JEE Main 2026 — Chemistry Physical Chemistry

Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of $\text{BaSO}_4$BaSO4​ is given by:

$$\Lambda^{\circ}_m(\text{BaSO}_4) = \Lambda^{\circ}_m(\text{BaCl}_2) + \Lambda^{\circ}_m(\text{H}_2\text{SO}_4) - 2\Lambda^{\circ}_m(\text{HCl})$$Λm∘​(BaSO4​)=Λm∘​(BaCl2​)+Λm∘​(H2​SO4​)−2Λm∘​(HCl)

Substituting the given values:

$$\Lambda^{\circ}_m(\text{BaSO}_4) = x_1 + x_2 - 2x_3$$Λm∘​(BaSO4​)=x1​+x2​−2x3​

For a sparingly soluble salt, the solubility $S$S (in mol L$^{-1}$−1) is related to its conductivity $\kappa$κ (in S cm$^{-1}$−1) and molar conductivity $\Lambda_m$Λm​ (in S cm$^2$2 mol$^{-1}$−1) by the relation:

$$\Lambda_m = \dfrac{\kappa \times 1000}{S}$$Λm​=Sκ×1000​

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

$$S = \dfrac{1000 x}{x_1 + x_2 - 2x_3}$$S=x1​+x2​−2x3​1000x​

The solubility product $K_{sp}$Ksp​ for $\text{BaSO}_4$BaSO4​ (which dissociates into $\text{Ba}^{2+}$Ba2+ and $\text{SO}_4^{2-}$SO42−​) is:

$$K_{sp} = S^2 = \left( \dfrac{1000 x}{x_1 + x_2 - 2x_3} \right)^2 = \dfrac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$$Ksp​=S2=(x1​+x2​−2x3​1000x​)2=(x1​+x2​−2x3​)2106x2​

Since $\alpha = 1$α=1, this expression is equivalent to $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$α2(x1​+x2​−2x3​)2106x2​.

Answer: $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$α2(x1​+x2​−2x3​)2106x2​