Organic Chemistry · Amines and Heterocycles

Biological Amines and the Henderson–Hasselbalch Equation

9 min read

pK_a / pK_aH values are approximate aqueous reference values from standard pharmacology and biochemistry references; verify against current sources before relying on them in assessments. Blood-pH and buffer discussion is a chemistry illustration only; no clinical guidance is implied.

On this page 9 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Biological systems are full of amines: the amino acid side chains of lysine and histidine, the neurotransmitters dopamine, serotonin, histamine, and epinephrine, the DNA/RNA bases adenine and guanine, and the plant-derived alkaloids such as morphine, nicotine, caffeine, and quinine. Every one of these molecules exists in an equilibrium between a neutral and a protonated ammonium form, and the position of that equilibrium is set by the local pH and the amine's pK_aH. The converts those two numbers into the ratio of protonated to unprotonated forms:

$pH = p K aH + log 10 [B] [BH +]$

The same equation, written with [A^-]/[HA], describes acids — including the that holds blood near pH 7.4. This topic derives the equation, shows how to use it, and applies it to drug absorption, isolation, and biological buffers.

Why this matters

Drug absorption and distribution: Only the uncharged free-base form of a drug crosses lipid membranes readily; the charged ammonium form is water-soluble but membrane-impermeable. Knowing pK_aH and the local pH tells you how much of a drug is where it needs to be.
Local anesthetics: Lidocaine (pK_aH ≈ 7.9) and procaine ( ≈ 8.9) work because a significant fraction exists as the free base at tissue pH, diffusing to the nerve, where lower local pH converts it to the active protonated form.
Alkaloid isolation: Natural-product chemists protonate alkaloids with acid to extract them into water, then deprotonate with base to recover the free amine — a direct application of the same equilibrium.
Blood pH homeostasis: The bicarbonate buffer, pH = 6.1 + log₁₀([HCO₃^-]/[H₂CO₃]), is the Henderson–Hasselbalch equation in action; its ~20:1 ratio at pH 7.4 is a standard chemistry illustration of buffering.
Protein chemistry: Histidine (pK_aH ≈ 6.0) and lysine ( ≈ 10.5) side chains change charge with local pH, controlling enzyme catalysis and protein folding.
Exams: Henderson–Hasselbalch calculations (fractions, ratios, required pH) are among the most tested quantitative skills in organic and biochemistry courses.

The college version

Core Concepts

Deriving the equation for an amine base

Start from the acid dissociation of the conjugate acid BH⁺:

$BH + ⇌ B + H + K a = [B][H +] [BH +]$

Solve for [H⁺]:

$[H +] = K a [BH +] [B]$

Take -log₁₀ of both sides. By the properties of logarithms, -log₁₀[H⁺] = pH, -log₁₀K_a = pK_aH, and the ratio term becomes a positive log:

$- log 10 [H +] = - log 10 K a - log 10 [BH +] [B] = p K aH + log 10 [B] [BH +]$

giving the working form:

$pH = p K aH + log 10 [B] [BH +]$

For an acid HA, the identical derivation with [A^-]/[HA] gives pH = pK_a + log₁₀([A^-]/[HA]). The only difference is which species is "unprotonated."

Reading the equation: three regimes

pH = pK_aH: log₁₀([B]/[BH⁺]) = 0, so [B] = [BH⁺] — 50% protonated.
pH < pK_aH: the log term is negative, so [BH⁺] > [B] — the protonated (ammonium) form dominates. Rule of thumb: pH below pK_aH means protonated.
pH > pK_aH: the free base dominates.

The ratio moves by factors of ten per pH unit: one unit away → 10:1, two units → 100:1. So an amine with pK_aH two units above the pH is ~99% protonated.

Biological amine pK_aH reference values

These are approximate aqueous values from standard pharmacology/biochemistry references — always check current sources for exact numbers:

Molecule	pK_aH (approx.)	Note
Amphetamine	≈ 9.9	Stimulant; almost fully protonated at pH 7.4
Ephedrine	≈ 9.6	Bronchodilator/decongestant
Atropine	≈ 9.4	Anticholinergic alkaloid
Morphine / codeine	≈ 8.2	Opiate analgesics
Procaine	≈ 8.9	Local anesthetic (ester)
Lidocaine	≈ 7.9	Local anesthetic (amide)
Nicotine (pyrrolidine N)	≈ 8.0	Alkaloid
Histidine side chain	≈ 6.0	Amino acid; titrates near physiological pH
Lysine side chain	≈ 10.5	Amino acid; protonated at pH 7.4

Alkaloid isolation by acid–base extraction

A typical isolation: treat plant material with dilute aqueous acid. The alkaloid free base accepts a proton, becomes a water-soluble ammonium salt, and leaves the organic tissue for the aqueous layer. Neutral plant fats and waxes stay behind. Then add base to the aqueous layer: pH rises above pK_aH, the ammonium ion loses its proton, and the neutral alkaloid precipitates or is extracted into an organic solvent. Each step is the same equilibrium pushed in a chosen direction by pH.

