The Ksp of AgCl is 1.8 × 10⁻¹⁰ mol² dm⁻⁶. What is its molar solubility in 0.10 mol dm⁻³ NaCl solution?
Correct answer A. 1.8 × 10⁻⁹ mol dm⁻³
Explanation
Here [Cl⁻] ≈ 0.10 mol dm⁻³ from NaCl, so s = Ksp / [Cl⁻] = 1.8 × 10⁻¹⁰ / 0.10 = 1.8 × 10⁻⁹ mol dm⁻³, far less than in pure water.
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