Problem 109 - Chemistry The Central Science In Si Units

Question

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14}\). (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?\) (b) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}{\underline{\phantom{xx}}}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) ?

Step-by-Step Solution

Answer

(a) The molar solubility of \(\mathrm{Cd(OH)}_2\) is approximately \(1.84 \times 10^{-5}\, \text{mol/L}\). (b) An initial concentration of NaBr of approximately \(3.16 \times 10^{-2}\, \text{mol/L}\) is needed.

Step 1: Write the Solubility Expression

For part (a), write the dissociation equation for \(\mathrm{Cd(OH)}_2\): \[\mathrm{Cd(OH)}_2(s) \rightleftharpoons \mathrm{Cd}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\]This yields the expression for the solubility product constant: \[K_{sp} = [\mathrm{Cd}^{2+}][\mathrm{OH}^{-}]^2\]

Step 2: Set Up the Solubility in Terms of 's'

Let the molar solubility of \(\mathrm{Cd(OH)}_2\) be \(s\). Then, \[[\mathrm{Cd}^{2+}] = s\] and \[[\mathrm{OH}^-] = 2s\]Substitute these into the \(K_{sp}\) expression, we get:\[K_{sp} = s\cdot (2s)^2\] which simplifies to:\[K_{sp} = 4s^3\]

Step 3: Solve for Molar Solubility 's'

Substitute \(K_{sp} = 2.5 \times 10^{-14}\) into the expression:\[4s^3 = 2.5 \times 10^{-14}\]Solve for \(s\):\[s^3 = \frac{2.5 \times 10^{-14}}{4} = 6.25 \times 10^{-15}\]\[s = \sqrt[3]{6.25 \times 10^{-15}} \approx 1.84 \times 10^{-5}\, \text{mol/L}\]

Step 4: Use Formation Constant of the Complex

For part (b), the new molar solubility is given as \(1.0 \times 10^{-3} \text{ mol/L}\).Consider the formation of \(\mathrm{CdBr}_4^{2-}\) with an equilibrium:\[\mathrm{Cd}^{2+} + 4\mathrm{Br}^- \rightleftharpoons \mathrm{CdBr}_4^{2-}\]The formation constant is:\[K_f = \frac{[\mathrm{CdBr}_4^{2-}]}{[\mathrm{Cd}^{2+}][\mathrm{Br}^-]^4} = 5 \times 10^3\]

Step 5: Setting Up the Relation with New Solubility

If \(s = 1.0 \times 10^{-3} \text{ mol/L}\), then the complex formation:\[[\mathrm{CdBr}_4^{2-}] = 1.0 \times 10^{-3}\]For equilibrium of \([\mathrm{Br}^-]\), from the relation: \[1.0 \times 10^{-3} = K_f \frac{s}{[\mathrm{Br}^-]^4}\] Substituting \(K_f = 5 \times 10^3\):\[[\mathrm{Br}^-]^4 = \mathbf{\frac{s}{(1.0 \times 10^{-3} \text{ mol/L})/5 \times 10^3}}\]

Step 6: Calculate the Initial Br^- Concentration

Rearrange and solve for \([\mathrm{Br}^-]\):\[[\mathrm{Br}^-]^4 = \frac{1.0 \times 10^{-3}}{5 \times 10^3} = 2 \times 10^{-7}\]\[[\mathrm{Br}^-] = (2 \times 10^{-7})^{1/4} \approx 3.16 \times 10^{-2}\, \text{mol/L}\]Thus, the initial concentration of NaBr needed is approximately \(3.16 \times 10^{-2}\, \text{mol/L}\).

Key Concepts

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution until the solution becomes saturated. For compounds like \(\text{Cd(OH)}_2\), understanding molar solubility is key to predicting how much of the compound will dissolve under equilibrium conditions.

In the case of \(\text{Cd(OH)}_2\), its solubility can be calculated by following few basic steps:

• Write the dissociation equation: \[\text{Cd(OH)}_2(s) \rightleftharpoons \text{Cd}^{2+}(aq) + 2\text{OH}^{-}(aq)\]
• Recognize that the equilibrium involves the formation of \(\text{Cd}^{2+}\) and \(\text{OH}^-\) ions.
• Use the solubility product constant \(K_{sp}\), a value given for this reaction, to find molar solubility.
• Let \(s\) represent the molar solubility, so that \([\text{Cd}^{2+}] = s\) and \([\text{OH}^-] = 2s\).
• Substitute these into the \(K_{sp}\) expression: \(K_{sp} = s\cdot (2s)^2 \)
• The expression simplifies to \(K_{sp} = 4s^3\). By finding \(s\) from this equation, we get the molar solubility.

This particular calculation for \(\text{Cd(OH)}_2\) yielded a molar solubility of approximately \(1.84 \times 10^{-5} \text{ mol/L}\), which describes the maximum concentration that can exist when \(\text{Cd(OH)}_2\) is dissolved in water.

Solubility Product Constant

The solubility product constant, \(K_{sp}\), is an equilibrium constant specific to dissolution reactions. It provides a quantitative measure of a sparingly soluble ionic compound's propensity to dissolve in a solvent, usually water.

For \(\text{Cd(OH)}_2\), the \(K_{sp}\) was given as \(2.5 \times 10^{-14}\). This low value indicates that the compound is not very soluble in water, emphasizing the need to be careful with quantities when dealing with slightly soluble substances.

• As the dissolution of \(\text{Cd(OH)}_2\) progresses, it reaches a dynamic equilibrium where the rate of dissolution equals the rate of precipitation.
• The expression for \(K_{sp}\) is created based on the equilibrium concentrations of the ions: \([\text{Cd}^{2+}][\text{OH}^-]^2\).
• The low \(K_{sp}\) value means that very few ions are actually present in solution at equilibrium.
• By calculating \(s\), a measure of molar solubility, we can relate this back to \(K_{sp}\). We determine \(s\) using the relationship \(4s^3 = K_{sp}\).

With this understanding, students can predict solubility and understand behaviors of ionic compounds in solution under various conditions.

Complex Ion Formation

Complex ion formation greatly enhances the solubility of some compounds, and it is a fascinating way to manipulate solubility dynamics through chemical equilibrium.

For example, when \(\text{Cd(OH)}_2\) is added to a solution containing \(\text{NaBr}\), the formation of the \(\text{CdBr}_4^{2-}\) complex ion can increase the solubility of cadmium hydroxide considerably.

• This occurs through an equilibrium reaction: \[\text{Cd}^{2+} + 4\text{Br}^- \rightleftharpoons \text{CdBr}_4^{2-}\]
• The stability of this complex is expressed in terms of a formation constant \(K_f\), which is given as \(5 \times 10^3\). This strong equilibrium favoring the complex ion formation means more \(\text{Cd}^{2+}\) ions are drawn into the solution.
• The key to leveraging this principle is determining the bromide ion concentration needed to achieve a target molar solubility, like \(1.0 \times 10^{-3}\, \text{mol/L}\).
• Rearranging the formula and substituting known values lets us solve for \([\text{Br}^-]\).
• In our example, an initial \([\text{Br}^-]\) concentration around \(3.16 \times 10^{-2}\, \text{mol/L}\) was found to be necessary.

By grasping the concept of complex ion formation, students can understand how chemists control solubility to refine processes and solutions in various applications.