Chapter F, Problem I.5E

(a)

Interpretation Introduction

Interpretation:

Balanced equation, complete ionic equation, and the net ionic equation has to be written for the given reaction.

    ${\text{BaBr}}_{\text{2}}(aq)+{\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{LiBr}(aq)$

Concept Introduction:

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.

Net ionic equation is the one that is obtained from the complete ionic equation by cancelling out the spectator ions.

Explanation of Solution

The chemical equation for the reaction is given as shown below;

    ${\text{BaBr}}_{\text{2}}(aq)+{\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{LiBr}(aq)$

Balancing barium atoms: In the reactant side, there is one barium atom while on the product side, there are three barium atoms.  Adding coefficient $3$ before ${\text{BaBr}}_{\text{2}}$ balances the barium atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{3BaBr}}_{\text{2}}(aq)+{\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{LiBr}(aq)$

Balancing bromine atoms:  In the above chemical equation, there are six bromine atoms on the left side of the equation, while in the product side, there is one bromine atom.  Adding coefficient $6$ before $\text{LiBr}$ in the product side balances the bromine atoms on both sides of the equation.  The chemical equation is given as shown below;

    ${\text{3BaBr}}_{\text{2}}(aq)+{\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6\text{LiBr}(aq)$

Balancing lithium atoms:  In the above chemical equation, there are three lithium atoms on the left side of the equation, while in the product side, there are six lithium atoms.  Adding coefficient $2$ before ${\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}$ in the reactant side balances the lithium atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    ${\text{3BaBr}}_{\text{2}}(aq)+2{\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6\text{LiBr}(aq)$

Complete ionic equation:

The complete ionic equation can be written considering the ionic compounds in aqueous medium to be written into respective ions.  Therefore, the complete ionic equation can be given as follows;

${\text{3Ba}}^{\text{2+}}(aq){\text{+6Br}}^{-}(aq){\text{+6Li}}^{\text{+}}(aq){\text{+2PO}}_{\text{4}}{}^{3-}(aq)\to {\text{Ba}}_{\text{3}}{({\text{PO}}_{\text{4}})}_2(s)+{\text{6Li}}^{\text{+}}(aq)+6{\text{Br}}^{-}(aq)$

Net ionic equation:

The net ionic equation can be obtained from the complete ionic equation by cancelling out the spectator ions on both sides of the equation.

${\text{3Ba}}^{\text{2+}}(aq)\text{+}\cancel{{\text{6Br}}^{-}(aq)}\text{+}\cancel{{\text{6Li}}^{\text{+}}(aq)}{\text{+2PO}}_{\text{4}}{}^{3-}(aq)\to {\text{Ba}}_{\text{3}}{({\text{PO}}_{\text{4}})}_2(s)+\cancel{{\text{6Li}}^{\text{+}}(aq)}+\cancel{{\text{6Br}}^{-}(aq)}$

Thus, the net ionic equation can be given as shown below;

  ${\text{3Ba}}^{\text{2+}}(aq)+2{\text{PO}}_{\text{4}}{}^{3-}(aq)\to {\text{Ba}}_{\text{3}}{({\text{PO}}_{\text{4}})}_2(s)$

(b)

Interpretation Introduction

Interpretation:

Balanced equation, complete ionic equation, and the net ionic equation has to be written for the given reaction.

    ${\text{NH}}_{\text{4}}\text{Cl}(aq)+{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to {\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(aq)+{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(s)$

Concept Introduction:

Refer part (a).

Explanation of Solution

The chemical equation for the reaction is given as shown below;

    ${\text{NH}}_{\text{4}}\text{Cl}(aq)+{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to {\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(aq)+{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(s)$

Balancing chlorine atoms: In the reactant side, there is one chlorine atom while on the product side, there are two chlorine atoms.  Adding coefficient $2$ before ${\text{NH}}_{\text{4}}\text{Cl}$ balances the chlorine atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{2NH}}_{\text{4}}\text{Cl}(aq)+{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to {\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(aq)+{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(s)$

Balancing nitrate ions:  In the above chemical equation, there are two nitrate ions on the left side of the equation, while in the product side, there is one nitrate ion.  Adding coefficient $2$ before ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ in the product side balances the nitrate ions on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    ${\text{2NH}}_{\text{4}}\text{Cl}(aq)+{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to 2{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(aq)+{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(s)$

