#KI solution

The solubility of PbI₂ in 0.05 M KI is 1.0 × 10⁻⁴ M . What is its Ksp ?

For PbI₂ Pb²⁺ + 2I- , [Pb²⁺] = 1.0 × 10⁻⁴ , [I-] = 0.05 + 2 × 1.0 × 10⁻⁴ ≈ 0.05 M . Ksp = [Pb²⁺][I-]² = (1.0 × 10⁻⁴) (0.05)² = 2.5 × 10⁻⁷ .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect