In each equation, label the acids, bases, and conjugate pairs: (a) \(\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) (b) \(\mathrm{HClO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons \mathrm{ClO}_{4}^{-}+\mathrm{H}_{3} \mathrm{SO}_{4}^{+}\) (c) \(\mathrm{HPO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{HSO}_{4}^{-}\)

Short Answer

Expert verified
In (a), HCl is an acid and Cl\(^{-}\) is its conjugate base; H\(_2\)O is a base and H\(_3\)O\(^+\) is its conjugate acid. In (b), HClO\(_4\) is an acid and ClO\(_4^-\) is its conjugate base; H\(_2\)SO\(_4\) is a base and H\(_3\)SO\(_4^+\) is its conjugate acid. In (c), H\(_2\)SO\(_4\) is an acid and HSO\(_4^-\) is its conjugate base; HPO\(_4^{2-}\) is a base and H\(_2\)PO\(_4^-\) is its conjugate acid.

Step by step solution

01

- Identify the acids and bases in equation (a)

In the reaction \(\text{HCl + H}_{2}\text{O} \rightleftharpoons \text{Cl}^{-}\text{ + H}_{3}\text{O}^{+}\), identify the substances that donate protons (acids) and those that accept protons (bases). HCl donates a proton to \(\text{H}_{2}\text{O}\), so HCl is an acid and \(\text{H}_{2}\text{O}\) is a base.
02

- Identify the conjugate pairs in equation (a)

After HCl donates a proton, it becomes \(\text{Cl}^{-}\), and after \(\text{H}_{2}\text{O}\) accepts a proton, it becomes \(\text{H}_{3}\text{O}^{+}\). Therefore, \(\text{Cl}^{-}\) is the conjugate base of HCl and \(\text{H}_{3}\text{O}^{+}\) is the conjugate acid of \(\text{H}_{2}\text{O}\).
03

- Identify the acids and bases in equation (b)

In the reaction \(\text{HClO}_{4}\text{ + H}_{2}\text{SO}_{4} \rightleftharpoons \text{ClO}_{4}^{-}\text{ + H}_{3}\text{SO}_{4}^{+}\), identify the substances that donate protons (acids) and those that accept protons (bases). HClO\text{4} donates a proton to H\(_2\)SO\(_4\), so HClO\(_4\) is an acid and H\(_2\)SO\(_4\) is a base.
04

- Identify the conjugate pairs in equation (b)

After HClO\text{4} donates a proton, it becomes ClO\(_4^{-}\), and after H\text{2}SO\text{4} accepts a proton, it becomes H\(_3\)SO\(_4^{+}\). Therefore, ClO\(_4^{-}\) is the conjugate base of HClO\text{4} and H\(_3\)SO\(_4^{+}\) is the conjugate acid of H\text{2}SO\text{4}.
05

- Identify the acids and bases in equation (c)

In the reaction \(\text{HPO}_{4} ^{2-}\text{ + H}_{2}\text{SO}_{4} \rightleftharpoons \text{H}_{2}\text{PO}_{4}^{-}\text{ + HSO}_{4}^{-}\), identify the substances that donate protons (acids) and those that accept protons (bases). H\text{2}SO\text{4} donates a proton to \(\text{HPO}_{4}^{2-}\), so H\text{2}SO\text{4} is an acid and \(\text{HPO}_{4}^{2-}\) is a base.
06

- Identify the conjugate pairs in equation (c)

After H\text{2}SO\text{4} donates a proton, it becomes HSO\(_{\text{4}}^{-}\), and after H\(\text{PO}_{4}^{2-}\) accepts a proton, it becomes H\(_2\)PO\(_4^{-}\). Therefore, HSO\(_4^{-}\) is the conjugate base of H\(_2\)SO\(_4\) and H\(_2\)PO\(_4^{-}\) is the conjugate acid of HPO\(_4^{2-}\).

