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JEE MainChemistryChemical Equilibrium

A weak electrolyte AB₂ (equilibrium constant K ) is dissolved in pure water to a concentration C , and its degree of dissociation is found to be ₁ . In a separate experiment, the same weak electrolyte AB₂ is dissolved to a concentration C in a solution already containing a strong electrolyte CB₂ at concentration C . The new degree of dissociation of AB₂ is ₂ . Assuming ₁ 1 and ₂ 1 , what is the ratio ₁ ₂ ?

Options

  1. A( 4C^2 K )^ 1 3
  2. B( K 4C^2 )^ 2 3
  3. C( 4C^2 K )^ 2 3
  4. D( C^2 K )^ 2 3

Correct answer

C. ( 4C^2 K )^ 2 3

Step-by-step solution

First, consider the dissociation of AB₂ in pure water: AB₂ A²⁺ + 2B^- The equilibrium concentrations are [AB₂] = C(1- ₁) C , [A²⁺] = C ₁ , and [B^-] = 2C ₁ . The equilibrium constant K is: K = [A²⁺][B^-]^2 [AB₂] = (C ₁)(2C ₁)^2 C = 4C^2 ₁^3 ₁ = ( K 4C^2 )^ 1 3 Next, consider the dissociation of AB₂ in the presence of the strong electrolyte CB₂ . CB₂ dissociates completely: CB₂ C²⁺ + 2B^- . Since the concentration of CB₂ is C , it provides an initial [B^-] = 2C . For the weak electrolyte AB₂ in this mixed solution:

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