Answer:
Option b. Sodium chloride.
Explanation:
Consider the dissociation equilibrium of [tex]\rm CaCl_2 \; (s)[/tex] in water:
[tex]\rm CaCl_2 \; (s) \rightleftharpoons Ca^{2+} \; (aq) + 2\; Cl^{-}\; (aq)[/tex].
By the Le Chatelier's Principle, increasing the concentration of either product will shift the equilibrium to the left. Some [tex]\rm CaCl_2 \; (s)[/tex] that was initially dissolved will precipitate out of the solution. This effect is called the common-ion effect.
For this [tex]\rm CaCl_2 \; (s)[/tex] solution, the products of dissociation are
- [tex]\rm Ca^{2+}[/tex] ions, and
- [tex]\rm Cl^{-}[/tex] ions.
Adding either to the solution will trigger the common-ion effect and reduce the solubility of [tex]\rm CaCl_2 \; (s)[/tex].
In the two choices,
- Sodium chloride [tex]\rm NaCl[/tex] will add [tex]\rm Na^{+}[/tex] and [tex]\rm Cl^{-}[/tex] ions to the solution.
- Sodium fluoride [tex]\rm NaF[/tex] will add [tex]\rm Na^{+}[/tex] and [tex]\rm F^{-}[/tex] ions to the solution.
[tex]\rm NaCl[/tex] contains [tex]\rm Cl^{-}[/tex] ions. It is capable of triggering the common-ion effect. However [tex]\rm NaF[/tex] contains neither [tex]\rm Ca^{2+}[/tex] ions nor [tex]\rm Cl^{-}[/tex] ions. It will not trigger the common-ion effect.
