Acetic acid and ethanol both give up a proton from an O–H bond. Yet acetic acid (pKa 4.76) is about 10¹¹ times stronger than ethanol (pKa 16). The two O–H bonds look much the same, so the difference must lie in what each acid leaves behind: its conjugate base.
That is the idea behind this whole page: the more stable the conjugate base, the stronger the acid. A stable conjugate base holds its new negative charge comfortably, so it has little pull to take the proton back. Conjugate acids & bases showed why a strong acid always has a weak conjugate base, and pKa gave the numbers. This page explains where those numbers come from, so that for any two hydrogens you can say which is more acidic and why.
Four questions about the conjugate base
Take four pairs of acids. In each pair, one feature of the conjugate base changes and the rest stay the same.
| Pair of acids | pKa | What differs in the conjugate base | Factor |
|---|---|---|---|
| water and HF | 15.7 and 3.2 | the charge sits on oxygen in HO⁻ and on fluorine in F⁻ | Atom |
| ethanol and acetic acid | 16 and 4.76 | ethoxide keeps the charge on one oxygen; acetate shares it between two | Resonance |
| acetic acid and chloroacetic acid | 4.76 and 2.86 | a chlorine near the charge pulls electron density away from it | Induction |
| ethane and ethyne | 50 and 25 | the lone pair sits in an sp³ orbital in one anion and an sp orbital in the other | Orbital |
Each row answers one question about the conjugate base. Together the four questions spell ARIO:
- A — Atom. Which atom carries the negative charge?
- R — Resonance. Can the charge spread over more than one atom?
- I — Induction. Does an electronegative atom nearby pull electron density away through the σ bonds (the single bonds of the framework)?
- O — Orbital. What is the hybridization of the atom that holds the charge?
Ask them in that order. Atom effects are usually the largest and resonance usually comes next, while induction and orbital effects most often break ties. So when you compare two acids, the first factor that differs usually settles the question.
The order is a rule of thumb, not a law. A carbanion (a carbon anion) with strong resonance can beat an alkoxide (the RO⁻ ion an alcohol leaves behind) that has none; the worked example "when two factors disagree" below shows one. Naming all four factors lets you see which ones are in play.
A — Atom: two trends, two different reasons
Start across the second row of the periodic table. CH₄ has pKa about 50, NH₃ about 38, water 15.7 and HF 3.2. Acidity rises from carbon to fluorine, and the step from nitrogen to oxygen alone is twenty-two powers of ten.
These four atoms are all about the same size, so electronegativity makes the difference. A more electronegative atom holds a negative charge more willingly. Its anion is more stable, and so its acid is stronger.
Now go down a group. HF has pKa 3.2, HCl −7, HBr −9 and HI −10. Acidity rises again, even though electronegativity falls from fluorine to iodine.
Here size wins. An iodide ion is far bigger than a fluoride ion, so its single negative charge is spread over a much larger volume. A large electron cloud is also more polarizable: it distorts easily, so the charge can shift to wherever it is least crowded. Spreading the charge out stabilizes the anion by far more than the lost electronegativity costs.
You will also see this trend explained by bond strength. The H–X bond gets longer and weaker down the group, from about 570 kJ/mol (136 kcal/mol) for H–F to about 300 (72 kcal/mol) for H–I, so there is less to break. That argument points the same way. This course leads with size and polarizability and treats bond strength as a second reason.
R — Resonance
Compare ethanol (pKa 16) with acetic acid (pKa 4.76) again. Both lose an O–H proton, and in both the charge lands on oxygen, so the Atom factor is the same. The difference appears once the proton has gone.
Ethoxide, CH₃CH₂O⁻, keeps its whole charge on one oxygen. Acetate can do better. Move a lone pair from the charged oxygen into the C–O bond, and move the C=O π bond onto the other oxygen, and you have drawn a second resonance structure (see Resonance). The two structures are equivalent, so the real ion carries half the charge on each oxygen, and both C–O bonds have the same length.
Spreading the charge over several atoms this way is called delocalization, and here it is worth eleven pKa units, a hundred billion to one. Among the four ARIO factors, resonance is usually the biggest after the atom.
