Chapter 13, Problem 10QAP

Interpretation Introduction

(a)

Interpretation:

The equation to show the acidic 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 10QAP

$\text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_3{\text{OH}}^{+}+{\text{H}}_{\text{2}}\text{O}\to \text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_3{\text{O+H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is $\text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_3{\text{OH}}^{+}$.

On reaction with water, it can act as an acid by donating hydrogen ion to the water. The reaction is shown as follows:

$\text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_3{\text{OH}}^{+}+{\text{H}}_{\text{2}}\text{O}\to \text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_3{\text{O+H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, $\text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_3{\text{OH}}^{+}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, $\text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_3\text{O}$ is a conjugate base and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is a conjugate acid.

Interpretation Introduction

(b)

Interpretation:

The equation to show the acidic 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 10QAP

${\text{HSO}}_4^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{SO}}_4^{2-}+{\text{H}}_3{\text{O}}^{+}$

Explanation of Solution

The given species is ${\text{HSO}}_4^{-}$.

On reaction with water, it can act as an acid by donating hydrogen ion to the water. The reaction is shown as follows:

${\text{HSO}}_4^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{SO}}_4^{2-}+{\text{H}}_3{\text{O}}^{+}$

In the above reaction, ${\text{HSO}}_4^{-}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, ${\text{SO}}_4^{2-}$ is a conjugate base and ${\text{H}}_3{\text{O}}^{+}$ is a conjugate acid.

Interpretation Introduction

(c)

Interpretation:

The equation to show the acidic 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 10QAP

${\text{HNO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{NO}}_2{}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is ${\text{HNO}}_{\text{2}}$.

On reaction with water, it can act as an acid by donating hydrogen ion to the water. The reaction is shown as follows:

${\text{HNO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{NO}}_2{}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, ${\text{HNO}}_{\text{2}}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, ${\text{NO}}_2{}^{-}$ is a conjugate base and ${\text{H}}_3{\text{O}}^{+}$ is a conjugate acid.

Interpretation Introduction

(d)

Interpretation:

The equation to show the acidic 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 10QAP

$\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{2+}{\text{+H}}_{\text{2}}\text{O}\to \text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_5^{+}\left(\text{OH}\right)+{\text{H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is as follows:

$\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{2+}$

On reaction with water, it can act as an acid by donating hydrogen ion to the water. The reaction is shown as follows:

$\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{2+}{\text{+H}}_{\text{2}}\text{O}\to \text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_5^{+}\left(\text{OH}\right)+{\text{H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, $\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{2+}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, $\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_5^{+}\left(\text{OH}\right)$ is a conjugate base and ${\text{H}}_3{\text{O}}^{+}$ is a conjugate acid.

Interpretation Introduction

(e)

Interpretation:

The equation to show the acidic 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 10QAP

${\text{H}}_2{\text{C}}_2{\text{H}}_3{\text{O}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}{}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is as follows:

${\text{H}}_2{\text{C}}_2{\text{H}}_3{\text{O}}_{\text{2}}$

On reaction with water, it can act as an acid by donating hydrogen ion to the water. The reaction is shown as follows:

${\text{H}}_2{\text{C}}_2{\text{H}}_3{\text{O}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}{}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, ${\text{H}}_2{\text{C}}_2{\text{H}}_3{\text{O}}_{\text{2}}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, ${\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}{}^{-}$ is a conjugate base and ${\text{H}}_3{\text{O}}^{+}$ is a conjugate acid.

Interpretation Introduction

(f)

Interpretation:

The equation to show the acidic 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 10QAP

${\text{H}}_{\text{2}}{\text{PO}}_4{}^{-}{\text{+H}}_{\text{2}}\text{O}\to {\text{HPO}}_4{}^{2-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is as follows:

${\text{H}}_{\text{2}}{\text{PO}}_4{}^{-}$

On reaction with water, it can act as an acid by donating hydrogen ion to the water. The reaction is shown as follows:

${\text{H}}_{\text{2}}{\text{PO}}_4{}^{-}{\text{+H}}_{\text{2}}\text{O}\to {\text{HPO}}_4{}^{2-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, ${\text{H}}_{\text{2}}{\text{PO}}_4{}^{-}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, ${\text{HPO}}_4{}^{2-}$ is a conjugate base and ${\text{H}}_3{\text{O}}^{+}$ is a conjugate acid.