Question

Is the total ionic equation the same as the net ionic equation when $\mathit{S}\mathit{r}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$and ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$react? Explain.

Short answer

Answer

Yes, the total ionic equation is the same as the net ionic equation when

$\mathit{S}\mathit{r}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$and ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$react.

Step-by-step solution

Determine difference between the total ionic equation and the net ionic equation

A complete ionic equation includes the spectator ions while the net ionic equation shows only the chemical species involved in a reaction.

Write the net ionic equation of the reaction

The balanced molecular reaction is:$\mathit{S}\mathit{r}{\left(OH\right)}_{\mathbf{2}}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\left(S{O}_4\right)\left(aq\right)\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(I\right)\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathit{S}\mathit{r}\mathit{S}{\mathit{O}}_{\mathbf{4}}\left(s\right)$

The total ionic equation is:$\mathit{S}{\mathit{r}}^{\mathbf{-}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}{\mathit{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{+}\mathit{S}\mathit{r}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

The net ionic equation is:

$\mathit{S}{\mathit{r}}^{\mathbf{-}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}{\mathit{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathit{S}{\mathit{O}}_{\mathbf{4}}^{\mathbf{-}\mathbf{2}}\mathbf{\to }\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{+}\mathit{S}\mathit{r}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

Since no ions are common in both sides of the total ionic equation, thus nothing is canceled out in the reaction.

Hence, the net ionic equation is the same as the total ionic equation.