Chemistry: Atoms First 2e · Acid-Base Equilibria

Buffers

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

A is a solution that resists large pH changes when small amounts of strong acid or strong base are added. A buffer contains a weak acid and its conjugate base — or a weak base and its conjugate acid — in significant, comparable amounts. The weak acid "absorbs" added hydroxide; the conjugate base "absorbs" added hydronium, so pH barely moves until the buffer is overwhelmed. The links buffer pH to the acid's pK_a and the ratio [A^-]/[HA]. Two limits matter: a buffer works only within about one pH unit of its pK_a (the ), and it neutralizes only a finite amount of added acid or base (the ).

Why this matters

Living systems run on buffers. Blood stays near pH 7.4 (normal arterial range roughly 7.35–7.45), mainly through the carbonic acid/bicarbonate pair working with the lungs and kidneys; a shift of a few tenths of a unit is a medical emergency. Cells use phosphate buffers, and enzymes work only in narrow pH windows. Pharmaceuticals are formulated at controlled pH; fermentation, pools, and aquariums rely on buffers. Buffers also explain the flat "buffer region" of weak-acid titration curves — the next topic.

The college version

Core Concepts

What a buffer is made of

Both members of a conjugate pair must be present at significant concentrations: acetic acid with sodium acetate provides HC₂H₃O₂ (HA) and C₂H₃O₂^- (A⁻); ammonium chloride with ammonia provides NH₄⁺ and NH₃. The pair acts as two sinks:

Added strong acid (H⁺) is consumed by the base: A^- + H⁺ → HA.
Added strong base (OH⁻) is consumed by the acid: HA + OH^- → A^- + H₂O.

A weak acid alone buffers poorly against added base, and a conjugate-base salt alone buffers poorly against added acid — you need both.

Why pH stays nearly constant

In HA ⇌ H⁺ + A^-, added H⁺ shifts the reaction left (combining with A⁻); added OH⁻ consumes H⁺, shifting it right (HA dissociates to replace the H⁺). Either way, the system restores its original ratio [A^-]/[HA], and since pH depends on that ratio, pH is nearly restored.

The Henderson–Hasselbalch equation

Start from the weak-acid equilibrium expression:

$K a = [H +][A -] [HA]$

Solve for [H⁺] and take -log of both sides:

$[H +] = K a \cdot [HA] [A -] \Rightarrow pH = p K a + log [A -] [HA]$

The key insight: buffer pH depends on the ratio [A^-]/[HA], not absolute concentrations — so diluting a buffer barely changes its pH but cuts its capacity. The equation assumes equilibrium concentrations equal the formal concentrations, which holds for normal buffers but not for strong acids or very dilute solutions.

Buffer range and buffer capacity

The buffer range is pK_a ± 1: outside it the ratio exceeds 10:1 (or falls below 1:10) and one member of the pair is nearly exhausted. Buffer capacity is the moles of strong acid or base per liter the buffer can neutralize before pH changes sharply. Capacity peaks when [A^-] = [HA] and grows with the total concentration of the pair.

Choosing and preparing a buffer

Pick a weak acid whose pK_a is within about one unit of the target pH, compute the required ratio from the Henderson–Hasselbalch equation, and mix the acid with a salt of its conjugate base. The HC₂H₃O₂/C₂H₃O₂^- pair (pK_a = 4.74) suits targets near pH 5; NH₄⁺/NH₃ (pK_a = 9.25) suits pH near 9; H₂PO₄^-/HPO₄^2- (pK_a ≈ 7.2) suits physiological pH. Real buffers are verified with a pH meter.

How It Works / Step-by-Step Process

Computing the pH of a buffer

Write the acid dissociation equilibrium; identify HA and A⁻.
Convert K_a to pK_a = -logK_a.
Determine the ratio [A^-]/[HA] from the concentrations mixed.
Substitute into the Henderson–Hasselbalch equation and solve.

Predicting the effect of added strong acid or base

Convert added moles of strong acid (or base) to moles of H⁺ (or OH⁻) by 1:1 neutralization stoichiometry.
Subtract from the species that consumes it; add to the other member of the pair.
Divide by the volume, recompute the ratio, and apply the Henderson–Hasselbalch equation.

Common Confusions

Do Not Confuse	With	The Difference
A buffer	Any weak acid solution	A weak acid alone resists added base poorly; you need the conjugate pair in comparable amounts
Buffer pH	Buffer capacity	pH depends on the ratio [A^-]/[HA]; capacity depends on the total amount of the pair
Adding acid	Adding base	H⁺ is consumed by A⁻ (pH dips slightly); OH⁻ is consumed by HA (pH rises slightly)
pK_a	pH	pKa is a property of the acid; pH is a property of the solution. Equal only when [A^-] = [HA]
Diluting a buffer	Destroying a buffer	Dilution barely changes pH (ratio unchanged) but sharply cuts capacity
Henderson–Hasselbalch equation	An exact law for any solution	Assumes equilibrium concentrations equal the formal concentrations — invalid for very dilute solutions and strong acids

Eli explains

The same idea, in plain words

Explain it like I’m 10

A buffer is like a seesaw with a heavy weight on each side: the weak acid and its conjugate base. When you push on one side — add a little acid or base — the weights slide a bit and the seesaw barely tips. That is why your blood keeps nearly the same pH whether you eat a lemon or exercise hard.

