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Organic Chemistry 1 · Acid-Base Chemistry

Brønsted-Lowry Acids and Bases

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

A is a proton (H⁺) donor; a is a proton acceptor. Losing a proton turns an acid into its , and gaining one turns a base into its . Acid strength is reported as = −log(Ka): the smaller the pKa, the stronger the acid. Every runs toward the weaker acid and weaker base, so comparing pKa values predicts reaction direction.

Why this matters

Blood pH is held near 7.4 by the amphiprotic carbonic-acid/bicarbonate conjugate pair, H₂CO₃ ⇌ H⁺ + HCO₃⁻. The same logic governs drug absorption: most drugs are weak acids or bases, and their ionized-versus-neutral form — set by the molecule's pKa relative to local pH — controls membrane crossing. A carboxylic-acid drug (pKa ≈ 4–5) is largely neutral and permeable in the acidic stomach but ionized in the more basic intestine.

The college version

1. Proton donors and acceptors define conjugate pairs

A Brønsted-Lowry acid donates a proton; a base accepts one. Every acid has a conjugate base (acid minus one H⁺), and every base has a conjugate acid (base plus one H⁺); the two differ by exactly one proton. In HCl + H₂O → H₃O⁺ + Cl⁻, HCl/Cl⁻ and H₂O/H₃O⁺ are the two conjugate pairs. A compound that can act as either an acid or a base is amphiprotic — water accepts a proton from HCl but donates one to NH₃.

2. Ka and pKa quantify acid strength

For a generic acid HA in water, HA + H₂O ⇌ H₃O⁺ + A⁻, with Ka = [H3O+] [A−]/[HA]. Because Ka spans many orders of magnitude, chemists report pKa = -log10 Ka, which inverts the intuition: large Ka = small pKa = strong acid. Acetic acid has Ka = 1.8 × 10⁻⁵, so pKa = 4.76; hydrochloric acid has pKa ≈ −7. One pKa unit equals a tenfold change in Ka.

3. Equilibrium favors the weaker acid and weaker base

An acid-base equilibrium is a competition for the proton between two bases. The proton ends up on the stronger base, so the equilibrium lies on the side containing the weaker acid and weaker base. A strong acid always has a weak conjugate base, and a weak acid has a strong conjugate base.

How it works

  1. Identify the two conjugate pairs: find the proton donor (acid) and acceptor (base), then add/subtract one H⁺ to name each product.
  2. Look up or estimate the pKa of the two acids present (one reactant, one product).
  3. Decide which acid is stronger (lower pKa); it donates, driving equilibrium away from itself.
  4. State the result: equilibrium favors the side with the weaker acid and weaker base.
  5. Check consistency: the strong acid's conjugate base is weak, and vice versa.

Common confusions

Do not confuseWithDifference
AcidConjugate baseThe acid has one extra proton; the conjugate base is what it becomes after donating
KapKaThey vary inversely — large Ka means small pKa and a stronger acid
Strong acidConcentrated acid"Strong" is about Ka (ionization); "concentrated" is about how much is dissolved
pKapHpKa is a property of the acid; pH is the proton level of a specific solution
Weaker conjugate baseWeaker acidThey go together — a strong acid always has a weak conjugate base

Memory aids

"DEA — a Donor is an Acid" fixes the direction. For equilibrium, "pKa low, proton flow": the proton flows from the low-pKa (strong) acid to the stronger base, ending on the side of the weak acid.

Quick review

Topic Recap

Brønsted-Lowry acids donate protons and bases accept them; each transfer forms a conjugate acid-base pair. Strength is measured by Ka and reported as pKa, with lower pKa meaning a stronger acid. Equilibrium always favors the weaker acid and weaker base, so comparing pKa values predicts proton-transfer direction in nearly any reaction.

Knowledge Check

  1. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, name the two conjugate acid-base pairs.
  2. Acid X has Ka = 1 × 10⁻³; acid Y has pKa = 6. Which is stronger?
  3. An acid with pKa = 4 reacts with the conjugate base of an acid with pKa = 9. Which way does equilibrium lie?
  4. Why is H₂O described as amphiprotic?
  5. Which is the stronger acid — ethanol (pKa ≈ 16) or acetic acid (pKa ≈ 4.8) — and why?

