Problem 3 - General Chemistry: Principles and Modern Applications

Question

For each of the following, identify the acids and bases involved in both the forward and reverse directions. (a) \(\mathrm{HOBr}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OBr}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) (b) \(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{SO}_{4}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{S}+\mathrm{OH}^{-}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

(a) \(\mathrm{HOBr}\) (acid), \(\mathrm{H}_{2} \mathrm{O}\) (base), \(\mathrm{OBr}^{-}\) (base), \(\mathrm{H}_{3} \mathrm{O}^{+}\) (acid); (b) \(\mathrm{HSO}_{4}^{-}\) (acid), \(\mathrm{H}_{2} \mathrm{O}\) (base), \(\mathrm{SO}_{4}^{2-}\) (base), \(\mathrm{H}_{3} \mathrm{O}^{+}\) (acid); (c) \(\mathrm{HS}^{-}\) (base), \(\mathrm{H}_{2} \mathrm{O}\) (acid), \(\mathrm{H}_{2} \mathrm{S}\) (acid), \(\mathrm{OH}^{-}\) (base); (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) (acid), \(\mathrm{OH}^{-}\) (base), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (base), \(\mathrm{H}_{2} \mathrm{O}\) (acid).

Step 1: Identify acids and bases for (a)

In the forward direction, \(\mathrm{HOBr}\) acts as the acid because it loses a proton to become \(\mathrm{OBr}^{-}\). \(\mathrm{H}_{2} \mathrm{O}\) is the base as it gains a proton from \(\mathrm{HOBr}\) to turn into \(\mathrm{H}_{3} \mathrm{O}^{+}\). In the reverse direction, \(\mathrm{OBr}^{-}\) gains a proton and thus acts as a base, and \(\mathrm{H}_{3} \mathrm{O}^{+}\) loses a proton to become \(\mathrm{H}_{2} \mathrm{O}\), acting as an acid.

Step 2: Identify acids and bases for (b)

For the reaction \(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\mathrm{SO}_{4}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}\), in the forward direction, \(\mathrm{HSO}_{4}^{-}\) is the acid as it donates a proton to \(\mathrm{H}_{2} \mathrm{O}\), thus forming \(\mathrm{SO}_{4}^{2-}\). \(\mathrm{H}_{2} \mathrm{O}\) is the base as it accepts this proton, turning into \(\mathrm{H}_{3} \mathrm{O}^{+}\). In the reverse direction, \(\mathrm{SO}_{4}^{2-}\) is the base and \(\mathrm{H}_{3} \mathrm{O}^{+}\) is the acid.

Step 3: Identify acids and bases for (c)

The equation for part (c) is \(\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{S}+\mathrm{OH}^{-}\). In the forward direction, \(\mathrm{HS}^{-}\) is a base, as it gains a proton from \(\mathrm{H}_{2} \mathrm{O}\) forming \(\mathrm{H}_{2} \mathrm{S}\). \(\mathrm{H}_{2} \mathrm{O}\) is an acid, as it loses a proton. In the reverse direction, \(\mathrm{OH}^{-}\) is a base and \(\mathrm{H}_{2} \mathrm{S}\) is the acid.

Step 4: Identify acids and bases for (d)

The reaction in part (d) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O}\). Forward direction: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) is an acid since it donates a proton to \(\mathrm{OH}^{-}\) turning into \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). \(\mathrm{OH}^{-}\) is a base in this case since it gains a proton. In the reverse direction: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) is a base, and \(\mathrm{H}_{2} \mathrm{O}\) is an acid.

Key Concepts

Acid-Base Reactions

Acid-base reactions are fascinating interactions where acids and bases exchange protons with each other. According to the Bronsted-Lowry theory:

• An acid is a substance that donates a proton (H⁺).
• A base is a substance that accepts a proton.

These reactions can occur in a number of directions, forward and reverse. Take the example of the reaction \[ \mathrm{HOBr} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OBr}^{-} + \mathrm{H}_{3} \mathrm{O}^{+} \]In the forward direction, \(\mathrm{HOBr}\) acts as an acid by giving up a proton to turn into \(\mathrm{OBr}^{-}\), while water, \(\mathrm{H}_{2} \mathrm{O}\), behaves as a base by gaining that proton to become \(\mathrm{H}_{3} \mathrm{O}^{+}\).
On the reverse side, the roles switch. \(\mathrm{OBr}^{-}\) accepts a proton becoming \(\mathrm{HOBr}\), and so it qualifies as a base. Meanwhile, \(\mathrm{H}_{3} \mathrm{O}^{+}\) donates a proton back to the cycle, functioning as an acid.
This ability to proceed in either direction allows such reactions to achieve equilibrium.

Chemical Equilibrium

Chemical equilibrium in reactions is that unique point where the forward and reverse reactions occur at the same rate.This state is dynamic, meaning molecules continuously react despite no net change in the concentration of reactants and products. Let's consider the chemical reaction:
\[ \mathrm{HSO}_{4}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{SO}_{4}^{2-} + \mathrm{H}_{3} \mathrm{O}^{+} \]
In equilibrium, the formation of \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) balances perfectly with the backward formation of \(\mathrm{HSO}_{4}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}\).

• At this point, the concentrations of reactants and products remain constant although the molecules interchange.
• The position of this equilibrium can be influenced by factors, such as concentration, temperature, and pressure according to Le Chatelier's Principle.

Thus, the reaction conserves balance, continually adjusting to maintain this equilibrium.

Proton Transfer

Proton transfer is a central theme in acid-base chemistry as it involves the exchange of hydrogen ions (protons) between substances.This exchange highlights the reversible nature of acid-base reactions. Consider the reaction:
\[ \mathrm{HS}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{S} + \mathrm{OH}^{-} \]
In this reaction, the \(\mathrm{HS}^{-}\) ion accepts a proton from \(\mathrm{H}_{2} \mathrm{O}\), producing \(\mathrm{H}_{2} \mathrm{S}\).In return, \(\mathrm{H}_{2} \mathrm{O}\) loses a proton, forming \(\mathrm{OH}^{-}\).

• This proton transfer rigorously characterizes the interaction between the acidic (proton donating) and basic (proton accepting) species.
• Through this process, the substances can transform back and forth, an essential aspect of their behavior in solutions.

Understanding this process helps explain not only the dynamic equilibrium in solutions but also the reactivity and strength of different acids and bases.