Henderson-Hasselbalch Equation — Use Henderson–Hasselbalch equation to compute pH of a buffer - Chemistry 2

SECTION 1

Historical Context & Motivation

In the early 1900s, scientists were wrestling with a very practical challenge: how do you keep the pH of a solution stable when small amounts of acid or base are added? Researchers in biochemistry and medicine needed reliable ways to control pH, since even small changes could ruin an experiment or endanger a patient. The mathematical tools for acid–base equilibria existed, but they were clunky—solving a full equilibrium expression every time you needed a pH was tedious. What chemists really wanted was a simple, direct formula that could predict the pH of a buffer at a glance.

1884 • Arrhenius Acid–Base Theory

Svante Arrhenius proposed that acids produce H⁺ ions and bases produce OH⁻ ions in water, laying the groundwork for quantitative acid–base chemistry.

1908 • Henderson's Equation

American biochemist Lawrence Joseph Henderson published a relationship linking hydrogen-ion concentration to the ratio of acid and conjugate base concentrations. His version used H⁺ directly, not pH.

1909 • Sørensen Introduces pH

Danish chemist Søren Sørensen defined the pH scale as the negative base-10 logarithm of the hydrogen-ion concentration, making it far easier to communicate acidity.

1917 • Hasselbalch Adds Logarithms

Danish physiologist Karl Albert Hasselbalch rewrote Henderson's equation using Sørensen's pH notation, creating the compact logarithmic form we use today.

The result of these combined efforts is the Henderson–Hasselbalch equation—a straightforward tool that answers a single question: If I know the pKₐ of my weak acid and the amounts of acid and conjugate base in my buffer, what is the pH? Let's explore how it works.

SECTION 2

Core Principles & Definitions

Before you can use the Henderson–Hasselbalch equation, you need to understand a few key ideas about buffers and weak-acid equilibria. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers are made from a weak acid paired with its conjugate base (or a weak base paired with its conjugate acid). The interplay between these two species is what gives the buffer its pH-stabilizing power.

Weak Acid (HA)

An acid that only partially ionizes in water. It donates H⁺ ions but not completely—some HA molecules remain intact. Examples include acetic acid (CH₃COOH) and carbonic acid (H₂CO₃).

Conjugate Base (A⁻)

The species that remains after the weak acid donates a proton. For acetic acid, the conjugate base is the acetate ion (CH₃COO⁻). It can accept H⁺ ions back.

Kₐ and pKₐ

Kₐ is the acid dissociation constant—it tells you how easily the acid donates H⁺. pKₐ = −log(Kₐ). A smaller pKₐ means a stronger weak acid.

Buffer Ratio

The mole ratio [A⁻]/[HA] determines how far the buffer's pH sits from the pKₐ. When [A⁻] = [HA], the ratio equals 1, and pH = pKₐ exactly.

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KEY TAKEAWAY

Think of a buffer like a temperature-controlled room. The thermostat (the weak acid/conjugate base pair) constantly adjusts: if someone opens a window and lets cold air in (adding acid), the heater kicks on (A⁻ neutralizes the H⁺). If someone turns on an oven (adding base), the air conditioner runs (HA donates H⁺ to neutralize OH⁻). The Henderson–Hasselbalch equation tells you the current "temperature" (pH) of that room based on how much heater fuel (A⁻) and coolant (HA) you have on hand.

SECTION 3

Visual Explanation — Buffer Equilibrium

This diagram shows the dynamic equilibrium inside a buffer solution. The weak acid (HA) and its conjugate base (A⁻) interconvert by donating and accepting protons. When acid or base is added, the buffer adjusts the ratio of A⁻ to HA, and the Henderson–Hasselbalch equation captures the resulting pH.

Notice how the two species act as a team. When extra H⁺ ions flood in, the conjugate base (A⁻) mops them up by reacting to form more HA. When OH⁻ ions are added, the weak acid (HA) donates protons to neutralize them, producing more A⁻ and water. In both cases the ratio [A⁻]/[HA] shifts only slightly, so the logarithm in the equation changes by a small amount and the pH barely moves. That's the magic of a buffer.

SECTION 4

Mathematical Framework

The Henderson–Hasselbalch equation is derived from the acid dissociation equilibrium expression for a weak acid. Let's walk through the algebra step by step so you can see where each part comes from.

