MHT CET202521 Apr 2025Morning ShiftChemistryIonic EquilibriumActual
The solubility product of PbI ₂ is 1.08 10⁻⁷ . Calculate its solubility in mol~ dm ⁻³ at 298 K ?
Options
- A2.018 10⁻³
- B2.011 10⁻⁹
- C1.259 10⁻⁹
- D3.0 10⁻³
Correct answer
D. 3.0 10⁻³
Step-by-step solution
The dissociation equilibrium for PbI ₂ is established as: PbI ₂(s) Pb ²⁺(aq) + 2 I ⁻(aq) Define s as the molar solubility in mol~dm ⁻³ , which yields [ Pb ²⁺] = s and [ I ⁻] = 2s . The solubility product is K_ sp = [ Pb ²⁺][ I ⁻]^2 = s (2s)^2 = 4s^3 . Given K_ sp = 1.08 10⁻⁷ , solve for s : 4s^3 = 1.08 10⁻⁷ s^3 = 2.7 10⁻⁸ s = [3] 2.7 10⁻⁸ 3 10⁻³~ mol~dm ⁻³ The calculated solubility corresponds to option D .