Topic 8.9 Notes – Henderson-Hasselbalch Equation

1. The Henderson-Hasselbalch Equation

A buffer contains a weak acid and its conjugate base (or weak base and its conjugate acid). The Henderson-Hasselbalch equation gives you the pH directly:

$$\textbf{pH = pKa + log}\left(\frac{[A^{-}]}{[HA]}\right)$$

• $HA$ = weak acid
• $A^{-}$ = conjugate base
• $pKa = -\log(Ka)$
• $pH = -\log[H^{+}]$

This comes from the equilibrium:

$$\ce{HA <=> H+ + A^{-}}$$

You do not need to derive it for AP. Just know how to use it and when it applies.

Big picture:
The pH of a buffer depends on:

• The strength of the acid (its pKa)
• The ratio $[A^{-}]/[HA]$

It depends on the ratio, not the absolute amounts.

2. What Each Part Means

pKa sets the baseline

Think of pKa as the “center point” of the buffer.

• Lower pKa → stronger acid → lower buffer pH
• Higher pKa → weaker acid → higher buffer pH
• If $[A^{-}] = [HA]$, then log(1) = 0 → pH = pKa

That equal-concentration point is where the buffer is most balanced.

The ratio controls direction

The log term tells you how far you move away from pKa.

• If $[A^{-}] > [HA]$ → log positive → pH > pKa
• If $[A^{-}] < [HA]$ → log negative → pH < pKa

Here’s the pattern that shows up on quizzes:

Every factor of 10 changes the pH by 1 unit. That shortcut saves time on multiple choice.

3. When You Can Use It

You must have both members of a conjugate pair present in significant amounts.

A. Direct buffer mixture

Example: mixing $\ce{HF}$ and $\ce{NaF}$.

If you’re given concentrations of the weak acid and its salt, plug directly into the equation. No ICE table needed.

B. During a titration before equivalence

If you titrate a weak acid with strong base:

$$\ce{HA + OH^{-} -> A^{-} + H2O}$$

Before the equivalence point:

• Some HA remains
• Some A⁻ has formed
• No excess OH⁻

That mixture is a buffer.

What you do:

1. Do stoichiometry first (subtract moles).
2. Find moles of HA and A⁻ after reaction.
3. Use the mole ratio directly in Henderson-Hasselbalch
   (since both are divided by the same total volume).

Students often forget step 1 and plug in initial amounts. That’s the most common mistake I see.

When NOT to use it

• At the start of a titration (no conjugate base yet)
• At equivalence point (only A⁻ present)
• After equivalence (excess strong base)
• Strong acid-strong base systems

If it’s not a buffer, this equation doesn’t apply.

4. Why Buffers Resist pH Change

The resistance comes from neutralization reactions.

Add strong acid:

$$\ce{A^{-} + H+ -> HA}$$

Add strong base:

$$\ce{HA + OH^{-} -> A^{-} + H2O}$$

In both cases:

• One component decreases slightly
• The other increases slightly
• The ratio changes only a little

Because pH depends on the log of the ratio, small ratio changes lead to small pH changes.

You are not expected to calculate the exact new pH after adding acid or base on the AP exam. You just need to explain why the change is small compared to pure water.

5. Buffer Effectiveness Range

Buffers work best when:

$$0.1 \le \frac{[A^{-}]}{[HA]} \le 10$$

That corresponds to:

$$\text{pH = pKa ± 1}$$

Outside that range, one component dominates and buffering weakens.

This range shows up in conceptual multiple-choice questions about choosing the best buffer for a target pH.