# Question #1a13e

Jun 20, 2015

Here's how the complete ionic equation and the net ionic equation will look like for this reaction.

#### Explanation:

You're mixing sulfuric acid, ${H}_{2} S {O}_{4}$, which is a strong acid, and sodium hydroxide, $N a O H$, which is a strong base.

This implies that the two compounds will be completely dissociated in aqueous solution. In the case of sulfuric acid, you have

${H}_{2} S {O}_{4 \left(a q\right)} \to 2 {H}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -}$

One mole of of sulfuric acid will produce 2 moles of ${H}^{+}$ and 1 mole of $S {O}_{4}^{2 -}$.

In the case of sodium hydroxide, you have

$N a O {H}_{\left(a q\right)} \to N {a}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

This time, one mole of sodium hydroxide will produce 1 mole of $N {a}^{+}$ and 1 mole of $O {H}^{-}$ in aqueous solution.

Sodium sulfate, $N {a}_{2} S {O}_{4}$, is soluble in aqueous solution, which implies that it too will exist as ions.

$N {a}_{2} S {O}_{4 \left(a q\right)} \to 2 N {a}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -}$

Your balanced chemical equation for this neutralization reaction looks like this

${H}_{2} S {O}_{4 \left(a q\right)} + \textcolor{red}{2} N a O {H}_{\left(a q\right)} \to N {a}_{2} S {O}_{4 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

The complete ionic equation will be

$2 {H}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -} + \textcolor{red}{2} N {a}_{\left(a q\right)}^{+} + \textcolor{red}{2} O {H}_{\left(a q\right)}^{-} \to 2 N {a}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -} + 2 {H}_{2} {O}_{\left(l\right)}$

To get the net ionic equation, eliminate the spectator ions, i.e. the ions that are on both sides of the equation. This will get you

$2 {H}_{\left(a q\right)}^{+} + \cancel{S {O}_{4 \left(a q\right)}^{2 -}} + \cancel{\textcolor{red}{2} N {a}_{\left(a q\right)}^{+}} + \textcolor{red}{2} O {H}_{\left(a q\right)}^{-} \to \cancel{2 N {a}_{\left(a q\right)}^{+}} + \cancel{S {O}_{4 \left(a q\right)}^{2 -}} + 2 {H}_{2} {O}_{\left(l\right)}$

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

You can simplify this by 2 to get

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