Chapter 5, Problem 5.31E

Reactions represented by the following equations take place in water solutions. Write each molecular equation in total ionic form, then identify spectator ions and write the equations in net ionic form. Solids that do not dissolve are designated by (s), gases that do not dissolve are designated by (g), and substances that dissolve but do not dissociate appear in blue.

a. ${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right)+{\text{SO}}_2\left(\text{aq}\right)\to {\text{2NaHSO}}_{\text{3}}\left(\text{aq}\right)$

b. $\begin{array}{l}\text{3Cu}\left(\text{s}\right)+{\text{8HNO}}_{\text{3}}\left(\text{aq}\right)\to \\ 3\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

c. $\text{2HCl}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{CaCl}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

d. ${\text{CaCO}}_3\left(\text{s}\right)+2\text{HCl}\left(\text{aq}\right)\to {\text{CaCl}}_2\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

e. ${\text{MnO}}_2\left(\text{s}\right)+4\text{HCl}\left(\text{aq}\right)\to {\text{MnCl}}_{\text{2}}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

f. ${\text{2AgNO}}_3\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to \text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$

Interpretation Introduction

(a)

Interpretation:

The given equation is to be written in total ionic form. The spectator ionsare to be identified. The net ionic equation is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The ionic compound conducts electricity in the molten state or in solution. The ionic compounds are soluble in water and dissociates into their respective ions.

Answer to Problem 5.31E

The total ionic form of given reaction is ${\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{Na}}^{+}\left(\text{aq}\right)+{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)+{\text{SO}}_2\left(\text{aq}\right)\to {\text{2Na}}^{+}\left(\text{aq}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)$. The spectator ions are ${\text{Na}}^{+}$ and ${\text{SO}}_{\text{3}}{}^{2-}$. The net ionic equation is ${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{SO}}_2\left(\text{aq}\right)\to 2{\text{H}}^{+}\left(\text{aq}\right)+{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}\left(\text{aq}\right)+{\text{SO}}_2\left(\text{aq}\right)\to {\text{2NaHSO}}_{\text{3}}\left(\text{aq}\right)$

The total ionic form of given reaction can be written as shown below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{Na}}^{+}\left(\text{aq}\right)+{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)+{\text{SO}}_2\left(\text{aq}\right)\to {\text{2Na}}^{+}\left(\text{aq}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)$

The spectator ions are those ions whose identity does not change after the reaction. Thus, the spectator ions are ${\text{Na}}^{+}$ and ${\text{SO}}_{\text{3}}{}^{2-}$.

The net ionic equation can be expressed by removing the spectator ions. Hence, the net ionic equation is shown below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{SO}}_2\left(\text{aq}\right)\to 2{\text{H}}^{+}\left(\text{aq}\right)+{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)$

Conclusion

The total ionic form of given reaction is

${\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{Na}}^{+}\left(\text{aq}\right)+{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)+{\text{SO}}_2\left(\text{aq}\right)\to {\text{2Na}}^{+}\left(\text{aq}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)$.

The spectator ions are ${\text{Na}}^{+}$ and ${\text{SO}}_{\text{3}}{}^{2-}$.

The net ionic equation is ${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{SO}}_2\left(\text{aq}\right)\to 2{\text{H}}^{+}\left(\text{aq}\right)+{\text{SO}}_{\text{3}}{}^{2-}\left(\text{aq}\right)$.

Interpretation Introduction

(b)

Interpretation:

The given equation is to be written in total ionic form. The spectator ions are to be identified. The net ionic equation is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The ionic compound conducts electricity in the molten state or in solution. The ionic compounds are soluble in water and dissociates into their respective ions.

Answer to Problem 5.31E

The total ionic form of given reaction is $\text{3Cu}\left(\text{s}\right)+{\text{8H}}^{+}\left(\text{aq}\right)+8{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to 3{\text{Cu}}^{2+}\left(\text{aq}\right)+6{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$. The spectator ion is ${\text{6NO}}_{\text{3}}{}^{-}$. The net ionic equation is $\text{3Cu}\left(\text{s}\right)+{\text{8H}}^{+}\left(\text{aq}\right)+2{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to 3{\text{Cu}}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

The given reaction is shown below.

$\text{3Cu}\left(\text{s}\right)+{\text{8HNO}}_{\text{3}}\left(\text{aq}\right)\to 3\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

The total ionic form of given reaction can be written as shown below.

$\text{3Cu}\left(\text{s}\right)+{\text{8H}}^{+}\left(\text{aq}\right)+8{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to 3{\text{Cu}}^{2+}\left(\text{aq}\right)+6{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

The spectator ions are those ions whose identity does not change after the reaction. Thus, the spectator ion is ${\text{6NO}}_{\text{3}}{}^{-}$.

The net ionic equation can be expressed by removing the spectator ions. Hence, the net ionic equation is shown below.

$\text{3Cu}\left(\text{s}\right)+{\text{8H}}^{+}\left(\text{aq}\right)+2{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to 3{\text{Cu}}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The total ionic form of given reaction is $\text{3Cu}\left(\text{s}\right)+{\text{8H}}^{+}\left(\text{aq}\right)+8{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to 3{\text{Cu}}^{2+}\left(\text{aq}\right)+6{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$. The spectator ion is ${\text{6NO}}_{\text{3}}{}^{-}$. The net ionic equation is $\text{3Cu}\left(\text{s}\right)+{\text{8H}}^{+}\left(\text{aq}\right)+2{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)\to 3{\text{Cu}}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$.

