StreakPeaked· Practice

ExamsJEE MainChemistry

The molar solubility (in mol L⁻¹) of a sparingly soluble salt MX4 is 's'. The corresponding solubility product is Ksp. 's' is given in term of Ksp by the relation

  1. s = (256 Ksp)^(1/5)
  2. s = (128 Ksp)^(1/4)
  3. s = (Ksp/128)^(1/4)
  4. s = (Ksp/256)^(1/5)

Correct answer: s = (Ksp/256)^(1/5)

Solution

The correct option relates the molar solubility 's' of the salt MX4 to its solubility product Ksp by recognizing that the dissolution of MX4 produces one mole of M and four moles of X, leading to the expression Ksp = [M][X]⁴. By substituting the concentrations in terms of 's', we derive the relationship s = (Ksp/256)^(1/5), which accurately reflects the stoichiometry of the dissolution process.

Related JEE Main Chemistry questions

⚔️ Practice JEE Main Chemistry free + battle 1v1 →