Problem 144 - Chemistry An Atoms First Approach

Question

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. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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.# \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\)

Step 3: 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.

Step 4: 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 \(\mathrm{H}_{2}\mathrm{SO}_{4}\).

Step 5: 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.# \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Step 6: 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.

Step 7: 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.# $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}$

Step 8: 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.

Step 9: 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.

Key Concepts

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.