Worked example

Example 1: pH of an acetic acid/acetate buffer

A solution is 0.50 M in acetic acid and 0.40 M in sodium acetate. Given K_a = 1.8 × 10^-5, find the pH.

First compute pK_a:

$p K a = - log (1.8 \times 10 -5) = 4.74$

Apply the Henderson–Hasselbalch equation with [A^-] = 0.40 M and [HA] = 0.50 M:

$pH = 4.74 + log 0.40 0.50 = 4.74 + log (0.80) = 4.74 - 0.097 = 4.64$

The pH is 4.64 — slightly below the pKa because the acid form is more concentrated. A 0.050 M/0.040 M buffer at the same ratio would have the same pH but far less capacity.

Example 2: Adding strong acid to a buffer versus to pure water

Add 0.010 mol of HCl to 1.0 L of the buffer from Example 1. The added acid reacts with the conjugate base:

$C 2 H 3 O 2 - + H + \to HC 2 H 3 O 2$

Convert the added HCl to moles of H⁺ by dimensional analysis:

$0.010 mol HCl \times 1 mol H + 1 mol HCl = 0.010 mol H +$

Each mole of H⁺ consumes one mole of acetate, so in 1.0 L:

$[C 2 H 3 O 2 -] = 0.40 - 0.010 mol 1.0 L = 0.39 M$

$[HC 2 H 3 O 2] = 0.50 + 0.010 mol 1.0 L = 0.51 M$

Now apply the Henderson–Hasselbalch equation:

$pH = 4.74 + log 0.39 0.51 = 4.74 - 0.117 = 4.62$

The pH dropped only 0.02 units (4.64 → 4.62). Put the same 0.010 mol of HCl into 1.0 L of pure water instead: [H⁺] = 0.010 mol/1.0 L = 0.010 M, so pH = -log(0.010) = 2.00 — a drop of five pH units. That contrast is the whole point of a buffer.

Buffer preparation in practice. To make a pH 5.00 buffer from 1.00 L of 0.100 M acetic acid, the required ratio is [A^-]/[HA] = 10^{5.00 - 4.74} = 1.8, so [A^-] = 0.18 M. That is 0.18 mol of sodium acetate — 0.18 mol × 82.03 gmol ≈ 15 g — dissolved in the acid solution.

Key takeaways

A buffer contains a weak acid and its conjugate base in significant amounts.
Added H⁺ is consumed by A⁻; added OH⁻ by HA; pH changes only slightly.
pH = pK_a + log[A^-][HA] — pH depends on the ratio, not absolute concentrations.
When [A^-] = [HA], pH = pK_a — the half-neutralization point and the ratio of maximum capacity.
Buffer range ≈ pK_a ± 1; capacity peaks at ratio 1:1 and grows with concentration.
Blood is buffered near pH 7.4 mainly by the H₂CO₃/HCO₃^- pair with the lungs and kidneys.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

What two species must a buffer contain, and what does each neutralize?
Show answer
A weak acid (HA) and its conjugate base (A⁻). HA neutralizes added OH⁻; A⁻ neutralizes added H⁺.
Write the Henderson–Hasselbalch equation. What is the pH when [A^-] = [HA]?
Show answer
pH = pK_a + log[A^-][HA]. When [A^-] = [HA], the log term is zero, so pH = pK_a.
A buffer made from an acid with pK_a = 4.74 has [A^-]/[HA] = 10. What is its pH?
Show answer
pH = 4.74 + log(10) = 5.74.
Which resists added base better: 0.10 M HA/0.10 M A⁻ or 0.50 M HA/0.50 M A⁻? Why?
Show answer
The 0.50 M/0.50 M buffer: same pH (ratio 1:1), but capacity scales with concentration, so it absorbs five times more added base.
Why is the bicarbonate buffer effective at blood pH 7.4 even though its pK_a is about 6.1?
Show answer
Blood is an open system: the lungs exhale CO₂ and the kidneys regulate HCO₃^-, keeping the ratio that gives pH 7.4.
What is the practical range of a buffer whose weak acid has K_a = 1.8 × 10^-5?
Show answer
pK_a = -log(1.8 × 10^-5) = 4.74, so the effective range is about 3.7 to 5.7.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

buffer: A weak acid + its conjugate base solution that resists pH change
conjugate acid–base pair: Two species differing by one proton, e.g., HC₂H₃O₂ and C₂H₃O₂^-
Henderson–Hasselbalch equation: Links buffer pH to pKa and the ratio [A^-]/[HA]
buffer range: The pH interval (pK_a ± 1) in which a buffer is effective
buffer capacity: Moles of strong acid or base per liter a buffer can absorb before pH changes sharply
pKₐ: -logK_a; the pH at which a weak acid is half dissociated

Sources & references

openstax.org — Chemistry Atoms First 2e

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