Question

If \(\mathrm{NaOH}\) is added to \(0.010 \mathrm{M} \mathrm{Al}^{3+},\) which will be the predominant species at equilibrium: \(\mathrm{Al}(\mathrm{OH})_{3}\) or \(\mathrm{Al}(\mathrm{OH})_{4}^{-} ?\) The \(\mathrm{pH}\) of the solution is \(14.00 .\left[K_{\mathrm{f}}\right.\) for \(\mathrm{Al}(\mathrm{OH})_{4}^{-}=2.0 \times 10^{33}\)

Short answer

The predominant species is \( \text{Al(OH)}_4^- \).

Step-by-step solution

Identify the relevant chemical reactions

The problem involves two possible species: \( \text{Al(OH)}_3 \) precipitate and \( \text{Al(OH)}_4^- \) complex ion. Therefore, we should consider these equilibria:\( \text{Al}^{3+} + 3 \text{OH}^- \rightleftharpoons \text{Al(OH)}_3(s) \) and \( \text{Al(OH)}_3(s) + \text{OH}^- \rightleftharpoons \text{Al(OH)}_4^{-} \).

Calculate the concentration of OH- at pH 14

At \( \text{pH} = 14 \), we can find the hydroxide ion concentration using \( \text{pOH} = 14 - \text{pH} = 0 \). Therefore, the concentration of \( \text{OH}^- \) is \( 10^{-\text{pOH}} = 10^{0} = 1 \text{ M} \).

Determine if Al(OH)3 precipitates at [OH-] = 1 M

The reaction of \( \text{Al}^{3+} \) with \( \text{OH}^- \) ions produces the solid \( \text{Al(OH)}_3 \). Given a high \( \text{OH}^- \) concentration (1 M), \( \text{Al(OH)}_3 \) might precipitate. However, since the concentration of \( \text{OH}^- \) is extremely high, it will also favor the formation of \( \text{Al(OH)}_4^- \).

Evaluate the formation constant of Al(OH)4-

The formation constant \( K_f \) for \( \text{Al(OH)}_4^- \) is \( 2.0 \times 10^{33} \), which is extremely large, indicating that the reaction strongly favors the formation of the \( \text{Al(OH)}_4^- \) complex in presence of excess \( \text{OH}^- \).

Conclude the predominant species

Given the extremely high formation constant \( K_f = 2.0 \times 10^{33} \), the equilibrium greatly favors the formation of \( \text{Al(OH)}_4^- \) in a solution containing \( \text{OH}^- \) at 1 M.

Key concepts

Precipitation Reactions

Precipitation reactions occur when two solutions combine to form an insoluble solid known as a precipitate. This process often involves the reaction of ions in a solution to form a compound that does not dissolve well in water. In the context of chemical equilibria, precipitation reactions help determine which substances will remain dissolved and which will form solids under given conditions.
In our specific exercise, we're looking at the potential formation of aluminum hydroxide, \( \text{Al(OH)}_3\), when sodium hydroxide \( \text{NaOH} \) is added to a solution of \( \text{Al}^{3+} \) ions. This reaction can lead to the formation of a solid precipitate if the conditions, such as ion concentrations, favor this process. However, whether the precipitate actually forms depends on other factors such as the presence of complex-forming ions, which can alter the equilibrium and result in different products.

Formation Constant

The formation constant, or stability constant \( K_f \), is a measure of the stability of a complex ion in solution. It indicates how favorably a complex ion is formed from its constituent ions. A high \( K_f \) value suggests that once the complex forms, it will stay intact rather than dissociate back into its components.
When considering the reaction between \( \text{Al(OH)}_3 \) and \( \text{OH}^- \), we look at the formation of \( \text{Al(OH)}_4^- \), which is heavily favored by a formation constant of \( 2.0 \times 10^{33} \). This suggests that once \( \text{OH}^- \) ions are added, the complex \( \text{Al(OH)}_4^- \) tends to form readily due to the stability conferred by such a large \( K_f \). Consequently, reaction equilibria will majorly shift towards the complex ion formation, overriding any tendency to form a solid precipitate.

Hydroxide Ion Concentration

The hydroxide ion concentration \( [\text{OH}^-] \) is crucial in influencing chemical equilibria and determining whether precipitation or complex formation occurs. It is a direct consequence of the pH level in a solution, where \text{pOH} = 14 - \text{pH}.
In this exercise, adding \( \text{NaOH} \) increases the \( [\text{OH}^-] \) concentration to 1 M when the \( \text{pH} \) is equal to 14. Such a high concentration of \( \text{OH}^- \) greatly affects the chemical equilibria, positively promoting the formation of complex ions like \( \text{Al(OH)}_4^- \) over the precipitation of \( \text{Al(OH)}_3 \). Thus, understanding and calculating the hydroxide concentration is essential when predicting the outcomes in complex equilibrium scenarios.

Complex Ion Formation

Complex ion formation involves the interaction between a central metal ion and surrounding ligands, which are molecules or ions that donate electron pairs to the metal. This process is crucial in altering the behavior and solubility of otherwise simple ions in solution.
In the given exercise, the central \( \text{Al}^{3+} \) ion and \( \text{OH}^- \) ions combine to form the complex ion \( \text{Al(OH)}_4^- \). The formation constant \( K_f = 2.0 \times 10^{33} \) implies that this complex ion is exceptionally stable and predominant at equilibrium when abundant \( \text{OH}^- \) ions are present.
This highlights how complex ion formation can shift equilibria away from precipitate formation in scenarios where both are possible, underscoring the impact of such ions in controlling solution characteristics.