Chapter 15, Problem 5PE

(a)

Interpretation Introduction

Interpretation:

Total and net ionic equation for reaction indicated has to be written.

  $\text{Zn}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{ZnCl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

Concept Introduction:

The ionic equation can be written based on rules summarized as follows:

• Strong electrolytes dissociate completely into ionic forms.
• Weak electrolytes remain in molecular form.
• Nonelectrolytes remain in molecular form.
• Insoluble precipitates, and gases remain in molecular forms.
• The net ionic equation includes only ions that have reacted and thus any spectator ions are usually omitted.

Explanation of Solution

The formula equation is given as follows:

  $\text{Zn}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{ZnCl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

Since ${\text{H}}_{\text{2}}$ is gas and $\text{Zn}$ is pure metal so the total ionic equation with all the ions in dissociated form gives total equation written as follows:

  $\text{Zn}\left(s\right)+2{\text{H}}^{+}+{\text{2Cl}}^{-}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{2Cl}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

The common ${\text{Cl}}^{-}$ ions are cancelled out from each side and final net ionic equation is written as follows:

  $\text{Zn}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

(b)

Interpretation Introduction

Interpretation:

Total and net ionic equation for reaction indicated has to be written.

  $\text{2Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{H}}_2{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formula equation is given as follows:

  $\text{2Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{H}}_2{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Since ${\text{H}}_{\text{2}}\text{O}$ is liquid and $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is solid so the total ionic equation with all the ions in dissociated form gives total equation written as follows:

  $\text{2Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{H}}^{+}+3{\text{SO}}_4^{2-}\left(aq\right)\to 2{\text{Al}}^{3+}\left(aq\right)+3{\text{SO}}_4^{2-}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The common ${\text{SO}}_4^{2-}$ ions are cancelled out from each side and thus net ionic equation is written as follows:

  $\text{2Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{H}}^{+}\to 2{\text{Al}}^{3+}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(c)

Interpretation Introduction

Interpretation:

Total and net ionic equation for reaction indicated has to be written.

  $\text{Ca}{\left({\text{HCO}}_3\right)}_2\left(s\right)+2\text{HBr}\left(aq\right)\to {\text{CaBr}}_2\left(aq\right)+2{\text{CO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formula equation is given as follows:

  $\text{Ca}{\left({\text{HCO}}_3\right)}_2\left(s\right)+2\text{HBr}\left(aq\right)\to {\text{CaBr}}_2\left(aq\right)+2{\text{CO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Since ${\text{CO}}_{\text{2}}$ is a gas, ${\text{H}}_{\text{2}}\text{O}$ is liquid and $\text{Ca}{\left({\text{HCO}}_3\right)}_2$ is solid so the total ionic equation with only ions in dissociated form gives total equation written as follows:

  $\text{Ca}{\left({\text{HCO}}_3\right)}_2+2{\text{H}}^{+}+{\text{2Br}}^{-}\to {\text{Ca}}^{2+}+2{\text{Br}}^{-}+2{\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

The common ${\text{Br}}^{-}$ ions are cancelled out from each side and thus net ionic equation is written as follows:

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

(c)

Interpretation Introduction

Interpretation:

Total and net ionic equations for reaction indicated have to be written.

  $\text{3KOH}\left(s\right)+{\text{H}}_3{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formula equation is given as follows:

  $\text{3KOH}\left(s\right)+{\text{H}}_3{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$

Since ${\text{H}}_{\text{2}}\text{O}$ is liquid and $\text{KOH}$ is solid so the total ionic equation with only ions in dissociated form is written as follows:

  $\text{3KOH}\left(s\right)+3{\text{H}}^{+}\left(aq\right)+{\text{PO}}_4^{3-}\left(aq\right)\to 3{\text{K}}^{+}\left(aq\right)+{\text{PO}}_4^{3-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$

The common ${\text{PO}}_4^{3-}$ ions are cancelled out from each side and thus net ionic equation is written as follows:

  $\text{3KOH}\left(s\right)+3{\text{H}}^{+}\left(aq\right)\to 3{\text{K}}^{+}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$