Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Short Answer

Expert verified
a. \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{OH}^-\) are a conjugate acid-base pair. b. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{SO}_{4}^{2-}\) are not a conjugate pair; the correct conjugate base for \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(\mathrm{HSO}_{4}^-\). c. \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^-\) are a conjugate acid-base pair. d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\) are a conjugate acid-base pair.

Step by step solution

01

Identify the acid and base

In this pair, \(\mathrm{H}_{2} \mathrm{O}\) (water) can act as an acid, and \(\mathrm{OH}^{-}\) (hydroxide) can act as a base.
02

Check for conjugate acid-base pair

Water can lose one proton (H+) to form \(\mathrm{OH}^{-}\): \[\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}^{+}\] This means that \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{OH}^{-}\) are a conjugate acid-base pair. #b.# \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\)
03

Identify the acid and base

In this pair, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (sulfuric acid) is the acid, and \(\mathrm{SO}_{4}^{2-}\) (sulfate) is the base.
04

Check for conjugate acid-base pair

We see that \(\mathrm{H}_{2} \mathrm{SO}_{4}\) can lose two protons (2 H+), but we need a species that only differs by one proton (H+), so sulfuric acid is not conjugate with sulfate. We must find the correct conjugate base for \(\mathrm{H}_{2}\mathrm{SO}_{4}\).
05

Find the correct conjugate base

\(\mathrm{H}_{2}\mathrm{SO}_{4}\) loses one proton to form \(\mathrm{HSO}_{4}^{-}\), so the correct conjugate base for \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(\mathrm{HSO}_{4}^{-}\). #c.# \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)
06

Identify the acid and base

In this pair, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (phosphoric acid) is the acid, and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (dihydrogen phosphate) is the base.
07

Check for conjugate acid-base pair

Phosphoric acid can lose one proton (H+) to form dihydrogen phosphate: \[\mathrm{H}_{3}\mathrm{PO}_{4} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{H}^{+}\] This means that \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) are a conjugate acid-base pair. #d.# $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}$
08

Identify the acid and base

In this pair, \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) (acetic acid) is the acid, and \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) (acetate) is the base.
09

Check for conjugate acid-base pair

Acetic acid can lose one proton (H+) to form acetate: \[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2} \rightleftharpoons \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-} + \mathrm{H}^{+}\] This means that \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) and \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) are a conjugate acid-base pair.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time. In acid-base reactions, equilibrium is essential for understanding how acids and bases interact in solution.
For example, in the reversible reaction of water losing a proton (\(\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}^{+}\)), both the forward and backward reactions continuously occur until equilibrium is reached.
At equilibrium:
  • The concentrations of \(\mathrm{H}_{2}\mathrm{O}\), \(\mathrm{OH}^{-}\), and \(\mathrm{H}^{+}\) remain constant.
  • The rate of \(\mathrm{H}_{2}\mathrm{O}\) turning into \(\mathrm{OH}^{-}\) and \(\mathrm{H}^{+}\) equals the rate of \(\mathrm{OH}^{-}\) and \(\mathrm{H}^{+}\) combining back into \(\mathrm{H}_{2}\mathrm{O}\).
Understanding equilibrium helps predict how changes in concentration, temperature, or pressure can shift the balance in an equilibrium reaction.
Acid-Base Reactions
Acid-base reactions involve the transfer of protons from one substance (the acid) to another (the base). These reactions can be seen as the movement of hydrogen ions (\(\mathrm{H}^{+}\)) between different species.
Conjugate acid-base pairs play a crucial role in these reactions. For instance:
  • \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{OH}^{-}\) form a conjugate acid-base pair, with water acting as an acid.
  • \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) and \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) also form a conjugate acid-base pair, where acetic acid donates a proton to form acetate.
This understanding helps in predicting the strength and direction of reactions, as stronger acids will more readily donate protons to become weaker conjugate bases.
Proton Transfer
Proton transfer is the fundamental mechanism in acid-base reactions. It involves an acid donating a proton (\(\mathrm{H}^{+}\)) to a base.
Here's how it works:
  • An acid like \(\mathrm{H}_{3}\mathrm{PO}_{4}\) donates a proton to become its conjugate base \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\).
  • A base like \(\mathrm{SO}_{4}^{2-}\) is derived from \(\mathrm{H}_{2}\mathrm{SO}_{4}\) after losing one or more protons.
In assessing reactions, the key is to identify the protons that can be transferred and the species that accept them. Recognizing these transfers designates the acid-base nature of each entity involved.
Acid and Base Definition
The acid and base definition according to the Bronsted-Lowry theory explains acids as proton donors and bases as proton acceptors. This theory is invaluable in understanding conjugate acid-base pairs.
For example:
  • In the pair \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{OH}^{-}\), water donates a proton to become hydroxide, thus acting as an acid.
  • \(\mathrm{H}_{2}\mathrm{SO}_{4}\), which can donate a proton, acts as an acid, but requires identification of the correct base (\(\mathrm{HSO}_{4}^{-}\) after losing one proton).
This definition is instrumental in predicting the directionality and outcome of acid-base reactions, helping to identify which species will react in a given chemical environment.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the species \(\mathrm{PO}_{4}^{3-}, \mathrm{HPO}_{4}^{2-},\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} .\) Each ion can act as a base in water. Determine the \(K_{\mathrm{b}}\) value for each of these species. Which species is the strongest base?

Calculate the pH of the following solutions: a. \(1.2 \mathrm{M} \mathrm{CaBr}_{2}\) b. \(0.84 M C_{6} H_{5} N H_{3} N O_{3}\left(K_{b} \text { for } C_{6} H_{5} N H_{2}=3.8 \times 10^{-10}\right)\) c. \(0.57 M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)\)

What are the major species present in the following mixtures of bases? a. \(0.050 M \mathrm{NaOH}\) and \(0.050 \mathrm{M} \mathrm{LiOH}\) b. \(0.0010 M \mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.020 \mathrm{M} \mathrm{RbOH}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Calculate the concentration of an aqueous HBr solution that has \(\mathrm{pH}=4.25 .\) HBr is a strong acid.

Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4},\) is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 146.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: \(\mathrm{CO}_{2}\) blood levels increase during cardiac arrest.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free