Practice question

MCQ

Question

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

Options

Choose one · Correct answer highlighted

1.0 × 10⁻⁴ 5.0 × 10⁻⁶ 2.5 × 10⁻⁷ Correct 4.0 × 10⁻⁸

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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