Carboxylic acids are weak organic acids. All acids are proton donors and the more stable the resulting anion (in this case the carboxylate ion), the greater the tendency for proton donation to occur and so the stronger the acid.
The negative log of the Ka in the pKa and this value is used to compare the strength of acids – the lower the pKa, the stronger the acid.
CH3COOH: pKa = – log Ka = 4.8
Carboxylic acids are stronger acids than phenol (pKa = 9.9) and alcohols (pKa ∼ 16) because they are able to stabilise the carboxylate anion by resonance.
Similarly, the phenoxide ion can be stabilised by delocalising the negative charge across the molecule.
Alcohols have neither a second oxygen atom or ring of delocalised electrons to help stabilise the alkoxide conjugate base so they are less likely to donate a proton and therefore less acidic.
Carboxylic acids are acidic enough to react with weak bases such as metal carbonates, giving a characteristic fizz as carbon dioxide is evolved. This is a useful way of distinguishing between carboxylic acids and phenols as phenols are not acidic enough to neutralise a metal carbonate.
2CH3COOH(aq) + Na2CO3(aq) ⇾ 2 CH3COO– Na+(aq) + CO2(g) + H2O(l)
As expected, carboxylic acids such as ethanoic acid and propanoic acid will also react with strong alkalis, metal oxides and metals.
CH3COOH(aq) + KOH(aq) ⇾ CH3COO–K+(aq) + H2O(l)
2CH3COOH(aq) + CaO(s) ⇾ (CH3COO–)2Ca2+(aq) + H2O(l)
2CH3COOH(aq) + Mg(s) ⇾ (CH3COO–)2Mg2+(aq) + H2(g)
Practice questions
- Write a balanced equation for the reaction between 2-methylpropanoic acid and
(a) calcium metal
(b) aqueous sodium carbonate
- Write a balanced equation for the reaction between aqueous sodium hydroxide and the molecule shown below, and explain the reasoning behind your answer.
Answers
- (a) 2 CH3CH(CH3)COOH + Ca(s) ⇾ (CH3CH(CH3)COO–)2 Ca2+ + H2(g)
(b) 2 CH3CH(CH3)COOH + Na2CO3(aq) ⇾ 2 CH3CH(CH3)COO– Na+ + CO2(g) + H2O(l)
- Carboxylic acids and phenols are acidic enough to react with a strong base but alcohols are not.
