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JIPMER2008ChemistryChemical Equilibrium

Solubility product of ( PbCl ₂ ) at 298 K is (1 10⁻⁶ ). At this temperature solubility of ( PbCl ₂ ) in ( mol / L ) is

Options

  1. A( (1 10⁻⁶ )^ 1 / 2 )
  2. B( (1 10⁻⁶ )^ 1 / 3 )
  3. C( (0.25 10⁻⁶ )^ 1 / 3 )
  4. D( (2.5 10⁻⁶ )^ 1 / 2 )

Correct answer

C. ( (0.25 10⁻⁶ )^ 1 / 3 )

Step-by-step solution

( PbCl ₂ s Pb ²⁺ + 2 s 2 Cl ⁻ ) ( aligned K_ S P & =(S)(2 S)^2 1 10⁻⁶ & =4 S^3 S & = ( 1 4 10⁻⁶ )^ 1 / 3 & = (0.25 10⁻⁶ )^ 1 / 3 aligned )

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