A student prepares a saturated sodium chloride solution represented by the equilibrium reaction: NaCl(s) ⇌ Na+(aq)+Cl−(aq) When the student adds concentrated sodium hydroxide (NaOH) to the solution, a precipitate forms. What happens to the equilibrium of the sodium chloride dissolution?

Question

A. The equilibrium shifts to the right, increasing the concentration of Na⁺ and Cl⁻ ions in solution.
B. The equilibrium shifts in both directions, creating more NaCl and NaOH in solution.
C. The equilibrium remains unchanged, as NaOH does not affect NaCl dissolution.
D. The equilibrium shifts to the left, resulting in the precipitation of solid NaCl.

Answer 1

When concentrated sodium hydroxide (NaOH) is added to the saturated sodium chloride solution, it introduces hydroxide ions (OH⁻) to the system. The presence of NaOH can lead to the formation of solid sodium chloride because it affects the equilibrium of the dissolution reaction.

The relevant reaction here is:

\[ \text{NaCl}(s) \rightleftharpoons \text{Na}^+(aq) + \text{Cl}^-(aq) \]

When NaOH dissolves in the solution, it dissociates into Na⁺ and OH⁻ ions. The addition of NaOH does not change the concentration of Na⁺ significantly (since it also adds Na⁺ ions), but the presence of OH⁻ ions can shift the equilibrium of the sodium chloride dissolution. According to Le Chatelier's principle, if we add more of a product (in this case, OH⁻ ions that can lead to the formation of the solid), the equilibrium will shift to counteract that change.

However, it's important to note that OH⁻ ions can react with dissolved Na⁺ ions to form undissolved sodium hydroxide or react with the water in the solution. But more importantly, the formation of OH⁻ can drive the equilibrium of NaCl to the left, promoting the precipitation of solid NaCl.

Thus, the correct option is:

D. The equilibrium shifts to the left, resulting in the precipitation of solid NaCl.