---
title: "AP Chemistry 8.9: Henderson-Hasselbalch Equation"
description: "Review the Henderson-Hasselbalch equation for AP Chemistry, including pH = pKa + log([A-]/[HA]), buffer pH, pKa, and conjugate acid-base ratios."
canonical: "https://fiveable.me/ap-chem/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW"
type: "study-guide"
subject: "AP Chemistry"
unit: "Unit 8 – Acids & Bases"
lastUpdated: "2026-06-09"
---

# AP Chemistry 8.9: Henderson-Hasselbalch Equation

## Summary

Review the Henderson-Hasselbalch equation for AP Chemistry, including pH = pKa + log([A-]/[HA]), buffer pH, pKa, and conjugate acid-base ratios.

## Guide

The Henderson-Hasselbalch equation, $\text{pH}=\text{p}K_a+\log\left(\frac{[A^-]}{[HA]}\right)$, lets you find the pH of a buffer from the $pK_a$ of the weak acid and the ratio of [conjugate base](/ap-chem/key-terms/conjugate-base "fv-autolink") to conjugate acid. When the conjugate acid and base concentrations are equal, the log term is zero and pH equals $pK_a$. For [AP Chemistry](/ap-chem "fv-autolink"), use this equation only when both buffer components are present.

## Why This Matters for the AP Chemistry Exam

Buffers and pH show up throughout [Unit 8](/ap-chem/unit-8 "fv-autolink"), and this equation is the fastest way to connect a buffer's composition to its pH. On the exam you may need to calculate or estimate the pH of a buffer from concentrations, reason about how the [A-]/[HA] ratio sets the pH, or interpret a [titration curve](/ap-chem/key-terms/titration-curve "fv-autolink") where a buffer region exists. The equation also reinforces a key idea: pH = pKa exactly at the point where conjugate acid and base concentrations match, which is the half-equivalence point in a weak acid titration.

Two things are good to know about limits. You will not be asked to derive the Henderson-Hasselbalch equation, and you will not have to compute the exact pH change after a small amount of acid or base is added to a buffer. You should still be able to explain why the pH barely shifts.

## Key Takeaways

- The equation is pH = pKa + log([A-]/[HA]), where A- is the conjugate base and HA is the conjugate acid.
- It comes from the [equilibrium expression](/ap-chem/key-terms/equilibrium-expression "fv-autolink") for a weak acid dissociating, so it only applies to buffer [solutions](/ap-chem/key-terms/solution "fv-autolink") with both members of a conjugate pair present.
- When [A-] = [HA], log(1) = 0, so pH = pKa. This is the most balanced buffer point.
- Because the concentrations sit inside a ratio, the same [volume](/ap-chem/key-terms/volume "fv-autolink") divides both, so you can often use moles or millimoles instead of [molarity](/ap-chem/key-terms/molarity "fv-autolink").
- Adding a small amount of acid or base barely changes the [A-]/[HA] ratio, so the pH stays nearly constant. That is what makes a buffer resist pH change.

## The Equation and What Each Part Means

The Henderson-Hasselbalch equation finds the pH of a buffer. From [Topic 8.8](/ap-chem/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD "fv-autolink"), a buffer is a solution that resists changes in pH and contains a weak acid with its conjugate base, or a [weak base](/ap-chem/key-terms/weak-base "fv-autolink") with its conjugate acid.

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

Here is what each piece does:

- **pH** is -log[[H3O+](/ap-chem/key-terms/h3o "fv-autolink")]. This is usually the unknown you solve for.
- **pKa** is -log(Ka), a logarithmic measure of [acid strength](/ap-chem/key-terms/acid-strength "fv-autolink"). A lower pKa means a stronger acid.
- **[A-]/[HA]** is the ratio of conjugate base concentration to conjugate acid concentration. HA is the weak acid, and A- is its conjugate base, so they are a [conjugate acid-base pair](/ap-chem/unit-4/intro-acid-base-reactions/study-guide/idvZ7Ve4pFo8gFIyYfMl "fv-autolink").

This is where the equation connects to buffers. A buffer always has some conjugate acid and some conjugate base present, so the ratio is always defined. It also explains the most balanced buffer: when [A-] = [HA], the ratio is 1, and log(1) = 0, so pH = pKa.

## Worked Examples

### Example 1: Buffer Given Directly

Find the pH of a buffer with 0.5 M CH3COOH mixed with 0.25 M CH3COONa (Ka = 1.8 x 10^-5).

The acetate ion (CH3COO-) is the conjugate base, and acetic acid (CH3COOH) is the conjugate acid. Plug straight into the equation.

$$pK_a = -\log(1.8 \times 10^{-5}) = 4.74$$

$$pH = 4.74 + \log\left(\frac{0.25}{0.5}\right) = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44$$

The pH is slightly below the pKa because there is more conjugate acid than conjugate base.

### Example 2: Using the Equation During a Titration

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH after adding 15.0 mL of 0.100 M NaOH.

First, write the [net ionic equation](/ap-chem/unit-4/net-ionic-equations/study-guide/VTCEO9cDrSHfPodoWGxS "fv-autolink") for the reaction:

$$CH_3COOH + OH^- \rightleftharpoons CH_3COO^- + H_2O$$

Next, use [stoichiometry](/ap-chem/unit-4/stoichiometry/study-guide/GjwCuhOQRvWLb4rKjYD2 "fv-autolink") to find the millimoles of each species after the reaction goes forward. Starting amounts: 2.5 mmol CH3COOH (25.0 mL x 0.100 M) and 1.5 mmol OH- (15.0 mL x 0.100 M).

| Species | CH3COOH | OH- | CH3COO- |
|:---|:---|:---|:---|
| Start (mmol) | 2.5 | 1.5 | 0 |
| End (mmol) | 1.0 | 0 | 1.5 |

Now both a weak acid and its conjugate base are present, so this is a buffer. Because both concentrations would be divided by the same total volume, the volume cancels and you can use millimoles directly in the ratio.

$$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) = 4.74 + \log\left(\frac{1.5}{1.0}\right) = 4.74 + 0.18 = 4.92$$

The pH is above the pKa here because the conjugate base now outweighs the conjugate acid.

