JEE Challenger
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ChemMedMain 2026 • 6th Apr - Shift 1

Solubility Product of Barium Sulfate from Electrolyte Conductivities

Consider the following data.

ElectrolyteΛm∘(S cm2 mol−1)BaCl2x1H2SO4x2HClx3\begin{array}{|c|c|} \hline \text{Electrolyte} & \Lambda^\circ_{\text{m}}(\text{S cm}^2\text{ mol}^{-1}) \\ \hline \text{BaCl}_2 & x_1 \\ \text{H}_2\text{SO}_4 & x_2 \\ \text{HCl} & x_3 \\ \hline \end{array}

BaSO4\text{BaSO}_4 is sparingly soluble in water. If the conductivity of the saturated BaSO4\text{BaSO}_4 solution is x S cm−1x\text{ S cm}^{-1} then the solubility product of BaSO4\text{BaSO}_4 can be given as (Here Λm=Λm∘\Lambda_{\text{m}} = \Lambda^\circ_{\text{m}})

Options

A

106x2α2(x1+x2−2x3)2\frac{10^6 x^2}{\alpha^2 \left(x_1 + x_2 - 2x_3\right)^2}

Correct
B

x2(x1+x2−2x3)2\frac{x^2}{\left(x_1 + x_2 - 2x_3\right)^2}

C

α2(x1+x2−2x3)2106x2\frac{\alpha^2 \left(x_1 + x_2 - 2x_3\right)^2}{10^6 x^2}

D

x2(x1+x2+2x3)2\frac{x^2}{\left(x_1 + x_2 + 2x_3\right)^2}

Topics & Concepts

ElectrochemistryElectrolysisSolubility Product

Step-by-Step Solution

To find the solubility product (KspK_{\text{sp}}) of BaSO4\text{BaSO}_4, we proceed step-by-step using Kohlrausch's Law of Independent Migration of Ions and basic principles of electrochemistry.

Step 1: Calculation of Limiting Molar Conductivity of BaSO4\text{BaSO}_4

According to Kohlrausch's Law, the molar conductivity at infinite dilution (Λm∘\Lambda^\circ_{\text{m}}) for an electrolyte is the sum of the individual ionic conductivities.

Given data:

  1. Λm∘(BaCl2)=λ∘(Ba2+)+2λ∘(Cl−)=x1\Lambda^\circ_{\text{m}}(\text{BaCl}_2) = \lambda^\circ(\text{Ba}^{2+}) + 2\lambda^\circ(\text{Cl}^-) = x_1
  2. Λm∘(H2SO4)=2λ∘(H+)+λ∘(SO42−)=x2\Lambda^\circ_{\text{m}}(\text{H}_2\text{SO}_4) = 2\lambda^\circ(\text{H}^+) + \lambda^\circ(\text{SO}_4^{2-}) = x_2
  3. Λm∘(HCl)=λ∘(H+)+λ∘(Cl−)=x3\Lambda^\circ_{\text{m}}(\text{HCl}) = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{Cl}^-) = x_3

We need to calculate Λm∘(BaSO4)\Lambda^\circ_{\text{m}}(\text{BaSO}_4): Λm∘(BaSO4)=λ∘(Ba2+)+λ∘(SO42−)\Lambda^\circ_{\text{m}}(\text{BaSO}_4) = \lambda^\circ(\text{Ba}^{2+}) + \lambda^\circ(\text{SO}_4^{2-})

By combining the given equations: Λm∘(BaSO4)=Λm∘(BaCl2)+Λm∘(H2SO4)−2Λm∘(HCl)\Lambda^\circ_{\text{m}}(\text{BaSO}_4) = \Lambda^\circ_{\text{m}}(\text{BaCl}_2) + \Lambda^\circ_{\text{m}}(\text{H}_2\text{SO}_4) - 2\Lambda^\circ_{\text{m}}(\text{HCl}) Λm∘(BaSO4)=x1+x2−2x3\Lambda^\circ_{\text{m}}(\text{BaSO}_4) = x_1 + x_2 - 2x_3

Step 2: Relation Between Molar Conductivity and Solubility

Let the solubility of BaSO4\text{BaSO}_4 in water be S mol L−1S \text{ mol L}^{-1}, and the conductivity of the saturated solution be κ=x S cm−1\kappa = x \text{ S cm}^{-1}.

The molar conductivity (Λm\Lambda_{\text{m}}) at concentration SS is given by: Λm=1000×κS=1000xS\Lambda_{\text{m}} = \frac{1000 \times \kappa}{S} = \frac{1000 x}{S}

The degree of dissociation (α\alpha) is defined as: α=ΛmΛm∘\alpha = \frac{\Lambda_{\text{m}}}{\Lambda^\circ_{\text{m}}}

Substituting the expressions for Λm\Lambda_{\text{m}} and Λm∘\Lambda^\circ_{\text{m}}: α=1000xSx1+x2−2x3=1000xS(x1+x2−2x3)\alpha = \frac{\frac{1000 x}{S}}{x_1 + x_2 - 2x_3} = \frac{1000 x}{S(x_1 + x_2 - 2x_3)}

Rearranging the above equation for the solubility SS: S=1000xα(x1+x2−2x3)S = \frac{1000 x}{\alpha (x_1 + x_2 - 2x_3)}

Step 3: Calculation of Solubility Product (KspK_{\text{sp}})

For the dissolution equilibrium of BaSO4\text{BaSO}_4: BaSO4(s)⇌Ba2+(aq)+SO42−(aq)\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)

The solubility product constant KspK_{\text{sp}} is given by: Ksp=[Ba2+][SO42−]=S×S=S2K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = S \times S = S^2

Substituting the value of SS into the expression for KspK_{\text{sp}}: Ksp=(1000xα(x1+x2−2x3))2=106x2α2(x1+x2−2x3)2K_{\text{sp}} = \left( \frac{1000 x}{\alpha (x_1 + x_2 - 2x_3)} \right)^2 = \frac{10^6 x^2}{\alpha^2 (x_1 + x_2 - 2x_3)^2}

Conclusion

The correct expression for the solubility product of BaSO4\text{BaSO}_4 is: 106x2α2(x1+x2−2x3)2\frac{10^6 x^2}{\alpha^2 (x_1 + x_2 - 2x_3)^2}

This corresponds to Option A.

Solubility Product of Barium Sulfate from Electrolyte Conductivities | Chemistry PYQ Solution - JEE Challenger