WEAK ACID EQUILIBRIUM

Kₐ = [H⁺][A⁻] / [HA]

Kₐ = acid dissociation constant, [H⁺] = hydrogen-ion concentration, [A⁻] = conjugate base concentration, [HA] = weak acid concentration.

To isolate [H⁺], rearrange the expression by multiplying both sides by [HA] and dividing both sides by [A⁻].

SOLVE FOR [H⁺]

[H⁺] = Kₐ × ([HA] / [A⁻])

This tells us that the hydrogen-ion concentration depends on Kₐ scaled by the ratio of acid to conjugate base.

Now take the negative log of both sides. Recall that pH = −log[H⁺] and pKₐ = −log(Kₐ). Also remember the log rule: log(a × b) = log(a) + log(b), and −log(HA/A⁻) = +log(A⁻/HA) because flipping the fraction reverses the sign.

HENDERSON–HASSELBALCH EQUATION

pH = pKₐ + log([A⁻] / [HA])

pH = acidity of the buffer, pKₐ = characteristic strength of the weak acid, [A⁻] = molar concentration of conjugate base, [HA] = molar concentration of weak acid. "log" means base-10 logarithm.

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Important Detail

You can substitute moles in place of molar concentrations in the ratio [A⁻]/[HA], because if both species are dissolved in the same volume of solution, the volume cancels out: (mol A⁻ / V) ÷ (mol HA / V) = mol A⁻ / mol HA. This shortcut saves you a calculation step.

SPECIAL CASE: EQUAL CONCENTRATIONS

When [A⁻] = [HA]: pH = pKₐ + log(1) = pKₐ

Since log(1) = 0, the pH of a buffer equals the pKₐ of the weak acid when the acid and its conjugate base are present in equal amounts. This is the midpoint of a titration curve.

SECTION 5

How the Buffer Ratio Affects pH

The Henderson–Hasselbalch equation reveals that the pH of a buffer depends on just two factors: the pKₐ of the weak acid (which is fixed for a given acid) and the ratio [A⁻]/[HA] (which you control). The table below illustrates how changing that ratio shifts the pH relative to pKₐ.

[A⁻] / [HA]	log([A⁻]/[HA])	pH relative to pKₐ	Buffer is…
0.10	−1.00	pKₐ − 1.00	More acidic (excess HA)
0.50	−0.30	pKₐ − 0.30	Slightly acidic
1.00	0.00	pKₐ exactly	Equal amounts—max buffering
2.00	+0.30	pKₐ + 0.30	Slightly basic
10.0	+1.00	pKₐ + 1.00	More basic (excess A⁻)

Effect of the [A⁻]/[HA] ratio on buffer pH

The graph shows a perfectly straight line with a slope of 1: for every factor-of-10 change in the [A⁻]/[HA] ratio, the pH shifts by exactly one unit from the pKₐ. When the ratio equals 1 (log = 0), the pH sits right at the pKₐ.

A useful rule of thumb: buffers work effectively within roughly ±1 pH unit of the pKₐ, which corresponds to a ratio of [A⁻]/[HA] between 0.1 and 10. Outside that range, one component is nearly used up and the buffer can no longer resist pH changes effectively.

SECTION 6

Worked Example

Let's put the Henderson–Hasselbalch equation to work with a concrete problem. Suppose you prepare a buffer by dissolving 0.40 mol of acetic acid (CH₃COOH) and 0.60 mol of sodium acetate (CH₃COONa) in enough water to make 1.00 L of solution. The Kₐ of acetic acid is 1.8 × 10⁻⁵. What is the pH of this buffer?

Finding the pH of an Acetic Acid / Acetate Buffer

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Step 1 — Identify the Weak Acid and Conjugate Base

Acetic acid (CH₃COOH) is the weak acid, abbreviated HA. Sodium acetate fully dissociates in water to give the acetate ion (CH₃COO⁻), which is the conjugate base A⁻. So [HA] = 0.40 M and [A⁻] = 0.60 M.

