Chapter 9, Problem 98A

Interpretation Introduction

Interpretation:

The complete ionic equation and net ionic equations for given chemical reaction needs to be determined.

${\text{CuCl}}_{\text{2}}\left(s\right)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to$

Concept introduction:

Chemical equation is the balanced chemical equation which shows the reactants and products using their chemical formulas. An ionic equation represents all the particles in a solution according to their existence and net ionic equations are the ionic equations that include only the particles participating in the reaction.

Answer to Problem 98A

Complete ionic equation for the reaction is,

${\text{CuCl}}_{\text{2}}\left(s\right)+2{\text{Na}}^{+}\text{(}aq\text{)}\text{+}{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)\to {\text{Cu}}^{+2}\left(aq\right)+{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)\text{+}{\text{2Cl}}^{-}\left(aq\right)$

Net ionic equation for the reaction is,

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

Explanation of Solution

The complete chemical equation for the given reaction is as follows:

${\text{CuCl}}_{\text{2}}\left(s\right)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{CuSO}}_{\text{4}}\left(aq\right)+2\text{NaCl}\left(aq\right)$

Complete ionic equation shows the ionic form of reactants. So, the complete ionic equation for the above reaction is as follows:

${\text{CuCl}}_{\text{2}}\left(s\right)+2{\text{Na}}^{+}\text{(}aq\text{)}\text{+}{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)\to {\text{Cu}}^{+2}\left(aq\right)+{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)\text{+}{\text{2Cl}}^{-}\left(aq\right)$

Net ionic equation cancels all the spectator ions of the reaction. So, the net ionic equation is as follows:

$\begin{array}{l}{\text{CuCl}}_{\text{2}}\left(s\right)+\cancel{2{\text{Na}}^{+}\text{(}aq\text{)}}\text{+}\cancel{{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)}\to {\text{Cu}}^{+2}\left(aq\right)+\cancel{{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)}+\cancel{2{\text{Na}}^{+}\left(aq\right)}\text{+}{\text{2Cl}}^{-}\left(aq\right)\\ {\text{CuCl}}_{\text{2}}\left(s\right)\to {\text{Cu}}^{+2}\left(aq\right)\text{+}{\text{2Cl}}^{-}\left(aq\right)\end{array}$

Conclusion

Net ionic equation contains fewer ions as it cancels the spectator ions of the reaction.