Chapter 4, Problem 39E

Write the balanced molecular, complete, and net ionicequations for the reaction, if any, that occurs when aqueoussolutions of the following are mixed.
a. ammonium sulfate and barium nitrate
b. lead(II) nitrate and sodium chloride
c. sodium phosphate and potassium nitrate
d. sodium bromide and rubidium chloride
e. copper(II) chloride and sodium hydroxide

Interpretation Introduction

Interpretation: The molecular equation, complete ionic equation and net ionic equation representing the formation of precipitate by the mixing of given solutions are to be predicted.

Concept introduction: The chemical equation in which molecules of reactants and products are present is termed as molecular equation. The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the chemical equation which contains ions only is termed as complete ionic equation. The equation obtained by the elimination of spectator ions from the complete ionic equation is known as net ionic equation.

Answer to Problem 39E

The molecular equation is shown below.

  ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4\left(aq\right)+\text{Ba}{\left({\text{NO}}_3\right)}_2\left(aq\right)\to 2{\text{NH}}_4{\text{NO}}_3\left(aq\right)+{\text{BaSO}}_4\left(aq\right)$

The complete ionic equation is shown below.

  $\begin{array}{l}{\text{2NH}}_4{}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)+{\text{Ba}}^{2+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)\to \\ {\text{2NH}}_4{}^{+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)+{\text{Ba}}^{2+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)\end{array}$

The net ionic equation does not exist.

Explanation of Solution

The given solutions are ammonium sulfate, ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ and barium nitrate, $\text{Ba}{\left({\text{NO}}_3\right)}_2$ .

The molecular equation representing the reaction between ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ and $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is shown below.

  ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4\left(aq\right)+\text{Ba}{\left({\text{NO}}_3\right)}_2\left(aq\right)\to 2{\text{NH}}_4{\text{NO}}_3\left(aq\right)+{\text{BaSO}}_4\left(aq\right)$

There is no precipitate formed in the reaction.

The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the complete ionic equation representing the reaction between ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ and $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is shown below.

  $\begin{array}{l}{\text{2NH}}_4{}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)+{\text{Ba}}^{2+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)\to \\ {\text{2NH}}_4{}^{+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)+{\text{Ba}}^{2+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)\end{array}$

The net ionic equation is obtained by eliminating the spectator ions present on left and right side of the complete ionic equation. As there is not any precipitate formed, therefore, the net ionic equation does not exist.

Interpretation Introduction

Interpretation: The molecular equation, complete ionic equation and net ionic equation representing the formation of precipitate by the mixing of given solutions are to be predicted.

Concept introduction: The chemical equation in which molecules of reactants and products are present is termed as molecular equation. The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the chemical equation which contains ions only is termed as complete ionic equation. The equation obtained by the elimination of spectator ions from the complete ionic equation is known as net ionic equation.

Answer to Problem 39E

The molecular equation is shown below.

  $\text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{NaCl}\left(aq\right)\to {\text{PbCl}}_2\left(s\right)+2{\text{NaNO}}_3\left(aq\right)$

The complete ionic equation is shown below.

  $\begin{array}{l}{\text{Pb}}^{2+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to \\ {\text{PbCl}}_2\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)\end{array}$

The net ionic equation is ${\text{Pb}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{PbCl}}_2\left(s\right)$ .

Explanation of Solution

The given solutions are lead (II) nitrate, $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and sodium chloride, $\text{NaCl}$ .

The molecular equation representing the reaction between $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and $\text{NaCl}$ is shown below.

  $\text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{NaCl}\left(aq\right)\to {\text{PbCl}}_2\left(s\right)+2{\text{NaNO}}_3\left(aq\right)$

The precipitate so formed in the reaction is of ${\text{PbCl}}_2$ .

The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the complete ionic equation representing the reaction between $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and $\text{NaCl}$ is shown below.

  $\begin{array}{l}{\text{Pb}}^{2+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to \\ {\text{PbCl}}_2\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)\end{array}$

The net ionic equation is obtained by eliminating the spectator ions present on left and right side of the complete ionic equation. Therefore, the net ionic equation representing the formation of precipitate is shown below.

  ${\text{Pb}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{PbCl}}_2\left(s\right)$

Interpretation Introduction

Interpretation: The molecular equation, complete ionic equation and net ionic equation representing the formation of precipitate by the mixing of given solutions are to be predicted.

Concept introduction: The chemical equation in which molecules of reactants and products are present is termed as molecular equation. The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the chemical equation which contains ions only is termed as complete ionic equation. The equation obtained by the elimination of spectator ions from the complete ionic equation is known as net ionic equation.

Answer to Problem 39E

The molecular equation is shown below.

  ${\text{Na}}_3{\text{PO}}_4\left(aq\right)+3{\text{KNO}}_3\left(aq\right)\to 3{\text{NaNO}}_3\left(aq\right)+{\text{K}}_3{\text{PO}}_4\left(aq\right)$

The complete ionic equation is shown below.

