Organic Chemistry · Carboxylic Acids and Nitriles
Biological Acids and the Henderson–Hasselbalch Equation
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In 30 seconds
Biology runs on weak acids. Lactic acid builds up in working muscle, pyruvic acid sits at the crossroads of metabolism, citric acid powers the Krebs cycle, and carbonic acid/bicarbonate keeps human blood near pH 7.4. Each exists in equilibrium with its conjugate base The species A- left after the acid loses H+. Full entry →, and the balance between acid and carboxylate is set by the pH of the surrounding fluid. The Henderson–Hasselbalch equation,
pH = pKa + log[A-][HA]
is a rearranged form of the acidity constant Ka that lets you compute what fraction of a weak acid An acid that only partially dissociates in water, with an equilibrium Ka. Full entry → is ionized at any pH. This topic derives it, shows how to use it with dimensional analysis, and applies it to biological buffers.
Why this matters
The Henderson–Hasselbalch equation is one of the few formulas from organic chemistry that appears daily in medicine, pharmacy, and biology. It explains why blood pH stays in a narrow band (about 7.35–7.45) despite constant production of carbon dioxide and lactic acid; why aspirin, a weak acid with pKa -logKa; smaller pKa means a stronger acid. Full entry → 3.5, is absorbed in the acidic stomach; why amino acids carry different charges at different pH; and why acidosis or alkalosis is dangerous. On exams, it is the reliable way to answer "what fraction is ionized?" questions.
The college version
Core Concepts
The weak-acid equilibrium and Ka
A weak acid HA (such as RCOOH) dissociates in water:
HA ⇌ H+ + A-
with acidity constant
Ka = [H+][A-][HA]
where brackets mean molar concentrations (mol/L). The pKa is -logKa, so a smaller pKa means a stronger acid. Simple carboxylic acids have pKa values near 4–5.
Deriving the Henderson–Hasselbalch equation
Take the negative logarithm of both sides of the Ka expression:
Ka = [H+][A-][HA]
-logKa = -log[H+] - log[A-][HA]
Substitute pKa = -logKa and pH = -log[H+], then rearrange:
pH = pKa + log[A-][HA]
The pH depends on the acid's pKa and the base/acid ratio. When [A-] = [HA], the log term is log1 = 0, so pH = pKa: at pH = pKa, a weak acid is exactly 50% ionized.
Reading the ratio: buffer region and beyond
When the pH is within about one unit of the pKa (the buffer A solution of a weak acid and its conjugate base that resists pH change. Full entry → region), both the acid and its conjugate base are present in significant amounts and the solution resists pH change — that is what a buffer is. If pH is more than about two units above pKa, the log ratio is ≥ 100 and the acid is essentially fully deprotonated; more than two units below, it is essentially fully protonated.
Biological acids and their pKa values
Standard reference pKa values: lactic acid 3.86, pyruvic acid 2.50, citric acid 3.13/4.76/6.39 (three successive protons), acetic acid 4.76, carbonic acid 6.35 (first proton). These explain biology: at blood pH 7.4, lactic acid (pKa 3.86) is mostly lactate anion plus protons; citric acid's three pKa values give it a mixture of charges in the cell; carbonic acid's 6.35 sits close enough to blood pH to buffer it.
The bicarbonate buffer of blood
Blood is buffered by the carbon dioxide/bicarbonate system:
CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3-
For this system, the apparent pKa used in practice is about 6.1. The working form is
pH = 6.1 + log[HCO3-](0.03 × PCO2)
where [HCO3-] is in mmol/L and PCO2 in mmHg. In healthy blood, bicarbonate is about 24 mmol/L and arterial PCO2 about 40 mmHg, giving the famous 20:1 base-to-acid ratio that holds pH near 7.4. This is why hyperventilation (lowering PCO2) raises blood pH and why CO₂ retention lowers it.
Predicting the charge on a biological acid
Compute pH - pKa: if it is positive by more than ~2 units, the acid is >99% in the A- form (negatively charged); if negative by more than ~2, it is >99% HA (neutral). This is how drug absorption is reasoned: aspirin (pKa 3.5) is neutral in the acidic stomach (pH ~2), so it crosses membranes as the uncharged form and is absorbed.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| pH | pKa | pH is the acidity of the solution; pKa is the intrinsic strength of the acid. They are equal only at 50% ionization. |
| Henderson–Hasselbalch for weak acids | Strong-acid pH calculations | Strong acids dissociate completely; the equilibrium ratio logic does not apply. |
| log([A-]/[HA]) | log([HA]/[A-]) | Inverting the ratio flips the sign; pH above pKa means more base, so the log term is positive. |
| Concentration units | Ratio | The log term needs the same units in numerator and denominator; mixing M with mM ruins the answer. |
| "Lactic acidosis" acid | Lactate | At pH 7.4, lactic acid (pKa 3.86) is >99% lactate; the problem is the co-produced protons, not the neutral acid. |
| Buffer region | Full neutralization | Buffering works within about ±1 pH unit of pKa; far outside, the solution behaves like a dilute strong acid/base. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A weak acid is like a see-saw between two teams: the "acid team" (HA) and the "base team" (A⁻). The pKa is the see-saw's balance point. If the pH is exactly at the pKa, both teams have the same number of players — 50/50. Move the pH one unit higher and the base team gets 10 times more players; the Henderson–Hasselbalch equation is just the scoreboard that tells you the team sizes.
