Q8 / 75 · JEE Main 2026 (06 April Shift 1) · Chemistry
Consider the following data.
Electrolyte $\Lambda^{\circ}_m$ (S cm$^2$ mol$^{-1}$) $\text{BaCl}_2$ $x_1$ $\text{H}_2\text{SO}_4$ $x_2$ HCl $x_3$
$\text{BaSO}_4$ is sparingly soluble in water. If the conductivity of the saturated $\text{BaSO}_4$ solution is $x$ S cm$^{-1}$ then the solubility product of $\text{BaSO}_4$ can be given as (Here $\Lambda_m = \Lambda^{\circ}_m$)
| Electrolyte | $\Lambda^{\circ}_m$ (S cm$^2$ mol$^{-1}$) |
|---|---|
| $\text{BaCl}_2$ | $x_1$ |
| $\text{H}_2\text{SO}_4$ | $x_2$ |
| HCl | $x_3$ |
$\text{BaSO}_4$ is sparingly soluble in water. If the conductivity of the saturated $\text{BaSO}_4$ solution is $x$ S cm$^{-1}$ then the solubility product of $\text{BaSO}_4$ can be given as (Here $\Lambda_m = \Lambda^{\circ}_m$)
- Correct answerA$\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$
- B$\dfrac{x^2}{(x_1 + x_2 - 2x_3)^2}$
- C$\dfrac{\alpha^2(x_1 + x_2 - 2x_3)^2}{10^6 x^2}$
- D$\dfrac{x^2}{(x_1 + x_2 + 2x_3)^2}$
Solution
Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of $\text{BaSO}_4$ is given by:
$\Lambda^{\circ}_m(\text{BaSO}_4) = \Lambda^{\circ}_m(\text{BaCl}_2) + \Lambda^{\circ}_m(\text{H}_2\text{SO}_4) - 2\Lambda^{\circ}_m(\text{HCl})$
Substituting the given values:
$\Lambda^{\circ}_m(\text{BaSO}_4) = x_1 + x_2 - 2x_3$
For a sparingly soluble salt, the solubility $S$ (in mol L$^{-1}$) is related to its conductivity $\kappa$ (in S cm$^{-1}$) and molar conductivity $\Lambda_m$ (in S cm$^2$ mol$^{-1}$) by the relation:
$\Lambda_m = \dfrac{\kappa \times 1000}{S}$
Given that $\kappa = x$ and $\Lambda_m = \Lambda^{\circ}_m$ (which implies complete dissociation, meaning the degree of dissociation $\alpha = 1$), we get:
$S = \dfrac{1000 x}{x_1 + x_2 - 2x_3}$
The solubility product $K_{sp}$ for $\text{BaSO}_4$ (which dissociates into $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$) is:
$K_{sp} = S^2 = \left( \dfrac{1000 x}{x_1 + x_2 - 2x_3} \right)^2 = \dfrac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$
Since $\alpha = 1$, this expression is equivalent to $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$.
Answer: $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$
$\Lambda^{\circ}_m(\text{BaSO}_4) = \Lambda^{\circ}_m(\text{BaCl}_2) + \Lambda^{\circ}_m(\text{H}_2\text{SO}_4) - 2\Lambda^{\circ}_m(\text{HCl})$
Substituting the given values:
$\Lambda^{\circ}_m(\text{BaSO}_4) = x_1 + x_2 - 2x_3$
For a sparingly soluble salt, the solubility $S$ (in mol L$^{-1}$) is related to its conductivity $\kappa$ (in S cm$^{-1}$) and molar conductivity $\Lambda_m$ (in S cm$^2$ mol$^{-1}$) by the relation:
$\Lambda_m = \dfrac{\kappa \times 1000}{S}$
Given that $\kappa = x$ and $\Lambda_m = \Lambda^{\circ}_m$ (which implies complete dissociation, meaning the degree of dissociation $\alpha = 1$), we get:
$S = \dfrac{1000 x}{x_1 + x_2 - 2x_3}$
The solubility product $K_{sp}$ for $\text{BaSO}_4$ (which dissociates into $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$) is:
$K_{sp} = S^2 = \left( \dfrac{1000 x}{x_1 + x_2 - 2x_3} \right)^2 = \dfrac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$
Since $\alpha = 1$, this expression is equivalent to $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$.
Answer: $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$
Hence, the correct answer is option: 'A'.