04 // Interactive Formula Compendium // CHEMISTRY-THERMOpH = pK_a + log₁₀([A⁻] / [HA])
Henderson-Hasselbalch Buffer pH Equation
The Henderson-Hasselbalch equation pH = pK_a + log₁₀([A⁻]/[HA]) relates solution pH to acid dissociation constant pK_a and the molar ratio of conjugate base [A⁻] to weak acid [HA].
Formula Reference
Main Formula
pH = pK_a + log₁₀([A⁻] / [HA])
Rearranged Forms
pK_a = −log₁₀(K_a)
[A⁻] / [HA] = 10^(pH − pK_a)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| pH | Buffer Solution pH | pH units |
| pKa | Acid Dissociation pK_a | dimensionless |
| base | Conjugate Base [A⁻] | mol/L |
| acid | Weak Acid [HA] | mol/L |
Inputs & Parameters
[dimensionless]
[mol/L]
[mol/L]
Calculated Result (Buffer Solution pH)
pH =4.936091pH units
Formula Notes & Background
- Derived directly by taking −log₁₀ of both sides of equilibrium expression K_a = [H⁺][A⁻]/[HA].
- Maximum buffer capacity β occurs when [A⁻] = [HA], where pH = pK_a exactly and effective buffering spans pK_a ± 1.
Step-by-Step Examples
#1: Acetate buffer prepared with 0.15 M sodium acetate ([A⁻]) and 0.10 M acetic acid ([HA], pK_a = 4.76).
pK_a = 4.76[A⁻] = 0.15 M[HA] = 0.10 M
pH = 4.76 + log₁₀(1.50) = 4.76 + 0.176
→ pH = 4.94
#2: Arterial blood bicarbonate buffer (pK_a = 6.10, [HCO₃⁻] = 24 mM, dissolved [CO₂] = 1.2 mM).
pK_a = 6.10Ratio = 24 / 1.2 = 20
pH = 6.10 + log₁₀(20) = 6.10 + 1.301
→ pH = 7.40 (Normal physiological arterial pH)
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Frequently Asked Questions
Why doesn't moderate dilution change a buffer's pH?+
Because both [A⁻] and [HA] share the same solution volume V, so the mole ratio n_base/n_acid stays invariant until ionic strength or water autoionization dominates.
What is the effective pH range of a buffer?+
Within pK_a ± 1 pH unit (corresponding to base-to-acid ratios from 1:10 to 10:1).
When does Henderson-Hasselbalch require exact quadratic correction?+
For very dilute solutions (< 1 mM) or very strong weak acids (pK_a < 2.5) where initial concentrations differ appreciably from equilibrium concentrations.