Acids & Bases · Section 19 of 64

pKa

Practice this — interactive lesson

pKa is the single most useful number in organic chemistry. It tells you whether a base is strong enough to remove a given proton, which direction a proton transfer runs, how good a leaving group something is, and how stable an anion is. Almost every "will this reaction work?" question in the first half of the course can be answered by comparing two pKa values.

Ka and pKa

For an acid HA dissociating in solution, HA ⇌ H⁺ + A⁻, the equilibrium constant is

Ka = [H⁺][A⁻] / [HA]

A larger Ka means the equilibrium sits further toward the dissociated side — a stronger acid. Because Ka values for real acids span something like sixty orders of magnitude, chemists take the negative logarithm instead, exactly as they do to turn [H⁺] into pH:

pKa = −log₁₀(Ka)

The negative sign flips the direction, and this is the source of endless confusion, so state it plainly and often: lower pKa means stronger acid. A pKa of −7 is a very strong acid. A pKa of 50 is not an acid in any practical sense.

Because pKa is logarithmic, every 1 unit is a tenfold change in acid strength. Acetic acid (pKa 4.76) is roughly 10¹¹ times more acidic than ethanol (pKa 16) — a hundred billion to one, from a gap that looks like eleven small steps on a table. When comparing two acids, always convert the gap into orders of magnitude before deciding whether it is "a big difference."

The table worth memorizing the shape of

You do not need exact values. You need the ordering and rough magnitudes, so that you can look at any two species and know which way a proton moves.

AcidpKaConjugate base
HI−10I⁻
HCl−7Cl⁻
H₃O⁺−1.7H₂O
HF3.2F⁻
CH₃COOH4.76CH₃COO⁻
H₂CO₃ / HCO₃⁻6.4 / 10.3bicarbonate / carbonate
1,3-diketone α-H9stabilized enolate
NH₄⁺9.2NH₃
phenol10phenoxide
water15.7HO⁻
ethanol16EtO⁻
ketone α-H19–20enolate
terminal alkyne C–H25acetylide
NH₃38⁻NH₂
alkene C–H44vinyl anion
alkane C–H50alkyl anion

A handful of anchors is enough to reconstruct the rest: carboxylic acid ≈ 5, ammonium ≈ 9, phenol ≈ 10, water ≈ 16, alcohol ≈ 16, ketone alpha-H ≈ 20, alkyne ≈ 25, amine N–H ≈ 38, alkane ≈ 50. Almost everything you meet sits near one of those.

HClpKa -7 · a strong mineral acidH₃O⁺pKa -1.7 · protonated watercarboxylic acid RCOOHpKa 4.8 · the O–H of a COOHammonium NH₄⁺pKa 9.2 · protonated ammoniaphenol PhOHpKa 10 · an aromatic O–Hwater H₂OpKa 15.7alcohol ROHpKa 16 · the O–H of an alcoholketone α-HpKa 20 · next to a carbonylterminal alkyne ≡C–HpKa 25 · sp carbonamine N–H RNH₂pKa 38alkane C–HpKa 50 · nothing touches thislower pKa · STRONGER acidhigher pKa · WEAKER acidevery step down is ten times stronger
The anchors, drawn to scale — which is the thing a table of numbers hides. The rungs are logarithmic: the gap from a carboxylic acid to an alcohol looks like eleven small steps and is a factor of a hundred billion. Almost every acidic hydrogen you meet in this course sits near one of these rungs, so learning the shape of the ladder is worth more than memorising any individual value.

Predicting which side an equilibrium favors

In any proton transfer, the equilibrium favors the side with the weaker acid — equivalently, the side whose acid has the higher pKa. The procedure is mechanical: identify the acid on the left, identify the acid that would be formed on the right (the conjugate acid of the base), compare their pKa values, and the side with the higher pKa wins.

Worked example — can hydroxide deprotonate a terminal alkyne?

