Chapter 16, Problem 1E

According to the Brønsted-Lowry theory, lebel each of the following as an acid of a base (a) ${\text{HNO}}_2$ , (b) ${\text{OCl}}^{\text{-}}$ , (C) ${\text{NH}}_2^{-}$ (d) ${\text{NH}}_4^{-}$ ; (e) ${\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}^{\text{-}}$

Interpretation Introduction

(a)

Interpretation:

Whether ${\text{HNO}}_{\text{2}}$ is an acid or base according to the Bronsted and Lowry should be determined.

Concept introduction:

Bronsted and Lowry state that an acid is the species which donates proton and base is the substance that accepts.

Species that vary from one another by one proton only are considered as an acid-base pair.

The acceptance of proton by the base results in the formation of conjugate acids whereas the loss of proton by the acids results in the formation of conjugate bases.

Answer to Problem 1E

According to Bronsted and Lowry ${\text{HNO}}_{\text{2}}$ is labeled as an acid.

Explanation of Solution

Nitrous acid ${\text{HNO}}_{\text{2}}$ is the parent acid. The nitrous acid ${\text{HNO}}_{\text{2}}$ loses one proton and it will form the conjugate base ${\text{NO}}_2^{-}$. The equation for the dissociation ${\text{HNO}}_{\text{2}}$ is as follows:

$\underset{\text{Acid}}{{\text{HNO}}_{\text{2}}}\text{ + }\underset{\text{base}}{{\text{H}}_{\text{2}}\text{O}}\to \underset{\text{Conjugate base}}{{\text{NO}}_2^{-}}+\underset{\text{Conjugate acid}}{{\text{H}}_{\text{3}}{\text{O}}^{+}}$

Therefore, ${\text{HNO}}_{\text{2}}$ is labeled as an acid because it donates a proton to water.

Interpretation Introduction

(b)

Interpretation:

Whether ${\text{OCl}}^{-}$ is an acid or base according to the Bronsted and Lowry should be determined.

Concept introduction:

Bronsted and Lowry state that an acid is the species which donates proton and base is the substance that accepts.

Species that vary from one another by one proton only are considered as an acid-base pair.

The acceptance of proton by the base results in the formation of conjugate acids whereas the loss of proton by the acids results in the formation of conjugate bases.

Answer to Problem 1E

According to Bronsted and Lowry $\text{HOCl}$ is labeled as an acid.

Explanation of Solution

Hypochlorous acid $\text{HOCl}$ is the parent acid. The hypochlorous acid $\text{HOCl}$ loses one proton and it will form the conjugate base ${\text{OCl}}^{-}$. The equation for the dissociation $\text{HOCl}$ is as follows:

$\underset{\text{Acid}}{\text{HOCl}}\text{ + }\underset{\text{base}}{{\text{H}}_{\text{2}}\text{O}}\to \underset{\text{Conjugate base}}{{\text{OCl}}^{-}}+\underset{\text{Conjugate acid}}{{\text{H}}_{\text{3}}{\text{O}}^{+}}$

Therefore, $\text{HOCl}$ is labeled as an acid because it donates a proton to water.

Interpretation Introduction

(c)

Interpretation:

Whether ${\text{NH}}_2^{-}$ is an acid or base according to the Bronsted and Lowry should be determined.

Concept introduction:

Bronsted and Lowry state that an acid is the species which donates proton and base is the substance that accepts.

Species that vary from one another by one proton only are considered as an acid-base pair.

The acceptance of proton by the base results in the formation of conjugate acids whereas the loss of proton by the acids results in the formation of conjugate bases.

Answer to Problem 1E

According to Bronsted and Lowry ${\text{NH}}_2^{-}$ is labeled as a base.

Explanation of Solution

The reaction of ${\text{NH}}_2^{-}$ and water is as follows:

$\underset{\text{Base}}{{\text{NH}}_2^{-}}\text{ + }\underset{\text{acid}}{{\text{H}}_{\text{2}}\text{O}}\to \underset{\text{Conjugate acid}}{{\text{NH}}_3}+\underset{\text{Conjugate base}}{{\text{OH}}^{-}}$

In the above reaction ${\text{NH}}_2^{-}$ accepts a proton from water and thus ${\text{NH}}_2^{-}$ is labeled as a base.

Interpretation Introduction

(d)

Interpretation:

Whether ${\text{NH}}_4^{+}$ is an acid or base according to the Bronsted and Lowry should be determined.

Concept introduction:

Bronsted and Lowry state that an acid is the species which donates proton and base is the substance that accepts.

Species that vary from one another by one proton only are considered as an acid-base pair.

The acceptance of proton by the base results in the formation of conjugate acids whereas the loss of proton by the acids results in the formation of conjugate bases.

Answer to Problem 1E

According to Bronsted and Lowry ${\text{NH}}_4^{+}$ is labeled as an acid.

Explanation of Solution

The reaction of ${\text{NH}}_4^{+}$ and water is as follows:

$\underset{\text{acid}}{{\text{NH}}_4^{+}}\text{ + }\underset{\text{base}}{{\text{H}}_{\text{2}}\text{O}}\to \underset{\text{Conjugate base}}{{\text{NH}}_3}+\underset{\text{Conjugate acid}}{{\text{H}}_{\text{3}}{\text{O}}^{+}}$

In the above reaction, ${\text{NH}}_4^{+}$ donates a proton to the water molecule and thus ${\text{NH}}_4^{+}$ is labeled as a base. Ammonia $\left({\text{NH}}_3\right)$ is its conjugate base.

Interpretation Introduction

(e)

Interpretation:

Whether ${\text{CH}}_3{\text{NH}}_3^{+}$ is an acid or base according to the Bronsted and Lowry should be determined.

Concept introduction:

Bronsted and Lowry state that an acid is the species which donates proton and base is the substance that accepts.

Species that vary from one another by one proton only are considered as an acid-base pair.

The acceptance of proton by the base results in the formation of conjugate acids whereas the loss of proton by the acids results in the formation of conjugate bases.

Answer to Problem 1E

According to Bronsted and Lowry ${\text{CH}}_3{\text{NH}}_3^{+}$ is labeled as an acid.

Explanation of Solution

The reaction of ${\text{CH}}_3{\text{NH}}_3^{+}$ and water is as follows:

$\underset{\text{acid}}{{\text{CH}}_3{\text{NH}}_3^{+}}\text{ + }\underset{\text{base}}{{\text{H}}_{\text{2}}\text{O}}\to \underset{\text{Conjugate base}}{{\text{CH}}_3{\text{NH}}_2}+\underset{\text{Conjugate acid}}{{\text{H}}_{\text{3}}{\text{O}}^{+}}$

In the above reaction ${\text{CH}}_3{\text{NH}}_3^{+}$ donates a proton to the water molecule and thus ${\text{CH}}_3{\text{NH}}_3^{+}$ is labeled as a base. Methylamine $\left({\text{CH}}_3{\text{NH}}_{\text{2}}\right)$ is its conjugate base.