COMEDK20269 May 2026Evening ShiftChemistryIonic EquilibriumActual
At 30°C the solubility of PbI₂ salt in 0.2 M KI solution will be X, if the solubility product of PbI₂ at 30°C is 2.4 10⁻⁸ . Identify the value of X.
Options
- A6.0 10⁻⁷ M
- B2.4 10⁻⁸ M
- C3.0 10⁻⁸ M
- D4.8 10⁻⁷ M
Correct answer
A. 6.0 10⁻⁷ M
Step-by-step solution
The dissociation of PbI₂ is given by: PbI₂(s) Pb²⁺(aq) + 2I^-(aq) Let the solubility of PbI₂ in 0.2 M KI solution be s . The concentration of I^- ions from KI is 0.2 M. Since KI is a strong electrolyte, it completely dissociates. The total concentration of I^- ions is (2s + 0.2) M. Because K_ sp is very small, 2s can be neglected compared to 0.2 . [I^-] 0.2 M The expression for the solubility product is: K_ sp = [Pb²⁺][I^-]^2 Substituting the given values: 2.4 10⁻⁸ = s (0.2)^2 2.4 10⁻⁸ = s 0.04 s = 2.4 10⁻⁸ 0.04 =