You have used pKa twice already. Resonance compared acetic acid with ethanol by it: a lower pKa means a stronger acid, and each unit is a factor of ten. Brønsted acids and bases used it to decide which side of a proton transfer wins. With acetic acid and ammonia, for example, the side with the weaker acid and the weaker base is favored by about 3 × 10⁴ to one.
This page explains where the number comes from, the acid dissociation constant Ka. It then lays out the scale, shows how to work out how far a proton transfer goes, and shows how to pick a base strong enough for a given proton. Later chapters lean on the same comparison to judge how easily a group leaves a molecule, so most “will this reaction work?” questions in the first half of the course come down to two pKa values. Why one acid is stronger than another is the subject of the next section, Factors affecting acidity.
Ka and pKa
Start with acetic acid dissolved in water. A small fraction of its molecules give up the proton on the O–H and become acetate ions. The rest stay whole. Written for any acid HA, the reaction is
Here H⁺ is shorthand. In water the proton actually sits on a water molecule as hydronium, H₃O⁺. Brønsted acids and bases showed this.
After a while the forward and reverse reactions run at the same rate, and the amounts stop changing. That balance is the equilibrium. The equilibrium constant is the ratio of product concentrations to reactant concentrations at equilibrium. Square brackets mean a concentration in moles per liter. For an acid, the constant is called the acid dissociation constant, Ka:
A larger Ka means more of the acid has given up its proton at equilibrium, so a larger Ka means a stronger acid. For acetic acid, Ka is 1.7 × 10⁻⁵.
Ka values for real acids span about sixty powers of ten, from around 10¹⁰ down to 10⁻⁵⁰. Numbers that far apart are awkward to compare, so chemists take the negative logarithm instead. It is the same step that turns [H⁺] into pH:
For acetic acid, −log₁₀(1.7 × 10⁻⁵) = 4.76. The minus sign is why a lower pKa means a stronger acid, the reading rule you already know: a larger Ka gives a smaller pKa. HCl, with a pKa of −7, is a very strong acid. An alkane C–H, with a pKa of 50, is not an acid in any practical sense.
The shape of the pKa table
You do not need every exact value. You need the order and the rough size, so that you can look at any two acids and tell which way a proton moves.
| Acid | pKa | Conjugate base |
|---|---|---|
| HI | −10 | I⁻ |
| HCl | −7 | Cl⁻ |
| H₃O⁺ | −1.7 | H₂O |
| HF | 3.2 | F⁻ |
| CH₃COOH | 4.76 | CH₃COO⁻ |
| H₂CO₃ / HCO₃⁻ | 6.4 / 10.3 | bicarbonate / carbonate |
| 1,3-diketone C–H between the two C=O | 9 | enolate |
| NH₄⁺ | 9.2 | NH₃ |
| phenol | 10 | phenoxide |
| water | 15.7 | HO⁻ |
| ethanol | 16 | CH₃CH₂O⁻ (ethoxide) |
| ketone α C–H | 19–20 | enolate |
| terminal alkyne C–H | 25 | acetylide |
| NH₃ | 38 | ⁻NH₂ (amide ion) |
| alkene C–H | 44 | vinyl anion |
| alkane C–H | 50 | alkyl anion |
Several rows name a hydrogen rather than a whole molecule. The figure below draws an example molecule for nine of the rows, with the hydrogen marked. An alpha (α) hydrogen sits on the carbon right next to a C=O carbon. A terminal alkyne has its triple bond at the end of the chain, so one of the triple-bond carbons carries a hydrogen.
The anions in the right-hand column have names you will meet again. An enolate is what is left when an α hydrogen is removed; Enolate Chemistry is built on them. An acetylide is what is left when a terminal alkyne loses its C–H proton. A vinyl anion and an alkyl anion are what an alkene and an alkane would leave behind after losing a C–H proton.
A handful of anchors is enough to rebuild the rest: carboxylic acid about 5, ammonium about 9, phenol about 10, water and alcohols about 16, ketone α C–H about 20, terminal alkyne about 25, amine N–H about 38, and alkane about 50. Almost every acidic hydrogen you meet sits near one of them.
Predicting which side an equilibrium favors
Take acetic acid and hydroxide. Hydroxide takes the O–H proton and becomes water, and acetic acid becomes acetate. The reaction can also run backward, with acetate taking a proton from water. So there is an acid on each side: acetic acid (pKa 4.76) on the left and water (pKa 15.7) on the right. Acetic acid is far stronger, so it is the one that ends up without its proton, and the equilibrium lies on the right.