Resonance is the standard explanation for this gap, and the one most courses grade. Some chemists argue that part of it comes from the carbonyl carbon itself, which is partly positive and sits right next to the charged oxygen. Carboxylic Acids & Derivatives returns to that view.
Phenol, an OH on a benzene ring, sits between the two at pKa 10. Its conjugate base, phenoxide, can push its charge into the ring, but only onto ring carbons. Carbon holds a negative charge far less willingly than oxygen does. So phenol gets real stabilization, but less than acetic acid: it is six units more acidic than ethanol and five units less acidic than acetic acid.
A preview. Carbon alone can still go a long way, as cyclopentadiene shows. It is a plain hydrocarbon with no oxygen or nitrogen anywhere, yet its CH₂ hydrogens have pKa 16, as acidic as an alcohol. Its conjugate base spreads the charge over all five ring carbons. The ring is also flat and holds six π electrons, which makes it aromatic, a special extra stability that Aromatic Chemistry explains. The result sits thirty-four pKa units below an ordinary alkane's 50. When a conjugate base turns out to be aromatic, expect its acid to be far stronger than the formula suggests.
Start with an ordinary ketone, such as acetone. A hydrogen on the carbon next to the C=O, the alpha carbon, has pKa about 20, which is remarkably low for a C–H. Its conjugate base is an enolate, met in Resonance: the charge is shared between the alpha carbon and the carbonyl oxygen.
Now put a C=O on each side of the same carbon, as in pentane-2,4-dione. The hydrogen between them drops to pKa 9. Its conjugate base shares the charge with two oxygens instead of one.
A pKa of 9 is about as acidic as phenol, and slightly more acidic than the ammonium ion (9.2), all from a hydrogen on carbon. An ordinary base such as ethoxide removes it completely. That is why 1,3-diketones and their relatives become key reagents in Enolate Chemistry.
I — Induction
Electronegativity introduced induction: an electronegative atom pulls electron density toward itself through the σ bonds. Near a negative charge, that pull spreads the charge out a little and steadies the anion. Chloroacetic acid shows the size of the effect. Put one chlorine on the carbon next to acetic acid's COOH group, and the pKa falls from 4.76 to 2.86.
The table adds more halogens and moves one further away.
| Acid | pKa | Effect |
|---|---|---|
| acetic acid, CH₃COOH | 4.76 | baseline |
| chloroacetic acid, ClCH₂COOH | 2.86 | one Cl, adjacent |
| dichloroacetic acid, Cl₂CHCOOH | 1.29 | two Cl |
| trichloroacetic acid, Cl₃CCOOH | 0.65 | three Cl |
| fluoroacetic acid, FCH₂COOH | 2.59 | F beats Cl |
| 4-chlorobutanoic acid, ClCH₂CH₂CH₂COOH | 4.52 | Cl three carbons away |
Induction adds up, but each extra chlorine adds less. The first chlorine is worth 1.90 pKa units, the second 1.57 and the third only 0.64, because each one finds less electron density left to pull on.
A more electronegative atom pulls harder. Fluoroacetic acid (2.59) is stronger than chloroacetic acid (2.86), because fluorine is more electronegative than chlorine.
Induction also dies off quickly with distance. Move a chlorine along butanoic acid (pKa 4.82), one carbon at a time. On the carbon next to the COOH group it lowers the pKa by nearly two units. One carbon further out it lowers it by about three quarters of a unit, and two carbons further out by only 0.3.
That fading is how you tell induction apart from resonance. Induction has almost vanished once the chlorine is two carbons further out. Resonance carries its effect along a chain of alternating double bonds and lone pairs with little loss. As a preview, a nitro group (–NO₂) on the far side of a benzene ring from phenol's OH still lowers the pKa by about three units, from 10 to 7.2.
Induction can also push electron density toward a charge. An alkyl group pushes slightly more electron density toward its neighbors than a hydrogen does, so it slightly destabilizes a nearby negative charge. Formic acid, HCOOH, has pKa 3.75, and acetic acid, CH₃COOH, has 4.76. Swapping one hydrogen for a methyl group costs a full pKa unit.
