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Consider the following data. Electrolyte ^ _m (S cm ^2 mol ⁻¹ ) BaCl ₂ x₁ H ₂ SO ₄ x₂ HCl x₃ 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 _m = ^ _m )

Options

  1. A10^6 x^2 ^2(x₁ + x₂ - 2x₃)^2
  2. Bx^2 (x₁ + x₂ - 2x₃)^2
  3. C^2(x₁ + x₂ - 2x₃)^2 10^6 x^2
  4. Dx^2 (x₁ + x₂ + 2x₃)^2

Correct answer

A. 10^6 x^2 ^2(x₁ + x₂ - 2x₃)^2

Step-by-step solution

Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of BaSO ₄ is given by: ^ _m( BaSO ₄) = ^ _m( BaCl ₂) + ^ _m( H ₂ SO ₄) - 2 ^ _m( HCl ) Substituting the given values: ^ _m( BaSO ₄) = x₁ + x₂ - 2x₃ For a sparingly soluble salt, the solubility S (in mol L ⁻¹ ) is related to its conductivity (in S cm ⁻¹ ) and molar conductivity _m (in S cm ^2 mol ⁻¹ ) by the relation: _m = 1000 S Given that = x and _m = ^ _m (which implies complete dissociation, meaning the degree of dissociatio

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