Problem 53 - Chemical Principles

Question

Determine \(K_{\mathrm{sp}}\) for each of the following sparingly soluble substances, given their molar solubilities: (a) \(\mathrm{AgBr}\), \(8.8 \times 10^{-7} \mathrm{~mol} \cdot \mathrm{L}^{-1}\), (b) \(\mathrm{PbCrO}_{4}, 1.3 \times 10^{-7} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) (c) \(\mathrm{Ba}(\mathrm{OH})_{2}, 0.11 \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (d) \(\mathrm{MgF}_{2}, 1.2 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1}\). For the purpose of this calculation, ignore any reaction of the anion with water and the autoprotolysis of water.

Step-by-Step Solution

Answer

The solubility product constants (\(K_{\mathrm{sp}}\)) are: (a) \(7.744 \times 10^{-13}\) for \(\mathrm{AgBr}\); (b) \(1.69 \times 10^{-14}\) for \(\mathrm{PbCrO4}\); (c) \(0.005346\) for \(\mathrm{Ba(OH)2}\); (d) \(6.912 \times 10^{-8}\) for \(\mathrm{MgF2}\).

Step 1: Write the Dissolution Equation for AgBr

For the solubility of silver bromide (AgBr), we write the dissolution equation which reveals the ions produced upon dissolution: \[\mathrm{AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq).}\] Each mole of \(\mathrm{AgBr}\) yields one mole of \(\mathrm{Ag}^{+}\) and one mole of \(\mathrm{Br}^{-}\).

Step 2: Calculate the Ksp for AgBr

Using the molar solubility of \(8.8 \times 10^{-7} \mathrm{mol/L}\) for both \(\mathrm{Ag}^{+}\) and \(\mathrm{Br}^{-}\), we substitute these concentrations into the solubility product expression: \[K_{\mathrm{sp}}= [\mathrm{Ag}^{+}][\mathrm{Br}^{-}] = (8.8 \times 10^{-7})(8.8 \times 10^{-7}).\] Solve for \(K_{\mathrm{sp}}\).

Step 3: Write the Dissolution Equation for PbCrO4

For lead(II) chromate (PbCrO4), the dissolution equation is: \[\mathrm{PbCrO4(s) \rightleftharpoons Pb^{2+}(aq) + CrO4^{2-}(aq).}\] Each mole of \(\mathrm{PbCrO4}\) gives one mole of \(\mathrm{Pb}^{2+}\) and one mole of \(\mathrm{CrO4}^{2-}\).

Step 4: Calculate the Ksp for PbCrO4

Substitute the molar solubility of \(1.3 \times 10^{-7} \mathrm{mol/L}\) into the solubility product expression: \[K_{\mathrm{sp}} = [\mathrm{Pb^{2+}}][\mathrm{CrO4^{2-}}] = (1.3 \times 10^{-7})(1.3 \times 10^{-7}).\] Then solve for \(K_{\mathrm{sp}}\).

Step 5: Write the Dissolution Equation for Ba(OH)2

For barium hydroxide (Ba(OH)2), the dissolution equation is: \[\mathrm{Ba(OH)2 (s) \rightleftharpoons Ba^{2+}(aq) + 2OH^{-}(aq).}\] Each mole of \(\mathrm{Ba(OH)2}\) produces one mole of \(\mathrm{Ba}^{2+}\) and two moles of \(\mathrm{OH}^{-}\).

Step 6: Calculate the Ksp for Ba(OH)2

Given the molar solubility of \(0.11 \mathrm{mol/L}\) for \(\mathrm{Ba}^{2+}\) and double that for \(\mathrm{OH}^{-}\), calculate \(K_{\mathrm{sp}}\) using the expression: \[K_{\mathrm{sp}} = [\mathrm{Ba}^{2+}][\mathrm{OH}^{-}]^2 = (0.11)(2 \times 0.11)^2 = (0.11)(0.22)^2.\] Solve for \(K_{\mathrm{sp}}\).

Step 7: Write the Dissolution Equation for MgF2

For magnesium fluoride (MgF2), the dissolution equation is: \[\mathrm{MgF2(s) \rightleftharpoons Mg^{2+}(aq) + 2F^{-}(aq).}\] One mole of \(\mathrm{MgF2}\) dissociates into one mole of \(\mathrm{Mg}^{2+}\) and two moles of \(\mathrm{F}^{-}\).

Step 8: Calculate the Ksp for MgF2

Use the molar solubility of \(1.2 \times 10^{-3} \mathrm{mol/L}\) to find \(K_{\mathrm{sp}}\) as follows: \[K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2 = (1.2 \times 10^{-3})(2 \times 1.2 \times 10^{-3})^2.\] Solve for \(K_{\mathrm{sp}}\).

Key Concepts

Molar Solubility

Molar solubility is a measure of the maximum amount of a substance that can dissolve in a liter of solvent to form a saturated solution. This term is particularly important in the context of sparingly soluble salts. These salts do not fully dissolve in water, resulting in a solution that is in equilibrium with the undissolved solid.

For instance, when we say the molar solubility of AgBr is \(8.8 \times 10^{-7} \mathrm{mol/L}\), it indicates the maximum number of moles of AgBr that can dissociate into its respective ions, \(\mathrm{Ag}^{+}\) and \(\mathrm{Br}^{-}\), in one liter of water at a given temperature. Understanding molar solubility is crucial for calculating the solubility product constant (Ksp), as it directly relates the concentration of the ions in a saturated solution.

Dissolution Equations

Dissolution equations represent the process of a solute dissolving in a solvent to form a solution. They are a key aspect of understanding how compounds dissociate into ions, especially for ionic solids such as salts. Each dissolution equation is specific to the compound in question and details which ions are formed upon dissolution.

For example, AgBr dissolves according to the equation \(\mathrm{AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)}\). This equation shows that solid silver bromide separates into silver ions (Ag+) and bromide ions (Br-) in water. The arrow pointing both ways indicates that the dissolution is reversible and, therefore, can achieve a state of chemical equilibrium.

Ksp Calculations

Calculating the solubility product constant (Ksp) is a method to quantify the solubility of sparingly soluble compounds. Ksp represents the equilibrium constant for the dissolution of a solid into its constituent ions. The value of Ksp is determined at a specific temperature and essentially indicates how soluble a compound is.

For calculation, one uses the molar solubility values of the ions produced and substitutes them into the solubility product expression, which is the product of the concentrations of the resulting ions, each raised to the power of their stoichiometric coefficient. For example, in the dissolution of MgF2, Ksp can be calculated as \(K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2\), where [Mg2+] and [F-] are the molar concentrations of the magnesium and fluoride ions, respectively, in a saturated solution.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a chemical process, resulting in no observable change in the amounts of reactants and products over time. This state is especially relevant in the context of dissolution reactions, where sparingly soluble compounds establish an equilibrium between their solid and dissolved ionic forms.

In the case of a solubility equilibrium, represented by the solubility product constant (Ksp), the equilibrium expresses the balance between the dissolved ions and the undissolved solid. Understanding chemical equilibrium helps us predict the extent to which a compound will dissolve and how the system will respond to changes in conditions, such as temperature or the presence of other ions.