Practice question
Question
The solubility product of AgCl is 1.8 × 10⁻¹⁰ . What is the solubility of AgCl in pure water in mol/L?
Explanation
Ksp = [Ag+][Cl-] = S² = 1.8 × 10⁻¹⁰ , so S = sqrt1.8 × 10⁻¹⁰ = 1.34 × 10⁻⁵ mol/L .
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