Equivalent conductance of saturated `BaSO_(4)` is 400 S `cm^(2) eq^(-1)` an d specific conductyance is 8 `xx10^(-5)S cm^(-1)`. Solubility product. `K_(sp)` of `BaSO_(4)` is

Options

A. `4xx10^(-8)`
B. `1xx10^(-8)`
C. `2xx10^(-4)`
D. `1xx10^(-4)`

Explanation

To find the solubility product \( K_{sp} \) of barium sulfate \( BaSO_4 \), we can follow these steps:

### Step 1: Identify Given Data
- Equivalent conductance of saturated \( BaSO_4 \) = 400 S cm² eq⁻¹
- Specific conductance \( \kappa \) = \( 8 \times 10^{-5} \) S cm⁻¹

### Step 2: Calculate Molar Conductance
Molar conductance \( \Lambda_m \) can be calculated using the formula:
\[ \Lambda_m = n \times \Lambda_{eq} \]
where \( n \) is the number of ions produced per formula unit of the solute (for \( BaSO_4 \), \( n = 2 \) because it dissociates into \( Ba^{2+} \) and \( SO_4^{2-} \)).

Substituting the values:
\[ \Lambda_m = 2 \times 400 \, \text{S cm}^2 \text{eq}^{-1} = 800 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 3: Calculate Solubility
The solubility \( S \) (in mol/L) can be calculated using the relationship:
\[ S = \frac{\kappa \times 1000}{\Lambda_m} \]
Substituting the values:
\[ S = \frac{8 \times 10^{-5} \, \text{S cm}^{-1} \times 1000}{800 \, \text{S cm}^2 \text{mol}^{-1}} \]
Calculating this gives:
\[ S = \frac{8 \times 10^{-2}}{800} = 1 \times 10^{-4} \, \text{mol/L} \]

### Step 4: Calculate the Solubility Product \( K_{sp} \)
The solubility product \( K_{sp} \) for \( BaSO_4 \) is given by:
\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]
Since \( BaSO_4 \) dissociates into one \( Ba^{2+} \) ion and one \( SO_4^{2-} \) ion, we can write:
\[ K_{sp} = S \times S = S^2 \]
Substituting the value of solubility:
\[ K_{sp} = (1 \times 10^{-4})^2 = 1 \times 10^{-8} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of \( BaSO_4 \) is:
\[ K_{sp} = 1 \times 10^{-8} \]

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