Chapter 10, Problem 10.42P

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for the neutralization reaction of ${\text{HClO}}_{\text{3}}{}_{\left(\text{aq}\right)}$ and ${\text{KOH}}_{\left(\text{aq}\right)}$ has to be written.

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 10.42P

The net ionic equation is,

  ${\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

Solutions of chloric acid and potassium hydroxide react to form potassium chlorate and water.  This is a neutralization reaction.  The complete equation is,

  ${\text{HClO}}_{\text{3(aq)}}{\text{+KOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{KClO}}_{\text{3}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The complete ionic equation is,

  ${\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+ClO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+ClO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The spectator ions, ${\text{K}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{ClO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$ 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.

The net ionic equation is written as,

  ${\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.42P

The net ionic equation is,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbCO}}_{\text{3}}{}_{\left(\text{s}\right)}$

Explanation of Solution

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

  $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{2NaNO}}_{\text{3(aq)}}{\text{+PbCO}}_{\text{3(s)}}$

The complete ionic equation is,

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

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.  Lead carbonate 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,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbCO}}_{\text{3}}{}_{\left(\text{s}\right)}$

(c)

Interpretation Introduction

The net ionic equation for the double replacement reaction of ${\text{AgClO}}_{\text{4(aq)}}$ and ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$ has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.42P

The net ionic equation is written as,

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

Explanation of Solution

Solutions of silver perchlorate and ammonium sulphate react to form silver sulphate and ammonium perchlorate.  This is a double displacement reaction.  The complete equation is,

  ${\text{2AgClO}}_{\text{4(aq)}}\text{+}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{2NH}}_{\text{4}}{\text{ClO}}_{\text{4(aq)}}{\text{+Ag}}_{\text{2}}{\text{SO}}_{\text{4(s)}}$

The complete ionic equation is,

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

The spectator ions, ${\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\left(\text{aq}\right)}$ and ${\text{ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$ 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 sulphate does not break into ions in this equation.

The net ionic equation is written as,

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

(d)

Interpretation Introduction

The net ionic equation for the reaction of ${\text{K}}_{\text{2}}{\text{S}}_{\left(\text{s}\right)}$ and $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}$ has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.42P

The net ionic equation is,

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

Explanation of Solution

Solutions of potassium sulphide and zinc nitrate react to form potassium nitrate and zinc sulphide.  The complete equation is,

  ${\text{K}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}\text{+Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }{\text{2KNO}}_{\text{3}}{}_{\left(\text{aq}\right)}{\text{+ZnS}}_{\left(\text{s}\right)}$

The complete ionic equation is,

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

The spectator ions, ${\text{K}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Zinc sulphide does not break up 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,

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