Problem 49 - Chemistry: The Central Science

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 \(1.1 \times 10^{-2} \mathrm{~g}\) of \(\mathrm{Sr} \mathrm{F}_{2}\) dissolves per \(100 \mathrm{~mL}\) of aqueous solution at \(25^{\circ} \mathrm{C}\). Calculate the solubility product for \(\mathrm{SrF}_{2}\). (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) The \(K_{sp}\) of CaF2 at \(35^{\circ} \mathrm{C}\) is \(3.87 \times 10^{-11}\). b) The solubility product for SrF2 is \(6.641 \times 10^{-10}\). c) The molar solubility of Ba(IO3)2 at \(25^{\circ} \mathrm{C}\) is \(1.54 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\).

Step 1: Write the balanced dissociation equation for CaF2

: CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)

Step 2: Obtain the molar solubility of CaF2

: We are given: Molar solubility of CaF2 = \(1.24 \times 10^{-3}\) mol/L

Step 3: Relate the concentrations of ions with their stoichiometric coefficients

: For CaF2 dissociation, \[ [Ca^{2+}] = 1.24 \times 10^{-3}\] \[ [F^-] = 2 \times (1.24 \times 10^{-3})\]

Step 4: Calculate the \(K_{sp}\) for CaF2

: \[ K_{sp} = [Ca^{2+}][F^-]^2 \] \[ K_{sp} = (1.24 \times 10^{-3}) \times (2 \times 1.24 \times 10^{-3})^2 \] \[ K_{sp} = 3.87 \times 10^{-11}\] b)

Step 1: Calculate molar solubility of SrF2

: Mass of SrF2 = \(1.1 \times 10^{-2}\) g Volume = 100 mL Molar mass of SrF2 = 125.62 g/mol \[ Molar \ solubility = \frac{1.1 \times 10^{-2}\ g} {125.62 \ g/mol \times 0.100 \ L} = 8.759 \times 10^{-4} \ mol/L \]

Step 2: Write the balanced dissociation equation for SrF2

: SrF2 (s) ⇌ Sr2+ (aq) + 2F- (aq)

Step 3: Relate the concentrations of ions with their stoichiometric coefficients

: For SrF2 dissociation, \[ [Sr^{2+}] = 8.759 \times 10^{-4}\] \[ [F^-] = 2 \times (8.759 \times 10^{-4})\]

Step 4: Calculate the \(K_{sp}\) for SrF2

: \[ K_{sp} = [Sr^{2+}][F^-]^2 \] \[ K_{sp} = (8.759 \times 10^{-4}) \times (2 \times 8.759 \times 10^{-4})^2 \] \[ K_{sp} = 6.641 \times 10^{-10}\] c)

Step 1: Write the balanced dissociation equation for Ba(IO3)2

: Ba(IO3)2 (s) ⇌ Ba2+ (aq) + 2IO3- (aq)

Step 2: Relate the molar solubility (x) to the concentrations with their stoichiometric coefficients

: For Ba(IO3)2 dissociation, \[ [Ba^{2+}] = x \] \[ [IO_3^-] = 2x \]

Step 3: Substitute concentrations into the given \(K_{sp}\) to find molar solubility

: Given \(K_{sp}\) of Ba(IO3)2 = \(6.0 \times 10^{-10}\), \[ K_{sp} = [Ba^{2+}][IO_3^-]^2 = x(2x)^2 \] \[ x(4x^2) = 6.0 \times 10^{-10} \] \[ 4x^3 = 6.0 \times 10^{-10} \] \[ x^3 = \frac{6.0 \times 10^{-10}}{4} \] \[ x^3 = 1.5 \times 10^{-10} \] \[ x = (1.5 \times 10^{-10})^{1/3} \] \[ Molar \ solubility = x = 1.54 \times 10^{-3}\ mol/L \]

Key Concepts

Molar Solubility

Understanding molar solubility is essential to solve solubility product constant problems in chemistry. Molar solubility refers to the number of moles of a substance that can dissolve to form a liter of solution before reaching saturation. In the exercises involving compounds like \( \text{CaF}_2 \) and \( \text{SrF}_2 \), we are provided with the molar solubility, or we determine it from given data. Once we know the molar solubility, we can relate it to the concentration of ions in the solution. For example, if the molar solubility of a compound is \( 1.24 \times 10^{-3} \text{ mol/L} \), this is the highest concentration at which the compound can dissolve under given conditions. This value provides a base number to express the concentration of the resultant ions after dissociation.

Dissociation Equation

A dissociation equation is a balanced chemical equation representing how an ionic compound dissolves into its constituent ions in water. For instance, the dissociation of \( \text{CaF}_2 \) is written as:

• \( \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \)

This equation shows that one mole of \( \text{CaF}_2 \) dissolves to produce one mole of \( \text{Ca}^{2+} \) ions and two moles of \( \text{F}^- \) ions. Understanding these equations is crucial because they help us identify the stoichiometric relationships between the dissolved solute and its ions. This relationship is the foundation to calculate ion concentrations, which will be further used to compute the solubility product constant \( K_{sp} \). By learning how to write these equations, you can better understand how changes in the system, such as concentration or temperature shifts, might affect the equilibrium position.

Ion Concentration

Ion concentration involves determining the concentration of each ion in a solution formed by the dissociation of the compound. During dissociation of \( \text{CaF}_2 \), if its molar solubility is known to be \( 1.24 \times 10^{-3} \text{ mol/L} \), then we can compute the ion concentrations:

• Concentration of \( \text{Ca}^{2+} \) ions is \( [\text{Ca}^{2+}] = 1.24 \times 10^{-3} \text{ mol/L} \).
• Concentration of \( \text{F}^- \) ions is twice that due to the stoichiometry \( [\text{F}^-] = 2 \times 1.24 \times 10^{-3} = 2.48 \times 10^{-3} \text{ mol/L} \).

Once these ion concentrations are known, they are substituted into the \( K_{sp} \) expression: \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]Understanding the relationships between ions during dissolution is crucial for calculating the solubility product and predicting solubility under various conditions.

Temperature Dependence

The solubility of compounds often increases or decreases with temperature, and this temperature dependence is important when dealing with the solubility product constant \( K_{sp} \). In general, solubility can change due to shifts in kinetic energy of molecules; however, the specific effect depends on whether the dissolution process is exothermic or endothermic. For the exercises given, different solubilities were observed for \( \text{CaF}_2 \) and \( \text{SrF}_2 \) at different temperatures. Understanding this, along with calculating \( K_{sp} \), requires noting that \( K_{sp} \) is temperature-specific—distinct values are accurate only for the given temperature during calculations. Awareness of temperature influence helps predict how solubility might be altered in real-life applications, such as in industrial processes or environmental conditions. Monitoring these changes allows chemists to control conditions for optimal dissolution or precipitation, aiding in desired phase changes or purity levels in chemical syntheses.