That is the rule from Brønsted acids and bases: a proton transfer favors the side with the weaker acid and the weaker base. In pKa terms, the weaker acid is the one with the higher pKa. To apply it, find the acid on the left. Then find the acid that the base turns into on the right, which is the base’s conjugate acid (see Conjugate acids and bases). Compare the two pKa values, and the side with the higher one is favored.
The same two numbers also give the size of the equilibrium constant:
Subtract in that order and the sign takes care of itself. A positive exponent means the reaction runs forward, and a negative one means it runs backward. For acetic acid and hydroxide, K ≈ 1015.7 − 4.76 ≈ 1011, so the reaction is complete.
Ethanol and hydroxide make a closer call. Ethanol (pKa 16) is on the left and water (pKa 15.7) is on the right, so K ≈ 1015.7 − 16 = 10−0.3, about one half. Neither side is clearly favored, so hydroxide cannot turn ethanol completely into ethoxide.
The table below turns a pKa gap into how much of the starting acid reacts, when the acid and the base start in equal amounts. By a gap of +4 the reaction is about 99% complete, and by −4 almost nothing happens. Both sides stay present in useful amounts only in the narrow band between.
| pKa gap (acid formed − acid used up) | Keq | Starting acid converted |
|---|---|---|
| −6 | 10⁻⁶ | about 0.1% |
| −2 | 10⁻² | about 9% |
| 0 | 1 | 50% |
| +2 | 10² | about 91% |
| +4 | 10⁴ | about 99% |
| +6 | 10⁶ | about 99.9% |
The alkyne C–H has a pKa of 25. Hydroxide would take that proton and become water, pKa 15.7.
Water is the stronger acid, so the equilibrium favors the left, the side with the alkyne. K ≈ 1015.7 − 25 ≈ 10−9. Hydroxide will not do this.
Now try sodium amide, NaNH₂. Its anion, the amide ion ⁻NH₂, would become ammonia, pKa 38. This time the alkyne is the stronger acid, so the equilibrium favors the right, with K ≈ 10¹³. That number is why NaNH₂, and not hydroxide, is the standard reagent for making acetylides in Alkenes & Alkynes.
A flask holds a carboxylic acid and a phenol, both dissolved in ether. To separate them, chemists use an extraction. The ether solution is shaken with water in a separatory funnel, a glass funnel with a tap at the bottom. Ether and water do not mix, so they settle into two layers. A charged compound moves into the water, and a neutral one stays in the ether.
Sodium hydroxide will not separate them. Its conjugate acid, water, has a pKa of 15.7, well above both compounds, so hydroxide deprotonates both and both move into the water. Sodium bicarbonate, NaHCO₃, works instead, because its conjugate acid, carbonic acid, has a pKa of 6.4.
The carboxylic acid (pKa 4.76) is the stronger acid of the two, so bicarbonate deprotonates it. It becomes a carboxylate salt and moves into the water. Chemists add bicarbonate in excess. The carbonic acid that forms breaks down to carbon dioxide gas, which bubbles out of the funnel, so the reaction goes to completion. The phenol (pKa 10) is the weaker acid, so it keeps its proton and stays in the ether.
Ether is less dense than water, so the ether layer is on top. You drain the lower, aqueous layer out through the tap. Adding acid to that layer afterward puts the proton back and recovers the carboxylic acid.
Every separation of two acids by extraction follows this pattern: choose a base whose conjugate acid’s pKa falls between the two compounds you want to separate.
The base's own pKa is its conjugate acid's
Hydroxide is used here as a base, not an acid, yet chemists quote 15.7 for it. That number belongs to water, its conjugate acid. A base is described by its conjugate acid’s pKa, often written pKaH: the pKa of the base with a proton added. Hydroxide’s pKaH is 15.7, and the amide ion’s is 38.
A base can fully remove a proton whose pKa is well below its own pKaH. Hydroxide fully deprotonates acids well below 15.7, gives a mixture with acids close to it such as ethanol, and does almost nothing to acids well above it. The amide ion, at 38, can remove every proton listed above ammonia’s own row in the table. So choosing a base is not memorization: find the pKa of the proton you want to remove, and pick a base whose pKaH is comfortably higher.