The bicarbonate buffer (chemistry illustration)

Carbon dioxide dissolves and hydrates to carbonic acid: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃^-. With the first pK_a ≈ 6.1:

$pH = 6.1 + log 10 [HCO 3 -] [H 2 CO 3]$

At blood pH 7.4 the ratio is about 20:1. This is a textbook chemistry model of a buffer; actual physiology involves the lungs and kidneys, which are beyond the scope of this chapter.

Common Confusions

Do Not Confuse	With	Difference
Using [A^-]/[HA] for an amine	[B]/[BH⁺]	For amines the unprotonated species is B and the protonated one is BH⁺; use pK_aH
"pH > pK_aH means protonated"	Free base dominates	pH above pK_aH → deprotonated (free base) dominates; pH below → protonated
pK_a and pK_aH interchangeable	Same symbol, different species	pK_aH always refers to the conjugate acid of the base in question
"Drugs must be fully uncharged to act"	Equilibrium balance	It's a ratio, not a switch; partial free-base fractions (like lidocaine's 24%) are functionally important
pH = pK_aH means fully protonated	50/50	At equality, [B] = [BH⁺] exactly
Henderson–Hasselbalch is exact in all systems	A good model in dilute solution	Proteins, membranes, and non-ideal media shift apparent pK_a values; treat results as approximations there

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a crowd of "base" molecules that can either carry a proton (charged, like a wet coat) or drop it (neutral, like a dry coat). Whether they carry it depends on the room's "protoniness" (pH) and each molecule's own preference number (pK_aH). The Henderson–Hasselbalch equation is the calculator: if the pH is below the molecule's number, most molecules carry the proton; if the pH is above, most drop it. At the exact number, half and half. That's why a drug like lidocaine is partly dry (free base, able to slip through cell membranes) and partly wet (protonated, active) inside your body at the same time.

Worked example

Example 1: Lidocaine at blood pH — what fraction is active?

Lidocaine has pK_aH ≈ 7.9. What fraction is protonated at blood pH 7.4?

Formula first:

$pH = p K aH + log 10 [B] [BH +]$

Substitute pH = 7.4, pK_aH = 7.9:

$7.4 = 7.9 + log 10 [B] [BH +] \Rightarrow log 10 [B] [BH +] = -0.5$

$[B] [BH +] = 10 -0.5 \approx 0.32$

Convert to fraction protonated (protonated = 1 part for every 0.32 parts free base):

$fraction protonated = 11 + 0.32 = 0.76$

Answer: about 76% of lidocaine is protonated and 24% is free base at pH 7.4. The ~24% free base crosses membranes to reach the nerve; inside the slightly acidic nerve environment the equilibrium shifts toward the protonated form that blocks the sodium channel.

Example 2: At what pH is procaine 90% protonated?

Procaine has pK_aH ≈ 8.9. Find the pH at which it is 90% protonated.

Set up the ratio first: 90% protonated means [B]/([B] + [BH⁺]) = 0.10, so [B]/[BH⁺] = 0.10/0.90 = 0.111.

Formula:

$pH = p K aH + log 10 [B] [BH +] = 8.9 + log 10 (0.111)$

Compute:

$log 10 (0.111) = -0.954 \Rightarrow pH = 8.9 - 0.95 = 7.95 \approx 8.0$

Answer: procaine is 90% protonated at pH ≈ 8.0. Since pK_aH - 1 = 7.9 gives 91% protonated, the "one pH unit ≈ 10:1" rule checks out.

Example 3: The bicarbonate buffer ratio at blood pH

Using pK_a ≈ 6.1 for carbonic acid, find [HCO₃^-]/[H₂CO₃] at pH 7.4.

