JEE Main 2026 — Electrochemistry Question with Solution

Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of ${\text{BaSO}}_4$ is given by:

${\Lambda }_m^{\circ }({\text{BaSO}}_4)={\Lambda }_m^{\circ }({\text{BaCl}}_2)+{\Lambda }_m^{\circ }({\text{H}}_2{\text{SO}}_4)-2{\Lambda }_m^{\circ }(\text{HCl})$

Substituting the given values:

${\Lambda }_m^{\circ }({\text{BaSO}}_4)=x_1+x_2-2x_3$

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:

${\Lambda }_m=\frac{\kappa \times 1000}{S}$

Given that $\kappa =x$ and ${\Lambda }_m={\Lambda }_m^{\circ }$ (which implies complete dissociation, meaning the degree of dissociation $\alpha =1$), we get:

$S=\frac{1000x}{x_1+x_2-2x_3}$

The solubility product $K_{sp}$ for ${\text{BaSO}}_4$ (which dissociates into ${\text{Ba}}^{2+}$ and ${\text{SO}}_4^{2-}$) is:

$K_{sp}=S^2={\left(\frac{1000x}{x_1+x_2-2x_3}\right)}^2=\frac{10^6{x}^2}{(x_1+x_2-2x_3{)}^2}$

Since $\alpha =1$, this expression is equivalent to $\frac{10^6{x}^2}{{\alpha }^2(x_1+x_2-2x_3{)}^2}$.

Answer: $\frac{10^6{x}^2}{{\alpha }^2(x_1+x_2-2x_3{)}^2}$