Problem 42 - Chemistry: Principles and Reactions

Question

Write a net ionic equation for the reaction with \(\mathrm{OH}^{-}\) by which (a) \(\mathrm{Ni}^{2+}\) forms a precipitate. (b) \(\mathrm{Sn}^{4+}\) forms a complex ion. (c) \(\mathrm{Al}(\mathrm{OH})_{3}\) dissolves.

Step-by-Step Solution

Answer

Question: Write the net ionic equation for the dissolution of aluminum hydroxide with the hydroxide ion. Answer: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Al}(\mathrm{OH})_{4}]^{-}(aq)\)

Step 1: (Net ionic equation for Nickel precipitation reaction)

For the precipitation of nickel ion (\(\mathrm{Ni}^{2+}\)) with hydroxide ion, the balanced molecular equation is: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)\) Now, split the reaction into its respective ions: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(s)\) The net ionic equation cancels out the spectator ions: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)\)

Step 2: (Net ionic equation for Tin complex-ion formation reaction)

When tin ion forms a complex ion with hydroxide ions, we can represent the balanced molecular equation as: \(\mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Sn}(\mathrm{OH})_{4}]^{2-}(aq)\) Split the reaction into its respective ions: \(\mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq)\) The net ionic equation cancels out the spectator ions: \(\mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Sn}(\mathrm{OH})_{4}]^{2-}(aq)\)

Step 3: (Net ionic equation for Aluminum hydroxide dissolution reaction)

To represent the dissolution of aluminum hydroxide (\(\mathrm{Al}(\mathrm{OH})_{3}\)) with hydroxide ion, the balanced molecular equation is: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Al}(\mathrm{OH})_{4}]^{-}(aq)\) Split the reaction into its respective ions: \(\mathrm{Al}^{3+}(aq) + 3\mathrm{OH}^{-}(s) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{Al}^{3+}(aq) + 4\mathrm{OH}^{-}(aq)\) The net ionic equation cancels out the spectator ions: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Al}(\mathrm{OH})_{4}]^{-}(aq)\)

Key Concepts

Precipitation Reaction

A precipitation reaction involves the formation of an insoluble solid, known as a precipitate, when two aqueous solutions are mixed. Let's explore the case with nickel ions (\(\mathrm{Ni}^{2+}\)). When these ions are combined with hydroxide ions (\(\mathrm{OH}^{-}\)), a solid, nickel hydroxide \( (\mathrm{Ni}(\mathrm{OH})_{2}) \), forms.
This can be observed as a sudden appearance of a solid in the solution. The balanced net ionic equation shows the direct transformation:

• \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)\)

No spectator ions (ions that don't change during the reaction) are involved directly, making it straightforward to write this net ionic equation. Precipitation reactions are crucial in processes like water purification, where unwanted ions are removed by forming solids.

Complex Ion Formation

Complex ion formation involves the assembly of metal ions with ligands to form a complex structure. Tin ions (\(\mathrm{Sn}^{4+}\)) bond with hydroxide ions in this case to form a complex ion \( [\mathrm{Sn}(\mathrm{OH})_{4}]^{2-} \).
In such reactions, hydroxide ions act as ligands, stabilizing the tin ion through a coordinate covalent bond.
Here's the net ionic equation:

• \(\mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Sn}(\mathrm{OH})_{4}]^{2-}(aq)\)

The result is not a precipitate but a dissolved complex ion. This differs from precipitation because the product remains in solution and is more stable due to the ligand interactions. Complex ions are used in extraction processes and analytical chemistry for detecting metals.

Aluminum Hydroxide Dissolution

Aluminum hydroxide dissolution is an intriguing process, showing how a solid can dissolve back into a solution by forming a soluble species. Starting with aluminum hydroxide \((\mathrm{Al}(\mathrm{OH})_{3})\), additional hydroxide ions make it soluble.
The process involves converting the solid into a complex ion \([\mathrm{Al}(\mathrm{OH})_{4}]^{-}\):

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

Here, hydroxide ions facilitate the breakdown of the solid, resulting in the more soluble complex form. This dissolution is a beautiful example of amphoterism, where aluminum hydroxide reacts with both acids and bases. It's important in various industrial applications, such as refining bauxite to aluminum.