Henderson-Hasselbalch Equation Explained- Khan Academy Guide

What the Henderson-Hasselbalch Equation Actually Is

Let's cut through the noise. The Henderson-Hasselbalch equation is a shortcut for calculating pH of buffer solutions. That's it. No mystical chemistry magic—just a rearranged version of the acid dissociation constant expression that saves you from messy logarithms.

If you're taking organic chemistry, biochemistry, or any lab science course, you'll encounter this equation constantly. Most students memorize it without understanding it. You won't be one of them after reading this.

The Formula

Here it is in its most common form:

pH = pKa + log([A⁻]/[HA])

That's the whole thing. Three variables. One equation. Here's what each piece means:

• pH — the acidity level of your solution (0-14 scale)
• pKa — the negative log of the acid dissociation constant (Ka). It's a measure of acid strength.
• [A⁻] — concentration of the conjugate base (the acid after it lost a proton)
• [HA] — concentration of the weak acid (the molecule still holding its proton)

Why This Equation Works

The equation comes from rearranging the Ka expression for weak acids:

Ka = [H⁺][A⁻]/[HA]

Take the negative log of both sides, and you get the Henderson-Hasselbalch equation. The pKa is just -log(Ka), and pH is -log[H⁺]. The math works itself out.

Understanding pKa and Why It Matters

pKa is the acid dissociation constant expressed on a logarithmic scale. Lower pKa means a stronger acid. Higher pKa means weaker.

When pH equals pKa, you're at the half-equivalence point. At this point, [A⁻] = [HA], so the log term becomes log(1) = 0. The pH equals the pKa. This is a useful reference point.

How to Use the Henderson-Hasselbalch Equation

Step-by-Step Calculation

Let's say you have a buffer with 0.1 M acetic acid (HA) and 0.1 M acetate (A⁻). The pKa of acetic acid is 4.76.

Step 1: Identify your values

• pKa = 4.76
• [A⁻] = 0.1 M
• [HA] = 0.1 M

Step 2: Calculate the ratio

[A⁻]/[HA] = 0.1/0.1 = 1

Step 3: Take the log

log(1) = 0

Step 4: Solve

pH = 4.76 + 0 = 4.76

When the concentrations are equal, pH equals pKa. This holds true every time.

A Different Example

Same buffer, but now you have 0.2 M acetate and 0.1 M acetic acid.

[A⁻]/[HA] = 0.2/0.1 = 2

log(2) = 0.301

pH = 4.76 + 0.301 = 5.06

The pH went up because you added more conjugate base. That's the buffer effect in action.

Common Mistakes to Avoid

• Using the wrong concentrations. Plug in molarity, not moles. If your solution volume changes, recalculate molarity.
• Confusing the conjugate base with the weak acid. [A⁻] is always the deprotonated form. [HA] is always the protonated form.
• Forgetting that this only works for buffer solutions. It gives approximate results for strong acid-strong base mixtures too, but the math breaks down when you're far from the buffer region.
• Misplacing the negative sign. pKa is already negative log. Don't add another negative sign somewhere else.
• Using molarity when you should use activity. In precise work, concentrations aren't enough. But for most undergraduate labs, concentration works fine.

When to Use This Equation

The Henderson-Hasselbalch equation is your go-to when:

• Designing buffer solutions for experiments
• Calculating pH changes during titrations
• Understanding drug absorption (pH partition hypothesis)
• Analyzing blood buffer systems
• Working with amino acids and proteins

Comparing pH Calculation Methods

Quick Reference

• pH = pKa + log([base]/[acid])
• When [base] = [acid], pH = pKa
• Base:acid ratio of 10:1 → pH = pKa + 1
• Base:acid ratio of 1:10 → pH = pKa - 1
• Buffer capacity is highest when [base] = [acid]

Bottom Line

The Henderson-Hasselbalch equation is a tool. It works when you understand what goes into it and when to apply it. Buffer solutions, titration curves, physiological pH—these are its domain.

Most students fail not because the equation is hard, but because they don't know what pKa represents or when the equation stops working. Now you do.