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

Option b. Sodium chloride.

Step-by-step 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.