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For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. a. \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\) b. \(\mathrm{H}_{2} \mathrm{O}+\mathrm{HONH}_{3}^{+} \rightleftharpoons \mathrm{HONH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+}\) c. \(\mathrm{HOCl}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightleftharpoons \mathrm{OCl}^{-}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\)

Short Answer

Expert verified
For each reaction: a. Acid: \(\mathrm{Al}(\mathrm{H}_{2}\mathrm{O})_{6}^{3+}\); Base: \(\mathrm{H}_{2}\mathrm{O}\); Conjugate Acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\); Conjugate Base: \(\mathrm{Al}(\mathrm{H}_{2}\mathrm{O})_{5}(\mathrm{OH})^{2+}\) b. Acid: \(\mathrm{HONH}_{3}^{+}\); Base: \(\mathrm{H}_{2}\mathrm{O}\); Conjugate Acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\); Conjugate Base: \(\mathrm{HONH}_{2}\) c. Acid: \(\mathrm{HOCl}\); Base: \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}\); Conjugate Acid: \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^{+}\); Conjugate Base: \(\mathrm{OCl}^{-}\)

Step by step solution

01

In the given reaction, \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) donates a proton (H+) to \(\mathrm{H}_{2} \mathrm{O}\). Therefore, the acid is \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). #Step 2: Identify the base#

In the given reaction, \(\mathrm{H}_{2}\mathrm{O}\) accepts a proton (H+) from \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). Therefore, the base is \(\mathrm{H}_{2}\mathrm{O}\). #Step 3: Identify the conjugate acid#
02

In the given reaction, \(\mathrm{H}_{2}\mathrm{O}\) gains a proton (H+) and forms \(\mathrm{H}_{3}\mathrm{O}^{+}\). Therefore, the conjugate acid is \(\mathrm{H}_{3}\mathrm{O}^{+}\). #Step 4: Identify the conjugate base#

In the given reaction, \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) loses a proton (H+) and forms \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\). Therefore, the conjugate base is \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\). b. \(\mathrm{H}_{2} \mathrm{O}+\mathrm{HONH}_{3}^{+} \rightleftharpoons \mathrm{HONH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+}\) #Step 1: Identify the acid#
03

In the given reaction, \(\mathrm{HONH}_{3}^{+}\) donates a proton (H+) to \(\mathrm{H}_{2}\mathrm{O}\). Therefore, the acid is \(\mathrm{HONH}_{3}^{+}\). #Step 2: Identify the base#

In the given reaction, \(\mathrm{H}_{2}\mathrm{O}\) accepts a proton (H+) from \(\mathrm{HONH}_{3}^{+}\). Therefore, the base is \(\mathrm{H}_{2}\mathrm{O}\). #Step 3: Identify the conjugate acid#
04

In the given reaction, \(\mathrm{H}_{2}\mathrm{O}\) gains a proton (H+) and forms \(\mathrm{H}_{3}\mathrm{O}^{+}\). Therefore, the conjugate acid is \(\mathrm{H}_{3}\mathrm{O}^{+}\). #Step 4: Identify the conjugate base#

In the given reaction, \(\mathrm{HONH}_{3}^{+}\) loses a proton (H+) and forms \(\mathrm{HONH}_{2}\). Therefore, the conjugate base is \(\mathrm{HONH}_{2}\). c. \(\mathrm{HOCl}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightleftharpoons \mathrm{OCl}^{-}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) #Step 1: Identify the acid#
05

In the given reaction, \(\mathrm{HOCl}\) donates a proton (H+) to \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). Therefore, the acid is \(\mathrm{HOCl}\). #Step 2: Identify the base#

In the given reaction, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) accepts a proton (H+) from \(\mathrm{HOCl}\). Therefore, the base is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). #Step 3: Identify the conjugate acid#
06

In the given reaction, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) gains a proton (H+) and forms \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\). Therefore, the conjugate acid is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\). #Step 4: Identify the conjugate base#

In the given reaction, \(\mathrm{HOCl}\) loses a proton (H+) and forms \(\mathrm{OCl}^{-}\). Therefore, the conjugate base is \(\mathrm{OCl}^{-}\).

