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Chapter 14: Problem 135

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

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

Step by step solution

01

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: \( \mathrm{Base} + \mathrm{H^+} \rightarrow \mathrm{Conjugate \space Acid} \) Now we can substitute \(\mathrm{Al(OH)_3}\) as the base to find the equation.
02

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: \( \mathrm{Al(OH)_3} + \mathrm{H^+} \rightarrow \mathrm{Al(OH)_4^-\space} \) In this reaction, aluminum hydroxide accepts one \(\mathrm{H^+}\), forming the tetrahydroxoaluminate(\(\mathrm{Al(OH)_4^-}\)) ion.
03

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: \( \mathrm{Lewis \space Acid} + :\mathrm{Lewis \space Base} \rightarrow \mathrm{Adduct} \) Now we can substitute \(\mathrm{Al(OH)_3}\) as the Lewis acid to find the equation.
04

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: \( \mathrm{Al(OH)_3} + :\mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^- \space} \) In this reaction, aluminum hydroxide accepts a lone pair of electrons from the hydroxide ion (\(\mathrm{OH^-}\)), forming the tetrahydroxoaluminate(\(\mathrm{Al(OH)_4^-}\)) ion. So, the two reactions are: 1. As a Brønsted-Lowry base: \( \mathrm{Al(OH)_3} + \mathrm{H^+} \rightarrow \mathrm{Al(OH)_4^- \space} \) 2. As a Lewis acid: \( \mathrm{Al(OH)_3} + :\mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^- \space} \)

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Most popular questions from this chapter

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Using your results from Exercise 129, place the species in each of the following groups in order of increasing base strength. a. \(\mathrm{OH}^{-}, \mathrm{SH}^{-}, \mathrm{SeH}^{-}\) b. \(\mathrm{NH}_{3}, \mathrm{PH}_{3}\) c. \(\mathrm{NH}_{3}, \mathrm{HONH}_{2}\)

Phosphoric acid is a common ingredient in traditional cola drinks. It is added to provide the drinks with a pleasantly tart taste. Although phosphoric acid is a triprotic acid, its protons are lost one at a time. Assuming that in cola drinks the concentration of phosphoric acid is \(0.007 M\), calculate the \(\mathrm{pH}\) in this solution.

A solution of formic acid \(\left(\mathrm{HCOOH}, K_{\mathrm{a}}=1.8 \times 10^{-4}\right)\) has a \(\mathrm{pH}\) of \(2.70 .\) Calculate the initial concentration of formic acid in this 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 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 Châtelier'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 148.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary?
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