Chapter 4, Problem 4.48QP

Complete the right side of each of the following molecular equations. Then write the net ionic equations. Assume all salts formed are soluble. Acid salts are possible.

1. a Ca(OH)₂(aq) + 2H₂SO₄(aq) →
2. b 2H₃PO₄(aq) + Ca(OH)₂(aq) →
3. c NaOH(aq) + H₂SO₄(aq) →
4. d Sr(OH)₂(aq) + 2H₂CO₃(aq) →

Interpretation Introduction

Interpretation:

The given set of molecular equations has to be completed and net ionic equation has to be found.

Concept introduction:

Neutralization Reaction:  In neutralization reactions acid and base reacts together to form water and an ionic compound(salt).

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides( Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.48QP

The complete molecular equation

${\text{2H}}_2{\text{SO}}_{\text{4(aq)}}{\text{ + Ca(OH)}}_{\text{2(aq)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + Ca(HSO}}_{\text{4}}{\text{)}}_{\text{2(aq)}}$

The net ionic equation

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

Explanation of Solution

The molecular equation for the reaction between sulfuric acid and calcium hydroxide is given below.

${\text{2H}}_2{\text{SO}}_{\text{4(aq)}}{\text{ + Ca(OH)}}_{\text{2(aq)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + Ca(HSO}}_{\text{4}}{\text{)}}_{\text{2(aq)}}$

In complete ionic equation the electrolytes are represented as its ions.  The complete ionic equation for the reaction is given below.

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

The ions common in the reactant and the product side are cancelled from the total ionic equation to get net ionic equation.

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

The entire equation is divided by two,

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

Interpretation Introduction

Interpretation:

The given set of molecular equations has to be completed and net ionic equation has to be found.

Concept introduction:

Neutralization Reaction:  In neutralization reactions acid and base reacts together to form water and an ionic compound(salt).

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides( Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.48QP

The complete molecular equation

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

The net ionic equation

${\text{H}}_3{\text{PO}}_{\text{4 }}{}_{\text{(aq)}}{\text{+ OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{O}}_{\text{(l)}}+{\text{H}}_2{\text{PO}}_{\text{4 }}{{}^{-}}_{\text{(aq)}}$

Explanation of Solution

The complete molecular equation for the reaction between calcium hydroxide and phosphoric acid is given below.  The salt formed is a soluble salt.

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

In complete ionic equation the electrolytes are represented as its ions.  The complete ionic equation for the reaction is given below.  As phosphoric acid is a weak acid its ionization is not included in the equation.

${\text{2H}}_3{\text{PO}}_{\text{4 }}{}_{\text{(aq)}}{\text{+ Ca}}^{2+}{}_{(aq)}{\text{+2OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}+{\text{Ca}}^{2+}{}_{(aq)}+2{\text{H}}_2{\text{PO}}_{\text{4 }}{{}^{-}}_{\text{(aq)}}$

The ions common in the reactant and the product side are cancelled from the total ionic equation to get net ionic equation.

${\text{2H}}_3{\text{PO}}_{\text{4 }}{}_{\text{(aq)}}{\text{+ +2OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}+2{\text{H}}_2{\text{PO}}_{\text{4 }}{{}^{-}}_{\text{(aq)}}$

The entire equation is divided by two,

${\text{H}}_3{\text{PO}}_{\text{4 }}{}_{\text{(aq)}}{\text{+ OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{O}}_{\text{(l)}}+{\text{H}}_2{\text{PO}}_{\text{4 }}{{}^{-}}_{\text{(aq)}}$

Interpretation Introduction

Interpretation:

The given set of molecular equations has to be completed and net ionic equation has to be found.

Concept introduction:

Neutralization Reaction:  In neutralization reactions acid and base reacts together to form water and an ionic compound(salt).

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides (Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.48QP

The complete molecular equation

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{ + NaOH}}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + NaHSO}}_4{}_{\text{(aq)}}$

The net ionic equation

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

Explanation of Solution

The complete molecular equation for the reaction between sulfuric acid and sodium hydroxide is given below.

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{ + NaOH}}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + NaHSO}}_4{}_{\text{(aq)}}$

In complete ionic equation the electrolytes are represented as its ions.  The complete ionic equation for the reaction is given below.

${\text{H}}^{+}{}_{(aq)}{\text{+HSO}}_4{{}^{-}}_{(aq)}{\text{ + Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+HSO}}_4{{}^{-}}_{(aq)}$

The ions common in the reactant and the product side are cancelled from the total ionic equation to get net ionic equation.

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

Interpretation Introduction

Interpretation:

The given set of molecular equations has to be completed and net ionic equation has to be found.

Concept introduction:

Neutralization Reaction:  In neutralization reactions acid and base reacts together to form water and an ionic compound (salt).

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides (Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.48QP

Molecular equation

${\text{2H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}{\text{ + Sr(OH)}}_2{}_{\text{(aq)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + Sr(HCO}}_3{\text{)}}_{\text{2(aq)}}$

The net ionic equation

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}{\text{ +OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ +HCO}}_3{{}^{-}}_{\text{(aq)}}$

Explanation of Solution

The molecular equation for the reaction between carbonic acid and strontium hydroxide is given below.

${\text{2H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}{\text{ + Sr(OH)}}_2{}_{\text{(aq)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + Sr(HCO}}_3{\text{)}}_{\text{2(aq)}}$

In complete ionic equation the electrolytes are represented as its ions.  The complete ionic equation for the reaction is given below.  Since, carbonic acid is a weak acid, its ionization is not included in the reactant side of net ionic equation.

${\text{2H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}{\text{ + Sr}}^{2+}{}_{(aq)}{\text{+2OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + Sr}}^{2+}{}_{(aq)}{\text{+2HCO}}_3{{}^{-}}_{\text{(aq)}}$

The ions common in the reactant and the product side are cancelled from the total ionic equation to get net ionic equation.

${\text{2H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}{\text{ +2OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ +2HCO}}_3{{}^{-}}_{\text{(aq)}}$

The entire equation is divided by two,

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}{\text{ +OH}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ +HCO}}_3{{}^{-}}_{\text{(aq)}}$