Practice question
Question
The Ksp of PbCl₂ is 1.7 × 10⁻⁵ . What is the solubility of PbCl₂ in a 0.1 M KCl solution?
Explanation
For PbCl₂ <=> Pb²⁺ + 2Cl- , Ksp = [Pb²⁺][Cl-]² = 1.7 × 10⁻⁵ . Let solubility = S , [Cl-] = 0.1 + 2S ≈ 0.1 , S (0.1)² = 1.7 × 10⁻⁵ , S = (1.7 × 10⁻⁵/0.01) = 1.7 × 10⁻³ M .
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