Carboxylic Acids

Carboxylic acids are the fourth homologous series you meet in organic chemistry, after alkanes, alkenes and alcohols. They all share the same functional group: the atoms that give the family its characteristic reactions.

(Separate Chemistry only) Carboxylic acids are assessed in GCSE Chemistry, not in Combined Science Trilogy.

The functional group of a carboxylic acid is –COOH, called the carboxyl group. It is a carbon atom joined to a double-bonded oxygen and to an –OH group. Because every member ends in the same group, they all behave in similar ways, which is exactly why chemists group them into a family.

The first four members follow the pattern methan-, ethan-, propan-, butan- with the ending -oic acid:

Name	Formula	Displayed structure of the acid end
Methanoic acid	HCOOH	H–COOH
Ethanoic acid	CH₃COOH	CH₃–COOH
Propanoic acid	CH₃CH₂COOH	CH₃CH₂–COOH
Butanoic acid	CH₃CH₂CH₂COOH	CH₃CH₂CH₂–COOH

Ethanoic acid is the acid in vinegar. You must be able to recognise a carboxylic acid from its name or its formula: look for a name ending in -oic acid, or the –COOH group in a structure.

Carboxylic acids are acids, so they behave like the acids you met earlier: they react with metal carbonates. This reaction is the classic acid + carbonate reaction and gives three products every time.

$\text{acid}+\text{carbonate}\to \text{salt}+\text{water}+\text{carbon dioxide}$

You see fizzing as carbon dioxide gas bubbles off. If you pass that gas through limewater, the limewater turns cloudy (milky), the standard test for carbon dioxide. This is useful in the lab: adding a carbonate to an unknown liquid and seeing effervescence that turns limewater cloudy is evidence the liquid is acidic.

The salt formed from a carboxylic acid is a carboxylate. For example, ethanoic acid produces salts called ethanoates, so ethanoic acid reacting with sodium carbonate gives sodium ethanoate, water and carbon dioxide.

The spec does not expect balanced symbol equations for carboxylic acid reactions. You do need the word equation, the products, and the observations (fizzing, limewater going cloudy).

When carboxylic acids dissolve in water they form acidic solutions, so the pH is below 7. This is what makes them behave as acids at all: dissolving releases hydrogen ions, H⁺, into the solution.

The key idea to hold on to for the next slide is that carboxylic acids do not produce as many H⁺ ions as you might expect. A solution of a carboxylic acid has a higher pH (it is less strongly acidic) than a solution of a strong acid such as hydrochloric acid at the same concentration.

Property	Carboxylic acid solution
pH	Less than 7 (acidic)
Compared with a strong acid of equal concentration	Higher pH (weaker)
Ions released in water	H⁺ ions, but only from a fraction of the molecules

You can recognise them as acids because they turn universal indicator or litmus the acidic colours and react with carbonates. The reason they are only weakly acidic is the Higher Tier explanation below.

(Higher Tier only) The following explanation of carboxylic acids as weak acids is assessed only at Higher Tier.

An acid ionises in water, splitting to release hydrogen ions (H⁺). A strong acid such as hydrochloric acid is fully ionised: essentially every molecule splits, so it releases the maximum number of H⁺ ions.

A carboxylic acid is a weak acid. It only partially ionises in water. At any moment most of the molecules stay whole and only a small fraction have split to release an H⁺ ion:

${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}+{\text{H}}^{+}$

The reversible arrow ($\rightleftharpoons$) shows the ionisation does not go to completion; an equilibrium is set up that lies well to the left. Because fewer H⁺ ions are released, the H⁺ concentration is lower, and since pH depends on H⁺ concentration, the pH is higher than for a strong acid of the same concentration.

Weak does not mean dilute. Strong/weak describes how far an acid ionises; concentrated/dilute describes how much acid is dissolved. A concentrated weak acid is still only partially ionised.

Carboxylic acids react with alcohols to produce a family of sweet-smelling compounds called esters. The reaction needs an acid catalyst (usually a small amount of concentrated sulfuric acid) and gentle warming.

$\text{carboxylic acid}+\text{alcohol}\stackrel{\text{acid catalyst}}{\to }\text{ester}+\text{water}$

The one ester you must name is ethyl ethanoate, made from ethanoic acid and ethanol:

$\text{ethanoic acid}+\text{ethanol}\stackrel{\text{acid catalyst}}{\to }\text{ethyl ethanoate}+\text{water}$

Notice how the name is built: the ethyl part comes from the alcohol (ethanol), and the ethanoate part comes from the acid (ethanoic acid). Esters have distinctive fruity, sweet smells, which is why they are used in perfumes, food flavourings and solvents.

You only need to name ethyl ethanoate. Names of other esters are not required by the spec, and balanced symbol equations for this reaction are not expected.

1. Writing the functional group as –OH

The –OH group belongs to alcohols. A carboxylic acid has the –COOH group, which contains a C=O as well as an –OH. Losing the C=O turns your answer into an alcohol and loses the mark.

2. Saying weak means dilute

A weak acid is one that only partially ionises, whatever its concentration. Do not confuse "weak" (few molecules ionised) with "dilute" (little acid dissolved). The Higher Tier answer must talk about partial ionisation, not the amount of water.

3. Forgetting a product of the carbonate reaction

Acid + carbonate gives three products: salt, water and carbon dioxide. Students often drop the water or forget the CO₂. The CO₂ is what causes the fizzing and turns limewater cloudy.

4. Naming the ester the wrong way round

In ethyl ethanoate, the first word (ethyl) comes from the alcohol and the second word (ethanoate) from the acid. Swapping them gives a name that does not exist.

5. Trying to write balanced symbol equations

The spec does not expect balanced symbol equations for carboxylic acid reactions (only word equations and observations). Do not waste time balancing formulae you were never asked for; give the word equation and the products instead.