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