Chapter 13, Problem 12QAP

Interpretation Introduction

(a)

Interpretation:

The equation to show the basic nature of the given species in water according to the Bronsted-Lowry model should be written.

Concept introduction:

According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

$\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 12QAP

${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+{\text{H}}_{\text{2}}\text{O}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}{\text{NH}}^{+}{\text{+OH}}^{-}$

Explanation of Solution

The given species is ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$.

On reaction with water, it can act as a base by accepting hydrogen ion from the water molecule. The reaction is shown as follows:

${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+{\text{H}}_{\text{2}}\text{O}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}{\text{NH}}^{+}{\text{+OH}}^{-}$

In the above reaction, ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ act as a base, ${\text{H}}_{\text{2}}\text{O}$ acts as an acid, ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}{\text{NH}}^{+}$ is a conjugate acid and ${\text{OH}}^{-}$ is a conjugate base.

Therefore, ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ results in the formation of a basic aqueous solution.

Interpretation Introduction

(b)

Interpretation:

The equation to show the basic nature of the given species in water according to the Bronsted-Lowry model should be written.

Concept introduction:

According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

$\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 12QAP

${\text{HC}}_{\text{2}}{\text{O}}_4^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_4^{}{\text{+OH}}^{-}$

Explanation of Solution

The given species is ${\text{HC}}_{\text{2}}{\text{O}}_4^{-}$.

On reaction with water, it can act as a base by accepting hydrogen ion from the water molecule. The reaction is shown as follows:

${\text{HC}}_{\text{2}}{\text{O}}_4^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_4^{}{\text{+OH}}^{-}$

In the above reaction, ${\text{HC}}_{\text{2}}{\text{O}}_4^{-}$ act as a base, ${\text{H}}_{\text{2}}\text{O}$ acts as an acid, ${\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_4^{}$ is a conjugate acid and ${\text{OH}}^{-}$ is a conjugate base.

Therefore, ${\text{HC}}_{\text{2}}{\text{O}}_4^{-}$ results in the formation of a basic aqueous solution.

Interpretation Introduction

(c)

Interpretation:

The equation to show the basic nature of the given species in water according to the Bronsted-Lowry model should be written.

Concept introduction:

According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

$\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 12QAP

${\text{CN}}^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{HCN+OH}}^{-}$

Explanation of Solution

The given species is ${\text{CN}}^{-}$.

On reaction with water, it can act as a base by accepting hydrogen ion from the water molecule. The reaction is shown as follows:

${\text{CN}}^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{HCN+OH}}^{-}$

In the above reaction, ${\text{CN}}^{-}$ act as a base, ${\text{H}}_{\text{2}}\text{O}$ acts as an acid, $\text{HCN}$ is a conjugate acid and ${\text{OH}}^{-}$ is a conjugate base.

Therefore, ${\text{CN}}^{-}$ results in the formation of a basic aqueous solution.

Interpretation Introduction

(d)

Interpretation:

The equation to show the basic nature of the given species in water according to the Bronsted-Lowry model should be written.

Concept introduction:

According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

$\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 12QAP

${\text{S}}^{2-}+{\text{H}}_{\text{2}}\text{O}\to {\text{HS}}^{-}{\text{+OH}}^{-}$

Explanation of Solution

The given species is ${\text{S}}^{2-}$.

On reaction with water, it can act as a base by accepting hydrogen ion from the water molecule. The reaction is shown as follows:

${\text{S}}^{2-}+{\text{H}}_{\text{2}}\text{O}\to {\text{HS}}^{-}{\text{+OH}}^{-}$

In the above reaction, ${\text{S}}^{2-}$ act as a base, ${\text{H}}_{\text{2}}\text{O}$ acts as an acid, ${\text{HS}}^{-}$ is a conjugate acid and ${\text{OH}}^{-}$ is a conjugate base.

Therefore, ${\text{S}}^{2-}$ results in the formation of a basic aqueous solution.

Interpretation Introduction

(e)

Interpretation:

The equation to show the basic nature of the given species in water according to the Bronsted-Lowry model should be written.

Concept introduction:

According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

$\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 12QAP

${\text{BrO}}_3{}^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{HBrO}}_3{\text{+OH}}^{-}$

Explanation of Solution

The given species is ${\text{BrO}}_3{}^{-}$.

On reaction with water, it can act as a base by accepting hydrogen ion from the water molecule. The reaction is shown as follows:

${\text{BrO}}_3{}^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{HBrO}}_3{\text{+OH}}^{-}$

In the above reaction, ${\text{BrO}}_3{}^{-}$ act as a base, ${\text{H}}_{\text{2}}\text{O}$ acts as an acid, ${\text{HBrO}}_3$ is a conjugate acid and ${\text{OH}}^{-}$ is a conjugate base.

Therefore, ${\text{BrO}}_3{}^{-}$ results in the formation of a basic aqueous solution.

Interpretation Introduction

(f)

Interpretation:

The equation to show the basic nature of the given species in water according to the Bronsted-Lowry model should be written.

Concept introduction:

According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

$\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 12QAP

${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}{\text{H}}_{\text{7}}{}^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_3{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}{\text{H}}_{\text{7}}{\text{+OH}}^{-}$

Explanation of Solution

The given species is ${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}{\text{H}}_{\text{7}}{}^{-}$.

On reaction with water, it can act as a base by accepting hydrogen ion from the water molecule. The reaction is shown as follows:

${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}{\text{H}}_{\text{7}}{}^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_3{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}{\text{H}}_{\text{7}}{\text{+OH}}^{-}$

In the above reaction, ${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}{\text{H}}_{\text{7}}{}^{-}$ act as a base, ${\text{H}}_{\text{2}}\text{O}$ acts as an acid, ${\text{H}}_3{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}\text{H}$ is a conjugate acid and ${\text{OH}}^{-}$ is a conjugate base.

Therefore, ${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{O}}_{\text{5}}{\text{H}}_{\text{7}}{}^{-}$ results in the formation of a basic aqueous solution.