Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is essential for calculating the pH of buffer solutions. It relates pH to pKa and the ratio of conjugate base to weak acid concentrations.

The Henderson-Hasselbalch Equation

pH = pKa + log([A⁻]/[HA])

Buffer pH from acid-base ratio

Variable Definitions

pH = Measure of Acidity

Units: Unitless

The pH of the buffer solution

pKa = Acid Dissociation Constant

Formula: pKa = -log(Ka)

Meaning: Strength of the weak acid

💡 Lower pKa = stronger acid

[A⁻] = Conjugate Base Concentration

Units: M (mol/L)

Concentration of the deprotonated form (base)

[HA] = Weak Acid Concentration

Units: M (mol/L)

Concentration of the protonated form (acid)

When to Use This Equation

✅ Good For:

  • Buffer solutions
  • Weak acid/conjugate base pairs
  • pH close to pKa (±1 unit)
  • Comparable amounts of acid and base

❌ Not For:

  • Strong acids or bases
  • Very dilute solutions
  • When [HA] or [A⁻] is very small
  • Non-buffer systems

Special Cases

When [A⁻] = [HA]

pH = pKa

log(1) = 0, so pH equals pKa when concentrations are equal

When [A⁻] > [HA]

pH > pKa

More base than acid → more basic solution

When [HA] > [A⁻]

pH < pKa

More acid than base → more acidic solution

Step-by-Step Example

Problem: Calculate pH of a buffer with 0.10 M acetic acid (pKa = 4.76) and 0.15 M acetate ion.

Given:

  • pKa = 4.76
  • [HA] = 0.10 M (acetic acid)
  • [A⁻] = 0.15 M (acetate)

Step 1: Write the equation

pH = pKa + log([A⁻]/[HA])

Step 2: Substitute values

pH = 4.76 + log(0.15/0.10)

Step 3: Calculate the ratio

pH = 4.76 + log(1.5) = 4.76 + 0.18 = 4.94

Answer: pH = 4.94

Since [A⁻] > [HA], pH is slightly above pKa ✓

Common Mistakes

❌ Flipping the ratio

It's [A⁻]/[HA] (base/acid), NOT [HA]/[A⁻]. Getting this backwards gives wrong pH.

❌ Using Ka instead of pKa

The equation requires pKa, not Ka. If given Ka, calculate pKa = -log(Ka) first.

❌ Using natural log (ln)

Use log (base 10), not ln (natural log). Make sure calculator is in log mode.

❌ Using with strong acids

This equation is ONLY for weak acid/conjugate base buffers. Don't use for HCl, NaOH, etc.

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Frequently Asked Questions

What is the Henderson-Hasselbalch equation?

pH = pKa + log([A⁻]/[HA]). It calculates buffer pH from the pKa and the ratio of conjugate base to weak acid.

When do I use Henderson-Hasselbalch?

Use it for buffer solutions containing a weak acid and its conjugate base, when pH is within ±1 of pKa.

What happens when pH equals pKa?

When pH = pKa, the concentrations of acid and conjugate base are equal ([A⁻] = [HA]). This is the optimal buffering point.

Can I use molar ratios instead of concentrations?

Yes! If the acid and base are in the same solution, you can use mole ratios: pH = pKa + log(moles A⁻/moles HA).

Why doesn't it work for strong acids?

Strong acids completely dissociate, so there's no equilibrium between HA and A⁻. The equation assumes weak acid equilibrium.

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