Chapter 10, Problem 10.44P

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for the double displacement of ${\text{AgNO}}_{\text{3}}{}_{\left(\text{aq}\right)}$ and ${\text{Na}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}$ has to be written and the equation has to be balanced.

Concept Introduction:

Net ionic equation:  A chemical equation, which lists only those species taking part in the reaction is called as net ionic equation.  The net ionic equation is used in acid-base neutralization reactions, double replacement reactions and redox reactions.

Spectator ions:  A spectator ion is one, which exists, in the same form on both the reactant and product sides of a chemical reaction.

Answer to Problem 10.44P

The net ionic equation is,

  ${\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ag}}_{\text{2}}{\text{S}}_{\left(\text{s}\right)}$

Explanation of Solution

Solutions of silver nitrate and sodium sulphide react to form silver sulphide and sodium nitrate.  This is a double replacement reaction.  The complete equation is,

  ${\text{AgNO}}_{\text{3(aq)}}{\text{+Na}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ag}}_{\text{2}}{\text{S}}_{\left(\text{s}\right)}{\text{+NaNO}}_{\text{3(aq)}}$

The balanced equation is,

  ${\text{2AgNO}}_{\text{3(aq)}}{\text{+Na}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ag}}_{\text{2}}{\text{S}}_{\left(\text{s}\right)}{\text{+2NaNO}}_{\text{3(aq)}}$

The complete ionic equation is,

  ${\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ag}}_{\text{2}}{\text{S}}_{\left(\text{s}\right)}{\text{+2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The spectator ions, ${\text{Na}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Because spectator ions appear on both sides of the complete ionic equation, the spectator ions can be cancelled.  Silver sulphide does not break into ions in this solution.

The net ionic equation is written as,

  ${\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ag}}_{\text{2}}{\text{S}}_{\left(\text{s}\right)}$

(b)

Interpretation Introduction

Interpretation:

The net ionic equation for the double replacement reaction of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$ and $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}$ has to be written and the equation has to be balanced.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.44P

The net ionic equation is,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}$

Explanation of Solution

Solutions of sulphuric acid and lead nitrate react to form lead sulphate and nitric acid.  This is a double replacement reaction.  The complete equation is,

  $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{PbSO}}_{\text{4(s)}}{\text{+HNO}}_{\text{3(aq)}}$

The balanced equation is,

  $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{PbSO}}_{\text{4(s)}}{\text{+2HNO}}_{\text{3(aq)}}$

The complete ionic equation is,

  ${\text{Pb}}^{}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbSO}}_{\text{4(s)}}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The spectator ions, ${\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Lead sulphate does not break into ions in this equation.  Because spectator ions  appear on both sides of the complete ionic equation, the spectator ions can be cancelled.

The net ionic equation is written as,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}$

(c)

Interpretation Introduction

The net ionic equation for the double displacement reaction of $\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}$ and ${\text{NaI}}_{\left(\text{aq}\right)}$ has to be written and the equation has to be balanced.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.44P

The net ionic equation is,

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

Explanation of Solution

Solutions of mercury (II) nitrate and sodium iodide react to form mercury iodide and sodium nitrate.  This is double replacement reaction.  The complete equation is,

  $\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+NaI}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{HgI}}_{\text{2(s)}}{\text{+NaNO}}_{\text{3(aq)}}$

The balanced equation is,

  $\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+2NaI}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{HgI}}_{\text{2(s)}}{\text{+2NaNO}}_{\text{3(aq)}}$

The complete ionic equation is,

  ${\text{Hg}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2I}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+HgI}}_{\text{2(s)}}$

The spectator ions, ${\text{Na}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Because spectator ions appear on both sides of the complete ionic equation, the spectator ions can be cancelled.  Mercury iodide does not break into ions in this solution.

The net ionic equation is written as,

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

(d)

Interpretation Introduction

The net ionic equation for the reaction of ${\text{CdCl}}_{\text{2(aq)}}$ and ${\text{AgClO}}_{\text{4(aq)}}$ has to be written and the equation has to be balanced.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.44P

The net ionic equation is,

  $\begin{array}{l}{\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2AgCl}}_{\left(\text{s}\right)}\\ {\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{AgCl}}_{\left(\text{s}\right)}\end{array}$

Explanation of Solution

Solutions of cadmium chloride and silver perchlorate react to form cadmium perchlorate and silver chloride.  The complete equation is,

  ${\text{CdCl}}_{\text{2(aq)}}{\text{+AgClO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{AgCl}}_{\left(\text{s}\right)}\text{+Cd}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2(sq)}}$

The balanced equation is,

  ${\text{CdCl}}_{\text{2(aq)}}{\text{+2AgClO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{2AgCl}}_{\left(\text{s}\right)}\text{+Cd}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2(sq)}}$

The complete ionic equation is,

  ${\text{Cd}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2AgCl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+Cd}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The spectator ions, ${\text{Cd}}^2{{}^{\text{+}}}_{\left(\text{aq}\right)}$ and ${\text{Cd}}^{\text{2+}}{}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Because spectator ions  appear on both sides of the complete ionic equation, the spectator ions can be cancelled.  Silver chloride does not break into ions in this solution.

The net ionic equation is,

  ${\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2AgCl}}_{\left(\text{s}\right)}$

The above net ionic equation is reduced to,

  ${\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{AgCl}}_{\left(\text{s}\right)}$