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NEET2018ChemistryIonic EquilibriumActual

The solubility of BaSO 4 in water is 2 .42 × 10 - 3   gL - 1 at 298   K . The value of its solubility product K sp will be (Given- the molar mass of BaSO 4 = 233   g   mol - 1 )

Options

  1. A1 .08 × 10 - 14   mol 2   L - 2
  2. B1 .08 × 10 - 12   mol 2   L - 2
  3. C1 .08 × 10 - 10   mol 2   L - 2
  4. D1 .08 × 10 - 8   mol 2   L - 2

Correct answer

C. 1 .08 × 10 - 10   mol 2   L - 2

Step-by-step solution

Solubility of BaSO 4 = 2 .42 × 10 - 3   gL - 1 ∴   s = 2 .42 × 10 - 3 233 = 1 .038 × 10 - 5   mol   L - 1 K sp = s 2 = 1 .038 × 10 - 5 2 = 1 .08 × 10 - 10   mol 2   L - 2

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