Question

Rewrite the following balanced equations as net ionic equations.

Short answer

A net ionic equation includes only the ions (and molecules) that actually take part in the reaction.

The net ionic equation provides the information on ions that exists in an aqueous medium.

The chemical species that change chemically are identified by the net ionic equation. The ions that appear on both sides of the equation remain constant and are therefore considered to be spectator ions.

Step-by-step solution

The concept of net ionic equations

A net ionic equation includes only the ions (and molecules) that actually take part in the reaction.

The net ionic equation provides the information on ions that exists in an aqueous medium.

The chemical species that change chemically are identified by the net ionic equation. The ions that appear on both sides of the equation remain constant and are therefore considered to be spectator ions.

Write the ionic net equation of the first part

a)

The balanced reaction is: ${\mathrm{Na}}_2{\mathrm{SO}}_4\left(aq\right)+{\mathrm{BaCl}}_2\left(aq\right)\to {\mathrm{BaSO}}_4\left(s\right)+2\mathrm{NaCl}\left(aq\right)$.

The ions that take part in this reaction are: ${\mathrm{Ba}}^{2+}$and $S{O^{2-}}_4$.

Therefore, the net ionic equation will be as follows: ${\mathrm{SO}}_4^{2-}\left(aq\right)+{\mathrm{Ba}}^{2+}\left(aq\right)\to {\mathrm{BaSO}}_4\left(s\right)$.

Write the ionic net equation of the second part

(b)

The balanced reaction is: .

The ions that take part in this reaction are: and .

Therefore, the net ionic equation will be as follows: .

Write the ionic net equation of third part

(c)

The balanced reaction is: ${\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2\left(aq\right)+2\mathrm{KI}\left(aq\right)\to {\mathrm{Hg}}_2{\mathrm{I}}_2\left(s\right)+2{\mathrm{KNO}}_3\left(aq\right)$.

The ions that take part in this reaction are: ${\mathrm{I}}^{-}$and ${\mathrm{Hg}}^{+}$.

Therefore, the net ionic equation will be as follows: $2{\mathrm{Hg}}_2\left(aq\right)+2{\mathrm{l}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Hg}}_2{\mathrm{l}}_2\left(s\right)$.

Write the ionic net equation of fourth part 

(d)

The balanced reaction is: $3\mathrm{NaOCl}\left(aq\right)+\mathrm{Kl}\left(aq\right)\to {\mathrm{NaIO}}_3\left(aq\right)+2\mathrm{NaCl}\left(aq\right)+\mathrm{KCl}\left(aq\right)$.

The ions that take part in this reaction are: ${\mathrm{I}}^{-}$and ${\mathrm{OCl}}^{-}$.

Therefore, the net ionic equation will be as follows: $3{\mathrm{OCl}}^{-}\left(aq\right)+{\mathrm{I}}^{-}\left(aq\right)\to {\mathrm{IO}}_3^{-}\left(aq\right)+3{\mathrm{Cl}}^{-}$.