Problem 17 - Chemistry: The Central Science

Question

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}{\underline{\phantom{xx}}}^{3-}(a q) \underset{\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)}{\rightleftharpoons}\)

Step-by-Step Solution

Answer

In the given reactions: (a) \(\mathrm{NH}_{4}^{+}\) is the Bronsted-Lowry acid, \(\mathrm{CN}^{-}\) is the Bronsted-Lowry base, \(\mathrm{HCN}\) is the conjugate acid, and \(\mathrm{NH}_{3}\) is the conjugate base. (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) is the Bronsted-Lowry acid, \(\mathrm{H}_{2} \mathrm{O}\) is the Bronsted-Lowry base, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\) is the conjugate acid, and \(\mathrm{OH}^{-}\) is the conjugate base. (c) \(\mathrm{HCOOH}\) is the Bronsted-Lowry acid, \(\mathrm{PO}_{4}^{3-}\) is the Bronsted-Lowry base, \(\mathrm{HPO}_{4}^{2-}\) is the conjugate acid, and \(\mathrm{HCOO}^{-}\) is the conjugate base.

Step 1: Identify the Bronsted-Lowry acid

The Bronsted-Lowry acid is the species that donates a proton (H+ ion). In this case, \(\mathrm{NH}_{4}^{+}\) donates a proton to \(\mathrm{CN}^{-}\).

Step 2: Identify the Bronsted-Lowry base

The Bronsted-Lowry base is the species that accepts a proton (H+ ion). In this case, \(\mathrm{CN}^{-}\) accepts a proton from \(\mathrm{NH}_{4}^{+}\).

Step 3: Identify the conjugate acid

The conjugate acid is the species formed after the base accepts a proton. In this case, \(\mathrm{HCN}\) is the conjugate acid.

Step 4: Identify the conjugate base

The conjugate base is the species formed after the acid donates a proton. In this case, \(\mathrm{NH}_{3}\) is the conjugate base. (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Step 1: Identify the Bronsted-Lowry acid

The Bronsted-Lowry acid is the species that donates a proton (H+ ion). In this case, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) donates a proton to \(\mathrm{H}_{2} \mathrm{O}\).

Step 2: Identify the Bronsted-Lowry base

The Bronsted-Lowry base is the species that accepts a proton (H+ ion). In this case, \(\mathrm{H}_{2} \mathrm{O}\) accepts a proton from \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\).

Step 3: Identify the conjugate acid

The conjugate acid is the species formed after the base accepts a proton. In this case, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\) is the conjugate acid.

Step 4: Identify the conjugate base

The conjugate base is the species formed after the acid donates a proton. In this case, \(\mathrm{OH}^{-}\) is the conjugate base. (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}{\underline{\phantom{xx}}}^{3-}(a q) \underset{\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)}{\rightleftharpoons}\)

Step 1: Identify the Bronsted-Lowry acid

The Bronsted-Lowry acid is the species that donates a proton (H+ ion). In this case, \(\mathrm{HCOOH}\) donates a proton to \(\mathrm{PO}_{4}^{3-}\).

Step 2: Identify the Bronsted-Lowry base

The Bronsted-Lowry base is the species that accepts a proton (H+ ion). In this case, \(\mathrm{PO}_{4}^{3-}\) accepts a proton from \(\mathrm{HCOOH}\).

Step 3: Identify the conjugate acid

The conjugate acid is the species formed after the base accepts a proton. In this case, \(\mathrm{HPO}_{4}^{2-}\) is the conjugate acid.

Step 4: Identify the conjugate base

The conjugate base is the species formed after the acid donates a proton. In this case, \(\mathrm{HCOO}^{-}\) is the conjugate base.

Key Concepts

Conjugate Acid-Base Pairs

Understanding the Bronsted-Lowry acid-base theory can be greatly enhanced by grasping the concept of conjugate acid-base pairs. This concept unveils the reciprocal relationship between acids and bases during chemical reactions.

To simplify, when an acid donates a proton during a reaction, it transforms into a conjugate base; conversely, when a base accepts a proton, it becomes a conjugate acid. Together, this acid and its corresponding base make up a conjugate acid-base pair.

In the exercise provided, for instance, when \( \mathrm{NH}_{4}^{+} \) donates a proton, it becomes \( \mathrm{NH}_{3} \), its conjugate base. Analogously, \( \mathrm{CN}^{-} \) accepts a proton to form its conjugate acid, \( \mathrm{HCN} \). These transformation processes are fundamental for the acid-base reaction to reach equilibrium.

Proton Transfer Reactions

Proton transfer reactions are at the core of the Bronsted-Lowry theory. They are essentially the back-and-forth exchange of a proton between an acid and a base.

Consider the reaction involving \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} \) and \( \mathrm{H}_{2} \mathrm{O} \). Here, \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} \) donates a proton to \( \mathrm{H}_{2} \mathrm{O} \) in a classic example of a proton transfer. This action results in a pair of new substances: the conjugate acid \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} \) and the conjugate base \( \mathrm{OH}^{-} \). The ease with which these proton transfers occur—and the relative strengths of the acids and bases involved—determine the reaction’s direction and position of equilibrium.

Acid-Base Equilibria

Acid-base equilibria are central to understanding reaction dynamics in chemistry. The idea here is that reactions don’t just go to completion but rather reach a state of balance where both reactants and products are present. The position where this balance occurs is known as the equilibrium position.

In the equation involving \( \mathrm{HCOOH} \) and \( \mathrm{PO}_{4}^{3-} \), equilibrium is achieved when the rate at which \( \mathrm{HCOOH} \) donates protons to \( \mathrm{PO}_{4}^{3-} \) is equal to the rate at which its conjugate base \( \mathrm{HCOO}^{-} \) re-accepts protons to form \( \mathrm{HCOOH} \) again. Factors such as concentration, temperature, and the strength of the acids and bases in question can shift the position of equilibrium, making it a crucial concept for predicting the outcome of reactions.