Consider the following data : $\mathrm{BaSO}_{4}$ is sparingly soluble in water. If the conductivity of the saturated $\mathrm{BaSO}_{4}$ solution is $\times \mathrm{S} \mathrm{cm}^{-1}$ then the solubility product of $\mathrm{BaSO}_{4}$ can be given as. (Here $\wedge_{\mathrm{m}}=\wedge_{\mathrm{m}}^{\mathrm{o}}$ )

(A) $\frac{10^{6} \mathrm{x}^{2}}{\alpha^{2}(\mathrm{x}_{1}+\mathrm{x}_{2}-2 \mathrm{x}_{3})^{2}}$

(B) $\frac{x^{2}}{(x_{1}+x_{2}-2 x_{3})^{2}}$

(C) $\frac{\alpha^{2}(x_{1}+x_{2}-2 x_{3})^{2}}{10^{6} x^{2}}$

(D) $\frac{x^{2}}{(x_{1}+x_{2}+2 x_{3})^{2}}$

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Redox Reactions and Electrochemistry

Consider the following data :

BaSO_{4}

is sparingly soluble in water. If the conductivity of the saturated

BaSO_{4}

solution is

\times S cm^{- 1}

then the solubility product of

BaSO_{4}

can be given as.
(Here

\land_{m} = \land_{m}^{o}

)

(A) $\frac{1 0 ^{6} x ^{2}}{α ^{2} ( x _{1} + x _{2} - 2 x _{3} ) ^{2}}$ ✓ Correct
(B) $\frac{x ^{2}}{( x _{1} + x _{2} - 2 x _{3} ) ^{2}}$
(C) $\frac{α ^{2} ( x _{1} + x _{2} - 2 x _{3} ) ^{2}}{1 0 ^{6} x ^{2}}$
(D) $\frac{x ^{2}}{( x _{1} + x _{2} + 2 x _{3} ) ^{2}}$

Answer: (A)

Solution

For

BaS O_{4} (s) ⇌ B a^{2 +} + S O_{4}^{2 -}

the solubility product is

K_{s p} = S^{2}

, where

S

is the molar solubility. Because the saturated solution is extremely dilute its molar conductivity equals the limiting value, built from the tabulated data as

Λ_{m}^{\circ} = x_{1} + x_{2} - 2 x_{3}

. From

Λ_{m} = \frac{κ \times 1000}{S}

with the measured conductivity

κ = x S c m^{- 1}

S = \frac{1000 x}{α ( x _{1} + x _{2} - 2 x _{3} )} mol L^{- 1} .

Squaring,

K_{s p} = S^{2} = \frac{1 0 ^{6} x ^{2}}{α ^{2} ( x _{1} + x _{2} - 2 x _{3} ) ^{2}} .

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Redox Reactions and Electrochemistry

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