8.9 Henderson-Hasselbalch Equation

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 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, 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, 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 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.

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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 for a weak acid dissociating, so it only applies to buffer solutions 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 divides both, so you can often use moles or millimoles instead of molarity.
• 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, a buffer is a solution that resists changes in pH and contains a weak acid with its conjugate base, or a weak base with its conjugate acid.

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

Here is what each piece does:

• pH is -log[H3O+]. This is usually the unknown you solve for.
• pKa is -log(Ka), a logarithmic measure of acid strength. 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.

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 for the reaction:

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

Next, use stoichiometry 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).

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 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
• 8.1 Introduction to Acids and Bases
• 8.2 pH and pOH of Strong Acids and Bases
• 8.4 Acid-Base Reactions and Buffers
• 8.3 Weak Acid and Base Equilibria
• 8.5 Acid-Base Titrations