Problem 17 - Chemistry The Central Science

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) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (c)\(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\)

Step-by-Step Solution

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 1: 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) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) #Step 1: Determine the Bronsted-Lowry acid and base#

Step 2: 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)\(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) \(\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\) #Step 1: Determine the Bronsted-Lowry acid and base#

Step 3: 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.