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If the solubility of AgCl is s mol/L in pure water, then its Ksp is:
AThe simple value $s$ (only one factor)
B$s^2$ ($[Ag^+] \times [Cl^-]$ = $s \times s$)
C$2s^2$ (incorrect stoichiometry coeff)
DJust $s^3$ (wrong dissociation count)
Answer & Solution
Correct answer: B. $s^2$ ($[Ag^+] \times [Cl^-]$ = $s \times s$)
AgCl(s) ⇌ Ag+(aq) + Cl-(aq). At saturation: [Ag+] = [Cl-] = s. Ksp = [Ag+][Cl-] = s × s = s². So solubility s = √Ksp.
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