Problem 53 - Chemistry The Central Science In Si Units

Question

(a) If the molar solubility of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(1.24 \times 10^{-3} \mathrm{~mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that \(1.1 \times 10^{-2} \mathrm{~g} \mathrm{SrF}_{2}\) dissolves per \(100 \mathrm{~mL}\) of aqueous solution at \(25^{\circ} \mathrm{C}\). Calculate the solubility product for \(\mathrm{SrF}_{2} .(\mathbf{c})\) The \(K_{s p}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10} .\) What is the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} ?\)

Step-by-Step Solution

Answer

(a) Ksp for CaF2 is 7.6 x 10^-9. (b) Ksp for SrF2 is 2.68 x 10^-10. (c) Molar solubility of Ba(IO3)2 is 1.14 x 10^-4 M.

Step 1: Write the Dissolution Reaction for Calcium Fluoride

The reaction for the dissolution of calcium fluoride, \( \text{CaF}_{2} \), in water is: \( \text{CaF}_{2} \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^{-} \). This shows that 1 mole of \( \text{CaF}_{2} \) produces 1 mole of \( \text{Ca}^{2+} \) ions and 2 moles of \( \text{F}^{-} \) ions in solution.

Step 2: Express the Solubility Product (Ksp) for Calcium Fluoride

The expression for the solubility product is derived from the dissolution reaction equilibrium: \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \]. Here, \([\text{Ca}^{2+}] = s\) and \([\text{F}^{-}] = 2s\), where \(s\) is the molar solubility.

Step 3: Calculate the Solubility Product (Ksp) for Calcium Fluoride

Given that the molar solubility \(s\) of \( \text{CaF}_{2} \) at 35°C is \(1.24 \times 10^{-3} \) mol/L, we substitute into the Ksp expression:\[ K_{sp} = (1.24 \times 10^{-3}) (2 \times 1.24 \times 10^{-3})^2 \]\[ K_{sp} = 1.24 \times 10^{-3} \times 4 \times (1.24 \times 10^{-3})^2 \]\[ K_{sp} = 7.6 \times 10^{-9} \].

Step 4: Convert Grams to Moles for Strontium Fluoride

The molar mass of \( \text{SrF}_{2} \) is calculated as \(87.62 + 2\times 18.998 \approx 125.62 \) g/mol. Therefore, the number of moles in 1.1 x 10^{-2} g of \( \text{SrF}_{2} \) is: \[ \frac{1.1 \times 10^{-2} \text{ g}}{125.62 \text{ g/mol}} \approx 8.756 \times 10^{-5} \text{ mol} \].

Step 5: Calculate Molarity and Solubility Product (Ksp) for Strontium Fluoride

Convert the moles to molarity since it is for 100 mL:\[ \text{Molarity} = \frac{8.756 \times 10^{-5} \text{ mol}}{0.1 \text{ L}} = 8.756 \times 10^{-4} \text{ M} \].The dissolution reaction: \( \text{SrF}_{2} \rightleftharpoons \text{Sr}^{2+} + 2\text{F}^{-} \) implies \([\text{SrF}^{2+}] = s \) and \([\text{F}^{-}] = 2s \). Thus,\[ K_{sp} = s(2s)^2 = 4s^3 = 2.68 \times 10^{-10} \].

Step 6: Use Ksp to Find Molar Solubility of Barium Iodate

From the dissolution reaction \( \text{Ba}(\text{IO}_3)_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{IO}_3^- \), we define \([\text{Ba}^{2+}] = s \) and \([\text{IO}_3^-] = 2s \). The expression for Ksp is:\[ K_{sp} = [\text{Ba}^{2+}][\text{IO}_3^-]^2 = s(2s)^2 = 4s^3 \].Given \( K_{sp} = 6.0 \times 10^{-10} \), solve for \( s \):\[ 4s^3 = 6.0 \times 10^{-10} \]\[ s^3 = \frac{6.0 \times 10^{-10}}{4} \approx 1.5 \times 10^{-10} \]\[ s = \sqrt[3]{1.5 \times 10^{-10}} \approx 1.14 \times 10^{-4} \text{ M} \].

