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If K_sp of AgCl is 1.6×10⁻¹⁰ at 25 °C, solubility of AgCl in pure water is approximately
A1.6×10⁻¹⁰ M
B1.26×10⁻⁵ M
C1.6×10⁻⁵ M
D1.26×10⁻¹⁰ M
Answer & Solution
Correct answer: B. 1.26×10⁻⁵ M
1. Let solubility s mol/L; then [Ag⁺] = [Cl⁻] = s.
2. K_sp = s² = 1.6×10⁻¹⁰.
3. s = √(1.6×10⁻¹⁰) ≈ 1.26×10⁻⁵ M.
_Source: Maharashtra Balbharati Std XII Chemistry, Ch 3 "Ionic Equilibria" §3.10_
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