Chapter F, Problem I.16E

(a)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{AgNO}}_{\text{3}}$ and ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{AgNO}}_{\text{3}}$ and ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$.  The reaction can be given as follows:

    ${\text{3AgNO}}_{\text{3}}(\text{aq})+{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})\to {\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}(s)+{\text{3NaNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{AgNO}}_{\text{3}}$ reacts with ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ resulting in formation of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{NaNO}}_{\text{3}}$.  The precipitate that is formed is ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$.  This is because the phosphate salts of ${\text{Ca}}^{\text{2+}}$, ${\text{Sr}}^{\text{2+}}$, ${\text{Ba}}^{\text{2+}}$, ${\text{Pb}}^{\text{2+}}$, ${\text{Hg}}_{\text{2}}{}^{\text{2+}}$, and ${\text{Ag}}^{\text{+}}$ are alone insoluble.

The balanced chemical equation for the reaction is given as follows:

    ${\text{3AgNO}}_{\text{3}}(\text{aq})+{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})\to {\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}(s)+{\text{3NaNO}}_{\text{3}}\text{(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{3Ag}}^{\text{+}}{\text{(aq)+3NO}}_{\text{3}}{}^{-}(\text{aq})+3{\text{Na}}^{\text{+}}{\text{(aq)+PO}}_{\text{4}}{}^{3-}(\text{aq})\to {\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}(s)+3{\text{Na}}^{\text{+}}{\text{(aq)+3NO}}_{\text{3}}{}^{-}\text{(aq)}$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{Na}}^{\text{+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

${\text{3Ag}}^{\text{+}}\text{(aq)+}\cancel{{\text{3NO}}_{\text{3}}{}^{-}(\text{aq})}+\cancel{3{\text{Na}}^{\text{+}}\text{(aq)}}{\text{+PO}}_{\text{4}}{}^{3-}(\text{aq})\to {\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}(s)+\cancel{3{\text{Na}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{3NO}}_{\text{3}}{}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

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

(b)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ and ${\text{NH}}_{\text{4}}\text{I}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ and ${\text{NH}}_{\text{4}}\text{I}$.  The reaction can be given as follows:

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

Aqueous solution of ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ reacts with ${\text{NH}}_{\text{4}}\text{I}$ resulting in formation of ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$ and ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.  The precipitate that is formed is ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$.

The balanced chemical equation for the reaction is given as follows:

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

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

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

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

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

Thus net ionic equation can be written as follows:

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

(c)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{BaCl}}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{BaCl}}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$.  The reaction can be given as follows:

    ${\text{BaCl}}_{\text{2}}(\text{aq})+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s)+2\text{NaCl(aq)}$

Aqueous solution of ${\text{BaCl}}_{\text{2}}$ reacts with ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ resulting in formation of ${\text{BaSO}}_{\text{4}}$ and $\text{NaCl}$.  The precipitate that is formed is ${\text{BaSO}}_{\text{4}}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{BaCl}}_{\text{2}}(\text{aq})+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s)+2\text{NaCl(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{Ba}}^{\text{2+}}{\text{(aq)+2Cl}}^{-}(\text{aq})+2{\text{Na}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}{}^{2-}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s){\text{+2Cl}}^{-}(\text{aq})+2{\text{Na}}^{\text{+}}\text{(aq)}$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{Na}}^{\text{+}}$ and ${\text{Cl}}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

  ${\text{Ba}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2Cl}}^{-}(\text{aq})}+\cancel{2{\text{Na}}^{\text{+}}\text{(aq)}}{\text{+SO}}_{\text{4}}{}^{2-}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s)\text{+}\cancel{{\text{2Cl}}^{-}(\text{aq})}+\cancel{2{\text{Na}}^{\text{+}}\text{(aq)}}$

Thus net ionic equation can be written as follows:

    ${\text{Ba}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}{}^{2-}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s)$

