Question

(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{s p^{2}}\) (b) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dissolves per liter of solution, calculate the solubilityproduct constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in grams per liter of solution.

Short answer

The Ksp of PbBr2 is \(4.0 \times 10^{-6}\), the Ksp of AgIO3 is \(4.37 \times 10^{-8}\), and the solubility of Cu(OH)2 can be calculated by solving the equation \(K_{sp} = 4s^{3}\) and converting the molar solubility to grams per liter using the molar mass of Cu(OH)2: Solubility in g/L = s (mol/L) * 97.57 g/mol.

Step-by-step solution

Understanding molar solubility and the solubility product constant (Ksp)

Molar solubility refers to the maximum number of moles of a solute that can dissolve in one liter of a solvent to form a saturated solution. The solubility product constant (Ksp) is an equilibrium constant that connects the equilibrium concentrations of the dissolved ions and the undissolved solid. For a general solute represented by "AaBb", the equilibrium reaction can be written as: \(A_{a}B_{b}(s) \rightleftharpoons aA_{(aq)}^{a+} + bB_{(aq)}^{b-} \) The solubility product constant expression is given by: \(K_{sp} = [A^{a+}]^{a} [B^{b-}]^{b}\) It's important to note that the concentration of the solid (AaBb) does not appear in the Ksp expression.

Calculate Ksp for PbBr2

Given the molar solubility of PbBr2, we can write the equilibrium reaction as: \(PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)\) The Ksp expression for this reaction is: \(K_{sp} = [Pb^{2+}] [Br^{-}]^{2}\) Since the molar solubility of PbBr2 is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\), we can assume that for every mole of PbBr2 dissolved, one mole of Pb2+ and two moles of Br- are formed. Therefore, \[ [Pb^{2+}] = 1.0 \times 10^{-2} \mathrm{M} \] \[ [Br^{-}] = 2 \times 1.0 \times 10^{-2} \mathrm{M} = 2.0 \times 10^{-2} \mathrm{M} \] Now we can plug these values into the Ksp expression: \(K_{sp} = (1.0 \times 10^{-2})(2.0 \times 10^{-2})^{2} = 4.0 \times 10^{-6}\)

Calculate Ksp for AgIO3

We know that \(0.0490 \mathrm{~g}\) of AgIO3 dissolves per liter of solution. To find the molar solubility, first, find the molar mass of AgIO3: Molar mass of AgIO3 = 107.87 (Ag) + 126.91 (IO3) = 234.78 g/mol Now convert the mass of AgIO3 dissolved to moles: \(0.0490 \mathrm{~g} \times \frac{1 \mathrm{~mol}}{234.78 \mathrm{~g}} = 2.09 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) For AgIO3, the equilibrium reaction is: \(AgIO_{3}(s) \rightleftharpoons Ag^{+}(aq) + IO_{3}^{-}(aq)\) The Ksp expression for this reaction is: \(K_{sp} = [Ag^{+}] [IO_{3}^{-}]\) Since the molar solubility of AgIO3 is \(2.09 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\), we can assume that for every mole of AgIO3 dissolved, one mole of Ag+ and one mole of IO3- are formed. Therefore, \[ [Ag^{+}] = 2.09 \times 10^{-4} \mathrm{M} \] \[ [IO_{3}^{-}] = 2.09 \times 10^{-4} \mathrm{M} \] Now we can plug these values into the Ksp expression: \(K_{sp} = (2.09 \times 10^{-4})(2.09 \times 10^{-4}) = 4.37 \times 10^{-8}\)