Interpretation Introduction

(c)

Interpretation:

The given equation is to be written in total ionic form. The spectator ions are to be identified. The net ionic equation is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The ionic compound conducts electricity in the molten state or in solution. The ionic compounds are soluble in water and dissociates into their respective ions.

Answer to Problem 5.31E

The total ionic form of given reaction is

${\text{2H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The spectator ion is ${\text{Cl}}^{-}$. The net ionic equation is

${\text{2H}}^{+}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

The given reaction is shown below.

$\text{2HCl}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{CaCl}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The total ionic form of given reaction can be written as shown below.

${\text{2H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The spectator ions are those ions whose identity does not change after the reaction. Thus, the spectator ion is ${\text{Cl}}^{-}$.

The net ionic equation can be expressed by removing the spectator ions. Hence, the net ionic equation is shown below.

${\text{2H}}^{+}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The total ionic form of given reaction is ${\text{2H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$. The spectator ion is ${\text{Cl}}^{-}$. The net ionic equation is ${\text{2H}}^{+}\left(\text{aq}\right)+\text{CaO}\left(\text{s}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Interpretation Introduction

(d)

Interpretation:

The given equation is to be written in total ionic form. The spectator ions are to be identified. The net ionic equation is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The ionic compound conducts electricity in the molten state or in solution. The ionic compounds are soluble in water and dissociates into their respective ions.

Answer to Problem 5.31E

The total ionic form of given reaction is ${\text{CaCO}}_3\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$. The spectator ion is ${\text{Cl}}^{-}$. The net ionic equation is ${\text{CaCO}}_3\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{CaCO}}_3\left(\text{s}\right)+2\text{HCl}\left(\text{aq}\right)\to {\text{CaCl}}_2\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The total ionic form of given reaction can be written as shown below.

${\text{CaCO}}_3\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The spectator ions are those ions whose identity does not change after the reaction. Thus, the spectator ion is ${\text{Cl}}^{-}$.

The net ionic equation can be expressed by removing the spectator ions. Hence, the net ionic equation is shown below.

${\text{CaCO}}_3\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The total ionic form of given reaction is ${\text{CaCO}}_3\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$. The spectator ion is ${\text{Cl}}^{-}$. The net ionic equation is ${\text{CaCO}}_3\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Ca}}^{2+}\left(\text{aq}\right)+{\text{CO}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Interpretation Introduction

(e)

Interpretation:

The given equation is to be written in total ionic form. The spectator ions are to be identified. The net ionic equation is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The ionic compound conducts electricity in the molten state or in solution. The ionic compounds are soluble in water and dissociates into their respective ions.

Answer to Problem 5.31E

The total ionic form of given reaction is ${\text{MnO}}_2\left(\text{s}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+4{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$. The spectator ion is ${\text{2Cl}}^{-}$. The net ionic equation is ${\text{MnO}}_2\left(\text{s}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{MnO}}_2\left(\text{s}\right)+4\text{HCl}\left(\text{aq}\right)\to {\text{MnCl}}_{\text{2}}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The total ionic form of given reaction can be written as shown below.

${\text{MnO}}_2\left(\text{s}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+4{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The spectator ions are those ions whose identity does not change after the reaction. Thus, the spectator ion is ${\text{2Cl}}^{-}$.

The net ionic equation can be expressed by removing the spectator ions. Hence, the net ionic equation is shown below.

${\text{MnO}}_2\left(\text{s}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The total ionic form of given reaction is ${\text{MnO}}_2\left(\text{s}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+4{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$. The spectator ion is ${\text{2Cl}}^{-}$. The net ionic equation is ${\text{MnO}}_2\left(\text{s}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right)+{\text{Cl}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Interpretation Introduction

(f)

Interpretation:

The given equation is to be written in total ionic form. The spectator ions are to be identified. The net ionic equation is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The ionic compound conducts electricity in the molten state or in solution. The ionic compounds are soluble in water and dissociates into their respective ions.

Answer to Problem 5.31E

The total ionic form of given reaction is ${\text{2Ag}}^{+}\left(\text{aq}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{aq}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$. The spectator ion is ${\text{NO}}_{\text{3}}{}^{-}$. The net ionic equation is ${\text{2Ag}}^{+}\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{2AgNO}}_3\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to \text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$

The total ionic form of given reaction can be written as shown below.

${\text{2Ag}}^{+}\left(\text{aq}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{aq}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$

The spectator ions are those ions whose identity does not change after the reaction. Thus, the spectator ion is ${\text{NO}}_{\text{3}}{}^{-}$.

The net ionic equation can be expressed by removing the spectator ions. Hence, the net ionic equation is shown below.

${\text{2Ag}}^{+}\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$

Conclusion

The total ionic form of given reaction is ${\text{2Ag}}^{+}\left(\text{aq}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{aq}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$. The spectator ion is ${\text{NO}}_{\text{3}}{}^{-}$. The net ionic equation is ${\text{2Ag}}^{+}\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{aq}\right)+\text{2Ag}\left(\text{s}\right)$.