Identify and label the Bronsted-Lowry acid, its conjugate base, the Bronsted- Lowry base, and its conjugate acid in each of the following equations: (a) \(\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{NO}_{3}^{-}\) (b) \(\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HCN}+\mathrm{OH}^{-}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{Cl}^{-} \longrightarrow \mathrm{HCl}+\mathrm{HSO}_{4}^{-}\) (d) \(\mathrm{HSO}_{4}^{-}+\mathrm{OH}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{O}^{2-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{OH}^{-}\) (f) \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}(\mathrm{OH})\right]^{+}+\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \longrightarrow\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}+\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}\) (g) \(\mathrm{H}_{2} \mathrm{S}+\mathrm{NH}_{2}^{-} \longrightarrow \mathrm{HS}^{-}+\mathrm{NH}_{3}\)

Step by step solution

01

- Identify the Acid and Base in (a)

In the equation \(\mathrm{HNO}_{3}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{H}_{3}\mathrm{O}^{+}+\mathrm{NO}_{3}^{-}\), \mathrm{HNO}_{3} donates a proton (\(\mathrm{H}^{+}\)) to \mathrm{H}_{2}\mathrm{O}, thus \mathrm{HNO}_{3} is the Bronsted-Lowry acid, and \mathrm{NO}_{3}^{-} is its conjugate base. \mathrm{H}_{2}\mathrm{O} is the Bronsted-Lowry base, and \mathrm{H}_{3}\mathrm{O}^{+} is its conjugate acid.

02

- Identify the Acid and Base in (b)

In the equation \(\mathrm{CN}^{-}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{HCN}+\mathrm{OH}^{-}\), \mathrm{H}_{2}\mathrm{O} donates a proton to \mathrm{CN}^{-}, making \mathrm{H}_{2}\mathrm{O} the Bronsted-Lowry acid and \mathrm{HCN} its conjugate base. \mathrm{CN}^{-} is the Bronsted-Lowry base, with \mathrm{OH}^{-} being its conjugate acid.

03

- Identify the Acid and Base in (c)

For \(\mathrm{H}_{2}\mathrm{SO}_{4}+\mathrm{Cl}^{-} \longrightarrow \mathrm{HCl}+\mathrm{HSO}_{4}^{-}\), \mathrm{H}_{2}\mathrm{SO}_{4} is the acid as it donates a proton to \mathrm{Cl}^{-}, and \mathrm{HSO}_{4}^{-} is its conjugate base. \mathrm{Cl}^{-} is the base with \mathrm{HCl} as its conjugate acid.

04

- Identify the Acid and Base in (d)

In the reaction \(\mathrm{HSO}_{4}^{-}+\mathrm{OH}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}+\mathrm{H}_{2}\mathrm{O}\), \mathrm{HSO}_{4}^{-} acts as the acid by donating a proton to \mathrm{OH}^{-}, and \mathrm{SO}_{4}^{2-} is its conjugate base. Conversely, \mathrm{OH}^{-} is the base and \mathrm{H}_{2}\mathrm{O} is its conjugate acid.

05

- Identify the Acid and Base in (e)

For the equation \(\mathrm{O}^{2-}+\mathrm{H}_{2}\mathrm{O} \longrightarrow 2\mathrm{OH}^{-}\), \mathrm{H}_{2}\mathrm{O} is the acid, donating a proton to \mathrm{O}^{2-}, making \mathrm{OH}^{-} the conjugate base of \mathrm{H}_{2}\mathrm{O}. \mathrm{O}^{2-} acts as the base, and two \mathrm{OH}^{-} ions are produced, indicating two instances of the conjugate acid-base pair.

