K_(sp)` of `BaSO_(4)` is

Options

A. `2.5xx10^(-10)M^(2)`
B. `2.5xx10^(-11)M^(2)`
C. `2.5xx10^(-20)M^(2)`
D. `2.5xx10^(-20)M^(2)`

Explanation

To find the solubility product (Ksp) of BaSO₄ from the given specific conductance and equivalent conductance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Specific conductance (κ) = \(8 \times 10^{-5} \, \Omega^{-1} \, \text{cm}^{-1}\)
- Equivalent conductance (Λ) = \(8000 \, \Omega^{-1} \, \text{cm}^2 \, \text{equiv}^{-1}\)

2. **Determine the N-factor:**
- For BaSO₄, barium (Ba) has a charge of +2 and sulfate (SO₄) has a charge of -2. Thus, the N-factor (number of moles of ions produced per mole of salt) is 2.

3. **Calculate Molar Conductivity (Λm):**
- Molar conductivity (Λm) can be calculated using the formula:
\[ \Lambda_m = N \times \Lambda \]
- Substituting the values:
\[ \Lambda_m = 2 \times 8000 \, \Omega^{-1} \, \text{cm}^2 \, \text{equiv}^{-1} = 16000 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

4. **Relate Molar Conductivity to Specific Conductance:**
- The relationship between specific conductance (κ) and molar conductivity (Λm) is given by:
\[ \kappa = \frac{\Lambda_m \times C}{1000} \]
- Where C is the molarity of the solution. Rearranging gives:
\[ C = \frac{1000 \times \kappa}{\Lambda_m} \]

5. **Calculate the Molarity (C):**
- Substitute the values:
\[ C = \frac{1000 \times 8 \times 10^{-5}}{16000} \]
- Performing the calculation:
\[ C = \frac{8 \times 10^{-2}}{16} = 5 \times 10^{-6} \, \text{mol/L} \]

6. **Calculate Ksp:**
- For BaSO₄ dissociating into Ba²⁺ and SO₄²⁻, the solubility product (Ksp) is given by:
\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] = S^2 \]
- Since both ions have the same concentration (S), we can substitute:
\[ K_{sp} = (5 \times 10^{-6})^2 = 25 \times 10^{-12} = 2.5 \times 10^{-11} \, \text{mol}^2/\text{L}^2 \]

### Final Answer:
The solubility product (Ksp) of BaSO₄ is \(2.5 \times 10^{-11} \, \text{mol}^2/\text{L}^2\).

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