Chapter 10, Problem 10.56P

(a)

Interpretation Introduction

Interpretation:

The equation has to be completed and balanced for the acid-base reaction of perchloric acid and calcium hydroxide.  The products have to be named and the corresponding net ionic equation has to be written.

Concept Introduction:

Double replacement reactions:  This type of reactions occurs when the cation and the anions switch between two reactants to form new products.  In order for the reaction to occur, one of the products is usually a solid precipitate, a gas or a molecular compound.

Acid-Base reactions:  A chemical reaction between an acid and a base is called as neutralization reaction.  The ionic compound is formed along with water in this reaction is called as salt.

Explanation of Solution

The product of the reaction between perchloric acid and calcium hydroxide is calcium perchlorate and water.  The completed equation is,

  $\begin{array}{l}{\text{HClO}}_{\text{4(aq)}}\text{+Ca}{\left(\text{OH}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }\text{Ca}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Perchloric}\text{Calcium}\text{Calcium}\text{Water}\\ \text{acid}\text{hydroxide}\text{perchlorate}\end{array}$

The balanced equation is,

  ${\text{2HClO}}_{\text{4(aq)}}\text{+Ca}{\left(\text{OH}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }\text{Ca}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2(aq)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The corresponding ionic equation is,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Cancellation of the spectator ions $\left({\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{,ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\right)$ yields the net ionic equation.  The net ionic equation is,

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

Dividing both the sides by two yields,

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

(b)

Interpretation Introduction

Interpretation:

The equation has to be completed and balanced for the acid-base reaction of hydrochloric acid and calcium carbonate.  The products have to be named and the corresponding net ionic equation has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between hydrochloric acid and calcium carbonate is calcium chloride, carbonic acid.  The completed equation is,

  $\begin{array}{l}{\text{HCl}}_{\left(\text{aq}\right)}\text{+}{\text{CaCO}}_{\text{3}}{}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{CaCl}}_{\text{2}\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}\\ \text{Hydrochloric}\text{Calcium}\text{Calicum}\text{Carbonic}\text{acid}\\ \text{acid}\text{carbonate}\text{chloride}\end{array}$

The balanced equation is,

  ${\text{2HCl}}_{\left(\text{aq}\right)}\text{+}{\text{CaCO}}_{\text{3}}{}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{CaCl}}_{\text{2}\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}$

The ionic equation is,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}$

Cancellation of the spectator ions yields the net ionic equation.  The net ionic equation is,

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

The decomposition of carbonic acid results in the production of water and carbon dioxide gas.  The equation is,

  ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+CO}}_{\text{2(g)}}$

The overall equation is,

  $\begin{array}{l}{\text{HCl}}_{\left(\text{aq}\right)}\text{+}{\text{CaCO}}_{\text{3}}{}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{CaCl}}_{\text{2}\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+CO}}_{\text{2(g)}}\\ \text{Hydrochloric}\text{Calcium}\text{Calicum}\text{Water}\text{Carbon}\\ \text{acid}\text{carbonate}\text{chloride}\text{dioxide}\end{array}$

The net ionic equation is,

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

(c)

Interpretation Introduction

The equation has to be completed and balanced for the acid-base reaction of nitric acid and aluminium oxide.  The products have to be named and the corresponding net ionic equation has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between nitric acid iodide and aluminium oxide is aluminium nitrate and water.  The completed equation is,

  $\begin{array}{l}{\text{HNO}}_{\text{3}}{}_{\left(\text{aq}\right)}{\text{+Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}\stackrel{}{\to }\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Nitric}\text{Aluminium}\text{Aluminium}\text{Water}\\ \text{acid}\text{oxide}\text{nitrate}\end{array}$

The balanced equation is,

  ${\text{6HNO}}_{\text{3}}{}_{\left(\text{aq}\right)}{\text{+Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}\stackrel{}{\to }\text{2Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3(aq)}}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Aluminium oxide is broken into aluminium (III) hydroxide and water.

The corresponding ionic equation is,

  ${\text{6H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+6NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Cancellation of the spectator ions $\left({\text{Al}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{,NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\right)$ yields the net ionic equation.  The net ionic equation is,

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

(d)

Interpretation Introduction

The equation has to be completed and balanced for the acid-base reaction of sulphuric acid and copper hydroxide.  The products have to be named and the corresponding net ionic equation has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between sulphuric acid and copper hydroxide is copper sulphate and water.  The completed equation is,

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\text{+Cu}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}\stackrel{}{\to }{\text{CuSO}}_{\text{4(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Sulphuric}\text{Copper}\text{Copper}\text{Water}\\ \text{acid}\text{hydroxide}\text{sulphate}\end{array}$

The balanced equation is,

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\text{+Cu}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}\stackrel{}{\to }{\text{CuSO}}_{\text{4(aq)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The corresponding ionic equation is,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Cancellation of the spectator ions $\left({\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{,SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\right)$ yields the net ionic equation.  The net ionic equation is,

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

Dividing both the sides by two yields,

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