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Key Concepts

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

Conjugate Acid
When discussing acids and bases in the context of the Br酶nsted-Lowry theory, it's crucial to understand the concept of conjugate acids. A conjugate acid is what forms when a base gains a proton (H鈦). This is a backward look from a dynamic balance which is usually called equilibrium. Let's take an example from the reaction between water (H鈧侽) and HONH鈧冣伜. In this interaction, water is acting as a base by accepting a proton from HONH鈧冣伜, thereby forming the hydronium ion (H鈧僌鈦), which is its conjugate acid.
This concept reveals that every base, when it gains a proton, becomes an acid. This is fundamental in understanding how substances interact in aqueous solutions. It helps to predict the outcome of reactions and the direction they will proceed toward equilibrium.
Conjugate Base
In any Br酶nsted-Lowry acid-base reaction, when an acid donates a proton, it forms its conjugate base. The conjugate base is what remains after the acid has given up its proton. This concept is easier to observe when examining a reaction such as the one between HOCl and C鈧咹鈧匩H鈧.
Here, HOCl donates a proton to C鈧咹鈧匩H鈧 and transforms into OCl鈦, its conjugate base. The importance of the conjugate base is that it shows the substance that can potentially accept a proton in the reverse reaction, illustrating the reversibility and the dynamic nature of aqueous reactions. By recognizing conjugate acid-base pairs, students can better predict and understand these interactions.
  • Acid donates a proton 鈫 Conjugate base forms
  • Base gains a proton 鈫 Conjugate acid forms
Proton Transfer
Proton transfer is the cornerstone of understanding Br酶nsted-Lowry acid and base reactions. It involves the movement of a proton (H鈦) from one molecule to another. This transfer could lead to the creation of a new molecule, changing the acidity or basicity of the environment, often under equilibrium conditions.
For instance, in the reaction of HONH鈧冣伜 with water: HONH鈧冣伜 donates a proton resulting in the formation of HONH鈧 and H鈧僌鈦. The movement of this proton indicates the dynamic nature of chemical reactions, as well as the fact that these interactions can run in both directions in a reversible and exchangeable pattern, essential for understanding complex chemical behavior in aqueous solutions.
Aqueous Reactions
Aqueous reactions are chemical reactions that occur in water. They are common in chemistry because water is an excellent solvent, and many substances dissolve well in it. In aqueous solutions, reactions proceed by the mechanisms of proton transfer between acids and bases.
Take the case of Al(H鈧侽)鈧喡斥伜 reacting with water. Al(H鈧侽)鈧喡斥伜 acts as an acid by transferring a proton to the water molecule, which acts as a base, forming hydronium ions (H鈧僌鈦). Such reactions are pivotal in many natural and industrial processes, like acid-base balancing in physiology or titration in laboratories.
Aqueous reactions exemplify chemical equilibria, where the forward and reverse reactions occur at the same rate, helping students understand how reactants convert to products and vice-versa in a balanced and dynamic system.

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Most popular questions from this chapter

Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\text {a }}\) value for hypochlorous acid is \(3.5 \times 10^{-8}\), which is the stronger base, \(\mathrm{OCl}^{-}\) or \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) ?

Calculate the \(\mathrm{pH}\) of the following solutions: a. \(1.2 M \mathrm{CaBr}_{2}\) b. \(0.84 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}}\right.\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\) ) c. \(0.57 \mathrm{M} \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)\)

Zinc hydroxide is an amphoteric substance. Write equations that describe \(\mathrm{Zn}(\mathrm{OH})_{2}\) acting as a Br酶nsted-Lowry base toward \(\mathrm{H}^{+}\) and as a Lewis acid toward \(\mathrm{OH}^{-}\).

The \(\mathrm{pH}\) of human blood is steady at a value of approximately \(7.4\) owing to the following equilibrium reactions: \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)\) Acids formed during normal cellular respiration react with the \(\mathrm{HCO}_{3}^{-}\) to form carbonic acid, which is in equilibrium with \(\mathrm{CO}_{2}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) During vigorous exercise, a person's \(\mathrm{H}_{2} \mathrm{CO}_{3}\) blood levels were \(26.3 \mathrm{~m} M\), whereas his \(\mathrm{CO}_{2}\) levels were \(1.63 \mathrm{~m} M .\) On resting, the \(\mathrm{H}_{2} \mathrm{CO}_{3}\) levels declined to \(24.9 \mathrm{~m} M\). What was the \(\mathrm{CO}_{2}\) blood level at rest?

A typical sample of vinegar has a pH of \(3.0\). Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.

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