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Chemistry 2e · Acid-Base Equilibria

Buffers

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On this page 8 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Check yourself
  7. Study tools
  8. Sources & references

In 30 seconds

A is a solution that resists changes in pH when small amounts of acid or base are added, or when it is diluted. Buffers work because they contain both members of a in appreciable concentrations — a weak acid with its conjugate base (e.g., acetic acid + acetate) or a weak base with its conjugate acid (e.g., ammonia + ammonium).

The chemistry is a direct application of the equilibrium ideas from earlier topics. The weak acid HA and its conjugate base A- are connected by

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

with

Ka = [H3O+][A−][HA]

Rearranging and taking -log of both sides gives the , the workhorse of buffer calculations:

pH = pKa + log([A−][HA])

Added H3O+ reacts with the base member (A-); added OH− reacts with the acid member (HA). As long as both members are present in quantity, the ratio [A−]/[HA] — and therefore the pH — changes only slightly.

Why this matters

  • The body runs on buffers. Blood is held near pH 7.4 by the bicarbonate buffer (H2CO3/HCO3−); cellular fluids use phosphate buffers; proteins themselves buffer. A shift of a few tenths of a pH unit in blood can be life-threatening (acidosis/alkalosis), so understanding buffers is understanding how the body stays stable.
  • Laboratory and industry: Buffer solutions calibrate pH meters, stabilize enzyme reactions, and control the pH of pharmaceuticals, food, and fermentation. Every biotech assay runs in a buffer.
  • Exams: Buffer pH calculations (Henderson–Hasselbalch), , and buffer preparation are among the highest-yield acid-base problems in general chemistry and nursing/med school prerequisites.

The college version

Core Concepts

What a buffer is made of

A buffer must contain a weak acid/base pair with both members present in comparable, non-trivial amounts:

  • Weak acid + its salt (e.g., CH3COOH + CH3COONa)
  • Weak base + its salt (e.g., NH3 + NH4Cl)

A strong acid cannot buffer: it fully ionizes, leaving no conjugate partner to react with added OH−. Similarly, a lone weak acid without added conjugate base resists base addition only weakly and is swamped by acid addition.

How a buffer resists pH change

Two reactions absorb the added stress (Le Châtelier's principle):

  • Added strong acid (H3O+): neutralized by the base member: A-(aq) + H3O+(aq) → HA(aq) + H2O(l)
  • Added strong base (OH−): neutralized by the acid member: HA(aq) + OH-(aq) → A-(aq) + H2O(l)

In both cases the added ions are converted into the other member of the pair — the total amount of acid+base is conserved, and the ratio [A−]/[HA] shifts only a little.

The Henderson–Hasselbalch equation

Derived from the Ka expression:

pH = pKa + log[A−][HA]

Three practical consequences:

  1. When [A−] = [HA], pH = pKa (the log term is zero). The best buffer for a target pH uses a weak acid whose pKa is close to that pH.
  2. Dilution does not change buffer pH (to a good approximation): both concentrations shrink by the same factor, so the ratio is unchanged.
  3. The equation is only valid within its assumptions — it works well when both concentrations are large compared with Ka-driven changes; it breaks down for very dilute or very strong buffers.

Buffer capacity and buffer range

  • Buffer capacity = the amount of strong acid or base a buffer can absorb before the pH changes significantly. Capacity is largest when both [HA] and [A−] are large, and it is maximal when the ratio is 1:1. Capacity is not the same as pH: a 1.0 M buffer and a 0.01 M buffer can have the same pH but wildly different capacities.
  • ≈ pKa ± 1. Outside this window the ratio becomes extreme (10:1 or worse), the log term dominates, and the buffer resists poorly. Choose the acid with pKa nearest the desired pH.

The bicarbonate buffer of blood

Blood uses H2CO3/HCO3− with pKa ≈ 6.3 for the first proton, yet blood pH is 7.4 — the buffer operates far from pKa because the body regulates it openly: lungs remove CO2 and kidneys manage HCO3−, continuously resupplying both members. This is a reminder that real biological buffers are dynamic systems, not just beakers.

How It Works / Step-by-Step Process

Worked example 1: pH of a buffer, then the effect of added acid

Problem. A buffer is made from 0.50 M CH3COOH and 0.50 M CH3COONa (Ka = 1.8 × 10-5, pKa = 4.74). (a) Find the initial pH. (b) Find the pH after adding 0.010 mol of HCl to 1.00 L of this buffer. (c) Compare with adding the same HCl to 1.00 L of pure water.

Solution (a):

  1. Write the Henderson–Hasselbalch equation: pH = pKa + log[A−][HA]
  2. Substitute: pH = 4.74 + log(0.50/0.50) = 4.74 + log(1) = 4.74.
  3. Equal concentrations → pH = pKa = 4.74.

Solution (b):

  1. Added H3O+ (0.010 mol) reacts with acetate: CH3COO− + H3O+ → CH3COOH + H2O.
  2. New amounts per liter: [CH3COOH] = 0.50 + 0.010 = 0.51 M; [CH3COO−] = 0.50 - 0.010 = 0.49 M.
  3. Substitute into the same formula: pH = 4.74 + log(0.490.51) = 4.74 + log(0.961) = 4.74 - 0.017 = 4.72
  4. The pH dropped by only 0.02 units.