O — Orbital
Compare three C–H bonds that differ only in the hybridization of the carbon: ethyne (sp) has pKa about 25, ethene (sp²) about 44 and ethane (sp³) about 50. When the proton leaves, the C–H bonding pair stays on carbon as a lone pair, in the same hybrid orbital that the C–H bond used.
The difference lies in that orbital. An sp orbital is 50% s-character, an sp² orbital 33% and an sp³ orbital 25% (Hybridization). An s orbital holds its electrons closer to the nucleus than a p orbital does. So the more s-character the orbital has, the closer to the nucleus the lone pair sits, the lower its energy, and the more stable the anion. That alone spans twenty-five powers of ten between ethyne and ethane. This is the largest orbital effect you will meet. Between sp² and sp³ the gap is only about six units (44 against 50), which is more typical, and it is why orbital effects usually just break ties.
In ethyne's anion, the lone pair points straight out along the C≡C axis, at right angles to both π bonds. It cannot spread into them, so resonance plays no part here.
The same factor works on bases. Pyridine, a benzene ring with one CH replaced by N, holds its nitrogen lone pair in an sp² orbital. An ordinary amine such as methylamine, CH₃NH₂, holds it in an sp³ orbital. The sp² pair is held more tightly, so pyridine is the weaker base. The pyridinium ion, pyridine's conjugate acid, has pKa 5.2, so pyridine's pKaH is 5.2 (a base's pKaH is the pKa of its conjugate acid, as pKa explains). The methylammonium ion has pKa 10.6.
Which is more acidic: ethanol, CH₃CH₂OH, or ethanethiol, CH₃CH₂SH?
A: the charge lands on O in one case and on S in the other. They are in the same group, so size dominates. Sulfur is bigger and more polarizable, so ethanethiol should be the stronger acid. R: neither anion has resonance. I: neither has an electronegative substituent. O: both O and S are sp³.
Only factor A differs, and it points to the thiol. The measured values agree: ethanol 16, ethanethiol 10.6, so the thiol is about 250,000 times more acidic. The framework got it right in one step.
The comparison above was the easy kind. Exams use harder ones, such as this: which is more acidic, ethanol or pentane-2,4-dione? The diketone's acidic hydrogen is on carbon.
A says ethanol. Ethanol's proton leaves an alkoxide, with the charge on oxygen. The diketone's proton leaves a carbanion, with the charge on carbon. Oxygen is far more electronegative, and A is usually the biggest factor, so the alcohol should win comfortably.
R says the diketone. The alkoxide has nowhere to put its charge. The diketone's anion sits between two carbonyls and delocalizes onto both oxygens, so most of its charge ends up on oxygen after all, and on two of them.
R wins, and not narrowly. Ethanol is 16 and pentane-2,4-dione is 9, so this C–H is ten million times more acidic than that O–H. Resonance did not merely offset the atom penalty; it overturned it, because sharing the charge between two oxygens is worth more than holding all of it on one.
This is what the ARIO order means: it is a search order, not a ranking of strengths. A comes first because it settles most comparisons quickly, not because it always wins. When the first factor and a later one point opposite ways, draw both anions and weigh them. A C–H that is unusually acidic is the tell that you are in that situation.
A common exam question gives you one molecule and asks which hydrogen comes off first. Take 4-hydroxybutan-2-one, CH₃–CO–CH₂–CH₂–OH, and go site by site. The figure numbers its carbons C1 to C4. There are four kinds of hydrogen on it.
The O–H. Its conjugate base is an ordinary alkoxide: the charge on oxygen and no resonance. The C=O group is three bonds away, too far for its induction to help much. That is an ordinary alcohol, pKa about 16.
The C3 hydrogens, between the C=O and the CH₂OH. C3 is an alpha carbon, so its conjugate base is an enolate, with the charge shared onto the carbonyl oxygen. That real stabilization makes this C–H thirty powers of ten more acidic than an alkane's: pKa about 20.
The C1 hydrogens, on the methyl group on the other side of the C=O. C1 is also an alpha carbon, so these are also about 20. Two alpha positions with similar acidity; neither changes the answer.