## How to Use This on the AP Chemistry Exam

### Problem Solving

- Identify the conjugate pair first. Decide which species is HA and which is A- before plugging in numbers.
- Find pKa from Ka using pKa = -log(Ka) if you are only given Ka.
- For titration problems, run the stoichiometry to get the leftover moles of weak acid and conjugate base, then use those in the ratio. Volume cancels, so moles or millimoles work.
- Check direction: if there is more conjugate base than acid, pH should be above pKa; if there is more acid, pH should be below pKa.

### Free Response

- Be ready to explain in words why a buffer resists pH change: adding a little acid or base barely shifts the [A-]/[HA] ratio, so pH stays close to pKa.
- Connect this to titration curves. The flattest part of a weak acid titration curve is the buffer region, and the half-equivalence point is where pH = pKa.

### Common Trap

- The exam will not ask you to derive this equation or to calculate the exact new pH after adding acid or base to a buffer. Focus on setting up the ratio and explaining the buffering behavior.

## Common Misconceptions

- **The equation works for any solution.** It only applies to buffers, where both the conjugate acid and conjugate base are present in meaningful amounts. It is not for a pure [strong acid](/ap-chem/key-terms/strong-acid "fv-autolink") or a single weak acid before any base is added.
- **You must convert to molarity first.** Since the concentrations are in a ratio with the same volume, moles or millimoles give the same answer. Converting to molarity is extra work that cancels out.
- **pKa and pH are the same thing.** They are equal only when [A-] = [HA]. The pKa is a fixed property of the acid, while the pH depends on the ratio of the conjugate pair.
- **A bigger ratio always means higher pH only by a lot.** Because the term is a base-10 log, the pH shifts gradually with the ratio. A ratio of 10 to 1 changes pH by just 1 unit from the pKa.
- **More conjugate base always means a stronger buffer.** A buffer resists pH change best when the conjugate acid and base concentrations are close to equal, not when one greatly outweighs the other.

## Related AP Chemistry Guides

- [Unit 8 Overview: Acids and Bases](/ap-chem/unit-8/review/study-guide/iJDbOp6pD1hPTfaU7AdM)
- [8.1 Introduction to Acids and Bases](/ap-chem/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj)
- [8.2 pH and pOH of Strong Acids and Bases](/ap-chem/unit-8/ph-poh-strong-acids-bases/study-guide/AhVlrEQS1kkfZGGWdFNT)
- [8.4 Acid-Base Reactions and Buffers ](/ap-chem/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh)
- [8.3 Weak Acid and Base Equilibria](/ap-chem/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw)
- [8.5 Acid-Base Titrations](/ap-chem/unit-8/acid-base-titrations/study-guide/5ugPp0ykDKthSY3MiYan)

## Vocabulary

- **buffer solution**: A solution containing a large concentration of both members of a conjugate acid-base pair that resists changes in pH when small amounts of acid or base are added.
- **concentration ratio**: The ratio of the concentration of the conjugate base to the concentration of the conjugate acid, [A-]/[HA], in a buffer solution.
- **conjugate acid-base pair**: Two species that differ by one proton, where one is the acid form and the other is the base form of the same substance.
- **dissociation**: The process by which a compound breaks apart into its constituent ions or molecules in solution.
- **equilibrium expression**: A mathematical equation that relates the concentrations or partial pressures of reactants and products at equilibrium, expressed as Kc or Kp.
- **pH**: A logarithmic scale used to express the concentration of hydronium ions in a solution, calculated as −log[H3O+].
- **pKa**: The negative logarithm of the acid dissociation constant (Ka); used to compare the relative strength of weak acids and predict protonation state at different pH values.
- **weak acid**: An acid that only partially ionizes in solution, establishing an equilibrium between the molecular form (HA) and its conjugate base (A-).

## FAQs

### What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is pH = pKa + log([A-]/[HA]). It relates buffer pH to the pKa of the weak acid and the concentration ratio of conjugate base to conjugate acid.

### When do you use the Henderson-Hasselbalch equation in AP Chemistry?

Use the Henderson-Hasselbalch equation when a buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid. It is used to find or interpret buffer pH from the conjugate pair ratio.

### What happens when [A-] equals [HA]?

When [A-] equals [HA], the ratio is 1 and log(1) equals 0, so pH = pKa. In a weak acid titration, this corresponds to the half-equivalence point.

### Can you use moles instead of molarity in the Henderson-Hasselbalch equation?

Yes, if both species are in the same final solution volume, the volume cancels in the ratio. That means moles or millimoles can often be used directly for [A-]/[HA].

### What is not assessed for AP Chem 8.9?

The AP Chemistry exam does not assess deriving the Henderson-Hasselbalch equation or computing the exact pH change after adding a small amount of acid or base to a buffer.

### How is the Henderson-Hasselbalch equation tested on the AP Chemistry exam?

AP Chemistry questions often ask you to identify the conjugate acid-base pair, calculate pKa from Ka, use pH = pKa + log([A-]/[HA]), or explain why a buffer resists large pH changes.

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