Chapter F, Problem J.10E

(a)

Interpretation Introduction

Interpretation:

The salt that is produced from sodium hydroxide and propanoic acid by means of neutralization reaction has to be given and also the complete ionic equation has to be written.

Concept Introduction:

Reaction between the acid and base is known as a neutralization reaction.  The product formed in neutralization reaction is salt and water.  The net reaction that takes place between the acid and base solutions is that the formation of water from hydrogen ions and hydroxide ions.  Neutralization reaction is the one that takes place between a strong acid and a metal hydroxide.

    $\text{Acid}+\text{Base}\to \text{Salt}+\text{Water}$

Complete ionic equation is the one that shows all the species that is present in the chemical reaction.  The dissolved ionic compounds are alone showed as ions.  Precipitates are not showed as ions.

Answer to Problem J.10E

The salt that is produced is sodium propanoate.  Complete ionic equation is ${\text{Na}}^{\text{+}}(aq){\text{+OH}}^{-}(aq)+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH(aq)}\to {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{-}{\text{(aq)+Na}}^{\text{+}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$.

Explanation of Solution

The acid and base given is propanoic acid and sodium hydroxide respectively.  The reaction between acid and base results in formation of salt and water.  Sodium hydroxide is a very strong base while propanoic acid is a weak acid.  The reaction between sodium hydroxide and propanoic acid results in formation of sodium propanoate.  Chemical equation can be given as shown below;

    $\text{NaOH}(aq)+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}(aq)\to {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COONa}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Therefore, the salt formed is sodium propanoate.  Complete ionic equation can be written by showing the ions explicitly that are in aqueous medium.  This can be written as follows;

${\text{Na}}^{\text{+}}(aq){\text{+OH}}^{-}(aq)+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH(aq)}\to {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{-}{\text{(aq)+Na}}^{\text{+}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

(b)

Interpretation Introduction

Interpretation:

The salt that is produced from ammonia and nitrous by means of neutralization reaction has to be given and also the complete ionic equation has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem J.10E

The salt that is produced is ammonium nitrite.  Complete ionic equation is ${\text{NH}}_{\text{4}}{}^{+}(aq)+{\text{OH}}^{\text{-}}(aq)+{\text{HNO}}_{\text{2}}(aq)\to {\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+NO}}_{\text{2}}{}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$.

Explanation of Solution

The acid and base given is nitrous acid and ammonia respectively.  The reaction between acid and base results in formation of salt and water.  Ammonia is a base while nitrous acid is a weak acid.  The reaction between ammonia and nitrous acid results in formation of ammonium nitrite.  Chemical equation can be given as shown below;

    ${\text{NH}}_{\text{4}}OH(aq)+{\text{HNO}}_{\text{2}}(aq)\to {\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Therefore, the salt formed is ammonium nitrite.  Complete ionic equation can be written by showing the ions explicitly that are in aqueous medium.  This can be written as follows;

    ${\text{NH}}_{\text{4}}{}^{+}(aq)+{\text{OH}}^{\text{-}}(aq)+{\text{HNO}}_{\text{2}}(aq)\to {\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+NO}}_{\text{2}}{}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

(c)

Interpretation Introduction

Interpretation:

The salt that is produced from cobalt(III) hydroxide and hydrosulfuric acid by means of neutralization reaction has to be given and also the complete ionic equation has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem J.10E

The salt that is produced is cobalt sulfide.  Complete ionic equation is ${\text{2Co}}^{\text{3+}}{\text{(aq)+3OH}}^{-}(aq)+3{\text{H}}_{\text{2}}\text{S}(aq)\to {\text{Co}}_{\text{2}}{\text{S}}_{\text{3}}(s)+6{\text{H}}_{\text{2}}\text{O}(l)$.

Explanation of Solution

The acid and base given is hydrosulfuric acid and cobalt(III) hydroxide respectively.  The reaction between acid and base results in formation of salt and water.  Cobalt(III) hydroxide is a strong base while hydrosulfuric acid is a weak acid.  The reaction between cobalt(III) hydroxide and hydrosulfuric acid results in formation of cobalt sulfide.  Chemical equation can be given as shown below;

    ${\text{2Co(OH)}}_{\text{3}}(aq)+3{\text{H}}_{\text{2}}\text{S}(aq)\to {\text{Co}}_{\text{2}}{\text{S}}_{\text{3}}(aq)+6{\text{H}}_{\text{2}}\text{O}(l)$

Therefore, the salt formed is cobalt(III) sulfide.  Complete ionic equation can be written by showing the ions explicitly that are in aqueous medium.  This can be written as follows;

    ${\text{2Co}}^{\text{3+}}{\text{(aq)+3OH}}^{-}(aq)+3{\text{H}}_{\text{2}}\text{S}(aq)\to {\text{Co}}_{\text{2}}{\text{S}}_{\text{3}}(s)+6{\text{H}}_{\text{2}}\text{O}(l)$

(d)

Interpretation Introduction

Interpretation:

The salt that is produced from barium hydroxide and chloric acid by means of neutralization reaction has to be given and also the complete ionic equation has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem J.10E

The salt that is produced is barium chlorite.  Complete ionic equation is ${\text{2Co}}^{\text{3+}}{\text{(aq)+3OH}}^{-}(aq)+3{\text{H}}_{\text{2}}\text{S}(aq)\to {\text{Co}}_{\text{2}}{\text{S}}_{\text{3}}(s)+6{\text{H}}_{\text{2}}\text{O}(l)$.

Explanation of Solution

The acid and base given is chloric acid and barium hydroxide respectively.  The reaction between acid and base results in formation of salt and water.  Barium hydroxide is a strong base and chloric acid is a strong acid.  The reaction between barium hydroxide and chloric acid results in formation of barium chlorite.  Chemical equation can be given as shown below;

    ${\text{Ba(OH)}}_{\text{2}}(aq)+2{\text{HClO}}_{\text{3}}(aq)\to {\text{Ba(ClO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$

Therefore, the salt formed is barium chlorite.  Complete ionic equation can be written by showing the ions explicitly that are in aqueous medium.  This can be written as follows;

    ${\text{Ba}}^{\text{2+}}{\text{(aq)+2OH}}^{-}(aq)+2{\text{H}}^{\text{+}}{\text{(aq)+2ClO}}_{\text{3}}{}^{-}(aq)\to {\text{Ba}}^{\text{2+}}{\text{(aq)+2ClO}}_{\text{3}}{}^{-}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$