The C4 hydrogens, on the carbon bearing the OH. Their conjugate base would be a bare carbanion next to an oxygen. It has no resonance, and induction from the oxygen is worth a unit or two at most against a starting point near 50, so this site is not a contender.
Verdict: the O–H wins, by about four pKa units, or ten thousand to one. Take this one slowly, because it corrects a strong intuition. A hydrogen next to a carbonyl feels "activated", and it is, by thirty powers of ten compared with an alkane. It is still not as acidic as an ordinary alcohol. Rank each site against the pKa table, not against how special it looks.
Two practical notes. If a question asks what one equivalent (one mole per mole of the compound) of sodium hydride (NaH, whose conjugate acid H₂ has pKa about 36) does to this molecule, the answer is that it removes the O–H proton and nothing else. And if it asks for the most basic site rather than the most acidic proton, the answer is the hydroxyl oxygen. A protonated alcohol has pKa about −2 and a protonated ketone about −7, so the alcohol oxygen holds the more available lone pair. (A preview: acid catalysis in Carbonyl Chemistry still works through the carbonyl oxygen, because that is the protonation that makes the carbonyl carbon more reactive, even though less of the compound is protonated there.)
Two more levers the four letters do not name
Charge. H₃O⁺ has pKa −1.7, while water has 15.7. NH₄⁺ has 9.2, while NH₃ has 38. In each pair the acidic proton sits on the same atom, with the same hybridization, and neither has resonance or induction. Yet the gaps are seventeen and twenty-nine pKa units.
The difference is where the charge starts and ends. A positively charged acid loses its proton and becomes neutral, so it creates no new charge at all. That makes a cationic acid far stronger than its neutral relative, and it is one of the largest effects in the pKa table. It is also why a protonated molecule gives its proton back so readily, which acid catalysis relies on (a preview of How Reactions Happen).
Solvation, and the bulk that blocks it. Methanol has pKa 15.5 and tert-butanol, (CH₃)₃COH, has pKa 18. Their conjugate bases differ only in bulk. Every pKa on this page was measured in a solvent, and water molecules crowd around an ion to share its charge. That is solvation. Water can crowd closely around small methoxide, but not around bulky tert-butoxide, so tert-butoxide is stabilized less and its acid is weaker.
Remove the solvent and the order reverses: in the gas phase, tert-butoxide is the more stable anion. Treat solvation as a reminder rather than a fifth factor. The four structural factors explain the anion, and the solvent has the last word on the number.
Running the framework on a base
Exams ask the mirror question just as often: which of two bases is stronger? No new theory is needed. The stronger acid has the weaker conjugate base (Conjugate acids & bases), so the stronger base is the one whose conjugate acid has the higher pKa, its pKaH. ARIO then explains the ranking: a base is strong when its lone pair or charge is poorly stabilized.
Take ethoxide against acetate. Their conjugate acids are ethanol (16) and acetic acid (4.76), so ethoxide is the stronger base by eleven units. The reason is the resonance argument read backwards: acetate's charge is already spread over two oxygens, so it has less need of a proton.
Ammonia against water works the same way. Their conjugate acids are NH₄⁺ (9.2) and H₃O⁺ (−1.7), so ammonia is the stronger base by about eleven units, and its lone pair is the one that gets protonated when both are present. The Atom factor explains it: nitrogen is less electronegative than oxygen, so it holds its lone pair less tightly.
The trap is picking the wrong conjugate acid. The conjugate acid of ammonia is ammonium, NH₄⁺, not ammonia itself. Confusing a base's own N–H pKa (38 for NH₃) with its pKaH (9.2) is the most common way to get these comparisons backwards.
What carries forward
These four factors are the general theory of anion stability, and anion stability sets leaving-group ability, nucleophile strength, enolate formation and carbanion chemistry. You will apply them to carboxylic acids (Carboxylic Acids & Derivatives), to alpha hydrogens (Enolate Chemistry), to amine basicity (Amines) and to the substituent effects that control aromatic substitution (Aromatic Chemistry). Very little in the second half of the course is new physics. Most of it is this page applied to a new structure.