[HA] = 0.40 M, [A⁻] = 0.60 M
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Step 2 — Calculate pKₐ

pKₐ = −log(Kₐ) = −log(1.8 × 10⁻⁵). Break this into parts: log(1.8) ≈ 0.26 and log(10⁻⁵) = −5. So −log(1.8 × 10⁻⁵) = −(0.26 + (−5)) = −0.26 + 5 = 4.74.

pKₐ = 4.74
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Step 3 — Calculate the Ratio [A⁻]/[HA]

[A⁻] / [HA] = 0.60 / 0.40 = 1.50. There is more conjugate base than acid, so we expect the pH to be slightly above the pKₐ.

Ratio = 1.50
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Step 4 — Apply the Henderson–Hasselbalch Equation

pH = pKₐ + log([A⁻] / [HA]) = 4.74 + log(1.50). Using a calculator or log table, log(1.50) ≈ 0.18.

pH = 4.74 + 0.18 = 4.92
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Step 5 — Check Reasonableness

Since there is more conjugate base than acid, the pH should be above pKₐ (4.74), and 4.92 satisfies that. The ratio of 1.50 is well within the effective buffer range (0.1 to 10), so this is a good buffer.

Final Answer: pH = 4.92

SECTION 7

Strengths & Limitations

The Henderson–Hasselbalch equation is incredibly useful, but like any shortcut, it comes with conditions. Understanding when it works well—and when it doesn't—will help you avoid common mistakes on exams and in the lab.

Strengths	Limitations
Quick calculation—just plug in pKₐ and the ratio to get pH instantly.	Assumes the concentrations of HA and A⁻ at equilibrium equal the initial concentrations. This breaks down at very low concentrations (below ~0.01 M).
You can use moles instead of molarity when both species share the same solution volume.	Only applies to buffer solutions containing a weak acid and its conjugate base (not strong acids or strong bases).
Easily rearranged to find [A⁻]/[HA] needed for a target pH, which is handy for designing buffers.	Ignores activity coefficients, so it becomes less accurate for solutions with high ionic strength.
Works at any point along a titration curve for a weak acid/strong base titration (before the equivalence point).	Not valid at the equivalence point of a titration, where all HA has been converted to A⁻ and there is no longer a buffer.

When the Henderson–Hasselbalch equation shines and where it falls short

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KEY TAKEAWAY

The Henderson–Hasselbalch equation is like a GPS shortcut through city streets—it gets you to the answer much faster than solving the full Kₐ equilibrium expression (the long route). But just like a GPS, it won't work if you're out of its service area: if the solution is extremely dilute, or if you're dealing with a strong acid instead of a weak one, you need to take the full route and solve the equilibrium expression from scratch.

SECTION 8

Connection to Advanced Concepts

The Henderson–Hasselbalch equation is your entry point into a family of more powerful acid–base tools. As you advance in chemistry, you'll encounter situations that require modifications or entirely different approaches. The table below compares the introductory form you've learned with some of its advanced relatives.

Feature	Henderson–Hasselbalch (This Lesson)	Full ICE-Table Approach
Setup required	Plug in pKₐ and ratio	Build Initial-Change-Equilibrium table, solve quadratic
Speed	Very fast—one equation	Slower—requires algebra
Accuracy at low concentration	Less accurate (approximation)	More accurate (no approximation)
Works for polyprotic acids?	One step at a time only	Yes, with multiple equilibria
Best for	Buffers, titration halfway points	Any weak acid/base equilibrium

Henderson–Hasselbalch vs. Full Equilibrium Methods

In AP Chemistry and college courses, you'll also see the Henderson–Hasselbalch equation applied to base buffers using pOH = pK_b + log([BH⁺]/[B]), where B is a weak base and BH⁺ is its conjugate acid. The same logic applies—you're just flipping the perspective from acid to base. In biochemistry, the equation is essential for understanding amino acid side-chain charges at physiological pH and for designing buffers in biological experiments.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A buffer is made from a weak acid with pKₐ = 6.35 and its conjugate base. Without doing any math, predict whether the pH will be above, below, or equal to 6.35 if the solution contains more conjugate base than weak acid. Explain your reasoning.

Show answer

The pH will be above 6.35. According to the Henderson–Hasselbalch equation, pH = pKₐ + log([A⁻]/[HA]). When [A⁻] > [HA], the ratio [A⁻]/[HA] is greater than 1, so log([A⁻]/[HA]) is positive. Adding a positive number to the pKₐ pushes the pH higher than 6.35. More conjugate base means the solution leans toward the basic side of the pKₐ.