Complete ionic equation:

The complete ionic equation can be written considering the ionic compounds in aqueous medium to be written into respective ions.  Therefore, the complete ionic equation can be given as follows;

${\text{2NH}}_{\text{4}}{}^{\text{+}}(aq){\text{+2Cl}}^{-}(aq){\text{+2Hg}}^{\text{+}}(aq){\text{+2NO}}_{\text{3}}{}^{-}(aq)\to {\text{2NH}}_{\text{4}}{}^{\text{+}}(aq)+{\text{2NO}}_{\text{3}}{}^{-}(aq)+{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(s)$

Net ionic equation:

The net ionic equation can be obtained from the complete ionic equation by cancelling out the spectator ions on both sides of the equation.

  $\begin{array}{l}{\text{2Cl}}^{-}(aq){\text{+2Hg}}^{\text{+}}(aq)\text{+}\cancel{{\text{2NH}}_{\text{4}}{}^{\text{+}}(aq)}\text{+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(aq)}\\ \to {\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(s)+\cancel{{\text{2NH}}_{\text{4}}{}^{\text{+}}(aq)}+\cancel{{\text{2NO}}_{\text{3}}{}^{-}(aq)}\end{array}$

Thus, the net ionic equation can be given as shown below;

  ${\text{2Cl}}^{-}(aq){\text{+2Hg}}^{\text{+}}(aq)\to {\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(s)$

(d)

Interpretation Introduction

Interpretation:

Balanced equation, complete ionic equation, and the net ionic equation has to be written for the given reaction.

    ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)+{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Co(OH)}}_{\text{3}}(s)+{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$

Concept Introduction:

Refer part (a).

Explanation of Solution

The chemical equation for the reaction is given as shown below;

    ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)+{\text{Ca(OH)}}_{\text{2}}(aq)\to {\text{Co(OH)}}_{\text{3}}(s)+{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$

Balancing hydroxide ions: In the reactant side, there are two hydroxide ion while on the product side, there are three hydroxide ions.  Adding coefficient $2$ before ${\text{Ca(OH)}}_{\text{2}}$ and $3$ before ${\text{Co(OH)}}_{\text{2}}$ balances the hydroxide ions on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)+3{\text{Ca(OH)}}_{\text{2}}(aq)\to 2{\text{Co(OH)}}_{\text{3}}(s)+{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$

Balancing nitrate ions:  In the above chemical equation, there are three nitrate ions on the left side of the equation, while in the product side, there are two nitrate ions.  Adding coefficient $2$ before ${\text{Co(NO}}_{\text{3}}{)}_3$ in the reactant side and $3$ before ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ in the product side balances the nitrate ions on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    ${\text{2Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)+3{\text{Ca(OH)}}_{\text{2}}(aq)\to 2{\text{Co(OH)}}_{\text{3}}(s)+3{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$

Complete ionic equation:

The complete ionic equation can be written considering the ionic compounds in aqueous medium to be written into respective ions.  Therefore, the complete ionic equation can be given as follows;

${\text{2Co}}^{\text{3+}}{\text{(aq)+6NO}}_{\text{3}}{}^{-}(aq)+3{\text{Ca}}^{\text{2+}}(aq){\text{+6OH}}^{-}(aq)\to 2{\text{Co(OH)}}_{\text{3}}(s)+3{\text{Ca}}^{\text{2+}}{\text{(aq)+6NO}}_{\text{3}}{}^{-}(aq)$

Net ionic equation:

The net ionic equation can be obtained from the complete ionic equation by cancelling out the spectator ions on both sides of the equation.

  $\begin{array}{l}{\text{2Co}}^{\text{3+}}\text{(aq)+}\cancel{{\text{6NO}}_{\text{3}}{}^{-}(aq)}\text{+}\cancel{3{\text{Ca}}^{\text{2+}}(aq)}{\text{+6OH}}^{-}(aq)\\ \to 2{\text{Co(OH)}}_{\text{3}}(s)+\cancel{3{\text{Ca}}^{\text{2+}}\text{(aq)}}+\cancel{{\text{6NO}}_{\text{3}}{}^{-}(aq)}\end{array}$

Thus, the net ionic equation can be given as shown below;

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