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Chapter 15: Problem 36

For the following reactions, label each species as an acid or a base. Indicate the species that are conjugates of one another. a) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{CO}_{3}\) \(\mathrm{b)} \mathrm{F}^{-}+\mathrm{HSO}_{4}^{-} \rightleftharpoons \mathrm{HF}+\mathrm{SO}_{4}^{2-}\) c) \(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{SO}_{4}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}\) d) \(\mathrm{H}_{2} \mathrm{~S}+\mathrm{CN}^{-} \rightleftharpoons \mathrm{HS}^{-}+\mathrm{HCN}\)

Short Answer

Expert verified
Label acids, bases and their conjugates for each reaction.

Step by step solution

01

Identifying Acids and Bases for Reaction a

In the reaction \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2} \mathrm{CO}_{3} \), identify the acids and bases:- \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) donates a proton to become \( \mathrm{HPO}_{4}^{2-} \), making it an acid.- \( \mathrm{HCO}_{3}^{-} \) accepts a proton to form \( \mathrm{H}_{2} \mathrm{CO}_{3} \), making it a base.- \( \mathrm{HPO}_{4}^{2-} \) is the conjugate base of \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \).- \( \mathrm{H}_{2} \mathrm{CO}_{3} \) is the conjugate acid of \( \mathrm{HCO}_{3}^{-} \).
02

Identifying Acids and Bases for Reaction b

In the reaction \( \mathrm{F}^{-} + \mathrm{HSO}_{4}^{-} \rightleftharpoons \mathrm{HF} + \mathrm{SO}_{4}^{2-} \), identify the acids and bases:- \( \mathrm{HSO}_{4}^{-} \) donates a proton to become \( \mathrm{SO}_{4}^{2-} \), making it an acid.- \( \mathrm{F}^{-} \) accepts a proton to form \( \mathrm{HF} \), making it a base.- \( \mathrm{HF} \) is the conjugate acid of \( \mathrm{F}^{-} \).- \( \mathrm{SO}_{4}^{2-} \) is the conjugate base of \( \mathrm{HSO}_{4}^{-} \).
03

Identifying Acids and Bases for Reaction c

In the reaction \( \mathrm{HSO}_{4}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{SO}_{4}^{2-} + \mathrm{H}_{3} \mathrm{O}^{+} \), identify the acids and bases:- \( \mathrm{HSO}_{4}^{-} \) donates a proton to become \( \mathrm{SO}_{4}^{2-} \), making it an acid.- \( \mathrm{H}_{2} \mathrm{O} \) accepts a proton to form \( \mathrm{H}_{3} \mathrm{O}^{+} \), making it a base.- \( \mathrm{SO}_{4}^{2-} \) is the conjugate base of \( \mathrm{HSO}_{4}^{-} \).- \( \mathrm{H}_{3} \mathrm{O}^{+} \) is the conjugate acid of \( \mathrm{H}_{2} \mathrm{O} \).
04

Identifying Acids and Bases for Reaction d

In the reaction \( \mathrm{H}_{2} \mathrm{~S} + \mathrm{CN}^{-} \rightleftharpoons \mathrm{HS}^{-} + \mathrm{HCN} \), identify the acids and bases:- \( \mathrm{H}_{2} \mathrm{~S} \) donates a proton to become \( \mathrm{HS}^{-} \), making it an acid.- \( \mathrm{CN}^{-} \) accepts a proton to form \( \mathrm{HCN} \), making it a base.- \( \mathrm{HS}^{-} \) is the conjugate base of \( \mathrm{H}_{2} \mathrm{~S} \).- \( \mathrm{HCN} \) is the conjugate acid of \( \mathrm{CN}^{-} \).