  $\begin{array}{l}{\text{3Na}}^{+}\left(aq\right)+{\text{PO}}_4{}^{3-}\left(aq\right)+{\text{3K}}^{+}\left(aq\right)+3{\text{NO}}_3{}^{-}\left(aq\right)\to \\ {\text{3Na}}^{+}\left(aq\right)+3{\text{NO}}_3{}^{-}\left(aq\right)+{\text{3K}}^{+}\left(aq\right)+{\text{PO}}_4{}^{3-}\left(aq\right)\end{array}$

There is nonet ionic equation.

Explanation of Solution

The given solutions are sodium phosphate, ${\text{Na}}_3{\text{PO}}_4$ and potassium nitrate, ${\text{KNO}}_3$ .

The molecular equation representing the reaction between ${\text{Na}}_3{\text{PO}}_4$ and ${\text{KNO}}_3$ is shown below.

  ${\text{Na}}_3{\text{PO}}_4\left(aq\right)+3{\text{KNO}}_3\left(aq\right)\to 3{\text{NaNO}}_3\left(aq\right)+{\text{K}}_3{\text{PO}}_4\left(aq\right)$

There is no precipitate formed in the reaction.

The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the complete ionic equation representing the reaction between ${\text{Na}}_3{\text{PO}}_4$ and ${\text{KNO}}_3$ is shown below.

  $\begin{array}{l}{\text{3Na}}^{+}\left(aq\right)+{\text{PO}}_4{}^{3-}\left(aq\right)+{\text{3K}}^{+}\left(aq\right)+3{\text{NO}}_3{}^{-}\left(aq\right)\to \\ {\text{3Na}}^{+}\left(aq\right)+3{\text{NO}}_3{}^{-}\left(aq\right)+{\text{3K}}^{+}\left(aq\right)+{\text{PO}}_4{}^{3-}\left(aq\right)\end{array}$

The net ionic equation is obtained by eliminating the spectator ions present on left and right side of the complete ionic equation. As there is not any precipitate formed, therefore, the net ionic equation does not exist.

Interpretation Introduction

Interpretation: The molecular equation, complete ionic equation and net ionic equation representing the formation of precipitate by the mixing of given solutions are to be predicted.

Concept introduction: The chemical equation in which molecules of reactants and products are present is termed as molecular equation. The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the chemical equation which contains ions only is termed as complete ionic equation. The equation obtained by the elimination of spectator ions from the complete ionic equation is known as net ionic equation.

Answer to Problem 39E

The molecular equation is shown below.

  $\text{NaBr}\left(aq\right)+\text{RbCl}\left(aq\right)\to \text{NaCl}\left(aq\right)+\text{RbBr}\left(aq\right)$

The complete ionic equation is shown below.

  $\begin{array}{l}{\text{Na}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)+{\text{Rb}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\to \\ {\text{Na}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+{\text{Rb}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\end{array}$

There is no net ionic equation.

Explanation of Solution

The given solutions are sodium bromide, $\text{NaBr}$ and rubidium chloride, $\text{RbCl}$ .

The molecular equation representing the reaction between $\text{NaBr}$ and $\text{RbCl}$ is shown below.

  $\text{NaBr}\left(aq\right)+\text{RbCl}\left(aq\right)\to \text{NaCl}\left(aq\right)+\text{RbBr}\left(aq\right)$

There is no precipitate formed in the reaction.

The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the complete ionic equation representing the reaction between $\text{NaBr}$ and $\text{RbCl}$ is shown below.

  $\begin{array}{l}{\text{Na}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)+{\text{Rb}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\to \\ {\text{Na}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+{\text{Rb}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\end{array}$

The net ionic equation is obtained by eliminating the spectator ions present on left and right side of the complete ionic equation. As there is not any precipitate formed, therefore, the net ionic equation does not exist.

Interpretation Introduction

Interpretation: The molecular equation, complete ionic equation and net ionic equation representing the formation of precipitate by the mixing of given solutions are to be predicted.

Concept introduction: The chemical equation in which molecules of reactants and products are present is termed as molecular equation. The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the chemical equation which contains ions only is termed as complete ionic equation. The equation obtained by the elimination of spectator ions from the complete ionic equation is known as net ionic equation.

Answer to Problem 39E

The molecular equation is shown below.

  ${\text{CuCl}}_{\text{2}}\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2\text{NaCl}\left(aq\right)$

The complete ionic equation is shown below.

  $\begin{array}{l}{\text{Cu}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to \\ \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\end{array}$

The net ionic equation is ${\text{Cu}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)$ .

Explanation of Solution

The given solutions are copper chloride, ${\text{CuCl}}_2$ and sodium hydroxide, $\text{NaOH}$ .

The molecular equation representing the reaction between ${\text{CuCl}}_2$ and $\text{NaOH}$ is shown below.

  ${\text{CuCl}}_{\text{2}}\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2\text{NaCl}\left(aq\right)$

The precipitate so formed in the reaction is of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ .

The substance that are strong electrolytes will get dissociate into their constituent ions. Therefore, the complete ionic equation representing the reaction between ${\text{CuCl}}_2$ and $\text{NaOH}$ is shown below.

  $\begin{array}{l}{\text{Cu}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to \\ \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\end{array}$

The net ionic equation is obtained by eliminating the spectator ions present on left and right side of the complete ionic equation. Therefore, the net ionic equation representing the formation of precipitate is shown below.

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