Problem 72 - Chemistry Matter and Change

Question

\(K_{\mathrm{sp}}\) for scandium fluoride \(\left(\mathrm{ScF}_{3}\right)\) at 298 \(\mathrm{K}\) is \(4.2 \times 10^{-18}\) Write the chemical equation for the solubility equilibrium of scandium fluoride in water. What concentration of \(\mathrm{Sc}^{3+}\) ions is required to cause a precipitate to form if the fluoride-ion concentration is 0.076 \(\mathrm{M}\) ?

Step-by-Step Solution

Answer

A concentration of \( 1.26 \times 10^{-16} \, M \) of \( Sc^{3+} \) is required to cause a precipitate to form when the fluoride-ion concentration is 0.076 M.

Step 1: 1. Write the Solubility Equilibrium equation for Scandium Fluoride

Given the compound scandium fluoride (\(ScF_{3}\)), it dissociates in water as follows: \[ ScF_{3(s)} \rightarrow Sc^{3+}_{(aq)} + 3F^-_{(aq)} \]

Step 2: 2. Write the Expression for the Solubility Product Constant

The expression for the solubility product constant \( K_{sp} \) is derived from the concentrations of the ions at equilibrium. Considering the stoichiometry of the reaction, the expression for \( K_{sp} \) for the reaction is: \[ K_{sp} = [Sc^{3+}][F^-]^3 \] Or in words: The solubility product constant is equal to the concentration of Scandium ions times the concentration of Fluoride ions cubed.

Step 3: 3. Calculation of Scandium Ion Concentration

We are asked to find the concentration of \(Sc^{3+}\) ions that will cause a precipitate to form given that the fluoride-ion concentration is 0.076 M. Precipitation would occur when the ionic product exceeds the \( K_{sp} \) value. Thus we can rearrange the solubility product equation to solve for the \(Sc^{3+}\) concentration: \[ [Sc^{3+}] = \frac{K_{sp}}{[F^-]^3} \] Substituting \( K_{sp} = 4.2 \times 10^{-18} \) and \( [F^-] = 0.076 \, M \) into the equation gives: \[ [Sc^{3+}] = \frac{4.2 \times 10^{-18}}{(0.076)^3} = 1.26 \times 10^{-16} \, M \] Therefore, a concentration of \( 1.26 \times 10^{-16} \, M \) of \( Sc^{3+} \) is required to cause a precipitate to form.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, commonly referred to as \( K_{sp} \), is a critical concept in chemistry, particularly when it comes to solubility equilibrium. This constant provides insight into the extent to which a solute will dissolve in a solution.
The \( K_{sp} \) is derived from the equilibrium concentrations of the ions in a saturated solution. It is specific to a certain solute at a given temperature and is calculated by multiplying the concentrations of the respective ions, each raised to the power of its coefficient in the balanced equation. For scandium fluoride \((\text{ScF}_3)\), the dissociation in water creates scandium and fluoride ions according to the equation:

• \( \text{ScF}_{3(s)} \rightarrow \text{Sc}^{3+}_{(aq)} + 3\text{F}^{-}_{(aq)} \)

The solubility product expression is then:

• \( K_{sp} = [\text{Sc}^{3+}][\text{F}^-]^3 \)

Understanding \( K_{sp} \) allows one to determine the maximum concentration of ions in solution before precipitation occurs, providing critical insight into the solubility behavior of various compounds.

Scandium Fluoride

Scandium fluoride, ScF\(_3\), is a chemical compound consisting of one scandium ion and three fluoride ions. It exhibits interesting solubility properties, particularly within aqueous solutions.
As a salt, scandium fluoride dissociates into its respective ions—\( \text{Sc}^{3+} \) and \( \text{F}^- \)—when dissolved in water. The specific dissociation reaction can be represented as:

• \( \text{ScF}_{3(s)} \rightarrow \text{Sc}^{3+}_{(aq)} + 3\text{F}^{-}_{(aq)}\)

Given its low solubility and small \( K_{sp} \) value of \( 4.2 \times 10^{-18} \), scandium fluoride remains mostly undissolved in solution, forming a precipitate under most conditions. The low solubility is what makes \( ScF_3 \) a good choice for exercises involving solubility equilibrium and \( K_{sp} \) calculations.
By understanding the behavior of \( ScF_3 \) in solution, students can gain an appreciation for how ionic compounds act in different conditions and the role of solubility product in predicting their behavior.

Precipitation Reaction

A precipitation reaction involves the formation of a solid in a solution during a chemical reaction. This solid, known as a precipitate, occurs when the ionic product of the reactants' solubility exceeds the \( K_{sp} \) of the compound formed.
To illustrate using the provided example, when scandium fluoride is added to water, its dissociation yields scandium ions and fluoride ions:

• \( \text{ScF}_{3(s)} \rightarrow \text{Sc}^{3+}_{(aq)} + 3\text{F}^-_{(aq)} \)

If the concentration of these ions in the solution becomes higher than its \( K_{sp} \), precipitation proceeds, forming more \( \text{ScF}_3 \) solid. The threshold concentration of \( \text{Sc}^{3+} \) ions required to initiate this process, given a fixed fluoride-ion concentration of 0.076 M, can be calculated by rearranging the \( K_{sp} \) equation:

• \( [\text{Sc}^{3+}] = \frac{K_{sp}}{[\text{F}^-]^3} \)
• For the given conditions: \( [\text{Sc}^{3+}] = \frac{4.2 \times 10^{-18}}{(0.076)^3} = 1.26 \times 10^{-16} \, M \)

The understanding of precipitation reactions is pivotal in fields like analytical chemistry and geochemistry, where control over the precipitation process is often required.