Chapter 4, Problem 32QP

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

$\begin{array}{l}\text{(a)}\\ \text{HBr}(aq){\text{+NH}}_{\text{3}}(aq)\to \\ \text{(b)}\\ {\text{Ba(OH)}}_{\text{2}}(aq){\text{+H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)\to \\ \text{(c)}\\ {\text{HClO}}_{\text{4}}(aq){\text{+Mg(OH)}}_{\text{2}}\text{(}s\text{)}\to \end{array}$

Interpretation Introduction

Interpretation:

The given equations are to be balanced and the corresponding ionic and net ionic equations are to be written.

Concept introduction:

The ionic and net ionic equations can be written using the following steps:

Write the balanced molecular equation along with the state of each substance.

Rewrite the equation as the complete ionic equation showing which compounds are dissociated into ions using the knowledge of solubility rules.

Cancel out the spectator ions on both sides of the equation to get the net ionic equation.

Answer to Problem 32QP

Solution:

(a) Balanced molecular equation:

$\text{HBr}\left(aq\right)+{\text{NH}}_3\left(aq\right)\to {\text{NH}}_{\text{4}}\text{Br}\left(aq\right)$

Ionic equation:

${\text{H}}^{+}\left(aq\right){\text{+ Br}}^{-}\left(aq\right)+{\text{NH}}_3\left(aq\right)\to {\text{NH}}_4^{+}\left(aq\right){\text{+ Br}}^{-}\left(aq\right)$

Net ionic equation:

${\text{H}}^{+}\left(aq\right)+{\text{NH}}_3\left(aq\right)\to {\text{NH}}_4^{+}\left(aq\right)$

(b) Balanced molecular equation:

$\text{3Ba}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{Ba}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{H}}_2\text{O}\left(l\right)$

Ionic equation:

${\text{3Ba}}^{2+}\left(aq\right)+6{\text{OH}}^{-}\left(aq\right)+2{\text{H}}_3{\text{PO}}_4\left(aq\right)\to {\text{Ba}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{H}}_2\text{O}\left(l\right)$

(c) Balanced molecular equation:

${\text{2HClO}}_{\text{4}}\left(aq\right)+\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to \text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Ionic equation:

${\text{2H}}^{+}\left(aq\right){\text{+ 2ClO}}_4^{-}\left(aq\right){\text{+ Mg}}^{2+}\left(aq\right){\text{+ 2OH}}^{-}\left(aq\right)\to {\text{Mg}}^{2+}\left(aq\right)+{\text{2ClO}}_4^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Net ionic equation:

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

Explanation of Solution

(a)

The balanced molecular equation for the given reaction is as follows:

$\text{HBr}\left(aq\right)+{\text{NH}}_3\left(aq\right)\to {\text{NH}}_{\text{4}}\text{Br}\left(aq\right)$

Write the complete ionic equation for the reaction showing all the dissociated ions.

${\text{H}}^{+}\left(aq\right){\text{+ Br}}^{-}\left(aq\right)+{\text{NH}}_3\left(aq\right)\to {\text{NH}}_4^{+}\left(aq\right){\text{+ Br}}^{-}\left(aq\right)$

The product formed, ${\text{NH}}_{\text{4}}\text{Br}$, is a soluble compound. Hence, it remains soluble in the solution. Cancel the spectator ion, ${\text{Br}}^{\text{-}}$, on both the sides of the equation.

Thus, we get the net ionic equation as

${\text{H}}^{+}\left(aq\right)+{\text{NH}}_3\left(aq\right)\to {\text{NH}}_4^{+}\left(aq\right)$

(b)

The balanced molecular equation for the given reaction is as follows:

$\text{3Ba}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to {\text{Ba}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{H}}_2\text{O}\left(l\right)$

Write the complete ionic equation for the reaction showing all the dissociated ions.

${\text{3Ba}}^{2+}\left(aq\right)+6{\text{OH}}^{-}\left(aq\right)+2{\text{H}}_3{\text{PO}}_4\left(aq\right)\to {\text{Ba}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{H}}_2\text{O}\left(l\right)$

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is a weak acid. The product formed, ${\text{Ba}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$, is an insoluble compound. There are no spectator ions, hence no net ionic equation.

(c)

The balanced molecular equation for the given reaction is as follows:

${\text{2HClO}}_{\text{4}}\left(aq\right)+\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to \text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Write the complete ionic equation for the reaction showing all the dissociated ions.

${\text{2H}}^{+}\left(aq\right){\text{+ 2ClO}}_4^{-}\left(aq\right){\text{+ Mg}}^{2+}\left(aq\right){\text{+ 2OH}}^{-}\left(aq\right)\to {\text{Mg}}^{2+}\left(aq\right)+{\text{2ClO}}_4^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

The product formed, $\text{Mg}{\left({\text{ClO}}_4\right)}_2$, is asoluble compound. Cancel the spectator ions, ${\text{Mg}}^{2+}{\text{and ClO}}_4^{-}$, on both the sides of the equation.

Thus, we get the net ionic equation as

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