Problem 121 - Chemistry

Question

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s),\) but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}.\) In terms of solubility, All \((\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the pH dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3},\) as a function of \(\left[\mathrm{H}^{+}\right],\) obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is 40.0 and \(K_{\mathrm{sp}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\) Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the ph range \(4-12.\)

Step-by-Step Solution

Answer

In acidic solutions, the reaction that increases the solubility of aluminum hydroxide is: \[ Al(OH)_3(s) + 3H^+(aq) \rightleftharpoons Al^{3+}(aq) + 3H_2O(l) \] In basic solutions, the reaction is: \[ Al(OH)_3(s) + OH^-(aq) \rightleftharpoons Al(OH)_4^-(aq) \] The solubility equation as a function of \(\left[ H^+ \right]\) is given by: \[ S = \frac{(([H^+])^3K_\mathrm{sp})} {(K_\mathrm{w})^3} + \frac{((K_\mathrm{w})K)}{[H^+]} \] Plotting the solubility over the pH range 4-12 using the given equilibrium constants will provide a graph displaying the solubility trend of aluminum hydroxide with respect to pH, showing that it increases as the acidity decreases or the basicity increases.

Step 1: Reaction of Aluminum hydroxide in acidic solution

In acidic solutions, the excess of protons (H+) reacts with the hydroxide ion (OH-) present in Aluminum hydroxide: \[ Al(OH)_3(s) + 3H^+(aq) \rightleftharpoons Al^{3+}(aq) + 3H_2O(l) \]

Step 2: Reaction of Aluminum hydroxide in basic solution

In basic solutions, the excess of hydroxide ions (OH-) reacts with Aluminum hydroxide to form the soluble complex ion: \[ Al(OH)_3(s) + OH^-(aq) \rightleftharpoons Al(OH)_4^-(aq) \] b) Deriving the solubility equation

Step 1: Write the equilibrium expressions

Write the expressions for the solubility product \(K_\mathrm{sp}\) and the equilibrium constant \(K\): \[ K_\mathrm{sp} = [Al^{3+}][OH^-]^3 \] \[ K = \frac{[Al(OH)_4^-]}{[Al(OH)_3][OH^-]} \]

Step 2: Relate the concentrations

Relate the concentrations of ions using the equilibrium constant for water \(K_\mathrm{w} = [H^+][OH^-]\): \[ [OH^-] = \frac{K_\mathrm{w}}{[H^+]} \]

Step 3: Substitute and Rearrange

Substitute the relationships into the first equation and rearrange: \[ S = [Al^{3+}] + [Al(OH)_4^-] \] \[ S = \frac{K_\mathrm{sp}} {[OH^-]^3} + \frac{K[OH^-]}{[Al(OH)_3]} \] \[ S = \frac{(([H^+])^3K_\mathrm{sp})} {(K_\mathrm{w})^3} + \frac{((K_\mathrm{w})K)}{[H^+]} \] c) Plotting the solubility Using a graphing software or calculator, plot the function: \[ S = \frac{(([H^+])^3(2\times10^{-32}))} {(K_\mathrm{w})^3} + \frac{((K_\mathrm{w})40.0)}{[H^+]} \] Over the pH range 4-12: remember that the concentration of \([H^+]\) ion is given by \([H^+] = 10^{-\mathrm{pH}}\). The graph will show a distinct trend in solubility as the pH varies, displaying the solubility of aluminum hydroxide increasing as the acidity decreases or as the basicity increases.

Key Concepts

Aluminum Hydroxide Reactions

Aluminum hydroxide, (\mathrm{Al}(\mathrm{OH})_{3}), is an amphoteric substance, meaning it can react both as an acid and a base. In acidic environments, aluminum hydroxide dissolves by reacting with excess hydrogen ions (H^+) present in the solution. This reaction leads to the formation of aluminum ions (\mathrm{Al}^{3+}) and water, as described by the equation:

• \( \mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{H}^{+}(aq) \rightleftharpoons \mathrm{Al}^{3+}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \)

On the other hand, in basic environments, where there is an excess of hydroxide ions (\mathrm{OH}^{-}), aluminum hydroxide reacts to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}\).This reaction is shown in the equation:

• \( \mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(aq) \)

These reactions of aluminum hydroxide with acids and bases demonstrate its dual capability of dissolving in both acidic and basic conditions, resulting in either free aluminum ions or complex ions formation. Understanding these reactions helps in predicting the solubility behavior of \(\mathrm{Al}(\mathrm{OH})_{3}\) across different pH levels.

pH Dependency

The solubility of aluminum hydroxide is highly dependent on the pH level of the solution. In extremely acidic conditions, the concentration of hydrogen ions (\(\mathrm{H}^{+}\)) is high, which shifts the equilibrium to produce more \(\mathrm{Al}^{3+}\) ions, thus increasing the compound's solubility.As the pH increases toward neutrality, the concentration of hydrogen ions decreases, resulting in lower solubility due to less pronounced dissolution into \(\mathrm{Al}^{3+}\). However, at higher pH levels (i.e., more basic conditions), the solubility again rises due to the formation of \(\mathrm{Al}(\mathrm{OH})_{4}^{-}\) complex ions. This peculiar behavior is explained by its amphoteric nature, which allows it to react differently at low and high pH values.

Complex Ion Formation

In basic solutions, aluminum hydroxide forms a complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}\) by associating with additional hydroxide ions. This process is crucial for enhancing the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in such environments. The formation of complex ions occurs because aluminum hydroxide tends to dissolve more as the hydroxide ion concentration increases.The complex ion formation equation is:

• \( \mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(aq) \)

This reaction indicates that the presence of excess \(\mathrm{OH}^{-}\) shifts the equilibrium towards the right, favoring the formation of the soluble complex ion. Such reactions are particularly important in various chemical processes and applications where precise control over solubility is required.

Equilibrium Constants

Equilibrium constants, such as the solubility product (K_{\mathrm{sp}}) and the formation constant (K), play significant roles in determining the solubility of compounds like aluminum hydroxide. The solubility product \(K_{\mathrm{sp}}\) is defined by the equilibrium concentration of the ions in a saturated solution, as given by:

• \( K_{\mathrm{sp}} = [\mathrm{Al}^{3+}][\mathrm{OH}^{-}]^3 \)

For aluminum hydroxide, the small value of \(K_{\mathrm{sp}} = 2 \times 10^{-32}\) indicates its limited solubility under normal conditions. Meanwhile, the equilibrium constant for the complex ion formation (K) is expressed as follows:

• \( K = \frac{[\mathrm{Al}(\mathrm{OH})_{4}^{-}]}{[\mathrm{Al}(\mathrm{OH})_{3}][\mathrm{OH}^{-}]} \)

With a given value of 40.0, this constant highlights the affinity of \(\mathrm{Al}(\mathrm{OH})_{3}\) to form the \(\mathrm{Al}(\mathrm{OH})_{4}^{-}\) ion in the presence of excess \(\mathrm{OH}^{-}\).Equilibrium constants help in calculating solubility across varying pH levels, allowing chemists to understand and manipulate reactions involving compounds like aluminum hydroxide.