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Chemistry · Electrochemistry

JEE Main 2026 — 6 April, Morning Shift — Question 52

Consider the following data :

ElectrolyteΛm∘ (S cm2 mol−1)BaClX2x1HX2SOX4x2HClx3\begin{array}{|c|c|} \hline \text{Electrolyte} & \Lambda_m^\circ\ (\text{S cm}^2\text{ mol}^{-1}) \\ \hline \ce{BaCl2} & x_1 \\ \hline \ce{H2SO4} & x_2 \\ \hline \ce{HCl} & x_3 \\ \hline \end{array}

BaSO4\mathrm{BaSO}_{4} is sparingly soluble in water. If the conductivity of the saturated BaSO4\mathrm{BaSO}_{4} solution is ×Scm−1\times \mathrm{S} \mathrm{cm}^{-1} then the solubility product of BaSO4\mathrm{BaSO}_{4} can be given as. (Here ∧m=∧m∘\wedge_{\mathrm{m}}=\wedge_{\mathrm{m}}^{\circ} )

  1. Option A:

    106x2α2(x1+x2−2x3)2\frac{10^{6} \mathrm{x}^{2}}{\alpha^{2}\left(\mathrm{x}_{1}+\mathrm{x}_{2}-2 \mathrm{x}_{3}\right)^{2}}

    Correct
  2. Option B:

    x2(x1+x2−2x3)2\frac{x^{2}}{\left(x_{1}+x_{2}-2 x_{3}\right)^{2}}

  3. Option C:

    α2(x1+x2−2x3)2106x2\frac{\alpha^{2}\left(x_{1}+x_{2}-2 x_{3}\right)^{2}}{10^{6} x^{2}}

  4. Option D:

    x2(x1+x2+2x3)2\frac{x^{2}}{\left(x_{1}+x_{2}+2 x_{3}\right)^{2}}

Answer: A

Step-by-step solution

BaSO4(s)⇌Ba(aq)2++SO4(aq)2−\mathrm{BaSO}_{4}(\mathrm{s}) \rightleftharpoons \mathrm{Ba}^{2+}_{(\mathrm{aq})} + \mathrm{SO}_{4(\mathrm{aq})}^{2-}

Ksp=S2K_{\mathrm{sp}} = S^{2}

KBaSO4=x S cm−1K_{\mathrm{BaSO}_{4}} = x\,\mathrm{S\,cm}^{-1}

Λm∘ of BaSO4=(x1+x2−2x3)Λm=αΛm∘=k×1000SS=x×1000α(x1+x2−2x3)Ksp=S2=1α2[x×1000x1+x2−2x3]2\begin{aligned} \Lambda_{\mathrm{m}}^{\circ}\text{ of }\mathrm{BaSO}_{4} &= \left(x_{1}+x_{2}-2x_{3}\right) \\ \Lambda_{\mathrm{m}} &= \alpha\Lambda_{\mathrm{m}}^{\circ} = \frac{k\times1000}{S} \\ S &= \frac{x\times1000} {\alpha\left(x_{1}+x_{2}-2x_{3}\right)} \\ K_{\mathrm{sp}} &= S^{2} = \frac{1}{\alpha^{2}} \left[ \frac{x\times1000} {x_{1}+x_{2}-2x_{3}} \right]^{2} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Chemistry
Chapter
Electrochemistry
Topic
Conductance of Solutions and Kohlrausch's Law