Chapter 4, Problem 4.33QP

Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate).

1. (a) $\text{HBr}(aq)+{\text{NH}}_3(aq)\stackrel{}{\to }$
2. (b) ${\text{Ba(OH)}}_2(aq)+{\text{H}}_3{\text{PO}}_4(aq)\stackrel{}{\to }$
3. (c) ${\text{HClO}}_4(aq)+\text{Mg(OH)}2(s)\stackrel{}{\to }$

Interpretation Introduction

Interpretation:

The balanced and ionic and net ionic equations for given reactions are should be written.

Concept introduction:

Balanced equation

• The balanced equation is nothing but the all ionic compounds are represented as neutral compounds using the molecular formula.
• In the balanced equation, the number of atoms for each element and total charge are equal in the reaction for in both reactants and the products sides.

Ionic equation:

• In the ionic reactions, all the species present in reaction are written with respective charges is called ionic equation.

Net ionic equation:

• The removal of spectator ions forms ionic equation to give net ionic equation.
• To write the balanced and ionic and net ionic equations for given reaction.

Answer to Problem 4.33QP

The balanced and ionic and net ionic equations for given reactions are,

The balanced equation is,

${\text{HBr}}_{\text{(aq)}}{\text{+NH}}_{\text{3}}{}_{\text{(aq) }}\to {\text{NH}}_{\text{4}}{\text{Br}}_{\text{(aq)}}$

The ionic equation is,

${\text{H}}^{+}{}_{\text{(aq)}}{\text{+Br}}^{-}{}_{\text{(aq)}}{\text{+NH}}_{\text{3}}{}_{\text{(aq) }}\to {\text{NH}}_{\text{4}}{{}^{+}}_{\text{(aq)}}{\text{+Br}}^{-}{}_{\text{(aq)}}$

The net ionic equation is,

${\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{ + NH}}_{\text{3}}{}_{\text{(aq) }}\to {\text{+4NH }}^{+}{}_{\text{(aq)}}$

Explanation of Solution

• The ionic compounds are dissolved in water to produced ions these ions are combined to give new product.
• The reaction is represent by the complex molecular formula means it is a balanced equation and it have equal number of atoms in both reactant and product sides.
• The ${\text{HBr}}_{\text{(aq)}}$ and ${\text{NH}}_{\text{3}}{}_{\text{(aq)}}$ are react to give the ${\text{NH}}_{\text{4}}{\text{Br}}_{\text{(aq)}}$ so the balanced equation for this reaction is,

  ${\text{HBr}}_{\text{(aq)}}{\text{+NH}}_{\text{3}}{}_{\text{(aq) }}\to {\text{NH}}_{\text{4}}{\text{Br}}_{\text{(aq)}}$

• All the species present in above equation is writing with respective charges to give ionic reaction.

  ${\text{H}}^{+}{}_{\text{(aq)}}{\text{+Br}}^{-}{}_{\text{(aq)}}{\text{+NH}}_{\text{3}}{}_{\text{(aq) }}\to {\text{NH}}_{\text{4}}{{}^{+}}_{\text{(aq)}}{\text{+Br}}^{-}{}_{\text{(aq)}}$

• The removal of common ions from the above equations to give net ionic equations.

  ${\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{ + NH}}_{\text{3}}{}_{\text{(aq) }}\to {\text{+4NH }}^{+}{}_{\text{(aq)}}$

Interpretation Introduction

Interpretation:

The balanced and ionic and net ionic equations for given reactions are should be written.

Concept introduction:

Balanced equation

• The balanced equation is nothing but the all ionic compounds are represented as neutral compounds using the molecular formula.
• In the balanced equation, the number of atoms for each element and total charge are equal in the reaction for in both reactants and the products sides.

Ionic equation:

• In the ionic reactions, all the species present in reaction are written with respective charges is called ionic equation.

Net ionic equation:

• The removal of spectator ions forms ionic equation to give net ionic equation.
• To write the balanced and ionic and net ionic equations for given reaction.

