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Chapter 15: Problem 27

For each of the following reactions, identify the Br酶nsted-Lowry acids and bases. What are the conjugate acid/base pairs? (a) \(\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCN}+\mathrm{OH}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HS}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{~S}\)

Short Answer

Expert verified
(a) Acid/Base: H2O/CN-, Conjugates: OH-/HCN. (b) Acid/Base: H3O+/HCO3-, Conjugates: H2O/H2CO3. (c) Acid/Base: CH3COOH/HS-, Conjugates: CH3COO-/H2S.

Step by step solution

01

Understand the Br酶nsted-Lowry Acid-Base Theory

The Br酶nsted-Lowry theory defines an acid as a substance that donates a proton (H+), while a base is a substance that accepts a proton. In a chemical reaction, acids and bases form conjugate base and acid pairs, respectively, through the donation and acceptance of protons.
02

Analyze Reaction (a)

In the reaction \ \( \mathrm{CN}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^{-} \ \), the cyanide ion \( \mathrm{CN}^{-} \) accepts a proton from water \( \mathrm{H}_{2} \mathrm{O} \) and forms \( \mathrm{HCN} \), making \( \mathrm{CN}^{-} \) a base and \( \mathrm{H}_{2} \mathrm{O} \) an acid. After donating a proton, \( \mathrm{H}_{2} \mathrm{O} \) becomes \( \mathrm{OH}^{-} \), its conjugate base, and the accepted proton transforms \( \mathrm{CN}^{-} \) into \( \mathrm{HCN} \), its conjugate acid.**Conjugate acid/base pairs are:** \( (\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}) \) and \( (\mathrm{CN}^{-}, \mathrm{HCN}) \).
03

Analyze Reaction (b)

For the reaction \( \mathrm{HCO}_{3}^{-} + \mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3} + \mathrm{H}_{2} \mathrm{O} \), bicarbonate \( \mathrm{HCO}_{3}^{-} \) accepts a proton from hydronium \( \mathrm{H}_{3} \mathrm{O}^{+} \), forming carbonic acid \( \mathrm{H}_{2} \mathrm{CO}_{3} \), making \( \mathrm{HCO}_{3}^{-} \) a base and \( \mathrm{H}_{3} \mathrm{O}^{+} \) an acid. Consequently, \( \mathrm{H}_{3} \mathrm{O}^{+} \) becomes \( \mathrm{H}_{2} \mathrm{O} \), its conjugate base, while \( \mathrm{HCO}_{3}^{-} \) turns into \( \mathrm{H}_{2} \mathrm{CO}_{3} \), its conjugate acid. **Conjugate acid/base pairs are:** \( (\mathrm{HCO}_{3}^{-}, \mathrm{H}_{2} \mathrm{CO}_{3}) \) and \( (\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{H}_{2} \mathrm{O}) \).
04

Analyze Reaction (c)

In the reaction \( \mathrm{CH}_{3} \mathrm{COOH} + \mathrm{HS}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-} + \mathrm{H}_{2} \mathrm{S} \), acetic acid \( \mathrm{CH}_{3} \mathrm{COOH} \) donates a proton to \( \mathrm{HS}^{-} \), forming acetate \( \mathrm{CH}_{3} \mathrm{COO}^{-} \) and \( \mathrm{H}_{2} \mathrm{S} \). Thus, \( \mathrm{CH}_{3} \mathrm{COOH} \) acts as an acid, and \( \mathrm{HS}^{-} \) acts as a base. After donating a proton, \( \mathrm{CH}_{3} \mathrm{COOH} \) becomes \( \mathrm{CH}_{3} \mathrm{COO}^{-} \), its conjugate base, and \( \mathrm{HS}^{-} \) becomes \( \mathrm{H}_{2} \mathrm{S} \), its conjugate acid. **Conjugate acid/base pairs are:** \( (\mathrm{CH}_{3} \mathrm{COOH}, \mathrm{CH}_{3} \mathrm{COO}^{-}) \) and \( (\mathrm{HS}^{-}, \mathrm{H}_{2} \mathrm{S}) \).

