STANDARD Chemistry Common Ion Effect MCQ Question
What will be the solubility of AgCl in 0.05 M NaCl aqueous solution if solubility product of AgCl is 1.5 × 10⁻¹⁰?
3 × 10⁻⁹ mol L⁻¹
0.05 mol L⁻¹
1.5 × 10⁻⁵ mol L⁻¹
3 × 10⁹ mol L⁻¹
Correct Answer
Detailed Explanation
In the presence of 0.05 M Cl⁻ from NaCl, the solubility of Ag⁺ is reduced. Kₛₚ = [Ag⁺][Cl⁻] = 1.5 × 10⁻¹⁰, solving gives [Ag⁺] = 3 × 10⁻⁹ mol L⁻¹.
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