Answer to Problem 4.33QP

The balanced and ionic and net ionic equations for given reactions are,

The balanced equation is,

${\text{Ba(OH)}}_{\text{2(aq)}}{\text{+ H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_2{}_{\text{(s)}}{\text{+ 6H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The ionic equation is,

${\text{3Ba}}^{2+}{}_{\text{(aq)}}{\text{+3OH}}^{-}{}_{\text{(aq)}}{\text{+ 3H}}^{+}{\text{+PO}}_{\text{4(aq)}}{}^{3-}\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_2{}_{\text{(s)}}{\text{+ 3OH}}^{-}{}_{\text{(aq)}}{\text{+ 3H}}^{+}{}_{\text{(aq)}}$

The net ionic equation is,

${\text{3Ba}}^{2+}{}_{\text{(aq)}}{\text{+2PO}}_{\text{4(aq)}}{}^{3-}\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_2{}_{\text{(s)}}$

Explanation of Solution

The ionic compounds are dissolved in water to produced ions these ions are combined to give new product.

• The reaction is represent by the complex molecular formula means it is a balanced equation and it have equal number of atoms in both reactant and product sides.
• The ${\text{Ba(OH)}}_{\text{2(aq)}}$ and ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ are react to give the ${\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{(s)}}$ so the balanced equation for this reaction is,

  ${\text{Ba(OH)}}_{\text{2(aq)}}{\text{+ H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_2{}_{\text{(s)}}{\text{+ 6H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

• All the species present in above equation is writing with respective charges to give ionic reaction.

  ${\text{3Ba}}^{2+}{}_{\text{(aq)}}{\text{+3OH}}^{-}{}_{\text{(aq)}}{\text{+ 3H}}^{+}{\text{+PO}}_{\text{4(aq)}}{}^{3-}\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_2{}_{\text{(s)}}{\text{+ 3OH}}^{-}{}_{\text{(aq)}}{\text{+ 3H}}^{+}{}_{\text{(aq)}}$

• The removal of common ions from the above equations to give net ionic equations.

  ${\text{3Ba}}^{2+}{}_{\text{(aq)}}{\text{+2PO}}_{\text{4(aq)}}{}^{3-}\to {\text{Ba}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_2{}_{\text{(s)}}$

Interpretation Introduction

Interpretation:

The balanced and ionic and net ionic equations for given reactions are should be written.

Concept introduction:

Balanced equation

• The balanced equation is nothing but the all ionic compounds are represented as neutral compounds using the molecular formula.
• In the balanced equation, the number of atoms for each element and total charge are equal in the reaction for in both reactants and the products sides.

Ionic equation:

• In the ionic reactions, all the species present in reaction are written with respective charges is called ionic equation.

Net ionic equation:

• The removal of spectator ions forms ionic equation to give net ionic equation.
• To write the balanced and ionic and net ionic equations for given reaction.

Answer to Problem 4.33QP

The balanced and ionic and net ionic equations for given reactions are,

The balanced equation is,

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

The ionic equation is,

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

The net ionic equation is,

${\text{2ClO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+Mg}}^{\text{2+}}{}_{\text{(aq)}}\to {\text{Mg(ClO}}_{\text{4}}{\text{)}}_{\text{2}\left(\text{aq}\right)}$

Explanation of Solution

• The ionic compounds are dissolved in water to produced ions these ions are combined to give new product.
• The reaction is represent by the complex molecular formula means it is a balanced equation and it have equal number of atoms in both reactant and product sides.
• The ${\text{HClO}}_{\text{4(aq)}}$ and ${\text{Mg(OH)}}_{\text{2(s)}}$ are react to give the ${\text{Mg(ClO}}_{\text{4}}{\text{)}}_{\text{2}\left(\text{aq}\right)}$ so the balanced equation for this reaction is,

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

• All the species present in above equation is writing with respective charges to give ionic reaction.

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

• The removal of common ions from the above equations to give net ionic equations.

  ${\text{2ClO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+Mg}}^{\text{2+}}{}_{\text{(aq)}}\to {\text{Mg(ClO}}_{\text{4}}{\text{)}}_{\text{2}\left(\text{aq}\right)}$