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Chapter 14: Problem 139

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

Short Answer

Expert verified
Aluminum hydroxide, Al(OH)₃, can act as a Bronsted-Lowry base by accepting a proton (H⁺) in the reaction: \( Al(OH)_{3} + H^{+} \rightarrow Al(OH)_{4}^{-} \). It can also act as a Lewis acid by accepting a hydroxide ion (OH⁻) in the reaction: \( Al(OH)_{3} + OH^{-} \rightarrow Al(OH)_{4}^{-} \).

Step by step solution

01

Reaction 1: Aluminum hydroxide as a Bronsted-Lowry base

To show aluminum hydroxide acting as a Bronsted-Lowry base, we need to show it accepting a proton (H⁺) forming Al(OH)₄⁻. The balanced reaction for Al(OH)₃ acting as a base towards H⁺ can be written as: \( Al(OH)_{3} + H^{+} \rightarrow Al(OH)_{4}^{-} \)
02

Reaction 2: Aluminum hydroxide as a Lewis acid

To show aluminum hydroxide acting as a Lewis acid, we need to demonstrate it accepting an OH⁻ ion to form Al(OH)₄⁻. The balanced reaction for Al(OH)₃ acting as a Lewis acid towards OH⁻ can be written as: \( Al(OH)_{3} + OH^{-} \rightarrow Al(OH)_{4}^{-} \) In these reactions, we have demonstrated that aluminum hydroxide can act as both a Bronsted-Lowry base and a Lewis acid by accepting a proton (H⁺) and a hydroxide ion (OH⁻). This confirms that Al(OH)₃ is an amphoteric substance.

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Most popular questions from this chapter

Calculate the percentage of pyridine $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)\( that forms pyridinium ion, \)\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+},\( in a \)0.10-M$ aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)

Arrange the following 0.10\(M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. $$ \mathrm{CaBr}_{2}, \mathrm{KNO}_{2}, \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4} $$

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables 14.2 and \(14.3 .\) $\begin{array}{ll}{\text { a. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}} & {\text { d. } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{ClO}_{2}} \\ {\text { b. } \mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}} & {\text { e. } \mathrm{NH}_{4} \mathrm{F}} \\ {\text { c. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{O} \mathrm{l}} & {\text { f. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}}\end{array}$

Calculate \(\left[\mathrm{OH}^{-}\right], \mathrm{pOH}\) , and pH for each of the following. a. 0.00040\(M \mathrm{Ca}(\mathrm{OH})_{2}\) b. a solution containing 25 \(\mathrm{g}\) KOH per liter c. a solution containing 150.0 \(\mathrm{g}\) NaOH per liter

An acid \(\mathrm{HX}\) is 25\(\%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M},\) calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\) .
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