Problem 55 - Chemistry The Molecular Science

Question

The water solubility of strontium fluoride, \(\mathrm{SrF}_{2}\), is \(0.011 \mathrm{~g} / 100 . \mathrm{mL} .\) Calculate its solubility product constant. Assume that there are no reactions other than the \(K_{\mathrm{sp}}\) reaction.

Step-by-Step Solution

Answer

The solubility product constant \(K_{sp}\) for \(\text{SrF}_2\) is \(2.7 \times 10^{-9}.\)

Step 1: Identify Dissolution Reaction

The dissolution reaction for strontium fluoride in water can be written as: \(\text{SrF}_2(s) \rightleftharpoons \text{Sr}^{2+}(aq) + 2\text{F}^- (aq)\). This equation represents the dissolution of strontium fluoride into its ions.

Step 2: Determine Molar Solubility

The given water solubility is \(0.011 \text{ g} / 100 \text{ mL.}\) To calculate the molar solubility, convert grams to moles using the molar mass of \(\text{SrF}_2\). The molar mass of \(\text{SrF}_2\) is approximately \(87.62 + 2 \times 18.998 = 125.62 \text{ g/mol.}\) Thus, the molar solubility is \(\dfrac{0.011 \text{ g}}{125.62 \text{ g/mol}} \times 10\) (since 100 mL is 0.1 L) = \(8.76 \times 10^{-4} \text{ mol/L.}\)

Step 3: Calculate Ion Concentrations

Using the molar solubility \(s = 8.76 \times 10^{-4} \text{ mol/L}\), the concentration of \(\text{Sr}^{2+}\) is \(s\), and the concentration of \(\text{F}^-\) is \(2s\). Therefore, \([\text{Sr}^{2+}] = 8.76 \times 10^{-4} \text{ M}\) and \([\text{F}^-] = 2 \times 8.76 \times 10^{-4} \text{ M} = 1.75 \times 10^{-3} \text{ M.}\)

Step 4: Calculate Solubility Product Constant \(K_{sp}\)

The solubility product \(K_{sp}\) is calculated using the expression: \(K_{sp} = [\text{Sr}^{2+}][\text{F}^-]^2\). Substitute the ion concentrations into this expression: \(K_{sp} = (8.76 \times 10^{-4}) (1.75 \times 10^{-3})^2\). After calculating, \(K_{sp} = 2.7 \times 10^{-9}.\)

Key Concepts

Molar Solubility

Molar solubility is a term used to describe the concentration of a solute in a solution at equilibrium, expressed in moles per liter. For strontium fluoride (\(\mathrm{SrF}_{2}\)), the exercise states that its solubility is initially given in grams per 100 mL. To find the molar solubility, you first need to convert these grams into moles. This requires the use of the molar mass of strontium fluoride.

• The molar mass of \(\mathrm{SrF}_2\) is calculated by adding the atomic masses: Sr (87.62 g/mol) and two F atoms (each 18.998 g/mol), resulting in 125.62 g/mol.
• The given solubility, 0.011 g/100 mL, must be converted to moles per liter (mol/L). Since 100 mL equals 0.1 L, the conversion results in a molar solubility of approximately \(8.76 \times 10^{-4} \text{ mol/L}\).

This molar solubility allows us to determine the concentrations of the ions in solution when \(\text{SrF}_2\) dissolves.

Strontium Fluoride

Strontium fluoride (\(\mathrm{SrF}_2\)) is an ionic compound made of strontium and fluoride ions. In solid form, it exists as a crystalline lattice. When you dissolve strontium fluoride in water, it separates into its constituent ions.

• This process is called dissolution, and for \(\mathrm{SrF}_2\), it leads to the formation of strontium ions (\(\mathrm{Sr}^{2+}\)) and fluoride ions (\(\mathrm{F}^{-}\)).
• Understanding this behavior is crucial for calculating its solubility product constant (\(K_{sp}\)), which measures the product of the molar concentrations of the dissociated ions when the solution is at maximum saturation.

Strontium fluoride is particularly significant in applications like optics and ceramics, where its solubility plays an essential role in determining its utility.

Ion Concentrations

When strontium fluoride dissolves in water, it dissociates into ions. These ions are strontium (\(\text{Sr}^{2+}\)) and fluoride (\(\text{F}^-\)).

• The concentration of strontium ions is the same as the molar solubility, because each formula unit of \(\mathrm{SrF}_2\) produces one \(\text{Sr}^{2+}\) ion.
• For fluoride ions, however, the concentration is twice that of the molar solubility, as each unit of \(\mathrm{SrF}_2\) dissociates to give two fluoride ions.
• In this case, with a molar solubility of \(8.76 \times 10^{-4} \text{ mol/L}\), the concentrations are \([\text{Sr}^{2+}] = 8.76 \times 10^{-4} \text{ M}\) and \([\text{F}^-] = 1.75 \times 10^{-3} \text{ M}\).

These concentrations are essential for calculating the solubility product constant, enabling us to predict how much more of the salt can dissolve in the solution.

Dissolution Reaction

The dissolution of strontium fluoride in water is governed by a chemical reaction. This reaction describes how the solid compound \(\mathrm{SrF}_2\) separates into its ions. The dissolution reaction is written as:\(\text{SrF}_2(s) \rightleftharpoons \text{Sr}^{2+}(aq) + 2\text{F}^- (aq)\).

• This equation shows that one formula unit of \(\mathrm{SrF}_2\) yields one strontium ion and two fluoride ions.
• Such a reaction is dynamic and reversible, meaning the ions can recombine to form \(\mathrm{SrF}_2\) solid again when the solution becomes saturated.

Understanding this equilibrium reaction is crucial as it allows us to calculate the solubility product constant (\(K_{sp}\)), which quantifies the solubility of \(\mathrm{SrF}_2\) and guides predictions about its behavior in different conditions.