Chapter F, Problem I.6E

(a)

Interpretation Introduction

Interpretation:

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

    ${\text{MgBr}}_{\text{2}}(aq)+{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{NaBr}(aq)$

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;

    ${\text{MgBr}}_{\text{2}}(aq)+{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{NaBr}(aq)$

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

    ${\text{3MgBr}}_{\text{2}}(aq)+{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{NaBr}(aq)$

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;

    ${\text{3MgBr}}_{\text{2}}(aq)+{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6\text{NaBr}(aq)$

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

    ${\text{3MgBr}}_{\text{2}}(aq)+2{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to {\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6\text{NaBr}(aq)$

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;

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

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.

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

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

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

(b)

Interpretation Introduction

Interpretation:

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

    $\text{CsI}(aq)+{\text{Hg}}_{\text{2}}{{\text{(HSO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}(s)+{\text{CsHSO}}_{\text{3}}(aq)$

Concept Introduction:

Refer part (a).

Explanation of Solution

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

    $\text{CsI}(aq)+{\text{Hg}}_{\text{2}}{{\text{(HSO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}(s)+{\text{CsHSO}}_{\text{3}}(aq)$

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

    $\text{2CsI}(aq)+{\text{Hg}}_{\text{2}}{{\text{(HSO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}(s)+{\text{CsHSO}}_{\text{3}}(aq)$

Balancing cesium atoms:  In the above chemical equation, there are two cesium atoms on the left side of the equation, while in the product side, there is one cesium atom.  Adding coefficient $2$ before ${\text{CsHSO}}_{\text{3}}$ in the product side balances the cesium atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    $\text{2CsI}(aq)+{\text{Hg}}_{\text{2}}{{\text{(HSO}}_{\text{3}}\text{)}}_{\text{2}}(aq)\to {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}(s)+2{\text{CsHSO}}_{\text{3}}(aq)$

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;

${\text{2Cs}}^{\text{+}}(aq){\text{+2I}}^{-}(aq){\text{+2Hg}}^{\text{+}}(aq){\text{+2HSO}}_{\text{3}}{}^{-}(aq)\to {\text{2Cs}}^{\text{+}}(aq)+{\text{2HSO}}_{\text{3}}{}^{-}(aq)+{\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}(s)$

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.

  $\begin{array}{l}\cancel{{\text{2Cs}}^{\text{+}}(aq)}{\text{+2I}}^{-}(aq){\text{+2Hg}}^{\text{+}}(aq)\text{+}\cancel{{\text{2HSO}}_{\text{3}}{}^{-}(aq)}\\ \to {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}(s)+\cancel{{\text{2Cs}}^{\text{+}}(aq)}+\cancel{{\text{2HSO}}_{\text{3}}{}^{-}(aq)}\end{array}$

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

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

(d)

Interpretation Introduction

Interpretation:

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

    ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(aq)+{\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)\to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}(s)+{\text{KNO}}_{\text{3}}(aq)$

Concept Introduction:

Refer part (a).

Explanation of Solution

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

    ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(aq)+{\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)\to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}(s)+{\text{KNO}}_{\text{3}}(aq)$

Balancing oxalate ions: In the reactant side, there are is one oxalate ion while on the product side, there are three oxalate ions.  Adding coefficient $3$ before ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ balances the oxalate ions on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{3K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(aq)+{\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)\to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}(s)+{\text{KNO}}_{\text{3}}(aq)$

Balancing potassium atoms:  In the above chemical equation, there are six potassium atoms on the left side of the equation, while in the product side, there is one potassium atom.  Adding coefficient $6$ before ${\text{KNO}}_{\text{3}}$ in the product side balances the potassium atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

  ${\text{3K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(aq)+{\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)\to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}(s)+6{\text{KNO}}_{\text{3}}(aq)$

Balancing cobalt atoms:  In the above chemical equation, there is one cobalt atom on the left side of the equation, while in the product side, there are two cobalt atoms.  Adding coefficient $2$ before ${\text{Co(NO}}_{\text{3}}{)}_2$ in the reactant side balances the cobalt atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    ${\text{3K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(aq)+2{\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{3}}(aq)\to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}(s)+6{\text{KNO}}_{\text{3}}(aq)$

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;

${\text{6K}}^{\text{+}}{\text{(aq)+3C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(aq)+2{\text{Co}}^{\text{3+}}(aq){\text{+6NO}}_{\text{3}}{}^{-}(aq)\to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}(s)+6{\text{K}}^{\text{+}}{\text{(aq)+6NO}}_{\text{3}}{}^{-}(aq)$

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.

  $\begin{array}{l}\cancel{{\text{6K}}^{\text{+}}\text{(aq)}}{\text{+3C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(aq)+{\text{2Co}}^{\text{3+}}\text{(aq)+}\cancel{{\text{6NO}}_{\text{3}}{}^{-}(aq)}\\ \to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}}_{{\text{O}}_{\text{4}}}(s)+\cancel{6{\text{K}}^{\text{+}}\text{(aq)}}+\cancel{{\text{6NO}}_{\text{3}}{}^{-}(aq)}\end{array}$

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

  ${\text{3C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(aq)+{\text{2Co}}^{\text{3+}}\text{(aq)}\to {\text{Co}}_{\text{2}}{{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}(s)$