Question

Question #1a13e

Answer 1

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_2SO_4`, which is a strong acid, and sodium hydroxide, `NaOH`, 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_2SO_(4(aq)) -> 2H_((aq))^(+) + SO_(4(aq))^(2-)`

One mole of of sulfuric acid will produce 2 moles of `H^(+)` and 1 mole of `SO_4^(2-)`.

In the case of sodium hydroxide, you have

`NaOH_((aq)) -> Na_((aq))^(+) + OH_((aq))^(-)`

This time, one mole of sodium hydroxide will produce 1 mole of `Na^(+)` and 1 mole of `OH^(-)` in aqueous solution.

Sodium sulfate, `Na_2SO_4`, is soluble in aqueous solution, which implies that it too will exist as ions.

`Na_2SO_(4(aq)) -> 2Na_((aq))^(+) + SO_(4(aq))^(2-)`

Your balanced chemical equation for this neutralization reaction looks like this

`H_2SO_(4(aq)) + color(red)(2)NaOH_((aq)) -> Na_2SO_(4(aq)) + 2H_2O_((l))`

The complete ionic equation will be

`2H_((aq))^(+) + SO_(4(aq))^(2-) + color(red)(2)Na_((aq))^(+) + color(red)(2)OH_((aq))^(-) -> 2Na_((aq))^(+) + SO_(4(aq))^(2-) + 2H_2O_((l))`

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

`2H_((aq))^(+) + cancel(SO_(4(aq))^(2-)) + cancel(color(red)(2)Na_((aq))^(+)) + color(red)(2)OH_((aq))^(-) -> cancel(2Na_((aq))^(+)) + cancel(SO_(4(aq))^(2-)) + 2H_2O_((l))`

`2H_((aq))^(+) + 2OH_((aq))^(-) -> 2H_2O_((l))`

You can simplify this by 2 to get

`H_((aq))^(+) + OH_((aq))^(-) -> H_2O_((l))`