Let the molar solubilities of solid in pure water, , , and be represented as , , , and respectively. — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The solubility of a sparingly soluble salt is inversely proportional to the concentration of the common ion. In pure water ($S_1$), common ion = 0. In $0.01 \text{ M } CaCl_2$, $[Cl^-] = 0.02 \text{ M}$. In $0.01 \text{ M } NaCl$, $[Cl^-] = 0.01 \text{ M}$. In $0.05 \text{ M } AgNO_3$, $[Ag^+] = 0.05 \text{ M}$. The common ion concentrations dictating suppression are: Water (0) < $NaCl$ (0.01) < $CaCl_2$ (0.02) < $AgNO_3$ (0.05). Therefore, the solubilities follow the exact reverse order: $S_1 > S_3 > S_2 > S_4$.