Question

The dissolution of calcium fluoride (CaF₂) in water can be represented by the equilibrium:

CaF₂​(s)⇌Ca₂+(aq)+2F−(aq)

The Ksp​ value for this equilibrium is given as 3.9×10⁻¹¹.

To determine in which solution the most calcium fluoride will dissolve, we need to consider the common ion effect. The common ion effect states that the presence of an ion in the solution can shift the equilibrium of a slightly soluble salt, decreasing its solubility. In this case, the common ion is the fluoride ion (F⁻).

Let's analyze the options:

A: Pure distilled water - No common ion is present.

B: 1 M HNO₃(aq) - Nitric acid (HNO₃​) does not contain fluoride ions, so it does not contribute to the common ion.

C: 1 M NaOH(aq) - Sodium hydroxide (NaOH) contains hydroxide ions (OH⁻), which do not share a common ion with calcium fluoride.

D: A saturated aqueous CaF₂ solution - This solution already contains fluoride ions, as it is saturated with calcium fluoride.

Answer 1

The solubility product of fluorite, CaF₂, can be calculated using the concentration of Ca²+ in a saturated solution and the stoichiometry of the dissolution equation. The solubility product expression is [Ca²+][F]² = 1.7 × 10-¹⁰.

Step-by-step explanation:

The dissolution of calcium fluoride (CaF₂) in water is represented by the equation:

CaF₂ (s) = Ca²+ (aq) + 2F¯(aq)

The concentration of Ca²+ in a saturated solution of CaF₂ is 2.15 x 10-4 M. The solubility product of fluorite, which is a measure of its solubility, can be calculated using the solubility product expression:

[Ca²+][F]² = 1.7 × 10-¹⁰

To calculate the solubility product, we need to find the concentration of fluoride ions ([F]). According to the stoichiometry of the dissolution equation, the fluoride ion molarity is equal to twice the calcium ion molarity:

[F] = (2 mol F / 1 mol Ca²+)

Substituting the concentration of Ca²+ (2.15 x 10-4 M) into the equation, we can calculate the concentration of fluoride ions:

[F] = (2) (2.15 × 10-4 M) = 4.30 × 10-4 M

Using this concentration, we can now calculate the solubility product:

[Ca²+][F]² = (2.15 x 10-4 M) * (4.30 x 10-4 M)² = 1.7 × 10-¹⁰