(d)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{K}}_{\text{2}}\text{S}$ and ${\text{Bi(NO}}_{\text{3}}{\text{)}}_{\text{3}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{K}}_{\text{2}}\text{S}$ and ${\text{Bi(NO}}_{\text{3}}{\text{)}}_{\text{3}}$.  The reaction can be given as follows:

    ${\text{3K}}_{\text{2}}\text{S}(\text{aq})+2{\text{Bi(NO}}_{\text{3}}{\text{)}}_{\text{3}}(\text{aq})\to {\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}(s)+{\text{6KNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{K}}_{\text{2}}\text{S}$ reacts with ${\text{Bi(NO}}_{\text{3}}{\text{)}}_{\text{3}}$ resulting in formation of ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}$ and ${\text{KNO}}_{\text{3}}$.  The precipitate that is formed is ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{3K}}_{\text{2}}\text{S}(\text{aq})+2{\text{Bi(NO}}_{\text{3}}{\text{)}}_{\text{3}}(\text{aq})\to {\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}(s)+{\text{6KNO}}_{\text{3}}\text{(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{6K}}^{\text{+}}{\text{(aq)+3S}}^{2-}(\text{aq})+2{\text{Bi}}^{\text{3+}}{\text{(aq)+6NO}}_{\text{3}}{}^{-}(\text{aq})\to {\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}(s)+{\text{6K}}^{\text{+}}{\text{(aq)+6NO}}_{\text{3}}{}^{-}(\text{aq})$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{K}}^{\text{+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

$\cancel{{\text{6K}}^{\text{+}}\text{(aq)}}{\text{+3S}}^{2-}(\text{aq})+2{\text{Bi}}^{\text{3+}}\text{(aq)+}\cancel{{\text{6NO}}_{\text{3}}{}^{-}(\text{aq})}\to {\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}(s)+\cancel{{\text{6K}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{6NO}}_{\text{3}}{}^{-}(\text{aq})}$

Thus net ionic equation can be written as follows:

    ${\text{3S}}^{2-}(\text{aq})+2{\text{Bi}}^{\text{3+}}\text{(aq)}\to {\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}(s)$

(e)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{Ba(CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{)}}_{\text{2}}$ and ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{Ba(CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{)}}_{\text{2}}$ and ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.  The reaction can be given as follows:

    ${\text{Ba(CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{)}}_{\text{2}}(\text{aq})+{\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}(\text{aq})\to {\text{BaCO}}_{\text{3}}(s)+2{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{Li(aq)}$

Aqueous solution of ${\text{Ba(CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{)}}_{\text{2}}$ reacts with ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ resulting in formation of ${\text{BaCO}}_{\text{3}}$ and ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{Li}$.  The precipitate that is formed is ${\text{BaCO}}_{\text{3}}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{Ba(CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{)}}_{\text{2}}(\text{aq})+{\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}(\text{aq})\to {\text{BaCO}}_{\text{3}}(s)+2{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{Li(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{Ba}}^{\text{2+}}{\text{(aq)+2CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{-}(\text{aq})+2{\text{Li}}^{\text{+}}{\text{(aq)+CO}}_{\text{3}}{}^{2-}(\text{aq})\to {\text{BaCO}}_{\text{3}}(s)+2{\text{Li}}^{\text{+}}{\text{(aq)+2CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{-}(\text{aq})$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{Li}}^{\text{+}}$ and ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

  $\begin{array}{r}{\text{Ba}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{-}(\text{aq})}+\cancel{2{\text{Li}}^{\text{+}}\text{(aq)}}{\text{+CO}}_{\text{3}}{}^{2-}(\text{aq})\to {\text{BaCO}}_{\text{3}}(s)+\cancel{2{\text{Li}}^{\text{+}}\text{(aq)}}\\ \text{+}\cancel{{\text{2CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{-}(\text{aq})}\end{array}$

Thus net ionic equation can be written as follows:

    ${\text{Ba}}^{\text{2+}}{\text{(aq)+CO}}_{\text{3}}{}^{2-}(\text{aq})\to {\text{BaCO}}_{\text{3}}(s)$