Chemistry Q58 – JEE-MAIN 2026

Home Chemistry ElectrochemistryQuestion #395

Consider the following data.

\begin{array}{cc} Electrolyte Λ_{m}^{\circ} {(S cm}^{2} {mol}^{- 1}) \\ {BaCl}_{2} & x_{1} \\ H_{2} {SO}_{4} & x_{2} \\ HCl & x_{3} \end{array}

BaSO

_{4}

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

_{4}

solution is

x

S cm

^{- 1}

then the solubility product of BaSO

_{4}

can be given as (Here

Λ_{m}^{\circ} = Λ_{m}^{\circ}

)

Clue 1 of 3: Kohlrausch's Law

Determine the limiting molar conductivity of BaSO $_{4}$ using the given data for strong electrolytes.

🥷

Ninja Strategy•Formula Structure and Kohlrausch's Law

First, correctly apply Kohlrausch's law to find $Λ_{m}^{\circ} ({BaSO}_{4})$ . Then, recognize that $K_{s p}$ is proportional to $S^{2}$ , implying $x^{2}$ must be in the numerator, and the $1000^{2} = 10^{6}$ factor should also be in the numerator due to unit conversion.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate

Λ_{m}^{\circ} ({BaSO}_{4})

using Kohlrausch's Law✦ Active

According to Kohlrausch's Law of independent migration of ions, the limiting molar conductivity of BaSO $_{4}$ can be expressed as:

Λ_{m}^{\circ} ({BaSO}_{4}) = λ^{\circ} ({Ba}^{2 +}) + λ^{\circ} ({SO}_{4}^{2 -})

Using the given data for strong electrolytes, we can derive this value:

Λ_{m}^{\circ} ({BaSO}_{4}) = Λ_{m}^{\circ} ({BaCl}_{2}) + Λ_{m}^{\circ} (H_{2} {SO}_{4}) - 2 Λ_{m}^{\circ} (HCl)

Substituting the given values:

Λ_{m}^{\circ} ({BaSO}_{4}) = x_{1} + x_{2} - 2 x_{3} {S cm}^{2} {mol}^{- 1}

Step 2: Relate

Λ_{m}^{\circ} ({BaSO}_{4})

to its solubility (

S

)○ Expand

For a sparingly soluble salt like BaSO $_{4}$ , the molar conductivity of its saturated solution is approximately equal to its limiting molar conductivity. The relationship between molar conductivity ( $Λ_{m}^{\circ}$ ), conductivity ( $κ$ ), and solubility ( $S$ ) is:

Λ_{m}^{\circ} ({BaSO}_{4}) = \frac{κ \times 1000}{S}

Where $κ$ is in S cm $^{- 1}$ , $S$ is in mol L $^{- 1}$ , and $1000$ is a conversion factor from L to cm $^{3}$ . Given $κ = x$ S cm $^{- 1}$ , we can find $S$ :

S = \frac{x \times 1000}{Λ_{m}^{\circ} ({BaSO}_{4})} = \frac{x \times 1000}{(x_{1} + x_{2} - 2 x_{3})} {mol L}^{- 1}

💡 Teacher's Secret Hint

Ensure correct unit conversion factor (1000) is used for S cm $^{- 1}$ and mol L $^{- 1}$ .

Step 3: Calculate the solubility product (

K_{s p}

)○ Expand

For the dissociation of BaSO $_{4}$ :

{BaSO}_{4} (s) ⇌ {Ba}^{2 +} (a q) + {SO}_{4}^{2 -} (a q)

If $S$ is the solubility, then the equilibrium concentrations are $[{Ba}^{2 +}] = S$ and $[{SO}_{4}^{2 -}] = S$ . The solubility product $K_{s p}$ is:

K_{s p} = [{Ba}^{2 +}] [{SO}_{4}^{2 -}] = S^{2}

Substituting the expression for $S$ from Step 2:

K_{s p} = {(\frac{x \times 1000}{(x_{1} + x_{2} - 2 x_{3})})}^{2} = \frac{x^{2} \times 10^{6}}{(x_{1} + x_{2} - 2 x_{3})^{2}}

Comparing this result with the given options, Option 1 matches if the variable 'a' is considered to be 1 (likely a typo in the option).

💡 Teacher's Secret Hint

Remember that $K_{s p}$ is $S^{2}$ for a 1:1 electrolyte like BaSO $_{4}$ .

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Chemistry Question 58 – JEE-MAIN 2026

Clue 1 of 3: Kohlrausch's Law

🧠 Full Solution Path