Three substrates each need to lose one proton completely: ethanol (O–H, pKa 16), a ketone α hydrogen (pKa 20) and a terminal alkyne (C–H, pKa 25). Four bases are on the shelf: sodium hydroxide, NaOH (pKaH 15.7); sodium ethoxide, NaOCH₂CH₃ (pKaH 16); sodium hydride, NaH, whose conjugate acid is H₂ (pKaH about 36); and sodium amide, NaNH₂ (pKaH 38). The table gives Keq for each pairing.
| Base | pKaH | Ethanol (16) | Ketone α C–H (20) | Terminal alkyne (25) |
|---|---|---|---|---|
| NaOH | 15.7 | 10−0.3 | 10−4.3 | 10−9.3 |
| NaOCH₂CH₃ | 16 | 1 | 10−4 | 10−9 |
| NaH | 36 | 1020 | 1016 | 1011 |
| NaNH₂ | 38 | 1022 | 1018 | 1013 |
Ethanol. NaOH leaves a mixture of ethanol and ethoxide. NaOCH₂CH₃ is ethanol’s own conjugate base, so K = 1 and nothing changes. NaH works, and the reaction cannot run backward, because its by-product, hydrogen gas, bubbles out of the flask. That is the standard way to make an alkoxide, the anion RO⁻ of an alcohol.
The ketone. NaOH and NaOCH₂CH₃ fall about four units short, so each gives only a trace of enolate at equilibrium. A trace is sometimes all a reaction needs, but it is useless when every ketone molecule has to be deprotonated. NaH and NaNH₂ are both 16 or more units above, so both are strong enough. Looking ahead: in practice chemists use LDA, a bulky nitrogen base (pKaH about 36) that removes the α hydrogen completely without reaching the C=O carbon; Enolate Chemistry covers it.
The alkyne. Hydroxide and ethoxide fall about nine units short and do almost nothing. NaNH₂ is 13 units above, so the reaction is complete.
The rule: pick a base whose pKaH is at least three or four units above the pKa of the proton you are removing. That gap already converts about 97 to 99% of the substrate. A gap of one or two units leaves a noticeable amount behind.
pH decides which form you are holding
Acetic acid in stomach acid, at about pH 2, is almost all neutral CH₃COOH. The same acid in blood, at pH 7.4, is almost all acetate, CH₃COO⁻. The molecule has not changed; the solution around it has.
pKa belongs to the molecule, and pH belongs to the solution. pH measures how much H₃O⁺ the solution holds, and a lower pH means more. Together they decide whether you have the acid or its conjugate base. Below its pKa, an acid is mostly protonated. Above its pKa, it is mostly deprotonated. At pH = pKa, the two forms are present in equal amounts.
The exact ratio comes from rearranging the Ka expression. The result is called the Henderson–Hasselbalch equation:
So one pH unit away from the pKa already gives a 10 : 1 ratio, and two units give 100 : 1. For acetic acid in blood, pH − pKa = 7.4 − 4.76 ≈ 2.6, so acetate outnumbers acetic acid by about 400 to one.
Amines follow the same rule through their conjugate acids. An amine whose ammonium ion, R–NH₃⁺, has a pKa of 10 stays mostly as that cation at any pH below 10. It is charged in the stomach and in blood alike.
This is why many drugs that contain an amine are sold as hydrochloride salts: the drug is charged at the body’s pH anyway, and the charged salt dissolves in water more readily than the neutral amine.
Three cautions on the numbers
pKa depends on the solvent. The values on this page are measured in water, or extrapolated to it. In other solvents they shift, sometimes by many units. Acetic acid, for example, has a pKa of about 12 in DMSO (dimethyl sulfoxide, a common organic solvent), against 4.76 in water. The order of the acids usually survives the change, and the order is what you rely on.
Water levels the strong acids. HCl, HBr, HI and the first proton of H₂SO₄ all dissociate completely in water, so water cannot tell them apart: each one simply becomes H₃O⁺. That makes H₃O⁺ the strongest acid that can exist in water, an effect called leveling. The quoted pKa values of those acids come from other solvents or from estimates.
The same limit applies to bases. Hydroxide is the strongest base that can exist in water, because any stronger base takes a proton from water. So NaNH₂ is used in liquid ammonia, the amide ion’s own conjugate acid, and NaH in THF (tetrahydrofuran), an ether with no O–H or N–H at all.
Two values are in use for H₃O⁺ and water. This course uses −1.7 for H₃O⁺ and 15.7 for water, the values most courses and exams use. Some sources give 0 and 14 instead. The two pairs differ in how they count the concentration of water when water is also the solvent. The choice rarely changes a conclusion, but it can in a close call such as water against methanol (pKa 15.5), so stick to one pair.
What carries forward
pKa reasoning runs through the whole course. It ranks leaving groups in How Reactions Happen and Substitution & Elimination. It decides whether a base will do the job in every elimination and enolate reaction. It explains the reactivity order of the acid derivatives in Carboxylic Acids & Derivatives and which nitrogen gets protonated in Amines. It is also the idea behind every extraction you will run in the lab. The next section, Factors affecting acidity, explains why the numbers fall where they do.