Henderson-Hasselbalch Equation- Khan Academy Guide

What Is the Henderson-Hasselbalch Equation?

The Henderson-Hasselbalch equation is a formula that connects pH, pKa, and the ratio of a weak acid to its conjugate base in solution. It's the workhorse of acid-base chemistry.

You use it when you need to:

• Calculate the pH of a buffer solution
• Determine how much acid or base to add to reach a target pH
• Understand why a buffer resists pH changes

That's it. No magic, no fluff. Just math.

The Equation

Here it is in its two common forms:

For acids:

pH = pKa + log([A⁻]/[HA])

For bases:

pOH = pKb + log([BH⁺]/[B])

In both cases, you're looking at the log of the ratio between the conjugate base form and the acid form.

Breaking Down the Components

pH — the measure of hydrogen ion concentration in solution

pKa — the negative log of the acid dissociation constant. Lower pKa means stronger acid.

[A⁻] — concentration of the conjugate base

[HA] — concentration of the weak acid

The ratio [A⁻]/[HA] tells you how much base you have relative to acid. When this ratio equals 1, log(1) = 0, and pH = pKa.

How to Actually Use This Thing

Step 1: Find Your pKa

Your pKa is either given in the problem or calculated from Ka:

pKa = -log(Ka)

If Ka = 1.8 × 10⁻⁵, then pKa = -log(1.8 × 10⁻⁵) = 4.74

Step 2: Plug In Your Concentrations

Let's say you have a buffer with 0.10 M acetic acid (HA) and 0.10 M acetate (A⁻). Ka for acetic acid is 1.8 × 10⁻⁵.

Step 1: pKa = -log(1.8 × 10⁻⁵) = 4.74

Step 2: Ratio = [A⁻]/[HA] = 0.10/0.10 = 1

Step 3: pH = 4.74 + log(1) = 4.74 + 0 = 4.74

When the concentrations are equal, pH equals pKa. This is a useful shortcut.

Step 3: Change the Ratio, Change the pH

What if you have more acetate than acetic acid? Say 0.15 M acetate and 0.10 M acetic acid.

Ratio = 0.15/0.10 = 1.5

pH = 4.74 + log(1.5) = 4.74 + 0.176 = 4.92

More base in the mix means higher pH. Makes sense.

pH vs. pKa: What the Numbers Mean

The relationship between pH and pKa determines which form dominates:

• pH < pKa — the solution is more acidic than the acid's pKa. The protonated form (HA) dominates.
• pH = pKa — equal amounts of HA and A⁻. This is the midpoint of a titration.
• pH > pKa — the solution is more basic. The deprotonated form (A⁻) dominates.

Buffers work best when pH is within ±1 of pKa. Outside that range, the buffer capacity drops fast.

Buffer Capacity: Why This Matters

A buffer resists pH changes, but it has limits. The buffer capacity depends on how much acid or base you can neutralize before the pH shifts significantly.

Two factors affect buffer capacity:

• Total concentration — higher concentrations = more buffering ability
• Ratio proximity to 1 — buffers work best when [A⁻] ≈ [HA]

A 1 M buffer can neutralize more acid than a 0.1 M buffer. Simple.

Common Mistakes to Avoid

• Using molarity instead of moles — the ratio matters, so you can use moles directly if the volume cancels out
• Forgetting the log — if you calculate the ratio but forget to take log₁₀, your answer will be way off
• Confusing Ka and pKa — Ka goes in the log, pKa replaces the log term entirely
• Using the wrong Ka for polyprotic acids — each proton has its own Ka and pKa. Use the one relevant to your titration stage.

Comparing Acid Strengths with Henderson-Hasselbalch

Here's how different acids behave at their pKa points:

The lower the pKa, the stronger the acid. Strong acids dissociate completely and don't follow Henderson-Hasselbalch — there's no meaningful [HA]/[A⁻] equilibrium to work with.

Getting Started: Worked Example

Problem: You need to prepare a buffer at pH 5.00 using acetic acid (pKa = 4.74). If you have 0.50 M acetic acid, how much sodium acetate do you need?

Step 1: Rearrange the equation to solve for the ratio

5.00 = 4.74 + log([A⁻]/[HA])

0.26 = log([A⁻]/[HA])

[A⁻]/[HA] = 10^0.26 = 1.82

Step 2: Solve for [A⁻]

[A⁻] = 1.82 × [HA] = 1.82 × 0.50 M = 0.91 M

You need 0.91 M sodium acetate in your buffer to hit pH 5.00.

When Henderson-Hasselbalch Breaks Down

This equation assumes:

• The acid is weak enough that you can ignore autoionization of water
• Concentrations are high enough that ionic strength effects are negligible
• Equilibrium concentrations ≈ initial concentrations (valid when the acid barely dissociates)

If you're working with very dilute solutions, very weak acids near their pKa, or strong acids, you need to solve the full equilibrium expression instead.

The Bottom Line

The Henderson-Hasselbalch equation is straightforward once you understand what each term represents. pKa is your reference point, the ratio tells you where you are relative to that point, and the log converts that ratio into pH units.

Practice with a few problems. The math is simple — the hard part is knowing when to apply it and which numbers to plug in. 📐