Chapter F, Problem I.13E

(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 ${\text{FeCl}}_{\text{2}}$ and ${\text{Na}}_{\text{2}}\text{S}$.

Explanation of Solution

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

    ${\text{FeCl}}_{\text{2}}(\text{aq})+{\text{Na}}_{\text{2}}\text{S}(\text{aq})\to \text{FeS}(s)+\text{2NaCl(aq)}$

Aqueous solution of ${\text{FeCl}}_{\text{2}}$ reacts with ${\text{Na}}_{\text{2}}\text{S}$ resulting in formation of $\text{FeS}$ and $\text{NaCl}$.  The precipitate that is formed is $\text{FeS}$.  This is because the sulfides are insoluble if the cations are apart from the elements of Group-1 and Group-2.

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

    ${\text{FeCl}}_{\text{2}}(\text{aq})+{\text{Na}}_{\text{2}}\text{S}(\text{aq})\to \text{FeS}(s)+\text{2NaCl(aq)}$

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

    ${\text{Fe}}^{\text{2+}}{\text{(aq)+2Cl}}^{-}(\text{aq})+2{\text{Na}}^{\text{+}}\text{(aq)}+{\text{S}}^{\text{2}-}(\text{aq})\to \text{FeS}(s)+{\text{2Na}}^{\text{+}}{\text{(aq)+2Cl}}^{-}\text{(aq)}$

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

    ${\text{Fe}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2Cl}}^{-}(\text{aq})}+\cancel{2{\text{Na}}^{\text{+}}\text{(aq)}}+{\text{S}}^{\text{2}-}(\text{aq})\to \text{FeS}(s)+\cancel{{\text{2Na}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{2Cl}}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

    ${\text{Fe}}^{\text{2+}}\text{(aq)}+{\text{S}}^{\text{2}-}(\text{aq})\to \text{FeS}(s)$

(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 ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ and $\text{KI}$.

Explanation of Solution

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

    ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})+2\text{KI}(\text{aq})\to {\text{PbI}}_{\text{2}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ reacts with $\text{KI}$ resulting in formation of ${\text{PbI}}_{\text{2}}$ and ${\text{KNO}}_{\text{3}}$.  The precipitate that is formed is ${\text{PbI}}_{\text{2}}$.  This is because the iodide salts of ${\text{Ag}}^{\text{+}}$, ${\text{Hg}}_{\text{2}}{}^{\text{2+}}$, and ${\text{Pb}}^{\text{2+}}$ are alone insoluble.

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

    ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})+2\text{KI}(\text{aq})\to {\text{PbI}}_{\text{2}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

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

    ${\text{Pb}}^{\text{2+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})+2{\text{K}}^{\text{+}}{\text{(aq)+2I}}^{-}(\text{aq})\to {\text{PbI}}_{\text{2}}(s)+{\text{2K}}^{\text{+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}\text{(aq)}$

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.

    ${\text{Pb}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(\text{aq})}+\cancel{2{\text{K}}^{\text{+}}\text{(aq)}}{\text{+2I}}^{-}(\text{aq})\to {\text{PbI}}_{\text{2}}(s)+\cancel{{\text{2K}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

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

(c)

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 ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ and ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

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

    ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})+{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{CaSO}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ reacts with ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ resulting in formation of ${\text{CaSO}}_{\text{4}}$ and ${\text{KNO}}_{\text{3}}$.  The precipitate that is formed is ${\text{CaSO}}_{\text{4}}$.  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:

    ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})+{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{CaSO}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

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

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

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.

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

Thus net ionic equation can be written as follows:

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

(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 ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}$ and ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.

Explanation of Solution

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

    ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}(\text{aq})+{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})\to {\text{PbCrO}}_{\text{4}}(s)+{\text{2NaNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}$ reacts with ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ resulting in formation of ${\text{PbCrO}}_{\text{4}}$ and ${\text{NaNO}}_{\text{3}}$.  The precipitate that is formed is ${\text{PbCrO}}_{\text{4}}$.  This is because the chromate 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:

    ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}(\text{aq})+{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})\to {\text{PbCrO}}_{\text{4}}(s)+{\text{2NaNO}}_{\text{3}}\text{(aq)}$

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

${\text{2Na}}^{\text{+}}\text{(aq)}+{\text{CrO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Pb}}^{\text{2+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})\to {\text{PbCrO}}_{\text{4}}(s)+{\text{2Na}}^{\text{+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}\text{(aq)}$

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

$\cancel{{\text{2Na}}^{\text{+}}\text{(aq)}}+{\text{CrO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Pb}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(\text{aq})}\to {\text{PbCrO}}_{\text{4}}(s)+\cancel{{\text{2Na}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

    ${\text{CrO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Pb}}^{\text{2+}}\text{(aq)}\to {\text{PbCrO}}_{\text{4}}(s)$

(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 ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ and ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

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

    ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}(\text{aq})+{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ reacts with ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ resulting in formation of ${\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{KNO}}_{\text{3}}$.  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:

    ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}(\text{aq})+{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}(\text{aq})\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

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

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

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.

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

Thus net ionic equation can be written as follows:

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