Henderson-Hasselbalch Calculator

Calculate buffer solution pH and understand acid-base equilibria

Henderson-Hasselbalch Calculator

Henderson-Hasselbalch Equation: pH = pKa + log([A⁻]/[HA])
Used to calculate the pH of buffer solutions and understand acid-base equilibria.

Notes:

  • • Buffer solutions resist pH changes when small amounts of acid or base are added
  • • Most effective buffering occurs within pH = pKa ± 1
  • • When pH = pKa, [A⁻] = [HA] (optimal buffer composition)
  • • Common buffers: acetate (pKa 4.76), phosphate (pKa 7.21), Tris (pKa 8.06)

Understanding the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental relationship in acid-base chemistry that relates the pH of a buffer solution to the pKa of the weak acid and the ratio of conjugate base to weak acid concentrations. It's one of the most important equations in chemistry and biochemistry.

The Henderson-Hasselbalch Equation

pH = pKa + log([A⁻]/[HA])

Buffer Solution pH Calculation

pH

Measure of acidity/basicity (−log[H⁺])

pKa

Acid dissociation constant (−log Ka), characteristic of each weak acid

[A⁻]

Concentration of conjugate base (deprotonated form)

[HA]

Concentration of weak acid (protonated form)

Derivation from Ka Expression

Starting with the acid dissociation equilibrium:

HA ⇌ H⁺ + A⁻

The equilibrium constant Ka:

Ka = [H⁺][A⁻] / [HA]

Rearranging for [H⁺]:

[H⁺] = Ka × [HA] / [A⁻]

Taking −log of both sides:

−log[H⁺] = −log(Ka) − log([HA]/[A⁻])

Since pH = −log[H⁺] and pKa = −log(Ka):

pH = pKa + log([A⁻]/[HA]) ✓

Key Relationships

When pH = pKa

[A⁻] = [HA]

Equal amounts of acid and conjugate base. Optimal buffer capacity.

When pH > pKa

[A⁻] > [HA]

More conjugate base present. Solution is more basic.

When pH < pKa

[HA] > [A⁻]

More weak acid present. Solution is more acidic.

Buffer Range

pH = pKa ± 1

Effective buffering region. Resists pH changes.

Practical Example: Acetate Buffer

Problem: What is the pH of a buffer containing 0.10 M acetic acid (CH₃COOH, pKa = 4.76) and 0.15 M sodium acetate (CH₃COO⁻Na⁺)?

Given:

  • pKa = 4.76
  • [HA] = [CH₃COOH] = 0.10 M
  • [A⁻] = [CH₃COO⁻] = 0.15 M

Solution:

pH = pKa + log([A⁻]/[HA])
pH = 4.76 + log(0.15/0.10)
pH = 4.76 + log(1.5)
pH = 4.76 + 0.18
pH = 4.94

Since pH > pKa, the solution contains more conjugate base than acid, making it slightly more basic than a 1:1 buffer would be.

Common Buffer Systems

BufferpKapH RangeUse
Acetate4.763.8 - 5.8Biochemistry
Phosphate7.216.2 - 8.2Physiological pH
Tris8.067.0 - 9.0Molecular biology
Carbonate10.339.3 - 11.3Basic solutions
Bicarbonate6.355.4 - 7.4Blood pH regulation

Applications

  • 🧬
    Biochemistry: Maintain optimal pH for enzyme activity and protein stability
  • 🩺
    Physiology: Blood pH regulation through bicarbonate-carbonic acid buffer system
  • ⚗️
    Analytical Chemistry: pH control for titrations, extractions, and chromatography
  • 💊
    Pharmaceuticals: Drug formulation and stability at specific pH values
  • 🧪
    Chemical Synthesis: Control reaction conditions and product selectivity
  • 🌊
    Environmental Science: Monitor and control water pH in ecosystems

📊Quick Reference

Equation:

pH = pKa + log([A⁻]/[HA])

Optimal Buffer:

pH = pKa ± 1 unit

Best Capacity:

When [A⁻] = [HA]

Common pKa:

Acetate: 4.76, Phosphate: 7.21

Level:

College General Chemistry

🔗Related Calculators

📐Related Formulas

🎯Where It's Used

  • 🧬

    Biochemistry

    Enzyme activity optimization

  • 🩺

    Medicine

    Blood pH regulation

  • 💊

    Pharmacology

    Drug formulation

  • ⚗️

    Analytical Chem

    pH-controlled reactions