The Ksp of PbCl₂ is 1.6 × 10⁻⁵ . What is its solubility in pure water in mol/L?
For PbCl₂ Pb²⁺ + 2Cl⁻ , Ksp = [Pb²⁺][Cl⁻]² = S (2S)² = 4S³ = 1.6 × 10⁻⁵ . Solving, S³ = 4 × 10⁻⁶ , S = (4 × 10⁻⁶)¹/³ ≈ 1.6 × 10⁻² .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
