Question

HCl was passed through a solution of $CaCl_2$,$MgCl_2$ and $NaCl$. Which of the following compound(s) crystallise(s)?

• A. Both $MgCl_2$ and $CaCl_2$
• B. Only $NaCl$
• C. Only $MgCl_2$
• D. $NaCl$, $MgCl_2$ and $CaCl_2$

Answer 1

The Correct Option is B

To determine which compound(s) crystallize when hydrochloric acid (HCl) is passed through a solution containing \(CaCl_2\), \(MgCl_2\), and \(NaCl\), we need to consider the solubility of these chlorides and the common ion effect.

1. Understanding the Solubility Principle: In a solution with common ions, the compound with lower solubility in the presence of its common ion is likely to crystallize first.
2. Solubility of the Chlorides:
   • \(CaCl_2\) and \(MgCl_2\) are highly soluble in water.
   • \(NaCl\) is also highly soluble, but when HCl is added, the solubility product (K_sp) of NaCl is exceeded, leading to crystallization due to the common ion effect (increased Cl^- concentration).
3. Effect of HCl Addition:
   • When HCl is added to the solution, the concentration of Cl^- ions increases.
   • The common ion effect decreases the solubility of NaCl. This is because NaCl has a lower solubility product threshold compared to CaCl₂ and MgCl₂ in the presence of chloride ions.
4. Conclusion: As a result of these considerations, \(NaCl\) will crystallize out of the solution when HCl is introduced, due to the common ion effect reducing its solubility.

Thus, the correct answer is: Only \(NaCl\).