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Key Concepts

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

Bronsted-Lowry acids and bases
In the realm of acid-base chemistry, the Bronsted-Lowry theory provides a detailed perspective. According to this theory, acids and bases are defined by their ability to donate or accept protons (hydrogen ions). In simple terms:
  • Acids are substances that donate protons (H+ ions).
  • Bases are substances that accept protons.
Let's explore equation (a) from our exercise: \( \text{HCl + H}_2\text{O} \rightleftharpoons \text{Cl}^{-}\text{ + H}_3\text{O}^{+} \) Here, HCl donates a proton to water (H2O), making HCl an acid and H2O a base.
By labeling the substances, we can understand how protons move from one molecule to another.
Did you know? The strength of an acid or base can be determined by its degree of proton donation or acceptance. Strong acids and bases completely dissociate in water, whereas weak ones do not.
In summary, the Bronsted-Lowry definition expands our understanding by focusing on proton transfer processes that occur during acid-base reactions.
Conjugate acid-base pairs
When an acid donates a proton, it becomes its conjugate base. Conversely, when a base accepts a proton, it becomes its conjugate acid. This transformation forms what we call a conjugate acid-base pair.
Using equation (b) from the exercise: \( \text{HClO}_4\text{ + H}_2\text{SO}_4 \rightleftharpoons \text{ClO}_4^{-}\text{ + H}_3\text{SO}_4^{+} \)
Here, HClO4 donates a proton to become ClO4-. Therefore, HClO4 is an acid, and ClO4- is its conjugate base.
Conversely, H2SO4 accepts a proton to become H3SO4+. Hence, H2SO4 is a base, and H3SO4+ is its conjugate acid. This elegant relationship simplifies complex reactions by grouping related substances.
It's fascinating to note that:
  • Every acid has a conjugate base formed after it donates a proton.
  • Every base has a conjugate acid formed after it accepts a proton.
Understanding these pairs helps predict the direction of acid-base reactions and determine their equilibrium states.
Proton transfer reactions
Proton transfer reactions are at the heart of acid-base chemistry. They involve the movement of protons from acids to bases during a chemical reaction. Let's examine equation (c) from our exercise: \( \text{HPO}_4^{2-}\text{ + H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{PO}_4^{-}\text{ + HSO}_4^{-} \)
In this reaction, H2SO4 donates a proton to HPO42- to form H2PO4- and HSO4-. This demonstrates a proton transfer from the acid (H2SO4) to the base (HPO42-).
Such reactions are characterized by:
  • A clear proton donor (acid).
  • A clear proton acceptor (base).
The direction and extent of these transfer reactions are influenced by the relative strengths of the acids and bases involved.
Always remember, in proton transfer reactionions:
  • Proton donors transform into their corresponding conjugate bases.
  • Proton acceptors transform into their corresponding conjugate acids.
These reactions are essential for many biological processes and industrial applications, illustrating the foundational role of acids and bases in chemistry.

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Most popular questions from this chapter

Many molecules with central atoms from Period 3 or higher take part in Lewis acid-base reactions in which the central atom expands its valence shell. \(\mathrm{SnCl}_{4}\) reacts with \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) as follows: (a) Identify the Lewis acid and the Lewis base in the reaction. (b) Give the \(n l\) designation of the sublevel of the central atom in the acid before it accepts the lone pair.

Methoxide ion, \(\mathrm{CH}_{3} \mathrm{O}^{-}\), and amide ion, \(\mathrm{NH}_{2}^{-}\), are very strong bases that are "leveled" by water. What does this mean? Write the reactions that occur in the leveling process. What species do the two leveled solutions have in common?

Tris(hydroxymethyl)aminomethane, known as TRIS or THAM, is a water-soluble base used in synthesizing surfactants and pharmaceuticals, as an emulsifying agent in cosmetics, and in cleaning mixtures for textiles and leather. In biomedical research, solutions of TRIS are used to maintain nearly constant pH for the study of enzymes and other cellular components. Given that the \(\mathrm{pK}_{\mathrm{b}}\) is \(5.91,\) calculate the \(\mathrm{pH}\) of \(0.075 \mathrm{M}\) TRIS.

Liquid ammonia autoionizes like water: $$ 2 \mathrm{NH}_{3}(l) \longrightarrow \mathrm{NH}_{4}^{+}(a m)+\mathrm{NH}_{2}^{-}(a m) $$ where \((a m)\) represents solvation by \(\mathrm{NH}_{3}\). (a) Write the ion-product constant expression, \(K_{\text {am }}\) (b) What are the strongest acid and base that can exist in \(\mathrm{NH}_{3}(l) ?\) (c) \(\mathrm{HNO}_{3}\) and \(\mathrm{HCOOH}\) are leveled in \(\mathrm{NH}_{3}(l) .\) Explain with equations. (d) At the boiling point of ammonia \(\left(-33^{\circ} \mathrm{C}\right), K_{\text {unt }}=5.1 \times 10^{-27}\) Calculate \(\left[\mathrm{NH}_{4}^{+}\right]\) at this temperature. (c) Pure sulfuric acid also autoionizes. Write the ion-product constant expression, \(K_{\text {sulf }}\), and find the concentration of the conjugate base at \(20^{\circ} \mathrm{C}\left(K_{\mathrm{sulf}}=2.7 \times 10^{-4} \mathrm{at} 20^{\circ} \mathrm{C}\right)\)

Write balanced net ionic equations for the following reactions, and label the conjugate acid-base pairs: (a) \(\mathrm{NaOH}(a q)+\mathrm{NaH}_{2} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{2} \mathrm{HPO}_{4}(a q)\) (b) \(\mathrm{KHSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{KHCO}_{3}(a q)\)

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