Henderson-Hasselbalch Equation

AP Chemistry· 12 min read

1. Derivation of the Henderson-Hasselbalch Equation★★☆☆☆⏱ 3 min

The equation is derived directly from the standard weak acid dissociation equilibrium expression, rearranged to isolate pH terms. No advanced math is required, and AP exam graders accept the full derivation for free-response points.

🔬 Derivation

Goal:

Derive Henderson-Hasselbalch from weak acid dissociation

Starting from:

Weak acid dissociation: $H A ⇌ H^{+} + A^{-}$

1
Write the Ka expression for the equilibrium:
2
$K_a = \frac{[H^+][A^-]}{[HA]}$
3
Rearrange to isolate the hydrogen ion concentration:
4
$[H^+] = K_a \times \frac{[HA]}{[A^-]}$
5
Take the negative base-10 logarithm of all terms:
6
$-\log[H^+] = -\log(K_a) - \log\left(\frac{[HA]}{[A^-]}\right)$
7
Substitute pH for $- lo g [H^{+}]$ and pKa for $- lo g K_{a}$ , then use logarithm rules to flip the fraction sign:

Result:

The final standard form of the equation for weak acid buffers is $p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$

📘 Definition

Henderson-Hasselbalch Equation

$p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$

An approximate formula that relates the pH of a weak acid buffer system to the pKa of the acid and the ratio of conjugate base to weak acid concentrations

✓ Quick check

Test your understanding of the derivation:

What mathematical operation is applied to both sides of the Ka expression to get the final pH form?
- Take natural logarithm
- Take negative base-10 logarithm
- Square both sides
- Multiply by [H+]
Reveal answer
Take negative base-10 logarithm —
This step converts the raw equilibrium constant and concentration terms to the familiar pH and pKa scales.

2. Calculating Buffer pH Using the Equation★★★☆☆⏱ 3 min

For most AP problems, you can use the initial molar concentrations of the weak acid and conjugate base directly in the ratio term, rather than calculating exact equilibrium concentrations. This approximation is valid for buffers where dissociation of HA is negligible.

📐 Worked Example

Calculate the pH of a buffer solution containing 0.25 M acetic acid (pKa = 4.76) and 0.15 M sodium acetate.

1
Identify the known values: pKa of acetic acid = 4.76, [HA] = 0.25 M, [A⁻] = 0.15 M
2
Substitute values directly into the Henderson-Hasselbalch equation:
3
$pH = 4.76 + \log\left(\frac{0.15}{0.25}\right)$
4
Calculate the log term: $lo g (0.6) = - 0.22$
5
Sum the terms to get final pH:
6
$pH = 4.76 - 0.22 = 4.54$

3. Calculating Target Buffer Ratios★★★☆☆⏱ 3 min

You can rearrange the Henderson-Hasselbalch equation to solve for the required ratio of conjugate base to weak acid if you are given a target pH and known pKa. This is the standard method for designing a buffer for a specific lab use case.

📐 Worked Example

What ratio of [A⁻] to [HA] is required to make a buffer with pH 5.2, using a weak acid with pKa = 4.9?

1
Rearrange the Henderson-Hasselbalch equation to isolate the log term:
2
$pH - pK_a = \log\left(\frac{[A^-]}{[HA]}\right)$
3
Substitute known values: $5.2 - 4.9 = 0.3$
4
Exponentiate both sides with base 10 to eliminate the log:
5
$10^{0.3} = \frac{[A^-]}{[HA]}$
6
Calculate the final ratio: $\frac{[ A ^{-} ]}{[ H A ]} = 2.0$

4. Limitations of the Henderson-Hasselbalch Equation★★★★☆⏱ 3 min

The equation is an approximation, and it fails under specific conditions that AP exam questions often test. You must be able to identify when the equation cannot be used to avoid losing points.

The equation is not valid for strong acid or strong base solutions, as no weak conjugate pair exists
The equation breaks down if the target pH is more than 1 unit away from the pKa of the weak acid, as dissociation of HA becomes significant
The equation cannot be used for very dilute buffer solutions, where autoionization of water contributes a large portion of total H⁺ concentration

5. Common Pitfalls

Wrong move:

Plugging in concentrations of strong acids or strong bases into the ratio term

Why:

Strong species fully dissociate, so no undissociated HA exists for the equilibrium expression

Correct move:

Only use concentrations of a matching weak acid-conjugate base pair in the ratio

Wrong move:

Flipping the ratio to [HA]/[A⁻] by accident

Why:

This reverses the log term sign and gives an incorrect pH value

Correct move:

Memorize the ratio order: conjugate base in the numerator, weak acid in the denominator

Wrong move:

Using natural log (ln) instead of base-10 log for the calculation

Why:

The derivation uses base-10 logarithm to match the pH scale definition

Correct move:

Always confirm your calculator is set to use base-10 log for these problems

Wrong move:

Attempting to use the equation for a buffer with pH 2 when the weak acid pKa is 5

Why:

At pH 3 units below pKa, the weak acid dissociation is no longer negligible, breaking the approximation

Correct move:

Only apply the equation for buffers where target pH is within ±1 of the acid pKa

Wrong move:

Forgetting to convert Ka to pKa before substituting into the equation

Why:

Using the raw Ka value (often ~10⁻⁵) will give a nonsensical negative pH

Correct move:

Always calculate pKa = -log(Ka) first before plugging values into the formula

6. Quick Reference Cheatsheet

Scenario	Henderson-Hasselbalch Form	Key AP Exam Note
Weak acid buffer	$p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$	Use for pH < 7 typical buffers
Weak base buffer	$pO H = p K_{b} + lo g (\frac{[ H B ^{+} ]}{[ B ]})$	Convert to pH via $14 - pO H$
1:1 conjugate pair ratio	$p H = p K_{a}$	Matches half-equivalence point of weak acid titration
pH within 1 unit of pKa	Equation fully valid	Approximation error is less than 5%

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 2
Buffer pH calculation problem
2021 · Paper 1
pKa from buffer ratio question
2019 · Paper 2
Buffer preparation design task

What's Next

Mastering the Henderson-Hasselbalch equation unlocks nearly all high-weight Unit 8 AP Chemistry points, as it is the foundation for buffer pH change calculations, titration curve analysis, and lab-based buffer design questions. You will now be able to quickly solve problems that previously required tedious full equilibrium ICE table calculations, saving critical time during the exam. The concepts you learned here also directly apply to acid-base titration half-equivalence points, a very common free-response topic that appears on almost every recent AP Chemistry exam.