Chapter 4, Problem 21QP

Write ionic and net ionic equations for the following reactions:

(a)

${\text{AgNO}}_3(aq)+N{a}_2{\text{SO}}_4\left(aq\right)\to$

(b)

${\text{BaCl}}_2\left(aq\right)+{\text{ZnSO}}_4(aq)\to$

(c)

${\left({\text{NH}}_4\right)}_2{\text{CO}}_3\left(aq\right)+{\text{CaCl}}_2\left(aq\right)\to$

Interpretation Introduction

Interpretation:

The ionic and net ionic equations are to be written for the given reactions.

Concept introduction:

The ionic and net ionic equations are written using the following steps:

Ionic equations show the dissolved ionic compounds as free ions. Net ionic equations show species that are actually participating in the reaction.

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

Rewrite the equation as a complete ionic equation that shows the compounds that are dissociated into ions, using the knowledge of the solubility rules.

The compounds containing alkali metal cation or ammonium ion are watersoluble.

The compounds containing nitrate ion, chlorate ion, perchlorate ion, or acetate ion are watersoluble.

The compounds containing chloride ion, bromide ion, or iodide ion are watersoluble with the exception of compounds containing ${\text{Ag}}^{\text{+}}$, ${\text{Hg}}_2^{2+}$, or ${\text{Pb}}^{\text{2+}}$ ions.

The compounds containing sulfate ion are soluble with the exception of compounds containing ${\text{Ag}}^{\text{+}}$, ${\text{Hg}}_2^{2+}$, ${\text{Pb}}^{\text{2+}}$, ${\text{Ca}}^{\text{2+}}$, ${\text{Sr}}^{\text{2+}}$, or ${\text{Ba}}^{2+}$ ions.

The compounds containing carbonate ion, phosphate ion, and sulfide ion, and most metal hydroxides are insoluble in water except the compounds containing alkali metal cation or ammonium ion.

Spectator ions are the ions that exist as a reactant and a product in a chemical reaction

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

Answer to Problem 21QP

Solution:

(a) Ionic equation:

${\text{2Ag}}^{+}\left(aq\right){\text{+ 2NO}}_3^{-}\left(\text{aq}\right){\text{+ 2Na}}^{+}\left(aq\right){\text{+ SO}}_4^{2-}\left(\text{aq}\right)\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+{\text{2Na}}^{+}\left(aq\right)+{\text{2NO}}_3^{-}\left(\text{aq}\right)$.

Net ionic equation:

${\text{2Ag}}^{+}\left(aq\right){\text{+ SO}}_4^{2-}\left(\text{aq}\right)\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)$.

(b) Ionic equation:

${\text{Ba}}^{2+}\left(aq\right){\text{+ 2Cl}}^{-}\left(aq\right){\text{+ Zn}}^{2+}\left(aq\right){\text{+ SO}}_4^{2-}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)+{\text{Zn}}^{2+}\left(aq\right)+{\text{2Cl}}^{-}\left(aq\right)$.

Net ionic equation:

${\text{Ba}}^{2+}\left(aq\right){\text{+ SO}}_4^{2-}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)$.

(c) Ionic equation:

${\text{2NH}}_4^{+}\left(aq\right){\text{+CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right){\text{+ 2Cl}}^{-}\left(aq\right)\to {\text{ CaCO}}_3\left(s\right)+{\text{2NH}}_4^{+}\left(aq\right)+{\text{2Cl}}^{-}\left(aq\right)$.

Net ionic equation:

${\text{CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{ CaCO}}_3\left(s\right)$.

Explanation of Solution

a) ${\text{AgNO}}_{\text{3}}\left(\text{aq}\right){\text{+ Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)$

The balanced molecular equation for the given reaction is:

${\text{2AgNO}}_{\text{3}}\left(\text{aq}\right){\text{+ Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+2{\text{NaNO}}_{\text{3}}\left(aq\right)$.

Write the complete ionic equation for the reaction that shows all the dissociated ions.

${\text{2Ag}}^{+}\left(aq\right){\text{+ 2NO}}_3^{-}\left(\text{aq}\right){\text{+ 2Na}}^{+}\left(aq\right){\text{+ SO}}_4^{2-}\left(\text{aq}\right)\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)+{\text{2Na}}^{+}\left(aq\right)+{\text{2NO}}_3^{-}\left(\text{aq}\right)$.

