Question

\(\begin{array}{llll}& \text { (a) If the molar solubility of } & \mathrm{CaF}_{2} & \text { at } & 35^{\circ} \mathrm{C} & \text { is }\end{array}\) \(1.24 \times 10^{-3} \mathrm{~mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that \(1.1 \times 10^{-2}\) of \(\mathrm{SrF}_{2}\) 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 \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) ?

Short answer

The solubility product constants (Ksp) of CaF₂ at 35°C is \(7.58 \times 10^{-9}\), the solubility product of SrF₂ at 25°C is \(6.08 \times 10^{-5}\), and the molar solubility of Ba(IO₃)₂ at 25°C is \(1.14 \times 10^{-3}\ mol/L\).

Step-by-step solution

(a) Identify the Dissociation Equation

For CaF₂, the dissociation equation in water is: \[CaF_2 \leftrightarrows Ca^{2+} + 2F^{-}\]

(a) Write the Solubility Product Formula

Based on the dissociation equation, we can write the solubility product constant (Ksp) formula of CaF₂ as: \[K_{sp} = [Ca^{2+}][F^-]^2\]

(a) Calculate the Concentrations of Ions

Given the molar solubility of CaF₂, we can calculate the concentration of ions using the stoichiometry of the dissociation equation: \[ [Ca^{2+}] = 1.24 \times 10^{-3} \ mol/L \] \[ [F^-] = 2 \times 1.24 \times 10^{-3} \ mol/L = 2.48 \times 10^{-3} \ mol/L \]

(a) Calculate Ksp

Substitute the concentrations of ions in the Ksp formula: \[K_{sp} = (1.24 \times 10^{-3})(2.48 \times 10^{-3})^2 = 7.58 \times 10^{-9}\] So, the solubility product constant of CaF₂ at 35°C is \(7.58 \times 10^{-9}\).

(b) Convert SrF₂ Solubility to Molar Solubility

Given that 1.1 g of SrF₂ dissolves per 100 mL of aqueous solution, we can find the molar solubility: \[ Molar\ Solubility\ = \frac{1.1\ g}{100\ mL} \times \frac{1\ mol}{126.62\ g} \times \frac{1000\ mL}{1\ L} \approx 8.69 \times 10^{-2} \ mol/L \]

(b) Identify the Dissociation Equation and Write Ksp Formula

For SrF₂, the dissociation equation and Ksp formula are: \[ SrF_2 \leftrightarrows Sr^{2+} + 2F^{-} \] \[ K_{sp}=[Sr^{2+}][F^-]^2 \]

(b) Calculate Ksp for SrF₂

Using the molar solubility of SrF₂, we can find the Ksp: \[ K_{sp}=(8.69 \times 10^{-2})(2 \times 8.69 \times 10^{-2})^2 = 6.08 \times 10^{-5} \] So, the solubility product of SrF₂ at 25°C is \(6.08 \times 10^{-5}\).

(c) Identify the Dissociation Equation and Write Ksp Formula

For Ba(IO₃)₂, the dissociation equation and Ksp formula are: \[ Ba(IO_3)_2 \leftrightarrows Ba^{2+} + 2IO_3^{-} \] \[ K_{sp}=[Ba^{2+}][IO_3^-]^2 \]

(c) Calculate the Molar Solubility of Ba(IO₃)₂

Given the Ksp of Ba(IO₃)₂, we can solve for the molar solubility (s) of Ba₂(IO₃)₂: \[ K_{sp} = 6.0 \times 10^{-10} = [Ba^{2+}][IO_3^-]^2 = (s)(2s)^2 = 4s^3 \] \[ s^3 = \frac{6.0 \times 10^{-10}}{4} = 1.5 \times 10^{-10} \] \[ s = \sqrt[3]{1.5 \times 10^{-10}} \approx 1.14 \times 10^{-3}\ mol/L \] So, the molar solubility of Ba(IO₃)₂ at 25°C is \(1.14 \times 10^{-3}\ mol/L\).

Key concepts

Ksp Calculation

Understanding the Solubility Product Constant, commonly referred to as Ksp, is pivotal for students studying chemical solubility equilibria. It's a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. Ksp calculation involves determining how much of a salt can dissolve in a solution to form a saturated solution. For this task, the steps are fairly straightforward: identify the dissociation equation, establish the Ksp expression from the stoichiometry of the dissociation, then calculate the concentrations of the ions involved and input these into the Ksp expression.

For example, in the case of calcium fluoride (CaF₂), the dissociation equation is \[CaF_2 \leftrightarrows Ca^{2+} + 2F^{-}\] and the resulting Ksp expression is \[K_{sp} = [Ca^{2+}][F^-]^2\].By plugging in the known molar solubility, we can then calculate the solubility product constant. This calculation helps us understand how salts interact with solvents and predict the point at which a solution becomes saturated, leading to precipitation.

Molar Solubility

Molar solubility is an expression of the maximum amount of a substance that can dissolve in a given volume of solvent to produce a saturated solution without any excess of the solute. It's usually expressed in moles per liter (mol/L). To calculate molar solubility from the mass of the substance and the volume of the solution, divide the mass (in grams) by the molar mass to convert to moles, then divide by the volume of the solution (in liters).

For instance, when finding the molar solubility of SrF₂, the process involves converting the grams of SrF₂ that can dissolve into moles using its molar mass and then adjusting for the volume of the solution. This figure is crucial for predicting how much of the compound can exist in a solution phase before solid forms, and it further helps in determining the concentration of the ions in the saturated solution.

Chemical Equilibrium

Chemical equilibrium refers to the state of a reaction in which the rates of the forward and reverse reactions are equal, leading to no overall change in the concentrations of reactants and products over time. This does not mean the amounts of reactants and products are equal, but that their ratios remain constant. In the context of solubility equilibria, the Ksp comes into play by representing the constant state at which a solid is in equilibrium with its constituent ions in solution.

Understanding equilibrium assists in grasping the solubility behavior of substances. The concept is applied in the dissolution process of sparingly soluble salts, such as in the example of Ba(IO₃)₂. By using the Ksp, one can establish the relationship between the concentrations of the ions in the saturated solution and calculate backward to find the molar solubility. This way, knowledge of the equilibrium concept and Ksp can predict the solubility of substances and aid in various applications, from chemical synthesis to environmental science.