The common ion effect occurs when an ion already present in solution suppresses the dissociation of a sparingly soluble salt. AgCl dissociates as AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq), and K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}. In 0.1 M NaCl, [Cl^-] \approx 0.1 M due to the common ion (much larger than the contribution from AgCl itself). Let the solubility of AgCl be s mol/L. Then [Ag^+] = s and [Cl^-] \approx 0.1 M. Thus K_{sp} = s \times 0.1 = 1.8 \times 10^{-10}, giving s = 1.8 \times 10^{-10} / 0.1 = 1.8 \times 10^{-9} mol L^{-1}. Option 1.34 \times 10^{-5} is the solubility in pure water (\sqrt{K_{sp}}), ignoring the common ion completely. Option 1.8 \times 10^{-10} is K_{sp} itself, not the solubility. Option 1.8 \times 10^{-11} would come from dividing by 10 instead of 0.1, an arithmetic error. This is a classic NCERT ionic equilibrium application demonstrating Le Chatelier's principle in the context of solubility. Plausibility check: solubility in NaCl (1.8 \times 10^{-9}) is about 10,000 times less than in pure water (1.34 \times 10^{-5}), confirming significant suppression due to the common ion effect.