Worked examples
A buffer is prepared with 0.10 M lactic acid and 0.25 M sodium lactate (the conjugate base). Lactic acid has pKa = 3.86. Write the Henderson–Hasselbalch equation, then substitute:
pH = pKa + log[A-][HA] = 3.86 + log0.250.10
The ratio is 0.25/0.10 = 2.5, and log2.5 = 0.40, so:
pH = 3.86 + 0.40 = 4.26
The buffer sits about 0.4 pH units above the pKa because the base form is 2.5 times more concentrated than the acid form. Note that only the ratio matters — diluting both by the same factor leaves the pH unchanged.
Acetic acid has pKa = 4.76, and the pH is exactly 1.00 unit above it. Substitute:
5.76 = 4.76 + log[A-][HA]
log[A-][HA] = 1.00 ⇒ [A-][HA] = 101 = 10
There is 10 times more acetate than acetic acid, so the fraction ionized is 10/(10+1) = 0.91, or 91%. At pH one unit below the pKa, the acid form dominates 10:1 and only about 9% is ionized; each pH unit above pKa adds one power of 10 to the ratio.
A patient's arterial blood has [HCO3-] = 24 mmol/L and PCO2 = 40 mmHg. First convert the CO₂ term to a concentration using the solubility factor 0.03 mmol/L per mmHg:
[dissolved CO2] = 0.03 mmol/LmmHg × 40 mmHg = 1.2 mmol/L
The units cancel (mmHg cancels mmHg), leaving mmol/L. Now substitute:
pH = 6.1 + log241.2 = 6.1 + log20 = 6.1 + 1.30 = 7.40
The healthy 20:1 ratio produces pH 7.40. If the patient hypoventilates and PCO2 rises to 60 mmHg, dissolved CO₂ becomes 1.8 mmol/L, the ratio drops to 24/1.8 = 13.3, log13.3 = 1.12, and pH falls to 7.22 — a respiratory acidosis. This is how a blood-gas report becomes a pH prediction.
Key takeaways
- Henderson–Hasselbalch: pH = pKa + log([A-]/[HA]); it is just Ka rearranged.
- At pH = pKa, the acid is 50% ionized; each pH unit away from pKa changes the base/acid ratio by a factor of 10.
- Simple carboxylic acids: pKa 4–5 (acetic 4.76, lactic 3.86, pyruvic 2.50); carbonic acid pKa₁ = 6.35.
- Blood buffer: pH = 6.1 + log([HCO₃⁻]/(0.03 × PCO₂)); healthy ratio ≈ 20:1 → pH ≈ 7.4.
- The equation applies only to weak acid/conjugate base pairs near equilibrium, not to strong acids.
- A weak-acid drug is absorbed best where it is neutral: aspirin (pKa 3.5) is neutral in the acidic stomach.
- Check units before substituting: molar concentrations for the general form; mmol/L and mmHg for the blood form.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Write the Henderson–Hasselbalch equation and state the condition under which pH = pKa.
Show answer
pH = pKa + log([A-]/[HA]); pH = pKa when [A-] = [HA] (the log term is zero), i.e., at 50% ionization.
A buffer contains 0.20 M propanoic acid (pKa 4.87) and 0.20 M sodium propanoate. What is its pH?
Show answer
Since the concentrations are equal, the log term is log1 = 0, so pH = pKa = 4.87.
At what pH is acetic acid (pKa 4.76) 50% ionized, and what fraction is ionized at pH 6.76?
Show answer
50% ionized at pH = pKa = 4.76. At pH 6.76 (two units above pKa), the base/acid ratio is 100, so the fraction ionized is 100/101 ≈ 0.99, or ~99%.
A patient has [HCO₃⁻] = 24 mmol/L and PCO₂ = 48 mmHg. Compute pH and state the direction of the disturbance.
Show answer
Dissolved CO₂ = 0.03 × 48 = 1.44 mmol/L; ratio = 24/1.44 = 16.7; log 16.7 = 1.22; pH = 6.1 + 1.22 = 7.32. This is below 7.35, consistent with a respiratory acidosis from CO₂ retention.
Why does the bicarbonate buffer The CO₂/HCO₃⁻ system that buffers blood at pH ~7.4. Full entry → use the factor 0.03, and what units must PCO₂ have?
Show answer
The factor 0.03 converts mmHg of CO₂ partial pressure into mmol/L of dissolved CO₂ (the solubility of CO₂ in plasma); PCO₂ must therefore be in mmHg for the units to cancel.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- weak acid
- An acid that only partially dissociates in water, with an equilibrium Ka.
- Kₐ
- The equilibrium constant for acid dissociation: [H+][A-]/[HA].
- pKa
- -logKa; smaller pKa means a stronger acid.
- buffer
- A solution of a weak acid and its conjugate base that resists pH change.
- conjugate base
- The species A- left after the acid loses H+.
- bicarbonate buffer
- The CO₂/HCO₃⁻ system that buffers blood at pH ~7.4.
Sources & references
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