Chapter 4, Problem 22QP

Write ionic and net ionic equations for the following reactions:

(a)

${\text{Na}}_{\text{2}}\text{S}\left(aq\right)+{\text{ZnCl}}_{\text{2}}\left(aq\right)\to$

(b)

${\text{K}}_3{\text{PO}}_{\text{4}}\left(aq\right)+3\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)\to$

(c)

$\text{Mg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+2\text{NaOH}\left(aq\right)\to$

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 can be 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 the complete ionic equation, showing the compounds thatare dissociated into ions using the knowledge of 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 sides of the equation in order to obtain the net ionic equation.

Answer to Problem 22QP

Solution:

(a) Ionic equation:

${\text{2Na}}^{+}\left(aq\right){\text{+ S}}^{2-}\left(aq\right){\text{+ Zn}}^{2+}\left(aq\right){\text{+2Cl}}^{-}\left(aq\right)\to \text{ZnS}\left(s\right)+2{\text{Na}}^{+}\left(aq\right){\text{+ 2Cl}}^{-}\left(aq\right)$

Net ionic equation:

${\text{Zn}}^{2+}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to \text{ZnS}\left(s\right)$

(b) Ionic equation:

${\text{6K}}^{+}\left(aq\right){\text{+ 2PO}}_4^{3-}\left(aq\right)+3{\text{Sr}}^{2+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)\to {\text{Sr}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{K}}^{+}\left(aq\right){\text{+ 6NO}}_3^{-}\left(aq\right)$

Net ionic equation:

$3{\text{Sr}}^{2+}\left(aq\right){\text{+ 2PO}}_4^{3-}\left(aq\right)\to {\text{Sr}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$

(c) Ionic equation:

${\text{Mg}}^{2+}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)+{\text{2Na}}^{+}\left(aq\right){\text{+ 2OH}}^{-}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)$

Net ionic equation:

${\text{Mg}}^{2+}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)$

Explanation of Solution

a) ${\text{Na}}_{\text{2}}\text{S}\left(aq\right)+{\text{ ZnCl}}_2\left(aq\right)$

The balanced molecular equation for the given reaction is:

${\text{Na}}_2\text{S}\left(\text{aq}\right){\text{+ ZnCl}}_2\left(\text{aq}\right)\to \text{ZnS}\left(s\right)+2\text{NaCl}\left(aq\right)$

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

${\text{2Na}}^{+}\left(aq\right){\text{+ S}}^{2+}\left(\text{aq}\right){\text{+ Zn}}^{2+}\left(aq\right){\text{+2Cl}}^{-}\left(\text{aq}\right)\to \text{ZnS}\left(s\right)+2{\text{Na}}^{+}\left(aq\right){\text{+ 2Cl}}^{-}\left(aq\right)$

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

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

Thus, we get the net ionic equation as:

${\text{Zn}}^{2+}\left(aq\right)+{\text{S}}^{2+}\left(\text{aq}\right)\to \text{ZnS}\left(s\right)$

b) ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+3\text{Sr}{\left({\text{NO}}_3\right)}_2\left(aq\right)$

The balanced molecular equation for the given reaction is:

${\text{2K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+3\text{Sr}{\left({\text{NO}}_3\right)}_2\left(aq\right)\to {\text{Sr}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{KNO}}_3\left(aq\right)$

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

${\text{6K}}^{+}\left(aq\right){\text{+ 2PO}}_4^{3-}\left(aq\right)+3{\text{Sr}}^{2+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)\to {\text{Sr}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{K}}^{+}\left(aq\right){\text{+ 6NO}}_3^{-}\left(aq\right)$

The product ${\text{Sr}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$ is an insoluble compound. The compounds containing phosphate ion are insoluble in water except the compounds containing alkali metal cation or ammonium ion. Potassium and nitrate are spectator ions. They exist as reactant and product on both sides of the equation.

The other product ${\text{KNO}}_{\text{3}}$ remains soluble in the solution. Cancel the spectator ions, ${\text{K}}^{+}{\text{ and NO}}_3^{-}$, from both sides of the equation.

Thus, we get the net ionic equation as:

$3{\text{Sr}}^{2+}\left(aq\right){\text{+ 2PO}}_4^{3-}\left(aq\right)\to {\text{Sr}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$

c) $\text{Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+\text{2NaOH}\left(aq\right)$

The balanced molecular equation for the given reaction is:

$\text{Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+\text{2NaOH}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{NaNO}}_3\left(aq\right)$

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

${\text{Mg}}^{2+}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)+{\text{2Na}}^{+}\left(aq\right){\text{+ 2OH}}^{-}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)$ The product $\text{Mg}{\left(\text{OH}\right)}_2\left(s\right)$ is an insoluble compound. 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. Sodium and nitrate are spectator ion. They exist as reactant and product.

The other product ${\text{NaNO}}_{\text{3}}$ remains soluble in the solution. Cancel the spectator ions, ${\text{Na}}^{+}{\text{ and NO}}_3^{-}$, from both sides of the equation.

Thus, we get the net ionic equation as:

${\text{Mg}}^{2+}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)$