Question

Identify the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also identify the conjugate acid and conjugate base of each on the right side: (a) ${\mathrm{NH}}_4^{+}(aq)+{\mathrm{CN}}^{-}(aq)\rightleftharpoons \mathrm{HCN}(aq)+{\mathrm{NH}}_3(aq)$ (b) ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons$ $$ ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}(aq)+{\mathrm{OH}}^{-}(aq)$ (c)$\mathrm{HCOOH}(aq)+{\mathrm{PO}}_4^{3-}(aq)\rightleftharpoons$ ${\mathrm{HCOO}}^{-}(aq)+{\mathrm{HPO}}_4^{2-}(aq)$

Short answer

In the given reactions, the Bronsted-Lowry acids, bases, conjugate acids, and conjugate bases are as follows: (a) ${\mathrm{NH}}_4^{+}$ is the acid, ${\mathrm{CN}}^{-}$ is the base, $\mathrm{HCN}$ is the conjugate acid, and ${\mathrm{NH}}_3$ is the conjugate base. (b) ${\mathrm{H}}_2\mathrm{O}$ is the acid, ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}$ is the 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 acid, ${\mathrm{PO}}_4^{3-}$ is the base, ${\mathrm{HPO}}_4^{2-}$ is the conjugate acid, and ${\mathrm{HCOO}}^{-}$ is the conjugate base.

Step-by-step solution

On the left side of the equation, we can see that ${\mathrm{NH}}_4^{+}$ donates a proton to ${\mathrm{CN}}^{-}$, which accepts the proton. Therefore, ${\mathrm{NH}}_4^{+}$ is the Bronsted-Lowry acid, and ${\mathrm{CN}}^{-}$ is the Bronsted-Lowry base. #Step 2: Determine the conjugate acid and conjugate base#

On the right side of the equation, the products after the reaction are $\mathrm{HCN}$ and ${\mathrm{NH}}_3$. As ${\mathrm{NH}}_4^{+}$ loses a proton to form ${\mathrm{NH}}_3$, ${\mathrm{NH}}_3$ is the conjugate base of the acid, while ${\mathrm{CN}}^{-}$ gains a proton to form $\mathrm{HCN}$, making $\mathrm{HCN}$ the conjugate acid of the base. (b) ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons$ ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}(aq)+{\mathrm{OH}}^{-}(aq)$ #Step 1: Determine the Bronsted-Lowry acid and base#

In this equation, the proton is being transferred from ${\mathrm{H}}_2\mathrm{O}$ to ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}$. So, the acid is ${\mathrm{H}}_2\mathrm{O}$, and the base is ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}$. #Step 2: Determine the conjugate acid and conjugate base#

As ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}$ gains a proton and forms ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}$, this species is the conjugate acid of the base. ${\mathrm{H}}_2\mathrm{O}$ loses a proton to form ${\mathrm{OH}}^{-}$, which is therefore the conjugate base of the acid. (c)$\mathrm{HCOOH}(aq)+{\mathrm{PO}}_4^{3-}(aq)\rightleftharpoons$ ${\mathrm{HCOO}}^{-}(aq)+{\mathrm{HPO}}_4^{2-}(aq)$ #Step 1: Determine the Bronsted-Lowry acid and base#

In this equation, the proton is being transferred from $\mathrm{HCOOH}$ to ${\mathrm{PO}}_4^{3-}$. Hence, the acid is $\mathrm{HCOOH}$, and the base is ${\mathrm{PO}}_4^{3-}$. #Step 2: Determine the conjugate acid and conjugate base#

The $\mathrm{HCOOH}$ loses a proton to form ${\mathrm{HCOO}}^{-}$, which is the conjugate base of the acid. Meanwhile, the ${\mathrm{PO}}_4^{3-}$ gains a proton to form ${\mathrm{HPO}}_4^{2-}$, which is the conjugate acid of the base.

Key concepts

Acid-Base Reactions

Acid-base reactions are fundamental chemical processes where an acid and a base interact to produce a conjugate base and a conjugate acid. According to the Bronsted-Lowry theory, an acid is a substance that can donate a proton (${\mathrm{H}}^{+}$), while a base is one that can accept a proton.

When these reactions occur, they typically result in the transfer of a proton from the acid to the base. This is the reason they are commonly referred to as proton transfer reactions. For example, in the reaction: ${\mathrm{NH}}_4^{+}+{\mathrm{CN}}^{-}\rightleftharpoons \mathrm{HCN}+{\mathrm{NH}}_3$, ${\mathrm{NH}}_4^{+}$ donates a proton to ${\mathrm{CN}}^{-}$.

This proton transfer showcases the key interaction in Bronsted-Lowry acid-base reactions. Understanding these interactions is essential for studying chemical processes that involve acids and bases, such as buffer solutions and metabolic pathways.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are related by the gain or loss of a proton. When an acid donates a proton, it becomes its conjugate base, and when a base accepts a proton, it turns into its conjugate acid. This relationship is crucial for balancing reactions and understanding equilibrium.

In the example reaction: ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}+{\mathrm{OH}}^{-}$,${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{OH}}^{-}$ form one conjugate pair, with ${\mathrm{OH}}^{-}$ being the conjugate base after the loss of a proton.

Meanwhile, ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}$ transitions into ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}$ on accepting a proton, thus forming the conjugate acid. Recognizing these pairs is straightforward; just look to see which species are related by a single proton difference.

Proton Transfer Reactions

Proton transfer reactions are at the heart of many acid-base interactions. They involve the movement of a proton from the acid to the base, facilitating the conversion of reactants to products. Because the transfer of a single proton can vastly change a molecule's structure and properties, these reactions are significant in both synthetic and natural processes.

For instance, in the equation: $\mathrm{HCOOH}+{\mathrm{PO}}_4^{3-}\rightleftharpoons {\mathrm{HCOO}}^{-}+{\mathrm{HPO}}_4^{2-}$, $\mathrm{HCOOH}$ donates a proton, becoming ${\mathrm{HCOO}}^{-}$, while ${\mathrm{PO}}_4^{3-}$ accepts that proton to form ${\mathrm{HPO}}_4^{2-}$.

These reactions can also delineate strong acids or bases from their weaker counterparts, as stronger acids have a higher tendency to release protons. This concept is applicable across chemistry, from buffering solutions in biochemistry to designing industrial catalysts.