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Key Concepts

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

Conjugate Acid-Base Pairs
In the Br酶nsted-Lowry framework, every acid has a conjugate base, and every base has a conjugate acid. These pairs are formed through the process of proton donation by an acid and proton acceptance by a base.
Whenever an acid donates a proton, it transforms into its conjugate base, and when a base accepts a proton, it becomes its conjugate acid.
This relationship is crucial for understanding acid-base reactivity.
  • Reaction (a): The pairs are
    • \( ( ext{H}_2 ext{O}, ext{OH}^-) \) 鈥� Water is the acid turning into hydroxide, its conjugate base.
    • \( ( ext{CN}^-, ext{HCN}) \) 鈥� Cyanide ion is the base turning into hydrogen cyanide, the conjugate acid.
  • Reaction (b): Bicarbonate and hydronium pair up as
    • \( ( ext{HCO}_3^-, ext{H}_2 ext{CO}_3) \) 鈥� Bicarbonate turns into carbonic acid, the conjugate acid.
    • \( ( ext{H}_3 ext{O}^+, ext{H}_2 ext{O}) \) 鈥� Hydronium turns into water, its conjugate base.
  • Reaction (c): Acetic acid partners with hydrogen sulfide
    • \( ( ext{CH}_3 ext{COOH}, ext{CH}_3 ext{COO}^-) \) 鈥� Acetic acid transforms into acetate, the conjugate base.
    • \( ( ext{HS}^-, ext{H}_2 ext{S}) \) 鈥� Bisulfide ion becomes hydrogen sulfide, the conjugate acid.
Proton Transfer
Proton transfer is at the heart of Br酶nsted-Lowry acid-base reactions. It involves the movement of a proton (\( ext{H}^+\)) from the acid to the base. This transfer is what forms conjugate acid-base pairs.
To visualize this, imagine the proton as a baton being passed in a relay race. The acid is passing the baton, and the base is grabbing it and changing its form in the process.
Here鈥檚 how this looks in each reaction:
  • In reaction (a), the proton is transferred from water (\( ext{H}_2 ext{O}\)) to the cyanide ion (\( ext{CN}^-\)). This action turns water into hydroxide (\( ext{OH}^-\)).
  • For reaction (b), hydronium (\( ext{H}_3 ext{O}^+\)) donates a proton to transform bicarbonate (\( ext{HCO}_3^-\)) into carbonic acid (\( ext{H}_2 ext{CO}_3\)).
  • In reaction (c), acetic acid (\( ext{CH}_3 ext{COOH}\)) gives up its proton to the bisulfide ion (\( ext{HS}^-\)), producing hydrogen sulfide (\( ext{H}_2 ext{S}\)).
Acid-Base Reactions
Acid-base reactions are a foundational concept in chemistry, characterized by the exchange of protons (\( ext{H}^+\)) between reactants.
These reactions are a special type of chemical reaction where the roles of acids and bases can be reversed, demonstrating the versatility of chemical processes.
Let鈥檚 break down their function in the given reactions:
  • In reaction (a), water acts as the acid, donating a proton to cyanide, which acts as the base. This allows us to see both substances switching between acidic and basic behavior.
  • For reaction (b), hydronium donates a proton to bicarbonate, showcasing a typical acid-base interaction with the common species in many aqueous solutions.
  • Finally, reaction (c) is another classic acid-base scenario where acetic acid is the proton donor and bisulfide ion is the proton acceptor.
Understanding these interactions helps in predicting the behavior of substances in different chemical environments. This knowledge also supports further learning in fields like biochemistry and environmental science.

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Most popular questions from this chapter

Fill in the following table, and indicate whether the solution is acidic, basic, or neutral. $$ \begin{array}{ccccc} \hline & \mathrm{pH} & {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], M} & \mathrm{pOH} & {\left[\mathrm{OH}^{-}\right], M} \\ \hline \text { (a) } & 10.34 & & & \\ \text { (b) } & & & 10.34 & \\ \text { (c) } & & 0.412 & & \\ \text { (d) } & & & & 11.2 \times 10^{-12} \\ \hline \end{array} $$

Identify the Lewis acid and the Lewis base in each reaction. (a) \(\mathrm{I}_{2}(\mathrm{~s})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_{3}^{-}(\mathrm{aq})\) (b) \(\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{BF}_{3}(\mathrm{~g}) \rightarrow \mathrm{O}_{2} \mathrm{SBF}_{3}(\mathrm{~s})\) (c) \(\mathrm{Au}^{+}(\mathrm{aq})+2 \mathrm{CN}^{-}(\mathrm{aq}) \rightarrow\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})\) (d) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(\) aq \()\)

Write the formula, and give the name of the conjugate acid of each of the following bases. (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{HCO}_{3}^{-}\) (c) \(\mathrm{Br}^{-}\)

Coniine ( 2 -propylpiperidine) is a weak base. It has the formula \(\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{~N}\). Calculate the \(\mathrm{pH}\) of a \(0.500 \mathrm{M}\) solution of coniine \(\left(\mathrm{p} K_{\mathrm{b}}=3.1\right)\) Coniine is extracted from the plant Conium maculatum, also called hemlock. This harmless-looking relative of the carrot produces a deadly poison that killed the Greek philosopher Socrates in 399 ?.c. Socrates was a gadfly. He set about demonstrating that many prominent Athenians were more concerned with their own self-interest than with the needs of the society as a whole. He was charged with impiety, corrupting the youth, and disturbing the society. Socrates defended himself and was found guilty by the other Athenians. They asked him to recommend his own punishment; he recommended that he be compensated for his work with young people because he had no other source of income. This suggestion angered his peers, who sentenced him to death. Socrates was not well guarded, and the Athenians hoped he would escape. But he felt a moral obligation to follow the edict of the state. So he drank hemlock and died. Three of Plato's dialogues speak of the events surrounding the death of Socrates. Apology depicts the trial, Crito includes his reasons for choosing death, and Phaedro comprises his musings as the coniine took effect.

What are the products of each of the following acidbase reactions? Indicate the acid and its conjugate base, and the base and its conjugate acid. (a) \(\mathrm{HClO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow\) (b) \(\mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \rightarrow\) (c) \(\mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-} \rightarrow\)
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