# Question #7cbb2

Jul 21, 2015

Here's how you can figure that out.

#### Explanation:

When doing molecular, ionic , and net ionic equations, it is very important to use state symbols to show that you understand which compound is ionic and which is covalent.

In your case, you're dealing with two ionic compounds, potassium bicarbonate, $K H C {O}_{3}$, and potassium sulfate, ${K}_{2} S {O}_{4}$.

The reaction takes place in aqueous solution, so you can expect these compounds to be broken up in cations and anions.

The other three species that take part in this reaction, sulfuric acid, ${H}_{2} S {O}_{4}$, water, ${H}_{2} O$, and carbon dioxide, $C {O}_{2}$, are all covalent compounds, so they will not exist as ions in solution.

So, let's take them one by one

• Molecular equation

The balanced chemical equation for this reaction looks like this

$\textcolor{red}{2} K H C {O}_{3 \left(a q\right)} + {H}_{2} S {O}_{4 \left(a q\right)} \to {K}_{2} S {O}_{4 \left(a q\right)} + \textcolor{b l u e}{2} C {O}_{2 \left(g\right)} + \textcolor{g r e e n}{2} {H}_{2} {O}_{\left(l\right)}$

• Complete ionic equation

This is where you need to show all the ions that are present in solution. Break up the aforementioned ionic compounds to get

$\textcolor{red}{2} {K}_{\left(a q\right)}^{+} + \textcolor{red}{2} H C {O}_{3 \left(a q\right)}^{2 -} + 2 {H}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -} \to 2 {K}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -} + \textcolor{b l u e}{2} C {O}_{2 \left(g\right)} + \textcolor{g r e e n}{2} {H}_{2} {O}_{\left(l\right)}$

The bicarbonate anion has a very, very small dissociation constant ($\approx 4.69 \cdot {10}^{- 11}$), so you can assume it to remain undissociated.

• Net ionic equation

This time you need to remove spectator ions, which are ions present on both sides of the equation. In your case, you need to remove the potassium cations and the sulfate anions

$\cancel{\textcolor{red}{2} {K}_{\left(a q\right)}^{+}} + \textcolor{red}{2} H C {O}_{3 \left(a q\right)}^{2 -} + 2 {H}_{\left(a q\right)}^{+} + \cancel{S {O}_{4 \left(a q\right)}^{2 -}} \to \cancel{2 {K}_{\left(a q\right)}^{+}} + \cancel{S {O}_{4 \left(a q\right)}^{2 -}} + \textcolor{b l u e}{2} C {O}_{2 \left(g\right)} + \textcolor{g r e e n}{2} {H}_{2} {O}_{\left(l\right)}$

This is equivalent to

$\textcolor{red}{2} H C {O}_{3 \left(a q\right)}^{2 -} + 2 {H}_{\left(a q\right)}^{+} \to \textcolor{b l u e}{2} C {O}_{2 \left(g\right)} + \textcolor{g r e e n}{2} {H}_{2} {O}_{\left(l\right)}$

You can divide all the stoichiometric coefficients by 2 to get

$H C {O}_{3 \left(a q\right)}^{2 -} + {H}_{\left(a q\right)}^{+} \to C {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(l\right)}$