Left-hand acid: the alkyne C–H, pKa ≈ 25. Right-hand acid: water, formed when hydroxide takes the proton, pKa 15.7.

Water is the stronger acid (lower pKa), so the equilibrium favors the left — the side with the weaker acid. The gap is about 9 units, so the equilibrium lies roughly 10⁹ to one against the product. Hydroxide will not do this.

Now try sodium amide, NaNH₂. Right-hand acid: ammonia, pKa 38. The alkyne at 25 is the stronger acid, so the equilibrium favors the right by about 10¹³. This works, and it is exactly why NaNH₂ is the reagent specified for generating acetylides in Module 7.

Can this base take that proton? Compare the acid on each side — the HIGHER pKa wins.alkyne ≡C–HpKa 25+ HO⁻water H–OHpKa 15.7NO — favours the leftwater is the stronger acid, by 10⁹alkyne ≡C–HpKa 25+ ⁻NH₂ammonia H–NH₂pKa 38YES — favours the rightthe alkyne is the stronger acid, by 10¹³
The whole procedure, mechanised. Name the acid on the left. Name the acid that would be made on the right — that is the conjugate acid of your base. Compare the two pKa values, and the equilibrium settles on whichever side holds the weaker acid, the one with the higher number. Do not reason about which base is stronger; convert everything to acids, where the number points the right way.This is exactly why sodium amide, and not hydroxide, is the reagent specified for making acetylides in Module 7. The number chose the reagent.
Worked example — separating a carboxylic acid from a phenol

Both dissolve in NaOH, so that does not separate them. Use sodium bicarbonate instead: the relevant acid on the right is carbonic acid, pKa 6.4.

The carboxylic acid (pKa 4.76) is the stronger acid, so it is deprotonated and dissolves into the aqueous layer as its carboxylate salt. The phenol (pKa 10) is the weaker acid, is not deprotonated, and stays in the organic layer.

This is a real laboratory separation, and it rests on nothing but two pKa comparisons.

Compare acids, not bases. The most common mistake is to reason about which base is stronger and get tangled in the double negative. Convert everything to acids and compare pKa values directly — it is the same question, asked in a form where the number points the right way.

The base's own pKa is its conjugate acid's

Bases do not have pKa values of their own; they inherit one from their conjugate acid, sometimes written pKaH. A base can remove a proton whose pKa is lower than its own pKaH. Hydroxide (pKaH 15.7) can deprotonate anything below about 15.7 and nothing much above it. Amide ion (pKaH 38) can deprotonate almost everything organic. LDA, a bulky lithium amide base (pKaH ≈ 36), is chosen precisely because it is strong enough for ketone alpha-hydrogens at pKa 20 while being too bulky to attack the carbonyl itself.

Once you see this, choosing a base stops being memorization. You look up what you need to deprotonate, and you pick a base whose pKaH is comfortably higher.

Two cautions on the numbers

pKa is solvent-dependent. The values in every table are measured in water (or extrapolated to it), and they shift in other solvents — sometimes by many units. The relative ordering usually survives, which is what you rely on, but absolute values in DMSO can differ substantially from the aqueous ones.

Strong acids are all levelled in water. HCl, HBr, HI and H₂SO₄ are all completely dissociated in aqueous solution, so water cannot distinguish between them — the strongest acid that can exist in water is H₃O⁺. Their quoted pKa values come from measurements in other solvents. The same levelling applies at the other end: the strongest base that survives in water is hydroxide, which is why reagents like NaNH₂ and LDA must be used in aprotic solvents.

What carries forward

pKa reasoning runs through the whole course. It ranks leaving groups (Module 2 and Module 6), it decides whether a base will do the job in every elimination and enolate reaction, it explains the relative reactivity of carboxylic acid derivatives in Module 10, it governs which nitrogen gets protonated in Module 12, and it is the framework behind every extraction you will run in the lab. The next two sections unpack where the numbers come from.