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Aluminum hydroxide is an amphoteric substance. It can act as either a Br酶nsted-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

Expert verified
Aluminum hydroxide acts as a Br酶nsted-Lowry base by accepting a proton (H+) in the following reaction: \[\mathrm{Al(OH)_{3}(s)} + \mathrm{H^{+}(aq)} \rightarrow \mathrm{Al(OH)_2(OH_2)^+ (aq)}\] It acts as a Lewis acid by accepting a hydroxide ion (OH-) in this reaction: \[\mathrm{Al(OH)_{3}(s)} + \mathrm{OH^{-}(aq)} \rightarrow \mathrm{Al(OH)_4^{-}(aq)}\]

Step by step solution

01

Reaction as a Br酶nsted-Lowry base

Since aluminum hydroxide can act as a Br酶nsted-Lowry base, it will be able to accept a proton (H+). We can write the reaction as follows: \[\mathrm{Al(OH)_{3}(s)} + \mathrm{H^{+}(aq)} \rightarrow \mathrm{Al(OH)_2(OH_2)^+ (aq)}\] The aluminum hydroxide, Al(OH)鈧, accepts a proton (H+) and forms the Al(OH)鈧(OH鈧)鈦 complex in an aqueous solution.
02

Reaction as a Lewis acid

In the second reaction, aluminum hydroxide acts as a Lewis acid, which means it can accept a pair of electrons. It will interact with the hydroxide ion (OH-) as a Lewis base. The reaction can be written as follows: \[\mathrm{Al(OH)_{3}(s)} + \mathrm{OH^{-}(aq)} \rightarrow \mathrm{Al(OH)_4^{-}(aq)}\] The aluminum hydroxide, Al(OH)鈧, accepts a hydroxide ion (OH-) and forms the Al(OH)鈧勨伝 complex in an aqueous solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Br酶nsted-Lowry Acid-Base Theory
The Br酶nsted-Lowry acid-base theory is fundamental in understanding how substances behave in chemical reactions. According to this theory, an acid is a substance that donates a proton (\(\text{H}^+\)) to another substance. Conversely, a base is defined as a substance that can accept a proton.

Aluminum hydroxide, \(\text{Al(OH)}_3\), is an excellent example of an amphoteric compound, meaning it can act as either a base or an acid. In its basic form, \(\text{Al(OH)}_3\) reacts with \(\text{H}^+\) ions (protons). Here, it accepts a proton, which adheres to the Br酶nsted-Lowry definition of a base. This reaction forms \(\text{Al(OH)}_2(\text{OH}_2)^+\) in an aqueous solution.

The ability of aluminum hydroxide to act both as an acid and a base makes it quite versatile in chemical reactions, paving the way for countless interactions in both industrial and biological processes.
Lewis Acid-Base Theory
The Lewis acid-base theory takes a broader perspective compared to the Br酶nsted-Lowry theory, by focusing on the role of electron pairs. According to Lewis, an acid is a substance that can accept an electron pair, while a base is one that donates an electron pair.

This expanded definition accounts for a wider range of reactions, often involving complex ions and coordination compounds. Aluminum hydroxide, \(\text{Al(OH)}_3\), can also function as a Lewis acid. In this role, it accepts an electron pair from the hydroxide ion, \(\text{OH}^-\). As a result, it forms a new species, \(\text{Al(OH)}_4^-\), illustrating its capability to act as an electron pair acceptor.

Understanding Lewis acid-base reactions is crucial in fields such as organic synthesis and biochemistry, where electron pair interactions play a major role in reaction mechanisms.
Aluminum Hydroxide Reactions
Aluminum hydroxide is a unique chemical that exhibits amphoteric properties, enabling it to participate in a variety of reactions by acting as either an acid or a base. This behavior is deeply rooted in both Br酶nsted-Lowry and Lewis theories.

When acting as a Br酶nsted-Lowry base, aluminum hydroxide captures protons to form complexes like \(\text{Al(OH)}_2(\text{OH}_2)^+\). This feature allows it to play a significant role in neutralizing acids and forming coordination compounds, which are pivotal in industrial and water treatment processes.

As a Lewis acid, aluminum hydroxide鈥檚 interaction with hydroxide ions, resulting in \(\text{Al(OH)}_4^-\), is crucial in understanding its behavior in different pH environments. This property contributes to its functionality in applications such as antacids and dye mordants.

These reactions underline the versatility of aluminum hydroxide and highlight the importance of understanding acid-base theories for practical and educational purposes.

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Most popular questions from this chapter

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{CaO}\) b. \(\mathrm{SO}_{2}\) c. \(\mathrm{Cl}_{2} \mathrm{O}\)

A \(0.050-M\) solution of the salt \(\mathrm{NaB}\) has a pH of \(9.00\). Calculate the \(\mathrm{pH}\) of a \(0.010-M\) solution of \(\mathrm{HB}\).

Calculate \(\left[\mathrm{CO}_{3}^{2-}\right]\) in a \(0.010-M\) solution of \(\mathrm{CO}_{2}\) in water (usually written as \(\mathrm{H}_{2} \mathrm{CO}_{3}\) ). If all the \(\mathrm{CO}_{3}{ }^{2-}\) in this solution comes from the reaction $$\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)$$ what percentage of the \(\mathrm{H}^{+}\) ions in the solution is a result of the dissociation of \(\mathrm{HCO}_{3}^{-} ?\) When acid is added to a solution of sodium hydrogen carbonate \(\left(\mathrm{NaHCO}_{3}\right)\), vigorous bubbling occurs. How is this reaction related to the existence of carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) molecules in aqueous solution?

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Ch芒telier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}{ }^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 146.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: \(\mathrm{CO}_{2}\) blood levels increase during cardiac arrest.)

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.10 \mathrm{MCH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.050 M \mathrm{NaCN}\)

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