Question

Adding which of the following substances will decrease the solubility of calcium chloride in water: CaCl2(s) Ca2+(aq) + 2 Cl–(aq)?
a. Sodium chloride
b. Sodium fluoride
c. No right choice.

Answer 1

Answer:

Option b. Sodium chloride.

Explanation:

Consider the dissociation equilibrium of $\rm CaCl_2 \; (s)$ in water:

$\rm CaCl_2 \; (s) \rightleftharpoons Ca^{2+} \; (aq) + 2\; Cl^{-}\; (aq)$.

By the Le Chatelier's Principle, increasing the concentration of either product will shift the equilibrium to the left. Some $\rm CaCl_2 \; (s)$ that was initially dissolved will precipitate out of the solution. This effect is called the common-ion effect.

For this $\rm CaCl_2 \; (s)$ solution, the products of dissociation are

• $\rm Ca^{2+}$ ions, and
• $\rm Cl^{-}$ ions.

Adding either to the solution will trigger the common-ion effect and reduce the solubility of $\rm CaCl_2 \; (s)$.

In the two choices,

• Sodium chloride $\rm NaCl$ will add $\rm Na^{+}$ and $\rm Cl^{-}$ ions to the solution.
• Sodium fluoride $\rm NaF$ will add $\rm Na^{+}$ and $\rm F^{-}$ ions to the solution.

$\rm NaCl$ contains $\rm Cl^{-}$ ions. It is capable of triggering the common-ion effect. However $\rm NaF$ contains neither $\rm Ca^{2+}$ ions nor $\rm Cl^{-}$ ions. It will not trigger the common-ion effect.