Equivalent condictance of `BaCI_(2), H_(2)SO_(4)` and `HCI` are `x_(1) , x_(2)` and `x_(3) S cm^(2) "equiv"^(-1)` at infinite dilution , if specific conductance of structured `BaSO_(4)` solution is of `y S cm^(-1)` then `K_(sp)` of `BaSO_(4)` is

Options

A. `(10^(3)y)/(2(x_(1) + x_(2)- 2x_(3)))`
B. `(10^(6)y^(2))/((x_(1) + x_(2)- 2x_(3))^(2))`
C. `(10^(6)y^(23))/(4(x_(1) x_(2)- 2x_(3))^(2))`
D. `(x_(1) x_(2)- 2x_(3))/(10^(3)y^(2))`

Explanation

To solve the problem, we need to find the solubility product constant (Ksp) of BaSO4 based on the given equivalent conductances at infinite dilution and the specific conductance of the BaSO4 solution.

### Step-by-Step Solution:

1. **Understanding Equivalent Conductance**:
The equivalent conductance (Λ) of a compound at infinite dilution can be expressed as the sum of the equivalent conductances of its constituent ions. For BaSO4, the ions are Ba²⁺ and SO₄²⁻.

\[ \Lambda_{\text{BaSO}_4} = \Lambda_{\text{Ba}^{2+}} + \Lambda_{\text{SO}_4^{2-}} \]

2. **Using Given Data**:
We have the equivalent conductances for BaCl2 (x₁), H₂SO₄ (x₂), and HCl (x₃). We need to express the equivalent conductance of BaSO4 in terms of these values.

From the dissociation of BaCl2:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \]
The equivalent conductance of BaCl2 can be represented as:
\[ \Lambda_{\text{BaCl}_2} = \Lambda_{\text{Ba}^{2+}} + 2\Lambda_{\text{Cl}^-} = x_1 \]

For H₂SO₄:
\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \]
The equivalent conductance of H₂SO₄ is:
\[ \Lambda_{\text{H}_2\text{SO}_4} = 2\Lambda_{\text{H}^+} + \Lambda_{\text{SO}_4^{2-}} = x_2 \]

For HCl:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
The equivalent conductance of HCl is:
\[ \Lambda_{\text{HCl}} = \Lambda_{\text{H}^+} + \Lambda_{\text{Cl}^-} = x_3 \]

3. **Finding the Equivalent Conductance of BaSO4**:
We can express the equivalent conductance of BaSO4 using the above equations:
\[ \Lambda_{\text{BaSO}_4} = \Lambda_{\text{Ba}^{2+}} + \Lambda_{\text{SO}_4^{2-}} = (x_1 - 2\Lambda_{\text{Cl}^-}) + \Lambda_{\text{SO}_4^{2-}} \]
From the H₂SO₄ equation, we can isolate \(\Lambda_{\text{SO}_4^{2-}}\):
\[ \Lambda_{\text{SO}_4^{2-}} = x_2 - 2\Lambda_{\text{H}^+} \]

Substituting back, we get:
\[ \Lambda_{\text{BaSO}_4} = x_1 - 2\Lambda_{\text{Cl}^-} + (x_2 - 2\Lambda_{\text{H}^+}) \]

4. **Relating Conductance to Solubility**:
The specific conductance (κ) of the BaSO4 solution is given as y S cm⁻¹. The relationship between specific conductance and solubility (S) can be expressed as:
\[ \kappa = S \cdot z \]
where z is the total charge carried by the ions in solution. For BaSO4, z = 2 (from Ba²⁺ and SO₄²⁻).

5. **Calculating Solubility**:
The solubility (S) of BaSO4 can be expressed as:
\[ S = \frac{y}{\Lambda_{\text{BaSO}_4}} = \frac{y}{x_1 + x_2 - 2x_3} \]

6. **Finding Ksp**:
The solubility product constant (Ksp) can be expressed as:
\[ K_{sp} = S^2 \]
Substituting the expression for S:
\[ K_{sp} = \left(\frac{y}{x_1 + x_2 - 2x_3}\right)^2 \]

### Final Expression for Ksp:
\[ K_{sp} = \frac{y^2}{(x_1 + x_2 - 2x_3)^2} \]