Question 2: BASIC CALCULATION

Calculate the pH of a buffer containing 0.25 M hydrofluoric acid (HF) and 0.25 M sodium fluoride (NaF). The Kₐ of HF is 6.8 × 10⁻⁴.

Show answer

First find pKₐ: pKₐ = −log(6.8 × 10⁻⁴) = −(log 6.8 + log 10⁻⁴) = −(0.83 + (−4)) = 3.17. Since [A⁻] = [HA] = 0.25 M, the ratio is 1.00 and log(1.00) = 0. Therefore pH = 3.17 + 0 = 3.17. When the acid and conjugate base concentrations are equal, the pH simply equals the pKₐ.

Question 3: INTERMEDIATE

A chemist mixes 0.30 mol of formic acid (HCOOH, Kₐ = 1.8 × 10⁻⁴) and 0.50 mol of sodium formate (HCOONa) in 2.0 L of water. What is the pH of the resulting buffer?

Show answer

Since both species are in the same 2.0 L volume, we can use moles directly in the ratio: [A⁻]/[HA] = 0.50/0.30 = 1.67. Next, pKₐ = −log(1.8 × 10⁻⁴) = −(0.26 + (−4)) = 3.74. Now apply Henderson–Hasselbalch: pH = 3.74 + log(1.67) = 3.74 + 0.22 = 3.96. The pH is above the pKₐ because there is more conjugate base than acid, which is consistent with our expectations.

Question 4: APPLIED

Blood is buffered primarily by the carbonic acid/bicarbonate system (H₂CO₃/HCO₃⁻). The pKₐ of carbonic acid is 6.35. If the concentration of HCO₃⁻ in blood is 24 mM and the concentration of H₂CO₃ is 1.2 mM, what is the predicted blood pH? Does this match the typical blood pH range of 7.35–7.45?

Show answer

The ratio [A⁻]/[HA] = 24 mM / 1.2 mM = 20. Apply the equation: pH = 6.35 + log(20) = 6.35 + 1.30 = 7.65. This is slightly higher than the typical blood pH of 7.35–7.45. The discrepancy occurs because the blood buffer system is more complex: CO₂ dissolved in blood contributes to the effective H₂CO₃ concentration, and the body uses respiration to adjust CO₂ levels. The Henderson–Hasselbalch equation gives a good first approximation, but biological systems layer additional regulatory mechanisms on top.

Question 5: CRITICAL THINKING

You need to prepare a buffer with pH = 5.00 using acetic acid (pKₐ = 4.74). What ratio of [CH₃COO⁻] to [CH₃COOH] is required? If you have 0.20 mol of acetic acid, how many moles of sodium acetate must you add? Explain whether this buffer would be more resistant to added acid or added base.

Show answer

Rearrange the Henderson–Hasselbalch equation: log([A⁻]/[HA]) = pH − pKₐ = 5.00 − 4.74 = 0.26. Therefore [A⁻]/[HA] = 10⁰·²⁶ ≈ 1.82. If you have 0.20 mol of acetic acid, you need 1.82 × 0.20 = 0.36 mol of sodium acetate. Because there is more conjugate base (A⁻) than acid (HA), the buffer has a larger reservoir of A⁻ available to neutralize added H⁺. This means the buffer is more resistant to added acid than to added base—adding base would consume the smaller HA supply more quickly.

SUMMARY

Lesson Summary

Summary

The Henderson–Hasselbalch equation, pH = pKₐ + log([A⁻]/[HA]), provides a fast and elegant way to calculate the pH of a buffer solution. It is derived from the Kₐ equilibrium expression by taking the negative logarithm of both sides. The equation shows that pH depends on two things: the inherent strength of the weak acid (pKₐ) and the ratio of conjugate base to acid.

Key points to remember: when [A⁻] = [HA], the pH equals the pKₐ exactly. Buffers are effective within approximately ±1 pH unit of the pKₐ. You can use moles instead of molarity when both species share the same volume. The equation works best at moderate concentrations of weak acid/conjugate base pairs and should not be applied to strong acids, strong bases, or extremely dilute solutions.