To compare two acids, deprotonate each on paper and decide which resulting anion is more stable.
Three O–H acids that differ by a factor of a billion in strength — the O–H bond is nearly identical, so the difference is entirely in the conjugate base.
1. A Lower pKa Means a Stronger Acid
Since pKa is a log scale, each unit is a factor of ten in Ka. Landmarks: HCl ≈ −7, carboxylic acids 4–5, water 15.7, alcohols 16–18, terminal alkynes ≈ 25, C–H bonds 45–50.
Left to right, acidity falls off a cliff — 57 pKa units separate HCl from ethane.
2. A More Stable Conjugate Base Means a Stronger Acid
Anything that stabilizes the negative charge on A− makes the parent acid stronger. Walk ARIO — Atom, Resonance, Induction, Orbital — and the first factor that differs usually settles the comparison.
Judge the acid on the right — the more stable the anion, the stronger the acid.
3. The Atom Holding the Charge Sets the Baseline
Two periodic-table trends govern which element holds the charge best:
- Across a period electronegativity rises: CH4 < NH3 < H2O < HF.
- Down a group size wins — a bigger atom spreads the charge: HF < HCl < HBr < HI.
O→F shows the across-a-period trend; F→Cl shows size winning down a group.
4. Resonance Spreads the Negative Charge and Lowers pKa
A conjugate base that delocalizes its charge by resonance is far more stable: acetate spreads charge over two oxygens (pKa 4.8), phenoxide into the ring's carbons (10), while ethoxide is stuck on one oxygen (16).
More (and better) resonance delocalization → more stable anion → stronger acid.
5. Nearby Electronegative Atoms Pull Charge Away by Induction
Electron-withdrawing groups pull charge through the sigma bonds, stabilizing a nearby anion — three chlorines drop acetic acid (pKa 4.76) to trichloroacetic acid (0.7). The effect fades with distance.
Three inductive chlorines stabilize the carboxylate and drop pKa by ~4 units.
6. More s-Character Stabilizes the Lone Pair (sp > sp2 > sp3)
More s character holds the lone pair closer to the nucleus, so C–H acidity follows sp > sp2 > sp3: alkyne (50% s, pKa 25) ≫ vinyl (44) ≫ alkane (50).
Deprotonating acetylene gives the acetylide anion [C⁻]#C, stabilized by 50% s character.
7. Equilibrium Always Favors the Weaker Acid
An acid–base equilibrium always runs toward the weaker acid (higher pKa): ethoxide readily deprotonates acetic acid (pKa 4.8) because the product ethanol (pKa 16) is the weaker acid.
Proton flows from the pKa-4.8 acid to give the pKa-16 acid — equilibrium favors the right.
8. Summary
Lower pKa = stronger acid · judge the conjugate base · walk ARIO: Atom · Resonance · Induction · Orbital · equilibrium favors the weaker acid.
Quiz yourself
Tap a question to reveal the answer — free, no login.
The pKa 4 acid is stronger — lower pKa = stronger acid. Since each unit is a factor of ten, a 6-unit gap means it ionizes about 106 (a million) times more readily.
Both lose a proton from O–H, but the acetate conjugate base delocalizes its negative charge by resonance across two equivalent oxygens, while ethoxide traps the charge on one oxygen. The more stable anion means the stronger acid — worth about 11 pKa units.
Orbital hybridization. The alkyne carbon is sp (50% s character), which holds the resulting acetylide lone pair closer to the nucleus and stabilizes it. The alkane carbon is sp3 (25% s), so its carbanion is far less stable. Order of s-character: sp > sp2 > sp3.
Yes. Equilibrium always favors forming the weaker acid (higher pKa). Removing acetic acid's proton produces ethanol (pKa 16), a much weaker acid than the acetic acid you started with, so the reaction runs strongly to the product side.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.