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JEE Main20262 April 2026Morning ShiftChemistryIonic EquilibriumActual

The solubility product constants of Ag₂CrO₄ and AgBr are 32x and 4y respectively at 298 K. The value of ( molarity of Ag₂CrO₄ molarity of AgBr ) can be expressed as :

Options

  1. A2 [3] x y
  2. B2 x y
  3. Cx y
  4. D[3] x y

Correct answer

D. [3] x y

Step-by-step solution

Let the solubility of Ag₂CrO₄ be S₁ . For Ag₂CrO₄ 2Ag^+ + CrO₄²⁻ K_ sp = (2S₁)^2(S₁) = 4S₁^3 4S₁^3 = 32x S₁^3 = 8x S₁ = 2 [3] x Let the solubility of AgBr be S₂ . For AgBr Ag^+ + Br^- K_ sp = S₂^2 S₂^2 = 4y S₂ = 2 y The ratio of their molarities is: S₁ S₂ = 2 [3] x 2 y = [3] x y Answer: [3] x y

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