NEET 2020 Chemistry Question 118 Solution

Crude sodium chloride obtained by evaporation of seawater contains impurities such as sodium sulphate ($$\text{Na}_2\text{SO}_4$$), calcium sulphate ($$\text{CaSO}_4$$), calcium chloride ($$\text{CaCl}_2$$), and magnesium chloride ($$\text{MgCl}_2$$). The chlorides of calcium and magnesium are deliquescent (they absorb moisture from the air).

To obtain pure $$\text{NaCl}$$, crude salt is dissolved in a minimum amount of water and filtered to remove insoluble impurities. Hydrogen chloride gas ($$\text{HCl}$$) is then passed through the saturated solution. The complete ionization of $$\text{HCl}$$ provides a high concentration of chloride ions ($$\text{Cl}^-$$).

Because $$\text{NaCl}$$ has a substantially lower solubility product than the extremely soluble $$\text{MgCl}_2$$ and $$\text{CaCl}_2$$, the common ion effect causes the ionic product $$[\text{Na}^+][\text{Cl}^-]$$ to exceed the solubility product ($$K_{sp}$$) of $$\text{NaCl}$$: $$[\text{Na}^+][\text{Cl}^-] > K_{sp}(\text{NaCl})$$ Pure sodium chloride crystallises out.

Calcium and magnesium chlorides remain in the mother liquor because they are vastly more soluble than sodium chloride and their ionic products do not exceed their large $$K_{sp}$$ values under these conditions. Thus, ONLY NaCl crystallises.