When HCl gas is passed through saturated solution of common salt, pure NaCl precipitated, because :

Question

Accepted Answer

When HCl gas is passed through a saturated solution of common salt (NaCl), pure NaCl precipitates due to the following reason:

### Explanation for Option 3:
**The ionic product $[Na^{+}][Cl^{-}]$ exceeds the solubility product of NaCl.**

When HCl gas is introduced into the saturated NaCl solution, it dissociates into H⁺ and Cl⁻ ions. The addition of Cl⁻ ions from HCl increases the concentration of chloride ions in the solution. As a result, the ionic product $[Na^{+}][Cl^{-}]$ becomes greater than the solubility product (Ksp) of NaCl. When the ionic product exceeds the Ksp, the solution becomes supersaturated with respect to NaCl, leading to the precipitation of pure NaCl to re-establish equilibrium.

### Explanation for why other options are incorrect:

1. **The impurities dissolve in HCl.**
   - This option is incorrect because the precipitation of NaCl is not primarily due to the dissolution of impurities. The process is driven by the increase in chloride ion concentration, which affects the ionic product relative to the solubility product of NaCl.

2. **HCl is slightly soluble in water.**
   - While HCl is indeed soluble in water, this fact alone does not explain the precipitation of NaCl. The key factor is the increase in chloride ion concentration, not the solubility of HCl itself.

4. **The solubility product of NaCl is lowered by Cl from aq. HCl.**
   - This option is incorrect because the solubility product (Ksp) of a compound is a constant at a given temperature and is not affected by the presence of additional ions. The precipitation occurs because the ionic product exceeds the Ksp, not because the Ksp itself is lowered.

In conclusion, the correct explanation for the precipitation of NaCl when HCl gas is passed through a saturated solution of common salt is that the ionic product $[Na^{+}][Cl^{-}]$ exceeds the solubility product of NaCl, as stated in Option 3.