Reaction Between Ethanoic Acid and Sodium Hydroxide

Carboxylic acids are those containing the “oic acid ending” and looking at these molecules you’ll realize a similarity in terms of carbon to oxygen and hydrogen bonds. All carboxylic have the “COOH” group attached to their molecules. One such good example of a carboxylic acid is ethanoic acid:

CH₃COOH

Sodium hydroxide is a metal hydroxide having the formula:

NaOH

If you have already done reactions involving acids and metal hydroxides you should probably already know the product that should be formed between these two.
When an acid reacts with a metal hydroxide the type of reaction that takes place is one of a neutralization reaction where the hydrogen ions are removed from the acid molecule and reacts with the hydroxide ions of the metal hydroxide. A salt and water will be the products of your reaction. The ionic equation below shows the reaction:

H⁺(aq) + OH^-(aq) ==> H₂O(l)

In this case the acid would be ethanoic acid and the metal hydroxide would be sodium hydroxide. Therefore it also follows that a neutralization reaction will take place between the ethanoic acid and the sodium hydroxide. The hydrogen ion will be removed from the ethanoic acid and replaced by the sodium metal forming a salt of a carboxylic acid known as sodium ethanoate. The hydrogen will then react with the hydroxide ions from the NaOH to form water. Since it’s a neutralization reaction no colour will be observed in the solution. The full equation for the reaction is:

CH₃COOH + NaOH ==> CH₃COONa + H₂O

(you could also use the ionic equation given above)

Or in words:

Ethanoic acid + Sodium Hydroxide ==> Sodium Ethanoate + Water

This equation will also be useful when testing for carboxylic acids using ethanoic acid as a typical one. You’ll realize that the strong smell/odor from the ethanoic acid will be gone when reacted with sodium hydroxide because it forms a neutrilisation reaction with the NaOH.