What Is a Buffer in Chemistry?

A buffer is a solution that resists changes in pH when small amounts of strong acid or strong base are added. Why does that matter? Many chemical reactions are pH-sensitive—they only proceed favorably within a narrow pH range, so they are often carried out in a buffered environment.

More importantly, most living organisms survive only within a very narrow pH window. Your blood, for example, maintains a pH of approximately 7.4—just on the basic side of neutral. If that pH drops below 7.1 or climbs above 7.8 for any significant period of time, the consequences are fatal. Your blood is therefore heavily buffered against pH changes.

Composition of a Buffer: Weak Acid and Conjugate Base

A buffer maintains pH by virtue of what it contains. Every buffer is composed of a weak acid paired with its conjugate base, or equivalently, a weak base paired with its conjugate acid. The key word here is weak—a strong acid with its conjugate base does not form a buffer, nor does a strong base with its conjugate acid.

Using Salts to Supply the Conjugate Ion

When you need to add ions like F− or NH₄+ to a solution, you don't just grab a bottle of bare ions off the shelf. Instead, these are added as part of a salt, as covered in chapter 16.

For a fluoride buffer, you want a cation that won't influence the solution—a negligible cation. Group 1 metal salts (lithium, sodium, potassium, rubidium, cesium) fit the bill perfectly: they're both negligible and soluble with fluoride. Group 2 metal fluorides tend to be insoluble, so they're ruled out. Among the Group 1 options, sodium fluoride (NaF) is by far the most common choice.

For an ammonia buffer, you need a negligible anion paired with the ammonium cation. The best candidates are conjugate bases of strong acids. Chloride (Cl−) is the most common, making ammonium chloride (NH₄Cl) the standard choice. Ammonium bromide, ammonium iodide, ammonium chlorate, and ammonium nitrate would also work, but NH₄Cl is what you'll see most often.

Buffer Ratios and Maximum Buffering Capacity

To function as a buffer, the weak acid and conjugate base must be mixed in similar amounts—ideally equimolar, meaning equal moles. When the ratio is exactly one-to-one, the buffer achieves its maximum buffering capacity, providing the greatest resistance to pH changes.

However, a solution still functions as a buffer if there is up to 10 times more acid than conjugate base, or up to 10 times more conjugate base than acid. Beyond that 10-fold ratio, the solution loses its buffering ability and will no longer resist pH changes effectively.

How a Buffer Protects Against pH Change

What causes pH to change in the first place? Typically, the addition of a strong acid or base. Think of the buffer as protecting the water molecules in the solution.

In unbuffered water, adding an acid protonates water to form H₃O+, while adding a base deprotonates water to form OH−—both of which directly shift the pH. A buffer prevents this by offering better alternatives:

• If a base is added: The weak acid (HF) is a better proton donor than water, so it donates H+ to the base instead.
• If an acid is added: The conjugate base (F−) is a better proton acceptor than water, so it absorbs the H+ instead.

The net result? Instead of converting water into hydronium or hydroxide—which would dramatically shift the pH—the buffer simply adjusts the ratio of HF to NaF in the solution. That produces only a small change in pH rather than a large one.

pK_a, Buffer Range, and the Logarithmic Scale

What Is pK_a?

When working with buffers, you'll encounter pK_a far more often than K_a. The pK_a is simply the negative logarithm of the K_a:

Buffer calculations are frequently more convenient using pK_a values. You may be given the K_a and need to convert, or the pK_a may be provided directly.

Buffer Range = pK_a ± 1

When weak acid and conjugate base are present in an exact one-to-one ratio, the pH equals the pK_a. Because the pH scale is logarithmic, shifting the ratio by a factor of 10 changes the pH by exactly one unit:

The buffer range is therefore pK_a ± 1. Within this window, the solution effectively resists pH changes—best at the pK_a itself, but functional throughout the range. Outside this window, the solution can no longer buffer effectively.

Acetic Acid/Sodium Acetate Buffer Range Example

Buffer Ranges for Polyprotic Acids

Polyprotic acids have more than one ionizable proton, which means they have multiple K_a values, multiple pK_a values, and multiple buffer ranges. This makes them remarkably versatile for buffering at different pH targets.

Carbonic Acid: 2 Buffer Ranges

Carbonic acid (H₂CO₃) is diprotic, with two acidic protons to donate. Each deprotonation has its own pK_a:

Notice that each successive pK_a is higher, reflecting the fact that removing additional protons becomes progressively more difficult. A lower K_a corresponds to a higher pK_a, indicating a weaker acid.

Phosphoric Acid: 3 pK_a Values and Physiological Buffering

Phosphoric acid (H₃PO₄) has three acidic protons and therefore 3 pK_a values: 2.1, 7.2, and 12.4. The second pK_a at 7.2 is especially useful because it falls right near physiological pH, making phosphate buffers extremely common for biological experiments, cell cultures, and biochemistry applications.

Carbonic Acid and Blood Buffering

The first equilibrium of carbonic acid is a major component of your blood's buffering system. In the body, dissolved CO₂—the carbon dioxide you exhale—combines with water to form carbonic acid. That carbonic acid then participates in the equilibrium with bicarbonate:

3 Ways to Prepare a Buffer Solution

Creating a buffer isn't about what you add—it's about what you end up with. The goal is always the same: a solution containing a roughly one-to-one ratio of weak acid and its conjugate base. There are 3 ways to prepare a buffer solution.

