Chapter 10, Problem 10.55P

(a)

Interpretation Introduction

Interpretation:

The equation has to be completed and balanced for the acid-base reaction of chloric acid and barium 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 chloric acid and barium hydroxide is barium chlorate and water.  The completed equation is,

  $\begin{array}{l}{\text{HClO}}_{\text{3(aq)}}\text{+Ba}{\left(\text{OH}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }\text{Ba}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Chloric}\text{Barium}\text{Barium}\text{Water}\\ \text{acid}\text{hydroxide}\text{chlorate}\end{array}$

The balanced equation is,

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

The corresponding ionic equation is,

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

Cancellation of the spectator ions $\left({\text{Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{,ClO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\right)$ yields the net ionic equation.  The net ionic equation is,

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

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

(b)

Interpretation Introduction

Interpretation:

The equation has to be completed and balanced for the acid-base reaction of acetic acid and potassium 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 acetic acid and potassium hydroxide is potassium acetate and water.  The completed equation is,

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+KOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{COOK}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Acetic}\text{acid}\text{Potassium}\text{Potassium}\text{Water}\\ \text{hydroxide}\text{acetate}\end{array}$

The above equation is self-balanced.  The corresponding ionic equation is,

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

Cancellation of the spectator ions $\left({\text{K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{,CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}\right)$ yields the net ionic equation.  The net ionic equation is,

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

(c)

Interpretation Introduction

The equation has to be completed and balanced for the acid-base reaction of hydrogen iodide and magnesium 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 hydrogen iodide and magnesium hydroxide is magnesium iodide and water.  The completed equation is,

  $\begin{array}{l}{\text{HI}}_{\left(\text{aq}\right)}\text{+Mg}{\left(\text{OH}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }{\text{MgI}}_{\text{2(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Hydrogen}\text{Magnesium}\text{Magnesium}\text{Water}\\ \text{iodide}\text{hydroxide}\text{iodide}\end{array}$

The balanced equation is,

  ${\text{2HI}}_{\left(\text{aq}\right)}\text{+Mg}{\left(\text{OH}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }{\text{MgI}}_{\text{2(aq)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The corresponding ionic equation is,

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

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

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

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

(d)

Interpretation Introduction

The equation has to be completed and balanced for the acid-base reaction of sulphuric acid and rubidium 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 rubidium hydroxide is rubidium sulphate and water.  The completed equation is,

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{+RbOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Sulphuric}\text{Rubidium}\text{Rubidium}\text{Water}\\ \text{acid}\text{hydroxide}\text{sulphate}\end{array}$

The balanced equation is,

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{+2RbOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The corresponding ionic equation is,

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

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

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

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