# What is the balanced equation for the reaction between acetic acid and NaOH?

Jun 8, 2017

Here's what I got.

#### Explanation:

Acetic acid, $\text{CH"_3"COOH}$, will react with sodium hydroxide, $\text{NaOH}$, to produce sodium acetate, $\text{CH"_3"COONa}$, and water.

The unbalanced chemical equation that describes this neutralization reaction looks like this

${\text{CH"_ 3"COOH"_ ((aq)) + "NaOH"_ ((aq)) -> "CH"_ 3"COONa"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

Now, you could check to see if this chemical equation is balanced by counting the number of atoms of each element present on both sides of the equation, or you could check by using the fact that sodium hydroxide is a strong base.

This implies that sodium hydroxide dissociates completely in aqueous solution to produce sodium cations and hydroxide anions

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

Sodium acetate is soluble in aqueous solution, so it will also exist as ions

${\text{CH"_ 3"COONa"_ ((aq)) -> "CH"_ 3"COO"_ ((aq))^(-) + "Na}}_{\left(a q\right)}^{+}$

Put this together to get

${\text{CH"_ 3"COOH"_ ((aq)) + color(blue)("Na"_ ((aq))^(+)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COO"_ ((aq))^(-) + color(blue)("Na"_ ((aq))^(+)) + "H"_ 2"O}}_{\left(l\right)}$

The sodium cations are spectator ions because they exist as ions on both sides of the equation. This means that you can rewrite the chemical equation as

"CH"_ 3"COO"color(red)("H")_ ((aq)) + color(red)("OH"_ ((aq))^(-)) -> "CH"_ 3"COO"_ ((aq))^(-) + color(red)("H"_ 2"O"_ ((l)))

Notice that the reactants' side has an undissociated acetate anion and the products' side has a dissociated acetate anion, so nothing to balance out here.

Similarly, the reactants' side has an undissociated proton and a dissociated hydroxide anion and the products' side has a water molecule, so once again, there's nothing to balance out here.

Therefore, you can say that the balanced chemical equation that describes this raection looks like this

${\text{CH"_ 3"COOH"_ ((aq)) + "NaOH"_ ((aq)) -> "CH"_ 3"COONa"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$