Question

Can someone explain what each of the answer choices will do to the equilibrium? Shiftwise?

Answer 1

Answer:

$\boxed{\text{(B)}}$

Explanation:

CaF₂(s) ⇌ Ca²⁺(aq) + F⁻(aq); ΔH > 0

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Let's consider each of the stresses in turn.

(A) Evaporating some of the water

The concentrations of the ions will increase, so calcium fluoride will precipitate out to remove the stress (the Ca²⁺ and F⁻ ions). The position of equilibrium does not shift, and [Ca²⁺] stays the same.

(B) Adding HNO₃

HF is a weak acid, so F⁻ is a relatively strong base. The added HNO₃ will convert the F⁻ ions to HF, removing them from solution. More CaF₂ will dissolve to replace the F⁻ ions, and this will add more Ca²⁺ ions as well. The position of equilibrium will shift to the right, and [Ca²⁺] will increase.

(C) Adding NaNO₃(aq)

There is no common ion, so NaNO₃ will have no effect. The added water will dilute the solution and decrease the concentrations of the ions. However, more CaF₂ will dissolve to increase the concentrations. The position of equilibrium does not shift, and [Ca²⁺] stays the same.

(D) Adding NaF

This is the common ion effect. F⁻ is the common ion. The added NaF will dissolve, increasing the concentration of F⁻ ions. More CaF₂ will precipitate to remove the added F⁻ ions, but it removes Ca²⁺ ions at the same time. The position of equilibrium shifts to the left, and [Ca²⁺] decreases.