Answers and Rationales

  1. NH₃/NH₄⁺ (base and its conjugate acid) and H₂O/OH⁻ (acid and its conjugate base); each pair differs by one proton.
  2. Acid X. Its Ka = 10⁻³ corresponds to pKa = 3, lower than 6, so it is the stronger acid.
  3. Toward products, away from the pKa = 4 acid. The pKa = 4 acid is stronger, so it donates; equilibrium favors the weaker acid (pKa = 9).
  4. Water can donate a proton (to NH₃, forming OH⁻) or accept one (from HCl, forming H₃O⁺), so it is amphiprotic.
  5. Acetic acid. Its pKa (4.8) is far lower than ethanol's (16); acetate is stabilized by resonance, ethoxide is not.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of a proton as a baton in a relay race. An acid holds it loosely, ready to hand it off; a base reaches out to take it. After the handoff, the first runner (now the conjugate base) is empty-handed and the second (now the conjugate acid) holds the baton. Whoever grips it more tightly is the weaker acid and keeps it.

Where it stops being exact: a proton is not an object passed by choice. The "handoff" is one bond breaking while another forms, driven by energy differences — acids do not "want" anything.

Simple Example

HCl + H₂O → H₃O⁺ + Cl⁻. HCl donates a proton (acid); H₂O accepts it (base). H₃O⁺ is the conjugate acid of water and Cl⁻ is the conjugate base of HCl.

Worked example

  1. Draw the proton transfer with curved arrows. A double-headed arrow starts at a lone pair on the base and points at the acidic proton; a second arrow starts at the H–A bond and points onto A, showing the bonding pair stays on the conjugate base. Arrows move electron pairs, never atoms: H2O: + H-Cl → H3O+ + :Cl−.
  2. Calculate pKa from Ka. For acetic acid, pKa = -log(1.8 × 10-5) = 4.76. Working backward, pKa = 10 means Ka = 10⁻¹⁰, a much weaker acid.
  3. Predict direction using pKa. Compare the acid on each side. In CH3CO2H + NH3 ⇌ CH3CO2− + NH4+, the acids are acetic acid (pKa ≈ 4.8) and ammonium (pKa ≈ 9.4). Acetic acid is stronger, so equilibrium lies to the right, favoring the weaker acid (NH₄⁺) and weaker base (CH₃CO₂⁻).

Key takeaways

  • High yield: A Brønsted-Lowry acid donates a proton; a base accepts one.
  • High yield: Conjugate pairs differ by exactly one proton (H⁺).
  • High yield: Lower pKa = stronger acid = weaker conjugate base.
  • High yield: pKa = −log(Ka), so each pKa unit is a tenfold change in Ka.
  • High yield: Equilibrium favors the weaker acid and weaker base.
  • Water is amphiprotic — the solvent base for most proton transfers you will draw.
  • pKa anchors: strong mineral acids (< 0), carboxylic acids (≈ 4–5), phenols (≈ 10), water (15.7), alcohols (≈ 16–18), terminal alkynes (≈ 25), amines (≈ 38), alkanes (≈ 50).
  • Quantitative acidity (Ka/pKa) differs from relative acidity (ranking); both appear on exams.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice Organic Chemistry 1

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Define Brønsted-Lowry acids and bases as proton donors and acceptors, and identify conjugate acid-base pairs in a reaction.
  • Write acid-dissociation equilibria and express both Ka and pKa, converting between them and interpreting their meaning.
  • Predict the direction of an acid-base equilibrium by comparing pKa values and applying the stronger-acid/weaker-conjugate-base relationship.
  • Use an approximate pKa ladder to rank common acids and apply it to unfamiliar molecules.

Key vocabulary

Brønsted-Lowry acid
A proton (H⁺) donor
Brønsted-Lowry base
A proton (H⁺) acceptor
Conjugate base
What remains after an acid loses a proton
Conjugate acid
What forms when a base gains a proton
Amphiprotic compound
Can donate or accept a proton
Acid-base equilibrium
Reversible proton transfer between two conjugate pairs
Ka
The acid-dissociation equilibrium constant
pKa
−log(Ka); smaller = stronger acid
Relative acidity
How strong an acid is compared with another

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