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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 a chemical reaction, acids and bases always come with their partners, known as conjugate acid-base pairs. These pairs play a crucial role in acid-base chemistry.
When an acid donates a proton (\(\text{H}^+\)), it transforms into its conjugate base. Conversely, when a base accepts a proton, it becomes its conjugate acid.
Considered as reversible processes, these transformations simply involve the exchange of a proton.
  • In the reaction \(\text{H}_{2} \text{PO}_{4}^{-} + \text{HCO}_{3}^{-} \rightleftharpoons \text{HPO}_{4}^{2-} + \text{H}_{2} \text{CO}_{3}\), \(\text{H}_{2} \text{PO}_{4}^{-}\) donates a proton to become \(\text{HPO}_{4}^{2-}\), illustrating an acid forming its conjugate base.
  • Simultaneously, \(\text{HCO}_{3}^{-}\) accepts a proton to form \(\text{H}_{2} \text{CO}_{3}\), a base becoming its conjugate acid.
Conjugate pairs highlight the reversible nature of proton exchange between acids and bases, and are key to understanding how these substances interact in water.
Proton Transfer
At the heart of acid-base reactions is the concept of proton transfer. This process is vital in determining whether a species will behave as an acid or a base.
An acid is a proton donor, while a base is a proton acceptor. This interaction is well exemplified by the reaction:
  • \(\text{HSO}_{4}^{-} + \text{H}_{2} \text{O} \rightleftharpoons \text{SO}_{4}^{2-} + \text{H}_{3} \text{O}^{+}\), here \(\text{HSO}_{4}^{-}\) acts as the acid by donating a proton, changing to \(\text{SO}_{4}^{2-}\).
  • \(\text{H}_{2} \text{O}\), meanwhile, accepts this proton to become \(\text{H}_{3} \text{O}^{+}\), functioning as a base.
Through proton transfer, we see how acids and bases interchange their roles depending on the direction of the reaction and their surroundings, demonstrating a dynamic equilibrium.
Chemical Equilibrium
Chemical reactions often don't proceed to completion but instead reach a balance called equilibrium. At this stage, both reactants and products are present in constant, unchanging concentrations.
In acid-base reactions, equilibrium is crucial as it dictates the extent of reaction between the acid and base. The reaction:
  • \(\text{H}_{2} \text{~S} + \text{CN}^{-} \rightleftharpoons \text{HS}^{-} + \text{HCN}\) rests at equilibrium with forward and reverse reactions occurring simultaneously.
  • This means the proton transfer can go both ways, with \(\text{H}_{2} \text{~S}\) donating a proton to \(\text{CN}^{-}\), forming \(\text{HS}^{-}\) and \(\text{HCN}\).
Chemical equilibrium ensures that the rate of proton transfer in one direction equals the rate in the opposite direction, maintaining balance between the acid and its conjugate base.

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Most popular questions from this chapter

Obtain the \(\mathrm{pH}\) corresponding to the following hydronium-ion concentrations a) \(1.0 \times 10^{-4} M\) b) \(3.2 \times 10^{-10} M\) c \(2.3 \times 10^{-5} M\) d) \(2.91 \times 10^{-11} M\)

A wine was tested for acidity, and its \(\mathrm{pH}\) was found to be 3.85 at \(25^{\circ} \mathrm{C}\). What is the hydronium-ion concentration?

What is the sum of the \(\mathrm{pH}\) and the \(\mathrm{pOH}\) for a solution at \(37^{\circ} \mathrm{C}\), where \(K_{w}\) equals \(2.5 \times 10^{-14}\) ?

The bicarbonate ion has the ability to act as an acid in the presence of a base or as a base in the presence of an acid, so it is said to be amphiprotic. Illustrate this behavior with water by writing Br酶nsted-Lowry acid- base reactions. Also illustrate this property by selecting a common strong acid and base to react with the bicarbonate ion.

Liquid ammonia undergoes autoionization similar to that of water: $$ \mathrm{NH}_{3}(l)+\mathrm{NH}_{3}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(l)+\mathrm{NH}_{2}^{-}(l) $$ How would you define an acid and a base similar to the way these terms are defined by the Br酶nsted-Lowry concept for aqueous solutions? Write the expression for the ionproduct constant, \(K_{a m},\) for this autoionization. The value of \(K_{a m}\) is \(5.1 \times 10^{-27}\). What is the concentration of \(\mathrm{NH}_{4}^{+}\) in a neutral solution of liquid ammonia? Suppose you dissolve \(\mathrm{NH}_{4} \mathrm{I}\) in liquid ammonia to give \(\mathrm{NH}_{4}{ }^{+}\) and \(\mathrm{I}^{-}\) ions. Similarly, you dissolve \(\mathrm{KNH}_{2}\) (potassium amide) in liquid ammonia to give a solution of \(\mathrm{K}^{+}\) and \(\mathrm{NH}_{2}^{-}\) ions. Which of these two solutions is acidic and which basic according to your definitions? Now, suppose you add the two solutions together. Write an equation for the neutralization reaction.
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