Formula (acid form):

$pH = p K a + log 10 [HCO 3 -] [H 2 CO 3]$

Substitute:

$7.4 = 6.1 + log 10 [HCO 3 -] [H 2 CO 3] \Rightarrow log 10 [HCO 3 -] [H 2 CO 3] = 1.3$

$[HCO 3 -] [H 2 CO 3] = 10 1.3 \approx 20$

Answer: about 20 bicarbonate ions for every carbonic acid molecule — the buffer is poised to absorb acid. This ratio is why a modest addition of H⁺ is resisted: HCO₃^- is the abundant partner ready to consume protons.

Key takeaways

Henderson–Hasselbalch for bases: pH = pK_aH + log₁₀([B]/[BH⁺]); for acids: pH = pK_a + log₁₀([A^-]/[HA]).
pH < pK_aH → protonated form dominates; pH > pK_aH → free base dominates; pH = pK_aH → 50/50.
Each 1.0 pH unit changes the ratio 10-fold (2 units → 100-fold).
At pH 7.4, amines with pK_aH ≳ 9 are >99% protonated; drugs like lidocaine (7.9) and procaine (8.9) exist in both forms — the balance is why they work.
Only the uncharged form crosses lipid membranes; charged salts are water-soluble and trapped.
Bicarbonate buffer: pH = 6.1 + log₁₀([HCO₃^-]/[H₂CO₃]); ratio ≈ 20:1 at pH 7.4.
Histidine (≈ 6.0) and lysine (≈ 10.5) side chains change charge with pH — central to protein behavior.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

Write the Henderson–Hasselbalch equation for an amine base and identify each symbol.
Show answer
pH = pK_aH + log₁₀([B]/[BH⁺]), where B is the free base, BH⁺ the protonated ammonium form, and pK_aH the pK_a of the conjugate acid.
A drug has pK_aH = 8.5. At pH 6.5, is it mostly protonated or mostly free base? By what ratio?
Show answer
Mostly protonated: pH (6.5) is 2 units below pK_aH (8.5), so [BH⁺]/[B] = 10² = 100:1 — about 99% protonated.
What fraction of a base with pK_aH = 6.0 is protonated at pH 7.0?
Show answer
pH 7.0 = 6.0 + log₁₀([B]/[BH⁺]) → log₁₀([B]/[BH⁺]) = 1.0 → [B]/[BH⁺] = 10, so fraction protonated = 1/(1+10) = 0.091 — about 9%.
Why must a local anesthetic exist partly as the free base to reach a nerve?
Show answer
The charged ammonium form cannot cross the lipid membrane; the neutral free base diffuses to the nerve, where protonation (favored at the lower local pH) regenerates the active form.
What is the [HCO₃^-]/[H₂CO₃] ratio at pH 7.4 (carbonic acid pK_a ≈ 6.1)?
Show answer
log₁₀([HCO₃^-]/[H₂CO₃]) = 7.4 - 6.1 = 1.3, so the ratio ≈ 20:1.
Lysine's side-chain pK_aH ≈ 10.5. At pH 7.4, is the side chain protonated or not?
Show answer
Protonated: pH 7.4 is ~3 units below 10.5, so the ammonium form dominates by ~1000:1.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Henderson–Hasselbalch equation: pH = pK_aH + log₁₀([B]/[BH⁺])
pKaH: pK_a of the conjugate acid of a base
Free base: Neutral, unprotonated amine (B)
Ammonium salt form: Protonated, charged form (BH⁺)
Alkaloid: Plant-derived basic nitrogen compound (morphine, nicotine, quinine)
Neurotransmitter: Signaling molecule, many are amines (dopamine, serotonin, histamine)
Bicarbonate buffer: H₂CO₃/HCO₃^- pair with pK_a ≈ 6.1
Membrane permeability: Ability to cross lipid bilayers

Sources & references

openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.

In 30 seconds

Why this matters

The college version

Core Concepts

Deriving the equation for an amine base

Reading the equation: three regimes

Biological amine pKaH reference values

Alkaloid isolation by acid–base extraction

The bicarbonate buffer (chemistry illustration)

Common Confusions

The same idea, in plain words

Worked example

Example 1: Lidocaine at blood pH — what fraction is active?

Example 2: At what pH is procaine 90% protonated?

Example 3: The bicarbonate buffer ratio at blood pH

Key takeaways

Check yourself

Keep learning

Key vocabulary

Related lessons

Sources & references

Biological amine pK_aH reference values