Chemistry · Acid-Base Chemistry

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation calculates the pH value of buffer solutions from the pKa of the acid and the concentration ratio.

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Formula

LaTeX: pH = pK_a + \log \frac{[A^-]}{[HA]}
pH dimensionless · pKa dimensionless · concentrations in mol/L

Variables & units – Henderson-Hasselbalch Equation

SymbolMeaningUnit
pHNegative base-10 logarithm of the H⁺ concentrationdimensionless
pKaNegative logarithm of the acid constantdimensionless
[A⁻]Concentration of the conjugate basemol/L
[HA]Concentration of the weak acidmol/L

Derivation & background – Henderson-Hasselbalch Equation

Derived from the law of mass action for weak acids: Ka = [H⁺][A⁻]/[HA] → [H⁺] = Ka × [HA]/[A⁻] → pH = pKa + log([A⁻]/[HA]).

Exam blueprint

Validity range

Applies to buffers of a weak acid and conjugate base, especially near pH ≈ pK_a.

Derivation steps

The equation follows directly from the acid constant and taking logarithms.

  1. 1K_a = [H⁺][A⁻]/[HA].
  2. 2Solve for [H⁺] and apply -log to get pH = pK_a + log([A⁻]/[HA]).

Rearrangements

Concentration ratio

At pH = pK_a, acid and base have equal concentration.

Task variant

pH is one unit above pK_a. What is [A⁻]/[HA]?

10^1 = 10; the base concentration is ten times the acid concentration.

Common mistakes

Swapping numerator and denominator.

The logarithm uses base/acid: [A⁻]/[HA].

Exam context

  • Typical in buffer preparation, dilution and blood-pH tasks.

These mistakes cost points in real exams. The set drills them until they stick.

Worked example

An acetic acid buffer (pKa = 4.75), [CH₃COO⁻] = 0.1 mol/L, [CH₃COOH] = 0.05 mol/L: pH = 4.75 + log(0.1/0.05) = 4.75 + 0.30 = 5.05.

Applications

Biochemistry (blood buffer pH 7.4), pharmacy (formulations), analytical chemistry, food chemistry

Quanta exam set

Curated exam set for "Henderson-Hasselbalch Equation":

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Which formula describes Henderson-Hasselbalch Equation?

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How do you rearrange pH = pKa + log([A⁻]/[HA]) for Concentration ratio?

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Which common mistake happens with Henderson-Hasselbalch Equation?

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Scientific sources

Common notations & search queries

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Frequently asked questions about Henderson-Hasselbalch Equation

How do you calculate the pH of a buffer with Henderson-Hasselbalch?+

Add to the pK_a of the acid the base-10 logarithm of the concentration ratio of conjugate base to acid: pH = pK_a + log([A⁻]/[HA]). Insert the concentrations in mol/L; only their ratio matters, not their absolute value. Example acetic acid buffer with pK_a = 4.75, [CH₃COO⁻] = 0.1 mol/L and [CH₃COOH] = 0.05 mol/L: pH = 4.75 + log(0.1/0.05) = 4.75 + 0.30 = 5.05. In the logarithm it is always base divided by acid, not the other way round. If both concentrations are equal, the logarithm is zero and the pH equals the pK_a.

Why does pH equal pK_a when acid and base have equal concentration?+

Because then the concentration ratio [A⁻]/[HA] is exactly one and the logarithm of one gives zero. In pH = pK_a + log([A⁻]/[HA]) the second term drops out, so pH = pK_a remains. This point is called the half-equivalence point and is especially important: there the buffering effect is greatest, because small additions of acid or base barely shift the pH. For this reason the pK_a of an acid is determined experimentally by reading the pH at the half-equivalence point of a titration. A good buffer works within about one pH unit around its pK_a.

When may you not use the Henderson-Hasselbalch equation?+

The equation is an approximation and works well near pH ≈ pK_a, that is in the actual buffer range, about one pH unit around the pK_a. It fails for very dilute solutions and at the edges, when one component is almost completely consumed. For strong acids or bases that dissociate completely it is also not applicable, because there is no equilibrium between acid and conjugate base. The self-ionization of water and the use of activity instead of concentration are also neglected. For standard buffer problems in chemistry class, however, it gives very reliable results.

How does the pH change when the base-to-acid ratio becomes ten times larger?+

Because the logarithm contains the ratio [A⁻]/[HA], a factor of ten shifts the pH up by exactly one unit, since log(10) = 1, so pH = pK_a + 1. If the ratio rises by a factor of 100, the pH rises by two units. Conversely a tenfold excess of acid lowers the pH by one unit below the pK_a. This logarithmic relationship means that large concentration changes cause only moderate pH changes, which explains the buffering effect. A buffer keeps the pH stable as long as both components are present in comparable amounts and neither dominates by more than a factor of ten.

How does a blood buffer work according to Henderson-Hasselbalch?+

Blood uses mainly the carbonic acid-bicarbonate system as a buffer. Here HA is carbonic acid H₂CO₃ and A⁻ is bicarbonate HCO₃⁻, with an effective pK_a of about 6.1. The physiological pH of 7.4 is reached because the ratio of bicarbonate to dissolved CO₂ is roughly 20 to 1: pH = 6.1 + log(20) ≈ 6.1 + 1.3 = 7.4. The body regulates this buffer dynamically: breathing controls the CO₂ content, the kidneys control the bicarbonate. As a result the blood pH stays within a very narrow, vital range despite metabolic fluctuations.

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How do you calculate with Henderson-Hasselbalch Equation?

Here is how to work through a typical Henderson-Hasselbalch Equation (pH = pKa + log([A⁻]/[HA])) task step by step:

  1. 1

    Task

    pH is one unit above pK_a. What is [A⁻]/[HA]?

    Solution path

    10^1 = 10; the base concentration is ten times the acid concentration.