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NTA Abhyas NEET2020ChemistryIonic EquilibriumPractice

The molar solubility [ i n m o l L - 1 ] of a sparingly soluble salt M X 4 is ‘s’. The corresponding solubility product is Ksp.s is given in terms of Ksp by the relation

Options

  1. As = K s p 256 1 / 5
  2. Bs = 128 K s p 1 / 4
  3. Cs = K s p 128 1 / 6
  4. Ds = 256 K s p 1 / 5

Correct answer

A. s = K s p 256 1 / 5

Step-by-step solution

For the solute AxBy ⇌ xA + yB K s p = x x y y s x + y MX 4 ⇌ M 4 + + 4 X − x = 1, y = 4 K s p = 4 4 1 1 s 5 = 256 s 5 s = K s p 256 1 / 5

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