Chapter 21, Problem 18QAP

Write a balanced net ionic equation for the disproportionation reaction of
(a) hypochlorous acid to chlorine gas and chlorous acid in acidic solution.
(b) chlorate ion to perchlorate and chlorite ions.

Interpretation Introduction

(a)

Interpretation:

The net ionic equation (balanced)for the following should be determined:

disproportionation reaction of hypochlorous acid to chlorine gas and chlorous acid in acidic solution

Concept introduction:

Net ionic equation is the ionic equation in which reactants are written in ionic form if they occur as ions in a reaction medium and product form are shown as combination of ions. The charges of ions and each atom in the reaction are balanced.

Answer to Problem 18QAP

Balanced net ionic equation for the disproportionation reaction of hypochlorous acidto chlorine gas and chlorous acid in acidic solution is,

$3\text{HClO}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{HClO}}_2\left(\text{g}\right)$

Explanation of Solution

The equation of hypochlorous acid changes to chlorine gas is,

$\text{HClO}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)$

Balance the above reaction.

1. Balance all atoms except O(Oxygen) and H(Hydrogen)

$\text{2HClO}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)$

2. Balance oxygen by adding water.

$\text{2HClO}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

3. Balance H by adding ${\text{H}}^{\text{+}}$ ion.

$\text{2HClO}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

4. Balance charges by adding electrons.

$\text{2HClO}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

5. Write in ionic form.

${\text{2H}}^{+}\left(\text{aq}\right)+2\text{HClO}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ - (1)

The equation of hypochlorous acid changes to chlorous acid.

$\text{HClO}\left(\text{aq}\right)\to {\text{HClO}}_2\left(\text{aq}\right)$

Balance the above reaction.

1. Balance all atoms except O(Oxygen) and H(Hydrogen)

$\text{HClO}\left(\text{aq}\right)\to {\text{HClO}}_2\left(\text{aq}\right)$

2. Balance oxygen by adding water.

$\text{HClO}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{HClO}}_2\left(\text{aq}\right)$

3. Balance H by adding ${\text{H}}^{\text{+}}$ ion.

$\text{HClO}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{HClO}}_2\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)$

4. Balance charges by adding electrons.

$\text{HClO}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{HClO}}_2\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}$ -(2)

Hence, net ionic equation is summation of equation (1) and (2).

$\begin{array}{l}{\text{2H}}^{+}\left(\text{aq}\right)+2\text{HClO}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \text{HClO}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{HClO}}_2\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\end{array}$

$3\text{HClO}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{HClO}}_2\left(\text{g}\right)$

So, the balanced net ionic equation is:

$3\text{HClO}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{HClO}}_2\left(\text{g}\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced net ionic equation for the following reaction should be determined:

disproportionation reaction of chlorate ion to perchlorate and chlorite ions.

Concept introduction:

Net ionic equation is the ionic equation in which reactants are written in ionic form if they occur as ions in a reaction medium and product form are shown as combination of ions. The charges of ions and each atom in the reaction are balanced.

Answer to Problem 18QAP

Balanced net ionic equation for the disproportionation reaction of chlorate ion to perchlorate and chlorite ions is,

$2{\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+{\text{ClO}}_2{}^{-}\left(\text{aq}\right)$

Explanation of Solution

Chlorate ion coverts to perchlorate ions.

$\underset{\text{Chlorate}\text{ion}}{\underbrace{{\text{ClO}}_{\text{3}}{}^{-}}}\left(\text{aq}\right)\to {\underset{\text{Perchlorate}}{\underbrace{{\text{ClO}}_{\text{4}}}}}^{-}\left(\text{aq}\right)$

Balance the above reaction.

1. Balance all atoms except O(Oxygen) and H(Hydrogen)

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)$

2. Balance oxygen by adding water.

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)$

3. Balance H by adding ${\text{H}}^{\text{+}}$ ion.

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)$

Hence, ionic form of reaction is,

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)$ …… (3)

The equation of conversion of chlorate ion to chlorite ion is,

$\underset{\text{Chlorate}\text{ion}}{\underbrace{{\text{ClO}}_{\text{3}}{}^{-}}}\left(\text{aq}\right)\to {\underset{\text{Chlorite}}{\underbrace{{\text{ClO}}_2}}}^{-}\left(\text{aq}\right)$

Balance the above reaction.

1. Balance all atoms except O(Oxygen) and H(Hydrogen)

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to {\text{ClO}}_2{}^{-}\left(\text{aq}\right)$

2. Balance oxygen by adding water.

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to {\text{ClO}}_2{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

3. Balance H by adding ${\text{H}}^{\text{+}}$ ion.

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)\to {\text{ClO}}_2{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Hence, ionic form of reaction is,

${\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)\to {\text{ClO}}_2{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ …… (4)

Hence, net ionic equation is summation of equation (3) and (4).

$\begin{array}{l}{\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)\\ {\text{ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)\to {\text{ClO}}_2{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\end{array}$

${\text{2ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+{\text{ClO}}_2{}^{-}\left(\text{aq}\right)$

Hence, net ionic equation is,

${\text{2ClO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to {\text{ClO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+{\text{ClO}}_2{}^{-}\left(\text{aq}\right)$