The solubility of AgCl in pure water at 25 °C is 1.3 × 10⁻⁵ mol dm⁻³. What is its solubility product?
Correct answer B. 1.7 × 10⁻¹⁰ mol² dm⁻⁶
Explanation
For AgCl, [Ag⁺] = [Cl⁻] = s, so Ksp = s² = (1.3 × 10⁻⁵)² = 1.69 × 10⁻¹⁰, which is 1.7 × 10⁻¹⁰ mol² dm⁻⁶.
Answer checked How we write and check questions
Discussion