Henderson-Hasselbalch Calculators

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The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of its conjugate base ([A⁻]) to the weak acid ([HA]). It is derived from the acid dissociation equilibrium and is the fundamental equation for designing buffers, calculating the degree of ionization of weak acids and bases at any pH, predicting the form (protonated or deprotonated) of amino acid side chains, and understanding blood pH regulation. The equation is most accurate when pH is within ±2 pH units of the pKa.

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Henderson-Hasselbalch Equation

pH = pKa + log([A⁻]/[HA])

Rearranged: [A⁻]/[HA] = 10^(pH − pKa)

At pH = pKa: log(1) = 0; [A⁻] = [HA] (50% dissociated). At pH = pKa + 1: [A⁻]/[HA] = 10 (91% dissociated). At pH = pKa − 1: [A⁻]/[HA] = 0.1 (9% dissociated).

Buffer Design Applications

Choose a buffer with pKa within ±1 pH unit of your target pH. Calculate ratio: [A⁻]/[HA] = 10^(pH − pKa). Example: prepare pH 6.0 buffer using acetate (pKa = 4.75): ratio = 10^(6.0 − 4.75) = 10^1.25 = 17.8. Mix sodium acetate:acetic acid = 17.8:1 → at this ratio, pH ≈ 6.0. Note: pH 6.0 is 1.25 units above the pKa — borderline of the effective buffering range (±1 unit). A better choice for pH 6.0 is MES (pKa 6.15) or citrate/phosphate.

Amino Acid Ionization

Predict ionization state at physiological pH 7.4: His side chain (imidazole, pKa ~6.0): [A⁻]/[HA] = 10^(7.4 − 6.0) = 10^1.4 = 25; 96% neutral, 4% protonated (cationic) at pH 7.4. Lys (pKa ~10.5): [A⁻]/[HA] = 10^(7.4 − 10.5) = 10^(−3.1) = 0.00079; >99.9% protonated (cationic) at physiological pH.

Limitations

Assumes ideal dilute solution; concentration effects at high ionic strength require activity corrections; less accurate when [A⁻] or [HA] is very low relative to Kw.

Glossary

Henderson-Hasselbalch Equation
pH = pKa + log([A⁻]/[HA]); relates buffer pH to pKa and the conjugate base:acid ratio; at pH = pKa, 50% is ionized; effective buffer range is pKa ± 1 pH unit.
Buffer Design
Select weak acid with pKa within ±1 pH unit of target pH; [A⁻]/[HA] = 10^(pH−pKa); mix acid and conjugate base in this ratio; verify and fine-tune with pH meter.
Ionization Fraction
Fraction of weak acid in deprotonated form: 1/(1 + 10^(pKa−pH)); at pH = pKa, 50% ionized; 1 pH unit above pKa = ~91% ionized; 1 pH unit below = ~9% ionized.

Frequently Asked Questions

Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). Derivation: from the acid dissociation equilibrium Ka = [H⁺][A⁻]/[HA]; solve for [H⁺]: [H⁺] = Ka × [HA]/[A⁻]; take −log of both sides: pH = −log(Ka) − log([HA]/[A⁻]) = pKa + log([A⁻]/[HA]). The equation is valid for any weak acid/conjugate base pair and tells you: (1) At what pH the acid is half-dissociated (pH = pKa). (2) What ratio of conjugate base to acid to use for a target pH. (3) What fraction of a weak acid is ionized at any given pH.

Steps: (1) Choose a weak acid with pKa within ±1 pH unit of target pH. (2) Calculate [A⁻]/[HA] ratio = 10^(target pH − pKa). (3) Mix the weak acid and its conjugate base (sodium salt) in this ratio at the desired total buffer concentration. (4) Check and adjust pH with calibrated pH meter; fine-tune with HCl or NaOH. Example: HEPES buffer at pH 7.5; pKa = 7.48. Ratio = 10^(7.5−7.48) = 10^0.02 = 1.047. Mix HEPES free acid : sodium HEPES ≈ 1:1.047 ≈ almost equal amounts. Total concentration 50 mM: ~24.4 mM acid + ~25.6 mM base.

Fraction ionized = [A⁻]/([A⁻] + [HA]) = 1/(1 + 10^(pKa−pH)). Fraction protonated = [HA]/([A⁻] + [HA]) = 1/(1 + 10^(pH−pKa)). Example: aspirin (acetylsalicylic acid, pKa = 3.5) in the stomach (pH 1.5): fraction ionized = 1/(1 + 10^(3.5−1.5)) = 1/(1 + 100) = 0.0099 ≈ 1%. So 99% is in the protonated (uncharged, lipid-soluble) form → absorbed across gastric mucosa. At intestinal pH 7.0: fraction ionized = 1/(1 + 10^(3.5−7.0)) = 1/(1 + 10^(−3.5)) = 1/1.000316 ≈ >99% ionized → poor absorption through lipid membranes.

Blood pH regulation uses the bicarbonate buffer system: CO₂(dissolved) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻. Henderson-Hasselbalch: pH = 6.1 + log([HCO₃⁻] / 0.03 × PCO₂). At normal blood values: [HCO₃⁻] = 24 mEq/L; PCO₂ = 40 mmHg; [CO₂]_dissolved = 0.03 × 40 = 1.2 mM. pH = 6.1 + log(24/1.2) = 6.1 + log(20) = 6.1 + 1.301 = 7.40. Lungs control PCO₂ (acid component); kidneys control [HCO₃⁻] (base component). In respiratory acidosis (elevated PCO₂): denominator increases → ratio decreases → pH falls. In metabolic alkalosis (elevated [HCO₃⁻]): numerator increases → ratio increases → pH rises.