Chapter 13, Problem 9QAP

Using the Brønsted-Lowry model, write equations to show why the following species behave as weak acids in water.

(a) Ni(H₂O)₅OH⁺(b) Al(H₂O)₆³⁺(c) H₂S

(d) HPO₄2-(e) HClO₂ (e) Cr(H₂O)₅(OH)⁺

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 9QAP

$\text{Ni}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\text{OH}}^{+}+{\text{H}}_{\text{2}}\text{O}\to \text{Ni}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\text{O+H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is $\text{Ni}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\text{OH}}^{+}$.

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

$\text{Ni}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\text{OH}}^{+}+{\text{H}}_{\text{2}}\text{O}\to \text{Ni}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\text{O+H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, $\text{Ni}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\text{OH}}^{+}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, $\text{Ni}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{O}$ 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 9QAP

$\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{3+}+{\text{H}}_{\text{2}}\text{O}\to \text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_5^{2+}\left(\text{OH}\right)+{\text{H}}_3{\text{O}}^{+}$

Explanation of Solution

The given species is $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{3+}$.

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

$\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{3+}+{\text{H}}_{\text{2}}\text{O}\to \text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_5^{2+}\left(\text{OH}\right)+{\text{H}}_3{\text{O}}^{+}$

In the above reaction, $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6^{3+}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_5^{2+}\left(\text{OH}\right)$ 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 9QAP

${\text{H}}_{\text{2}}{\text{S+H}}_{\text{2}}\text{O}\to {\text{HS}}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

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

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

${\text{H}}_{\text{2}}{\text{S+H}}_{\text{2}}\text{O}\to {\text{HS}}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, ${\text{H}}_{\text{2}}\text{S}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, ${\text{HS}}^{-}$ 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 9QAP

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

Explanation of Solution

The given species is as follows:

${\text{HPO}}_4^{2-}$

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

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

In the above reaction, ${\text{HPO}}_4^{2-}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, ${\text{PO}}_4^{3-}$ 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 9QAP

${\text{HClO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{ClO}}_{\text{2}}{}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is as follows:

${\text{HClO}}_{\text{2}}$

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

${\text{HClO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{ClO}}_{\text{2}}{}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, ${\text{HClO}}_{\text{2}}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, ${\text{ClO}}_{\text{2}}{}^{-}$ 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 9QAP

$\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{+}{\text{+H}}_{\text{2}}\text{O}\to \text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{O}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

Explanation of Solution

The given species is as follows:

$\text{Cr}{\left(\text{H2O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{+}$

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

$\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{+}{\text{+H}}_{\text{2}}\text{O}\to \text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{O}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

In the above reaction, $\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{+}$ acts as an acid, ${\text{H}}_{\text{2}}\text{O}$ acts as a base, $\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{O}$ is a conjugate base and ${\text{H}}_3{\text{O}}^{+}$ is a conjugate acid.