Chapter F, Problem I.14E

(a)

Interpretation Introduction

Interpretation:

Net ionic equation has to be written for the formation of precipitate and the spectator ions has to be identified in the reaction between ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{AgNO}}_{\text{3}}$.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{AgNO}}_{\text{3}}$.  The reaction can be given as follows:

    ${\text{2AgNO}}_{\text{3}}(\text{aq})+{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{AgNO}}_{\text{3}}$ reacts with ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ resulting in formation of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{KNO}}_{\text{3}}$.  The precipitate that is formed is ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$.  This is because the sulfate 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:

    ${\text{2AgNO}}_{\text{3}}(\text{aq})+{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

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

    ${\text{2Ag}}^{\text{+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})+2{\text{K}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}{}^{2-}(\text{aq})\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+{\text{2K}}^{\text{+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}\text{(aq)}$

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 ${\text{K}}^{\text{+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

${\text{2Ag}}^{\text{+}}\text{(aq)+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(\text{aq})}+\cancel{2{\text{K}}^{\text{+}}\text{(aq)}}{\text{+SO}}_{\text{4}}{}^{2-}(\text{aq})\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+\cancel{{\text{2K}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

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

(b)

Interpretation Introduction

Interpretation:

Net ionic equation has to be written for the formation of precipitate and the spectator ions has to be identified in the reaction between ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{SrBr}}_{\text{2}}$.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{SrBr}}_{\text{2}}$.  The reaction can be given as follows:

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})+3{\text{SrBr}}_{\text{2}}(\text{aq})\to {\text{Sr}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{6HBr(aq)}$

Aqueous solution of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ reacts with ${\text{SrBr}}_{\text{2}}$ resulting in formation of ${\text{Sr}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ and $\text{HBr}$.  The precipitate that is formed is ${\text{Sr}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.  This is because the phosphates are insoluble if the cations are apart from the elements of Group-1 and ammonium ion.

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

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})+3{\text{SrBr}}_{\text{2}}(\text{aq})\to {\text{Sr}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\text{6HBr(aq)}$

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

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

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 ${\text{H}}^{\text{+}}$ and ${\text{Br}}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

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

Thus net ionic equation can be written as follows:

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

(b)

Interpretation Introduction

Interpretation:

Net ionic equation has to be written for the formation of precipitate and the spectator ions has to be identified in the reaction between ${\text{Na}}_{\text{2}}\text{S}$ and ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.

Explanation of Solution

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

    ${\text{Na}}_{\text{2}}\text{S}(\text{aq})+2{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(\text{aq})\to {({\text{NH}}_{\text{4}}\text{)}}_{\text{2}}\text{S}(\text{aq})+{\text{2NaNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{Na}}_{\text{2}}\text{S}$ reacts with ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ resulting in formation of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}\text{S}$ and ${\text{NaNO}}_{\text{3}}$.  There is no precipitate formed in this reaction because all salts in the reaction are soluble.

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

    ${\text{Na}}_{\text{2}}\text{S}(\text{aq})+2{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}(\text{aq})\to {({\text{NH}}_{\text{4}}\text{)}}_{\text{2}}\text{S}(\text{aq})+{\text{2NaNO}}_{\text{3}}\text{(aq)}$

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

  $\begin{array}{r}{\text{S}}^{2-}(\text{aq})+2{\text{Na}}^{\text{+}}\text{(aq)}+2{\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})\to {\text{S}}^{2-}(\text{aq})+2{\text{Na}}^{\text{+}}\text{(aq)}+2{\text{NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}\\ {\text{+2NO}}_{\text{3}}{}^{-}(\text{aq})\end{array}$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus all are spectator ions in the given equation.

As all the ions present in the ionic equation, there is no net ionic equation.

(d)

Interpretation Introduction

Interpretation:

Net ionic equation has to be written for the formation of precipitate and the spectator ions has to be identified in the reaction between ${\text{CdSO}}_{\text{4}}$ and ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}$.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{CdSO}}_{\text{4}}$ and ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}$.  The reaction can be given as follows:

    ${\text{CdSO}}_{\text{4}}(\text{aq})+{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}(\text{aq})\to {\text{CdCO}}_{\text{3}}(s)+{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

Aqueous solution of ${\text{CdSO}}_{\text{4}}$ reacts with ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}$ resulting in formation of ${\text{CdCO}}_{\text{3}}$ and ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$.  The precipitate that is formed is ${\text{CdCO}}_{\text{3}}$.  This is because the carbonate salts of cations except that is from Group-1 elements and ammonium ion are insoluble.

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

    ${\text{CdSO}}_{\text{4}}(\text{aq})+{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}(\text{aq})\to {\text{CdCO}}_{\text{3}}(s)+{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

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

${\text{Cd}}^{\text{2+}}\text{(aq)}+{\text{SO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{2NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+CO}}_{\text{3}}{}^{2-}(\text{aq})\to {\text{CdCO}}_{\text{3}}(s)+{\text{SO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{2NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}$

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 ${\text{SO}}_{\text{4}}{}^{2-}$ and ${\text{NH}}_{\text{4}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

${\text{Cd}}^{\text{2+}}\text{(aq)}+\cancel{{\text{SO}}_{\text{4}}{}^{2-}(\text{aq})}+\cancel{{\text{2NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}{\text{+CO}}_{\text{3}}{}^{2-}(\text{aq})\to {\text{CdCO}}_{\text{3}}(s)+\cancel{{\text{SO}}_{\text{4}}{}^{2-}(\text{aq})}+\cancel{{\text{2NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}$

Thus net ionic equation can be written as follows:

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

(e)

Interpretation Introduction

Interpretation:

Net ionic equation has to be written for the formation of precipitate and the spectator ions has to be identified in the reaction between ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

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

    ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(\text{aq})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+\text{2HCl(aq)}$

Aqueous solution of ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ reacts with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ resulting in formation of ${\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{HCl}$.  The precipitate that is formed is ${\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}$.  This is because the sulfate 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:

    ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}(\text{aq})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+\text{2HCl(aq)}$

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

    ${\text{Hg}}_{\text{2}}{}^{\text{2+}}{\text{(aq)+2Cl}}^{-}(\text{aq})+2{\text{H}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}{}^{2-}(\text{aq})\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+{\text{2H}}^{\text{+}}{\text{(aq)+2Cl}}^{-}\text{(aq)}$

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 ${\text{H}}^{\text{+}}$ and ${\text{Cl}}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

${\text{Hg}}_{\text{2}}{}^{\text{2+}}\text{(aq)+}\cancel{{\text{2Cl}}^{-}(\text{aq})}+\cancel{2{\text{H}}^{\text{+}}\text{(aq)}}{\text{+SO}}_{\text{4}}{}^{2-}(\text{aq})\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+\cancel{{\text{2H}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{2Cl}}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

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