Key Concepts

Molar Solubility

Molar solubility is a key concept in understanding how substances dissolve in water. It tells us the number of moles of a compound that can dissolve in a liter of solution. This is crucial because when a compound dissolves, it splits into its ionic components, each contributing to the solution's overall concentration. When we know the molar solubility, we can determine the concentration of ions in the solution. This is particularly useful for calculating the solubility product constant, or \(K_{sp}\). The \(K_{sp}\) provides insight into the solubility of sparingly soluble salts in water.
The higher the molar solubility, the more of the compound can dissolve, indicating a state of equilibrium between the solid and its ions in the solution. Understanding this helps us predict how substances will behave in various chemical environments.

Dissolution Reactions

In dissolution reactions, ionic compounds break apart in water to form ions. This process is vital for understanding how solids dissolve and how their concentrations affect chemical equilibria. For instance, when calcium fluoride (\( \text{CaF}_{2} \)) dissolves, the equation is \( \text{CaF}_{2} \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^{-} \). Such reactions illustrate the balance that exists between the undissolved solid and the dissolved ions in solution. The equation's double arrows signify that this is a dynamic equilibrium, where dissolution and precipitation occur at the same rate.
The \(K_{sp}\) expression, derived from these equilibrium concentrations, helps quantify how a particular compound behaves in solution. These reactions are foundational in chemical studies, providing insights into the solubility and behavior of compounds.

Calcium Fluoride

Calcium fluoride (\( \text{CaF}_{2} \)) is a sparingly soluble salt found in numerous chemical reactions. When it dissolves in water, it dissociates into calcium ions (\( \text{Ca}^{2+} \)) and fluoride ions (\( \text{F}^{-} \)). This reaction allows us to calculate its \(K_{sp}\), which signifies its solubility equilibrium.The dissolution of \( \text{CaF}_{2} \) involves the equation \( \text{CaF}_{2} \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^{-} \). Using the molar solubility data, we can substitute into the \(K_{sp}\) expression: \( K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \). Knowing the molarity of dissolved ions helps predict the conditions under which \( \text{CaF}_{2} \) forms or dissolves further.
This understanding is crucial in fields like geology and biology, where calcium minerals play vital roles.

Strontium Fluoride

Strontium fluoride (\( \text{SrF}_{2} \)) shares similarities with calcium fluoride in its dissolution process. It dissolves into strontium ions (\( \text{Sr}^{2+} \)) and fluoride ions (\( \text{F}^{-} \)). The equation \( \text{SrF}_{2} \rightleftharpoons \text{Sr}^{2+} + 2\text{F}^{-} \) indicates the stoichiometry of the dissolution.Calculating the \(K_{sp}\) for \( \text{SrF}_{2} \) involves understanding the concentration of ions as they form. After converting from grams to moles and calculating molarity, the solubility product gives insight into how much \( \text{SrF}_{2} \) can dissolve in a given amount of water at specific temperatures.
These calculations are important in applications such as materials science, where solidity and solubility impact the quality of materials.

Barium Iodate

Barium iodate (\( \text{Ba}(\text{IO}_3)_2 \)) is another sparingly soluble compound that provides a deeper understanding of solubility dynamics. Its dissolution reaction is \( \text{Ba}(\text{IO}_3)_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{IO}_3^- \).To determine its molar solubility, you use the \(K_{sp}\) expression \( K_{sp} = [\text{Ba}^{2+}][\text{IO}_3^-]^2 \), where each ion's concentration is considered. Knowing the \(K_{sp}\) of \( \text{Ba}(\text{IO}_3)_2 \) at a certain temperature allows for calculating the concentration of barium and iodate ions in solution.
This knowledge is significant in environmental chemistry and analytical applications, where predicting ion concentrations in aqueous solutions is necessary.