06

- Identify the Acid and Base in (f)

In the complex ion exchange reaction \(\left[\mathrm{Cu}\left(\mathrm{H}_{2}\mathrm{O}\right)_{3}(\mathrm{OH})\right]^{+}+\left[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3+} \longrightarrow \left[\mathrm{Cu}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+}+\left[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}\), the \left[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3+} species donates a proton to \left[\mathrm{Cu}\left(\mathrm{H}_{2}\mathrm{O}\right)_{3}(\mathrm{OH})\right]^{+}, making it the acid with \left[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+} as its conjugate base. Conversely, \left[\mathrm{Cu}\left(\mathrm{H}_{2}\mathrm{O}\right)_{3}(\mathrm{OH})\right]^{+} is the base, and \left[\mathrm{Cu}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+} is its conjugate acid.

07

- Identify the Acid and Base in (g)

In the reaction \(\mathrm{H}_{2}\mathrm{S}+\mathrm{NH}_{2}^{-} \longrightarrow \mathrm{HS}^{-}+\mathrm{NH}_{3}\), \mathrm{H}_{2}\mathrm{S} is the Bronsted-Lowry acid as it donates a proton to \mathrm{NH}_{2}^{-}, and \mathrm{HS}^{-} is its conjugate base. \mathrm{NH}_{2}^{-} is the Bronsted-Lowry base, and \mathrm{NH}_{3} is its conjugate acid.

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Key Concepts

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

Acid-Base Reactions

Understanding acid-base reactions is fundamental in chemistry, as they are widespread in both nature and laboratory settings. An acid-base reaction involves the transfer of a proton (hydrogen ion, \(H^+\)) from one substance to another. According to the Bronsted-Lowry theory, an acid is a substance that can donate a proton, while a base is a substance that can accept a proton.

Using the reactions provided in the exercise, we can see that when \(HNO_3\) reacts with \(H_2O\), \(HNO_3\) donates a proton to the \(H_2O\), forming \(H_3O^+\) and \(NO_3^-\). This transfer of a proton is what characterizes the acid-base reaction. After identifying the acid and base in such reactions, students can better understand the behavior of substances when mixed together and predict the outcomes of similar reactions.

Conjugate Acid-Base Pairs

In each acid-base reaction, there's a pair of two species that differ only by a proton: these are the conjugate acid-base pairs. A conjugate acid-base pair consists of a Bronsted-Lowry acid and its corresponding base after a proton has been transferred. For example, in the equation involving \(CN^-\) and \(H_2O\), \(H_2O\) donates a proton to become \(OH^-\), which is the base. The \(CN^-\), upon accepting a proton, becomes \(HCN\), the conjugate acid.

An important note for students is that recognizing these pairs helps in understanding the reaction's direction and identifying the strength of acids and bases. Stronger acids have weaker conjugate bases, and vice versa.

Proton Transfer Reactions

Proton transfer reactions are the heart of acid-base chemistry. They revolve around the movement of protons from one species to another. A simple tip to remember is that the substance that loses a proton becomes a conjugate base, while the one gaining a proton becomes a conjugate acid. The example of \(HSO_4^-\) donating a proton to \(OH^-\) showcases this proton transfer where \(OH^-\) becomes \(H_2O\) and \(HSO_4^-\) becomes \(SO_4^{2-}\).

Understanding the movement of protons can assist students in visualizing and memorizing reactions. Additionally, it highlights the fact that these reactions can proceed forwards and backwards, implying a state of equilibrium can be attained.

Chemical Equilibrium

Chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. This means the rate of the forward reaction (acid donating a proton) is equal to the rate of the reverse reaction (base receiving a proton). For instance, in the reaction between \(H_2S\) and \(NH_2^-\), equilibrium would be reached when the rate of formation of \(HS^-\) and \(NH_3\) equals the rate at which they reform \(H_2S\) and \(NH_2^-\).

Equilibrium is a dynamic process, and understanding it is essential for predicting the extent of a reaction as well as its spontaneity. Students should also understand that equilibrium can be influenced by various factors, such as concentration, temperature, and pressure, which are governed by Le Chatelier's principle.