Question

Aluminum hydroxide is an amphoteric substance. It can act as either a Brønsted-Lowry base or a Lewis acid. Write a reaction showing $\mathrm{Al}(\mathrm{OH}{)}_3$ acting as a base toward ${\mathrm{H}}^{+}$ and as an acid toward ${\mathrm{OH}}^{-}$.

Short answer

When aluminum hydroxide $\mathrm{Al}(\mathrm{OH}{)}_3$ acts as a Brønsted-Lowry base, it accepts a proton (${\mathrm{H}}^{+}$) in the reaction: $\mathrm{Al}(\mathrm{OH}{)}_3+{\mathrm{H}}^{+}\to \mathrm{Al}(\mathrm{OH}{)}_4^{-}$. When it acts as a Lewis acid, it accepts a lone pair of electrons from the hydroxide ion (${\mathrm{OH}}^{-}$) in the reaction: $\mathrm{Al}(\mathrm{OH}{)}_3+:{\mathrm{OH}}^{-}\to \mathrm{Al}(\mathrm{OH}{)}_4^{-}$.

Step-by-step solution

Understanding Brønsted-Lowry base

A Brønsted-Lowry base is a substance that can accept a proton (hydrogen ion, ${\mathrm{H}}^{+}$) in a reaction. In a reaction with a Brønsted-Lowry base, the base will typically take a ${\mathrm{H}}^{+}$ from an acid, and the complete reaction looks like this: $\mathrm{Base}+{\mathrm{H}}^{+}\to \mathrm{Conjugate}\mathrm{Acid}$ Now we can substitute $\mathrm{Al}(\mathrm{OH}{)}_3$ as the base to find the equation.

Writing the equation for aluminum hydroxide as a Brønsted-Lowry base

With aluminum hydroxide as the base and knowing the reaction, we can write the equation: $\mathrm{Al}(\mathrm{OH}{)}_3+{\mathrm{H}}^{+}\to \mathrm{Al}(\mathrm{OH}{)}_4^{-}$ In this reaction, aluminum hydroxide accepts one ${\mathrm{H}}^{+}$, forming the tetrahydroxoaluminate($\mathrm{Al}(\mathrm{OH}{)}_4^{-}$) ion.

Understanding Lewis acid

A Lewis acid is a substance that can accept a lone pair of electrons in a reaction. In a reaction with a Lewis acid, the acid will typically accept a lone pair of electrons from a Lewis base, forming a coordinated covalent bond. The general reaction looks like this: $\mathrm{Lewis}\mathrm{Acid}+:\mathrm{Lewis}\mathrm{Base}\to \mathrm{Adduct}$ Now we can substitute $\mathrm{Al}(\mathrm{OH}{)}_3$ as the Lewis acid to find the equation.

Writing the equation for aluminum hydroxide as a Lewis acid

With aluminum hydroxide as the Lewis acid and knowing the reaction, we can write the equation: $\mathrm{Al}(\mathrm{OH}{)}_3+:{\mathrm{OH}}^{-}\to \mathrm{Al}(\mathrm{OH}{)}_4^{-}$ In this reaction, aluminum hydroxide accepts a lone pair of electrons from the hydroxide ion (${\mathrm{OH}}^{-}$), forming the tetrahydroxoaluminate($\mathrm{Al}(\mathrm{OH}{)}_4^{-}$) ion. So, the two reactions are: 1. As a Brønsted-Lowry base: $\mathrm{Al}(\mathrm{OH}{)}_3+{\mathrm{H}}^{+}\to \mathrm{Al}(\mathrm{OH}{)}_4^{-}$ 2. As a Lewis acid: $\mathrm{Al}(\mathrm{OH}{)}_3+:{\mathrm{OH}}^{-}\to \mathrm{Al}(\mathrm{OH}{)}_4^{-}$