The product ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is an insoluble compound. The compounds containing sulfate ion are soluble with the exception of compounds containing ${\text{Ag}}^{\text{+}}$, ${\text{Hg}}_2^{2+}$, ${\text{Pb}}^{\text{2+}}$, ${\text{Ca}}^{\text{2+}}$, ${\text{Sr}}^{\text{2+}}$, or ${\text{Ba}}^{2+}$ ions. Sodium and nitrate are spectator ion. They exist as reactant and as product

The other product, ${\text{NaNO}}_{\text{3}}$, remains soluble in the solution. Cancel the spectator ions, ${\text{Na}}^{\text{+}}{\text{and NO}}_{\text{3}}^{\text{-}}$, from both the sides of the equation.

Thus, we get the net ionic equation as:

${\text{2Ag}}^{+}\left(aq\right){\text{+ SO}}_4^{2-}\left(\text{aq}\right)\to {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)$.

b) ${\text{BaCl}}_2\left(\text{aq}\right){\text{+ ZnSO}}_{\text{4}}\left(\text{aq}\right)$

The balanced molecular equation for the given reaction is:

${\text{BaCl}}_2\left(\text{aq}\right){\text{+ ZnSO}}_{\text{4}}\left(\text{aq}\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)+{\text{ZnCl}}_{\text{2}}\left(aq\right)$.

Write the complete ionic equation for the reaction that shows all the dissociated ions.

${\text{Ba}}^{+}\left(aq\right){\text{+ 2Cl}}^{-}\left(\text{aq}\right){\text{+ Zn}}^{2+}\left(aq\right){\text{+ SO}}_4^{2-}\left(\text{aq}\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)+{\text{Zn}}^{2+}\left(aq\right)+{\text{2Cl}}^{-}\left(\text{aq}\right)$.

The product ${\text{BaSO}}_{\text{4}}$ is an insoluble compound. The compounds containing sulfate ion are soluble with the exception of compounds containing ${\text{Ag}}^{\text{+}}$, ${\text{Hg}}_2^{2+}$, ${\text{Pb}}^{\text{2+}}$, ${\text{Ca}}^{\text{2+}}$, ${\text{Sr}}^{\text{2+}}$, or ${\text{Ba}}^{2+}$ ions. Zinc and chloride are spectator ion. They exist as reactant and as product

The other product, ${\text{ZnCl}}_{\text{2}}$, remains soluble in the solution. Cancel the spectator ions, ${\text{Zn}}^{\text{2+}}{\text{and Cl}}^{\text{-}}$, from both the sides of the equation.

Thus, we get the net ionic equation as:

${\text{Ba}}^{+}\left(aq\right){\text{+ SO}}_4^{2-}\left(\text{aq}\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)$.

c) ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+{\text{CaCl}}_{\text{2}}\left(aq\right)$

The balanced molecular equation for the given reaction is:

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+{\text{CaCl}}_{\text{2}}\left(aq\right)\to {\text{ CaCO}}_3\left(s\right)+2{\text{NH}}_{\text{4}}\text{Cl}\left(aq\right)$.

Write the complete ionic equation for the reaction that shows all the dissociated ions.

${\text{2NH}}_4^{+}\left(aq\right){\text{+CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right){\text{+ 2Cl}}^{-}\left(aq\right)\to {\text{ CaCO}}_3\left(s\right)+{\text{2NH}}_4^{+}\left(aq\right)+{\text{2Cl}}^{-}\left(aq\right)$.

The product ${\text{CaCO}}_{\text{3}}$ is an insoluble compound. The compounds containing carbonate ion are insoluble in water except the compounds containing alkali metal cation or ammonium ion. Ammonium and chloride are spectator ion. They exist as reactant and as product

The other product, ${\text{NH}}_4\text{Cl}$, remains soluble in the solution. Cancel the spectator ions, ${\text{NH}}_4^{+}{\text{ and Cl}}^{\text{-}}$, from both the sides of the equation.

Thus, we get the net ionic equation as:

${\text{CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{ CaCO}}_3\left(s\right)$.