Chapter F, Problem J.6E

(a)

Interpretation Introduction

Interpretation:

Overall equation, complete ionic equation, and the net ionic equation for the given reaction has to be written.

    ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+NaOH(aq)\to$

Concept Introduction:

Reaction between the acid and base is known as a neutralization reaction.  The product formed in neutralization reaction is salt and water.  The net reaction that takes place between the acid and base solutions is that the formation of water from hydrogen ions and hydroxide ions.  Neutralization reaction is the one that takes place between a strong acid and a metal hydroxide.

    $\text{Acid}+\text{Base}\to \text{Salt}+\text{Water}$

Complete ionic equation is the one that shows all the species that is present in the chemical reaction.  The dissolved ionic compounds are alone showed as ions.  Precipitates are not showed as ions.

Net ionic equation is the one that is obtained from the complete ionic equation by cancelling out the spectator ions.

Explanation of Solution

Overall equation:

The given incomplete equation is written as follows;

    ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+NaOH(aq)\to$

Arsenic acid is a weak triprotic acid.  Hence, it is kept in the molecular form itself.  The product formed in the reaction is water and trisodium arsenate.  The overall equation can be given as shown below;

    ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+3NaOH(aq)\to {\text{Na}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+3{\text{H}}_{\text{2}}\text{O}(l)$

Complete ionic equation:

The complete ionic equation can be written considering the ionic compounds in aqueous medium to be written into respective ions.  Therefore, the complete ionic equation can be given as follows;

  ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+3N{a}^{+}(aq)+3O{H}^{-}(aq)\to 3{\text{Na}}^{\text{+}}{\text{(aq)+AsO}}_{\text{4}}{}^{3-}(aq)+3{\text{H}}_{\text{2}}\text{O}(l)$

Net ionic equation:

The net ionic equation can be obtained from the complete ionic equation by cancelling out the spectator ions on both sides of the equation.

  ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+\cancel{3N{a}^{+}(aq)}+3O{H}^{-}(aq)\to \cancel{3{\text{Na}}^{\text{+}}\text{(aq)}}{\text{+AsO}}_{\text{4}}{}^{3-}(aq)+3{\text{H}}_{\text{2}}\text{O}(l)$

Thus, the net ionic equation can be given as shown below;

  ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+3O{H}^{-}(aq)\to {\text{AsO}}_{\text{4}}{}^{3-}(aq)+3{\text{H}}_{\text{2}}\text{O}(l)$

(b)

Interpretation Introduction

Interpretation:

Overall equation, complete ionic equation, and the net ionic equation for the given reaction has to be written.

    ${\text{Sr(OH)}}_{\text{2}}(aq)+{\text{HClO}}_{\text{4}}(aq)\to$

Concept Introduction:

Refer part (a).

Explanation of Solution

Overall equation:

The given incomplete equation is written as follows;

    ${\text{Sr(OH)}}_{\text{2}}(aq)+{\text{HClO}}_{\text{4}}(aq)\to$

Strontium hydroxide is a strong base and perchloric acid is a strong acid.  The product formed in the reaction is ${\text{Sr(ClO}}_{\text{4}}{\text{)}}_{\text{2}}$ and water.  The overall equation can be given as shown below;

    ${\text{Sr(OH)}}_{\text{2}}(aq)+2{\text{HClO}}_{\text{4}}(aq)\to {\text{Sr(ClO}}_{\text{4}}{\text{)}}_{\text{2}}(\text{aq})+2{\text{H}}_{\text{2}}\text{O}(l)$

Complete ionic equation:

The complete ionic equation can be written considering the ionic compounds in aqueous medium to be written into respective ions.  Therefore, the complete ionic equation can be given as follows;

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

Net ionic equation:

The net ionic equation can be obtained from the complete ionic equation by cancelling out the spectator ions on both sides of the equation.

$\cancel{{\text{Sr}}^{\text{2+}}\text{(aq)}}{\text{+2OH}}^{-}(aq)+2{\text{H}}^{\text{+}}\text{(aq)+}\cancel{{\text{2ClO}}_{\text{4}}{}^{2-}(aq)}\to \cancel{{\text{Sr}}^{\text{2+}}\text{(aq)}}\text{+}\cancel{{\text{2ClO}}_{\text{4}}{}^{2-}(\text{aq})}+2{\text{H}}_{\text{2}}\text{O}(l)$

Thus, the net ionic equation can be given as shown below;

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

(c)

Interpretation Introduction

Interpretation:

Overall equation, complete ionic equation, and the net ionic equation for the given reaction has to be written.

    ${\text{Ca(OH)}}_{\text{2}}(\text{s})+\text{HBrO}(aq)\to$

Concept Introduction:

Refer part (a).

Explanation of Solution

Overall equation:

The given incomplete equation is written as follows;

    ${\text{Ca(OH)}}_{\text{2}}(\text{s})+\text{HBrO}(aq)\to$

Calcium hydroxide is a strong base.  As it is in solid state, the molecular form is retained as such.  Hypobromous acid is a weak acid and hence it is also kept in molecular form.  The product formed in the reaction is ${\text{Ca(BrO)}}_{\text{2}}$ and water.  The overall equation can be given as shown below;

    ${\text{Ca(OH)}}_{\text{2}}(\text{s})+2\text{HBrO}(aq)\to {\text{Ca(BrO)}}_{\text{2}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$

Complete ionic equation:

The complete ionic equation can be written considering the ionic compounds in aqueous medium to be written into respective ions.  Therefore, the complete ionic equation can be given as follows;

  ${\text{Ca(OH)}}_{\text{2}}(\text{s})+2\text{HBrO}(aq)\to {\text{Ca}}^{\text{2+}}{\text{(aq)+2BrO}}^{-}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$

Net ionic equation:

The net ionic equation can be obtained from the complete ionic equation by cancelling out the spectator ions on both sides of the equation.  In the complete ionic equation, there are no spectator ions.  Hence, the net ionic equation is the complete ionic equation and it is given as follows;

  ${\text{Ca(OH)}}_{\text{2}}(\text{s})+2\text{HBrO}(aq)\to {\text{Ca}}^{\text{2+}}{\text{(aq)+2BrO}}^{-}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$