Method 1: Mix Weak Acid Directly with Its Conjugate Base Salt

The most straightforward approach to buffer preparation: combine the weak acid and a salt of its conjugate base in approximately equal amounts.

Method 2: Mix a Weak Acid with a Strong Base (2:1 Ratio)

Instead of adding the conjugate base directly, you can create it by partially neutralizing a weak acid with a strong base. The strong base converts some of the weak acid into its conjugate base. The trick is using a two-to-one ratio—two parts weak acid to one part strong base—so that exactly half the weak acid gets neutralized.

Method 3: Mix a Weak Base with a Strong Acid (2:1 Ratio)

This works exactly like Method 2, but in reverse. Partially neutralize a weak base with a strong acid using a two-to-one ratio (two parts weak base to one part strong acid). Half the weak base gets converted into its conjugate acid, leaving a 1:1 mixture.

2 Common Wrong Answers to Watch For

Exam questions love to test whether you truly understand buffer composition. Watch out for these traps:

Buffer pH Calculations: The Henderson-Hasselbalch Equation

Why Use the Henderson-Hasselbalch Equation?

Calculating the pH of a buffer is the fifth type of pH calculation you'll encounter in general chemistry, following strong acid, strong base, weak acid, and weak base calculations from previous chapters. For buffers, the Henderson-Hasselbalch equation provides a more convenient approach than traditional ICE tables.

This equation applies only when both a weak acid and its conjugate base are present in solution. If either one is missing (concentration of zero), the equation breaks down—you'd be taking the log of zero, which is undefined.

Deriving the Henderson-Hasselbalch Equation

Starting from the K_a expression and taking the negative logarithm of both sides:

1. Begin with: K_a = [H⁺][A⁻] / [HA]
2. Take the log of both sides: log(K_a) = log([H⁺]) + log([A⁻]/[HA])
3. Multiply through by −1: −log(K_a) = −log([H⁺]) − log([A⁻]/[HA])
4. Recognize that −log(K_a) = pK_a and −log([H⁺]) = pH
5. Rearrange to arrive at the final form:

4 Versions of the Equation (and Which One to Use)

There are actually 4 versions of this equation floating around, which can be a source of confusion. Since log(x/y) = −log(y/x), you can flip the ratio and change the sign. Additionally, there are pOH/pK_b versions. To keep things simple, stick with this one form:

pH = pK_a + log([A⁻] / [HA])

This is the most commonly presented version, and any buffer pH calculation works exactly the same way using this form alone.

Worked Example: Finding the pH of a Buffer Solution

First, recognize that this is a buffer: you have a weak acid and its conjugate base present together in similar amounts. With a buffer, any ±x terms in an ICE table would be negligible, so skip the ICE table entirely and go straight to Henderson-Hasselbalch.

Step 1: Convert K_a to pK_a:

pK_a = −log(1.76 × 10⁻⁵) = 4.75

Step 2: Apply the Henderson-Hasselbalch equation:

pH = 4.75 + log(0.080 / 0.10) = 4.75 + log(0.80) = 4.75 + (−0.10) = 4.65

Worked Example: pH After Adding a Strong Acid to a Buffer

Setting Up the Problem

Because this is a buffer, the pH should change only slightly. We expect it to decrease a little (we're adding acid), but not by much.

Neutralization Reaction First

When HCl (a strong acid) enters the buffer, it neutralizes the conjugate base. Strong acids neutralize all bases—strong, weak, everything. The base in this solution is the acetate ion (CH₃COO−):

HCl + CH₃COO− → CH₃COOH + Cl−

This reaction goes to completion because HCl fully dissociates. A reaction that goes to completion requires a limiting reagent calculation, which means we need moles, not molarities.

Converting to Moles

Using moles = molarity × volume (in liters):

• Conjugate base (CH₃COO−): 0.100 L × 0.080 mol/L = 0.0080 mol
• Conjugate acid (CH₃COOH): 0.100 L × 0.10 mol/L = 0.010 mol
• HCl added: 0.0050 L × 0.10 mol/L = 0.00050 mol

Limiting Reagent Stoichiometry

The HCl is the limiting reagent (0.00050 mol). It reacts in a 1:1 ratio with the conjugate base, consuming 0.00050 mol of acetate while producing 0.00050 mol of acetic acid:

• Conjugate base remaining: 0.0080 − 0.00050 = 0.0075 mol
• HCl remaining: 0 mol (completely consumed)
• Conjugate acid total: 0.010 + 0.00050 = 0.0105 mol

Both weak acid and conjugate base are still present in reasonable amounts, so the solution is still a buffer and Henderson-Hasselbalch still applies.

Why You Can Use Moles Instead of Molarities in Henderson-Hasselbalch

Here's a useful shortcut. In Henderson-Hasselbalch, the ratio [A⁻]/[HA] can be written as (moles of A⁻ / V) / (moles of HA / V), where V is the total solution volume. Since V appears in both numerator and denominator, it cancels:

pH = pK_a + log(moles of A⁻ / moles of HA)

Final pH Calculation

pH = 4.75 + log(0.0075 / 0.0105) = 4.75 + log(0.714) = 4.75 + (−0.15) = 4.60

The pH dropped from 4.65 to 4.60—a change of only 0.05 units. That's the buffer doing its job. For comparison, adding the same amount of HCl to 100 mL of pure water would have dropped the pH by several units.

The Pattern: What to Add or Subtract

Once you set up Henderson-Hasselbalch in terms of moles, the pattern is simple: