Chapter 10, Problem 10.43P

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for the double displacement of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{}_{\left(\text{aq}\right)}$ and ${\text{NaOH}}_{\left(\text{aq}\right)}$ 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 10.43P

The net ionic equation is,

  ${\text{Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3(s)}}$

Explanation of Solution

Solutions of iron (III) nitrate and sodium hydroxide react to form iron (III) hydroxide and sodium nitrate.  This is a double replacement reaction.  The complete equation is,

  $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3(aq)}}{\text{+NaOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3(s)}}{\text{+NaNO}}_{\text{3(aq)}}$

The balanced equation is,

  $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3(aq)}}{\text{+3NaOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3(s)}}{\text{+3NaNO}}_{\text{3(aq)}}$

The complete ionic equation is,

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

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.  Iron (III) hydroxide does not break into ions in this equation.

The net ionic equation is written as,

  ${\text{Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3(s)}}$

(b)

Interpretation Introduction

Interpretation:

The net ionic equation for the double replacement reaction of $\text{Zn}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}\left(\text{aq}\right)}$ and ${\text{K}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}$ has to be written and the equation has to be balanced.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.43P

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 perchlorate react to form zinc sulphide and potassium perchlorate.  This is a double replacement reaction.  The complete equation is,

  $\text{Zn}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2(aq)}}{\text{+K}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{ZnS}}_{\left(\text{s}\right)}{\text{+KClO}}_{\text{4(aq)}}$

The balanced equation is,

  $\text{Zn}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2(aq)}}{\text{+K}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{ZnS}}_{\left(\text{s}\right)}{\text{+2KClO}}_{\text{4(aq)}}$

The complete ionic equation is,

  ${\text{Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+ZnS}}_{\left(\text{s}\right)}$

The spectator ions, ${\text{K}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Zinc sulphide 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.  Zinc sulphide does not break into ions in this equation.

The net ionic equation is written as,

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

(c)

Interpretation Introduction

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.43P

The net ionic equation is,

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

Explanation of Solution

Solutions of lead nitrate and potassium hydroxide react to form potassium nitrate and lead (III) hydroxide.  This is a double replacement reaction.  The complete equation is,

  $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+KOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }\text{Pb}{\left(\text{OH}\right)}_{\text{2(s)}}{\text{+KNO}}_{\text{3(aq)}}$

The balanced equation is,

  $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+2KOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }\text{Pb}{\left(\text{OH}\right)}_{\text{2(s)}}{\text{+2KNO}}_{\text{3(aq)}}$

The complete ionic equation is,

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

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.  Because spectator ions appear on both sides of the complete ionic equation, the spectator ions can be cancelled.  Lead (III) hydroxide does not break into ions in this equation.

The net ionic equation is written as,

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

(d)

Interpretation Introduction

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.43P

The net ionic equation is,

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

Explanation of Solution

Solutions of zinc nitrate and sodium carbonate react to form zinc carbonate and sodium nitrate.  The complete equation is,

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

The balanced equation is,

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

The complete ionic equation is,

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

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.  Because spectator ions appear on both sides of the complete ionic equation, the spectator ions can be cancelled.  Zinc carbonate does not break into ions in this equation.

The net ionic equation is,

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