Question

Consider a beaker containing a saturated solution of \(\mathrm{CaF}_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s)\). Solid \(\mathrm{CaCl}_{2}\) is then added to the solution. (a) Will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(\mathbf{F}^{-}\) ions in solution increase or decrease?

Short answer

(a) The amount of solid CaF₂ will increase due to the common ion effect and equilibrium shift. (b) The concentration of Ca²⁺ ions will remain constant as the equilibrium counteracts the additional ions from CaCl₂. (c) The concentration of F⁻ ions will decrease due to the equilibrium shift favoring the formation of solid CaF₂.

Step-by-step solution

Identify the solubility product constant expression for CaF₂

CaF₂ has the formula unit: Ca²⁺ + 2F⁻. Its solubility product constant expression is given by: \[K_{sp} = [Ca^{2+}][F^{-}]^2\]

Write the equilibrium expressions for the addition of CaCl₂

When CaCl₂ is added to the solution, it will dissociate into its ions: Ca²⁺ and 2Cl⁻. We will not consider Cl⁻ ions in our analysis, as they don't affect the CaF₂ equilibrium. The equilibrium expressions for the addition of CaCl₂ are: Before CaCl₂ addition: \[K_{sp} = [Ca^{2+}][F^{-}]^2\] After CaCl₂ addition: \[K_{sp} = ([Ca^{2+}] + x)[F^{-}]^2\] where x is the increase in concentration of Ca²⁺ ions due to the addition of CaCl₂.

Consider the common ion effect on the equilibrium

Since adding CaCl₂ increases the concentration of the common ion Ca²⁺, the equilibrium will shift to the left, favoring the formation of solid CaF₂ according to Le Châtelier's principle. This means the amount of solid CaF₂ will increase, the concentration of Ca²⁺ ions in solution will remain constant, and the concentration of F⁻ ions will decrease.

Answer the exercise questions

(a) The amount of solid CaF₂ at the bottom of the beaker will increase due to the common ion effect and the shift in the equilibrium. (b) The concentration of Ca²⁺ ions in the solution will remain constant because the equilibrium will shift to counteract the addition of more Ca²⁺ ions from CaCl₂. (c) The concentration of F⁻ ions in the solution will decrease, as the equilibrium shift to the left due to the common ion effect will favor the formation of solid CaF₂.

Key concepts

Common Ion Effect

When we talk about the common ion effect, we are delving into a principle that is crucial in understanding solubility and equilibrium in chemistry. Imagine a solution where a compound is dissolved, and it's sitting at equilibrium. Now, let's say we add another compound that shares a common ion with the first one. This additional amount of the ion shifts the equilibrium position.
In the case of calcium fluoride (\(\mathrm{CaF}_{2}\)) in our beaker, when \(\mathrm{CaCl}_{2}\) is added, it dissociates into calcium ions (\(\mathrm{Ca}^{2+}\)) and chloride ions (\(\mathrm{Cl}^{-}\)). These extra calcium ions are the common ion because \(\mathrm{CaF}_{2}\) also dissolves to give calcium ions.

• This presence of more \(\mathrm{Ca}^{2+}\) shifts the equilibrium because there are more calcium ions than before.
• According to the common ion effect, the system will want to counteract this change.
• As a result, it will favor the reverse reaction — leading to more solid \(\mathrm{CaF}_{2}\) forming.

In simpler terms, adding more of the same type of ion makes the solution more likely to revert back to forming the solid, decreasing its solubility.

Le Châtelier's Principle

Le Châtelier's Principle is an important concept in predicting how changes in conditions can affect chemical equilibria. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts itself to counteract the change and tries to re-establish equilibrium.
In our example with calcium fluoride (\(\mathrm{CaF}_{2}\)), the disturbance is caused by adding more \(\mathrm{Ca}^{2+}\) ions through the addition of \(\mathrm{CaCl}_{2}\). Here's how the system responds:

• Le Châtelier's principle predicts that the system will shift to minimize the effect of the added \(\mathrm{Ca}^{2+}\) ions.
• Therefore, the reaction where solid \(\mathrm{CaF}_{2}\) reforms from \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\) is favored.
• This results in increased precipitation of \(\mathrm{CaF}_{2}\) which reduces the concentration of fluoride ions (\(\mathrm{F}^{-}\)) in the solution.

This principle shows how increases in ion concentrations can make a solution less soluble by encouraging the backward reaction, solidifying more of the dissolved substance.

Calcium Fluoride Equilibrium

Calcium fluoride equilibrium refers to the balance state of \(\mathrm{CaF}_{2}\) in an aqueous solution, where its dissolved ions coexist with undissolved solid. At this point, \(\mathrm{CaF}_{2}\) has established a saturated solution, denoted by the equation:\[\mathrm{CaF}_{2}(s) \leftrightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^{-}(aq)\]The solubility product (\(K_{sp}\)) for \(\mathrm{CaF}_{2}\) is expressed as:\[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\]Here's what happens when \(\mathrm{CaCl}_{2}\) dissolves and increases the amount of \(\mathrm{Ca}^{2+}\):

• The equation strives to maintain its established \(K_{sp}\) value.
• Extra \(\mathrm{Ca}^{2+}\) causes a shift, leading to the precipitation of more \(\mathrm{CaF}_{2}\) solid.
• This reduces the fluoride ion concentration in solution, as part of the equilibrium adjustment.

Calcium fluoride equilibrium is a delicate balance and provides a great example of how chemical systems respond and shift under various conditions, such as changes in ion concentrations.