Calculate the solubility of Cu(OH)2

We know the Ksp value for Cu(OH)2 from Appendix D (check the table provided by your textbook or teacher). Let's assume the Ksp value is x. The equilibrium reaction for Cu(OH)2 is: \(Cu(OH)_{2}(s) \rightleftharpoons Cu^{2+}(aq) + 2OH^{-}(aq)\) The Ksp expression for this reaction is: \(K_{sp} = [Cu^{2+}] [OH^{-}]^{2}\) Let the molar solubility of Cu(OH)2 be s. Then, \[ [Cu^{2+}] = s \] \[ [OH^{-}] = 2s \] Now we can plug these values into the Ksp expression and solve for s: \(K_{sp} = s(2s)^{2}\) \(x = 4s^{3}\) Now, find s by solving this equation. Once you have the molar solubility s, convert it to grams per liter using the molar mass of Cu(OH)2: Molar mass of Cu(OH)2 = 63.55 (Cu) + 2(16.00 + 1.01) (OH) = 97.57 g/mol Solubility in g/L = s (mol/L) * 97.57 g/mol

Key concepts

Molar Solubility

Molar solubility is an essential concept in chemistry when it comes to understanding how substances dissolve in solvents. It tells us the number of moles of a solute that can be dissolved in one liter of solution until the solution becomes saturated. When a solution is saturated, it means no more solute can dissolve without changing the temperature or the surrounding conditions. This maximum capacity is critical for predicting how much of a compound will dissolve in a given solvent.

• Molar solubility is often expressed in \( \text{mol/L} \) or M (molarity).
• The concept applies primarily to sparingly soluble compounds, like \( \text{PbBr}_2 \) and other salts.

To calculate molar solubility, you'll often start with known data like the mass of solute that dissolves or the equilibrium concentrations of ions in the solution. For example, if \( 0.0490 \, \text{g} \) of \( \text{AgIO}_3 \) dissolves in one liter of solution, you would first convert this mass into moles using its molar mass, and then that number becomes its molar solubility.
The molar solubility becomes foundational when solving for equilibrium constants like the solubility product constant \( K_{sp} \).

Equilibrium Constant

The equilibrium constant, specifically the solubility product constant (\( K_{sp} \)), pertains to the equilibria of sparingly soluble salts as they dissolve and reach saturation in a solution. These constants help chemists understand and predict the behavior of ionic compounds in an aqueous environment.

• \( K_{sp} \) is unique to every salt at a given temperature and represents the extent of its dissolution.
• While the solids aren't included, the ions into which they dissociate are pivotal.

The general representation of a dissolution reaction shows how a compound breaks into its ionic components, as seen with \( A_aB_b(s) \rightleftharpoons aA^{a+}_{(aq)} + bB^{b-}_{(aq)} \). The \( K_{sp} \) expression emerges from this balance and is devoid of any influence from the undissolved solid itself.
As you calculate \( K_{sp} \):

• Recognize that it captures the concentrations of the dissociated ions in a saturated solution.
• Note: Larger \( K_{sp} \) values indicate higher solubility, and vice versa.

For instance, if you know the molar solubility of \( \text{PbBr}_2 \) is \( 1.0 \times 10^{-2} \text{ mol/L} \), you can determine \( K_{sp} \) by plugging the ion concentrations into its expression.

Dissolution Reactions

Dissolution reactions are processes where solids separate into their constituent ions in a solvent, often leading to the formation of a solution. These reactions are crucial because they provide insight into how compounds behave in various environments. When a compound like \( \text{Cu(OH)}_2 \) dissolves, it creates a saturated solution in equilibrium with the undissolved solid.
How do dissolution reactions work?

• The compound dissociates into its ions, following the general pattern \( A_aB_b(s) \rightarrow aA^{a+}(aq) + bB^{b-}(aq) \).
• Some substances dissolve fully, while others achieve equilibrium, hence the concept of molar solubility.

In dissolution, the dynamic balance between the dissolved and undissolved phases is reflected in the equilibrium expression derived for \( K_{sp} \). It involves

• Balancing the rate of dissolution and precipitation of ions.
• Quantifying how much of these ions exist in solution at equilibrium.

The dissolution of \( \text{AgIO}_3 \) involves a simple dissociation into \( \text{Ag}^+ \) and \( \text{IO}_3^- \) ions, represented by its \( K_{sp} \), while compounds like \( \text{Cu(OH)}_2 \) involve more complex ion equilibria due to their stoichiometry, with additional tasks like adjusting \( [\text{OH}^-] \) when calculating molar solubility.