Chapter 14, Problem 1QAP

Write a net ionic equation for the reaction between aqueous solutions of

(a) ammonia and hydrofluoric acid.

(b) perchloric acid and rubidium hydroxide.

(c) sodium sulfite and hydriodic acid.

(d) nitric acid and calcium hydroxide.

Interpretation Introduction

Interpretation:

The net ionic equation needs to be determined for Ammonium and hydrofluoric acid.

Concept Introduction:

In the net ionic equation, the species that participates in the reaction are present. It does not include the ions which are not participating in a chemical reaction. It is commonly used in the neutralization reactions of acid and base.

Answer to Problem 1QAP

  ${\text{NH}}_{\text{3}}\left(\text{aq}\right)\text{ + HF }\left(\text{aq}\right)\to {\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)$

Explanation of Solution

The reaction between ammonium and hydrofluoric is as follows:

  ${\text{NH}}_{\text{3}}\left(\text{aq}\right)\text{ + HF }\left(\text{aq}\right)\to {\text{NH}}_{\text{4}}\text{F }\left(\text{aq}\right)$

As the reaction is occurring in solution, the complete ionic form can be written as follows:

  ${\text{NH}}_{\text{3}}\left(\text{aq}\right)\text{ + HF }\left(\text{aq}\right)\to {\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)$

So, the net ionic equation is shown as follows:

  ${\text{NH}}_{\text{3}}\left(\text{aq}\right)\text{ + HF }\left(\text{aq}\right)\to {\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)$

Interpretation Introduction

Interpretation:

The net ionic equation needs to be determined for Per chloric acid and rubidium hydroxide.

Concept Introduction:

In the net ionic equation, the species that participates in the reaction are present. It does not include the ions which are not participating in a chemical reaction. It is commonly used in the neutralization reactions of acid and base.

Answer to Problem 1QAP

  ${\text{H}}^{\text{+}}\left(\text{aq}\right)\text{ + RbOH }\left(\text{s}\right)\to {\text{Rb}}^{\text{+}}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\text{O }\left(\text{l}\right)$

Explanation of Solution

The reaction between per chloric acid and rubidium hydroxide is as follows:

  ${\text{HClO}}_{\text{4}}\left(\text{aq}\right)\text{ + RbOH }\left(\text{s}\right)\to {\text{RbClO}}_{\text{4}}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\text{O }\left(\text{l}\right)$

As the reaction is occurring in solution, one can write the equation in its complete ionic form. The complete ionic equation will be as follows:

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

Cancel the spectator ions to get the net ionic equation as follows:

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

So, the net ionic equation is:

  ${\text{H}}^{\text{+}}\left(\text{aq}\right)\text{ + RbOH }\left(\text{s}\right)\to {\text{Rb}}^{\text{+}}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\text{O }\left(\text{l}\right)$

Interpretation Introduction

Interpretation:

The net ionic equation needs to be determined for Sodium sulfite and hydro iodic acid.

Concept Introduction:

In the net ionic equation, the species that participates in the reaction are present. It does not include the ions which are not participating in a chemical reaction. It is commonly used in the neutralization reactions of acid and base.

Answer to Problem 1QAP

  ${\text{SO}}_{\text{3}}{}^{\text{2-}}\left(\text{aq}\right){\text{ + 2H}}^{\text{+}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right)$

Explanation of Solution

The reaction between sodium sulfite and hydro iodic acid is as follows:

  ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right)\text{ + 2HI }\left(\text{aq}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right)\text{ + 2 NaI }\left(\text{aq}\right)$

As the reaction is occurring in solution, one can write the equation in its complete ionic form. The compete ionic equation will be as follows:

  ${\text{2Na}}^{\text{+}}\left(\text{aq}\right){\text{ + SO}}_{\text{3}}{}^{\text{2-}}\left(\text{aq}\right){\text{ + 2H}}^{\text{+}}\left(\text{aq}\right){\text{ + 2I}}^{\text{-}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right){\text{ + 2 Na}}^{\text{+}}\left(\text{aq}\right){\text{ + 2I}}^{\text{-}}\left(\text{aq}\right)$

Cancel the spectator ions to get the net ionic equation as follows:

  $\cancel{{\text{2Na}}^{+}\left(\text{aq}\right)}+{\text{ SO}}_{\text{3}}{}^{\text{2}-}\left(\text{aq}\right)+{\text{ 2H}}^{+}\left(\text{aq}\right)+\cancel{{\text{ 2I}}^{-}\left(\text{aq}\right)}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right)+\cancel{{\text{2 Na}}^{+}\left(\text{aq}\right)}+\cancel{{\text{ 2I}}^{-}\left(\text{aq}\right)}$

So, the net ionic equation is:

  ${\text{SO}}_{\text{3}}{}^{\text{2-}}\left(\text{aq}\right){\text{ + 2H}}^{\text{+}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right)$

Interpretation Introduction

Interpretation:

The net ionic equation needs to be determined for Nitric acid and calcium hydroxide.

Concept Introduction:

In the net ionic equation, the species that participates in the reaction are present. It does not include the ions which are not participating in a chemical reaction. It is commonly used in the neutralization reactions of acid and base.

Answer to Problem 1QAP

  ${\text{2H}}^{\text{+}}\left(\text{aq}\right)\text{ + Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\left(\text{aq}\right){\text{ + 2H}}_{\text{2}}\text{O }\left(\text{l}\right)$

Explanation of Solution

The reaction between Nitric acid and calcium hydroxide is as follows:

  ${\text{2HNO}}_{\text{3}}\left(\text{aq}\right)\text{ + Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{aq}\right)\to \text{Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{ + 2 H}}_{\text{2}}\text{O }\left(\text{l}\right)$

As the reaction is occurring in solution, one can write the equation in its complete ionic form. The complete ionic equation will be as follows:

  ${\text{2H}}^{\text{+}}\left(\text{aq}\right){\text{ + 2NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)\text{ + Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\left(\text{aq}\right){\text{ + 2NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right){\text{ + 2H}}_{\text{2}}\text{O }\left(\text{l}\right)$

Cancel the spectator ions to get the net ionic equation as follows:

  ${\text{2H}}^{+}\left(\text{aq}\right)+\cancel{{\text{2NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)}+\text{ Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\to {\text{Ca}}^{\text{2}+}\left(\text{aq}\right)+\cancel{{\text{2NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)}+{\text{ 2H}}_{\text{2}}\text{O }\left(\text{l}\right)$

So, the net ionic equation is:

  ${\text{2H}}^{\text{+}}\left(\text{aq}\right)\text{ + Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\left(\text{aq}\right){\text{ + 2H}}_{\text{2}}\text{O }\left(\text{l}\right)$