The solubility product constant (Ksp) of silver chloride (AgCl) in pure water at 25 °C is 1.8 x 10^-10. What is the molar solubility of AgCl in pure water?
Answer and explanation
Correct answer: B. 1.34 x 10^-5 M
The dissolution equilibrium is AgCl(s) <=> Ag+(aq) + Cl-(aq). If s is molar solubility, [Ag+] = s and [Cl-] = s. Ksp = [Ag+][Cl-] = s^2. Therefore, s = sqrt(Ksp) = sqrt(1.8 x 10^-10) ≈ 1.34 x 10^-5 M.
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