Chapter 5, Problem 111QP

Predict whether reactions should occur between aqueous solutions of the following compounds. If so, write balanced molecular and net ionic equations for the reactions. Include physical states.

$\begin{array}{l}\left(\text{a}\right)\text{ Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{\text{ and H}}_{\text{2}}{\text{SO}}_{\text{4}}\hfill \\ \left(\text{b}\right)\text{ Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{\text{ and Na}}_{\text{2}}{\text{SO}}_{\text{4}}\hfill \\ \left(\text{c}\right){\text{ CuSO}}_{\text{4}}\text{ and BaS}\hfill \\ \left(\text{d}\right){\text{ NaHCO}}_{\text{3}}{\text{ and CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\hfill \end{array}$

Interpretation Introduction

Interpretation:

The balanced molecular and net ionic equations are to be written for the reaction of strontium nitrate with sulfuric acid.

Explanation of Solution

Strontium nitrate and sulfuric acid are electrolytes.Strontium ion in a solution reacts with sulfate ion to form strontium sulfate, which is insoluble in an aqueous solution. The given reaction is an example of a precipitation reaction. The balanced molecular equation is as follows:

${\text{Sr(NO}}_{\text{3}}{\text{)}}_2(aq)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}^{}(aq)\to {\text{SrSO}}_{\text{4}}^{}(s)+2{\text{HNO}}_3(aq)$

The balanced ionic equation of the given reaction is as follows:

${\text{Sr}}^{2+}(aq)+2{\text{NO}}_3^{-}(aq)+2{\text{H}}^{+}(aq)+{\text{SO}}_{\text{4}}^{2-}(aq)\to {\text{SrSO}}_{\text{4}}^{}(s)+2{\text{NO}}_3^{-}(aq)+2{\text{H}}^{+}(aq)$

As nitrate and hydrogen ion are spectator ions, the net ionic equation is as follows:

${\text{Sr}}^{2+}(aq)+{\text{SO}}_{\text{4}}^{2-}(aq)\to {\text{SrSO}}_{\text{4}}^{}(s)$

Interpretation Introduction

Interpretation:

The balanced molecular and net ionic equations are to be written for the reaction of zinc nitrate with sodium sulfate in an aqueous solution.

Explanation of Solution

No reaction takes place on mixing zinc nitrate and sodium sulfate in an aqueous solution. For instance, if product forms, the possible products that can be formed are zinc sulfate and sodium nitrate. Both the products and reactants are electrolytes and remain in a dissociated form. Since all the ions remain as such, no new products are formed in the given reaction.

Interpretation Introduction

Interpretation:

The balanced molecular and net ionic equations are to be written for the reaction of copper sulfate with barium sulfide in an aqueous solution.

Explanation of Solution

Except for copper sulfate, all other substances are non-electrolytes. The given reaction is a double displacement reaction as well as a precipitation reaction. The balanced molecular equation is as follows:

$\text{BaS}(s)+{\text{CuSO}}_{\text{4}}^{}(aq)\to {\text{BaSO}}_{\text{4}}^{}(s)+\text{CuS}(s)$

The balanced ionic equation of the given reaction is as follows:

$\text{BaS}(s)+{\text{Cu}}^{2+}(aq)+{\text{SO}}_{\text{4}}^{2-}(aq)\to {\text{BaSO}}_{\text{4}}^{}(s)+\text{CuS}(s)$

As no ion is a spectator ion, the net ionic equation is same as the ionic equation.

Interpretation Introduction

Interpretation:

The balanced molecular and net ionic equations are to be written for the reaction of sodium bicarbonate with acetic acid.

Explanation of Solution

Acetic acid is a weak electrolyte and sodium bicarbonate is an electrolyte. As the given reaction is an example of acid and base neutralization reaction, acetic acid reacts slowly with bicarbonate to form carbonic acid which in turn decomposes to water and carbon dioxide. The balanced molecular equation is as follows:

${\text{NaHCO}}_3(aq)+{\text{CH}}_{\text{3}}\text{COOH}(aq)\to {\text{CH}}_{\text{3}}\text{COONa}(aq)+{\text{H}}_{\text{2}}\text{O}(l)+{\text{CO}}_2(g)$

Since most of the acetic acid molecules remain undissociated, the molecular form is used in writing the ionic equation.The balanced ionic equation of the given reaction is as follows:

${\text{Na}}^{+}(aq)+{\text{HCO}}_3^{-}(aq)+{\text{CH}}_{\text{3}}\text{COOH}(aq)\to {\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{Na}}^{+}(aq)+{\text{H}}_{\text{2}}\text{O}(l)+{\text{CO}}_2(g)$

As sodium ion isa spectator ion, the net ionic equation is as follows:

${\text{HCO}}_3^{-}(aq)+{\text{CH}}_{\text{3}}\text{COOH}(aq)\to {\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)+{\text{CO}}_2(g)$