Solution (c): In pure water, 0.010 mol HCl in 1.00 L gives [H3O+] = 0.010 M and pH = -log(0.010) = 2.00 — a change of 5 units. The buffer absorbed the same acid with a 250× smaller pH change.

Worked example 2: preparing a buffer at a target pH

Problem. You need 1.00 L of buffer at pH 5.00 using acetic acid (pKa = 4.74) and sodium acetate. What ratio of [CH3COO−]/[CH3COOH] is required?

Solution.

  1. Solve Henderson–Hasselbalch for the ratio: log[A−][HA] = pH - pKa = 5.00 - 4.74 = 0.26
  2. Exponentiate both sides (base 10): [A−][HA] = 100.26 = 1.8
  3. So use about 1.8 mol of acetate per 1.0 mol of acetic acid (e.g., 0.64 M acetate + 0.36 M acetic acid gives the ratio 1.8 and pH 5.00). The target pH sits within pKa ± 1, so this buffer will resist well.

Worked example 3: ammonia buffer (weak base system)

Problem. A buffer contains 0.20 M NH3 and 0.30 M NH4Cl. For the ammonium ion, Ka = 5.6 × 10-10, pKa = 9.25. Find the pH.

Solution.

  1. Treat NH4+ as the weak acid HA and NH3 as its conjugate base A-. Use the same formula: pH = pKa + log[NH3][NH4+]
  2. Substitute: pH = 9.25 + log(0.20/0.30) = 9.25 + log(0.667) = 9.25 - 0.18 = 9.07.
  3. The buffer is basic (pH 9.07), as expected for an ammonia-based system; note the pKa used is for NH4+, not Kb of NH3 — a classic trap.

Common Confusions

Do Not ConfuseWithDifference
Strong acid + its saltBufferStrong acids fully ionize — no conjugate partner remains to absorb base; buffers require a weak pair
pH of bufferBuffer capacitypH depends on the ratio [A−]/[HA]; capacity depends on the total amount of both members
Adding acid vs adding baseSame reactionAdded H3O+ is consumed by A-; added OH− is consumed by HA — always identify which member reacts
Kb of NH3Ka of NH4+For an ammonia buffer use pKa of the conjugate acid NH4+ (9.25), not pKb of NH3 (4.75)
Buffer rangeExact fixed pHThe buffer holds pH near pKa but the pH value depends on the ratio; range is roughly pKa ± 1
Biological buffer (blood)Closed beaker bufferBlood's bicarbonate buffer operates away from pKa (6.3 vs pH 7.4) because lungs/kidneys continuously replenish both members
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

A buffer is like a seesaw with a grown-up on each side. If a small kid jumps on one side, the grown-ups shift their weight and the seesaw barely moves. One grown-up catches "sour" (acid) pushes, and the other catches "bitter" (base) pushes. When the grown-ups are equal in size, the seesaw stays level — that's when the buffer is best. If a huge truck jumps on instead of a kid, the seesaw tips — that's the buffer getting used up.

Key takeaways

  • Buffer = weak acid + conjugate base (or weak base + conjugate acid), both at significant concentration.
  • Henderson–Hasselbalch: pH = pKa + log([A−]/[HA]).
  • At [A−] = [HA], pH = pKa; this is where capacity is greatest.
  • Added H3O+ is consumed by A-; added OH− is consumed by HA — the ratio barely changes.
  • Buffer range ≈ pKa ± 1; choose an acid with pKa within ~1 unit of the target pH.
  • Dilution changes concentrations but not the ratio, so buffer pH is dilution-resistant (ratio unchanged).
  • Capacity depends on the amounts of both members, not on the ratio alone.
  • Blood's H2CO3/HCO3− buffer (pKa ≈ 6.3) holds pH near 7.4 because lungs and kidneys continuously regenerate both members.
  • pKa = -logKa; acetic acid pKa = 4.74 (Ka = 1.8 × 10-5); ammonium pKa = 9.25.

Check yourself

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

  1. What two components must a buffer contain?

    Show answer

    A weak acid and its conjugate base (or a weak base and its conjugate acid), both at significant concentration.

  2. Write the Henderson–Hasselbalch equation and state the condition under which pH = pKa.

    Show answer

    pH = pKa + log([A−]/[HA]); pH = pKa when [A−] = [HA].

  3. A buffer of 0.40 M CH3COOH and 0.40 M CH3COONa (pKa 4.74): what is its pH?

    Show answer

    pH = 4.74 + log(0.40/0.40) = 4.74.

  4. Which member of the pair consumes added OH−?

    Show answer

    The weak acid HA reacts with OH− to form A- and water.

  5. Why is a 1.0 M buffer more effective than a 0.01 M buffer even at the same pH?

    Show answer

    Capacity depends on the amounts of both members: the 1.0 M buffer can neutralize 100× more added acid or base before the ratio (and pH) shifts appreciably.

  6. What happens to a buffer's pH when it is diluted 10-fold? Why?

    Show answer

    Essentially unchanged: both [HA] and [A−] decrease by the same factor, so their ratio — and the pH — stay the same (to a good approximation).

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

buffer
Solution containing a weak acid/base pair that resists pH change
conjugate pair
Acid/base pair differing by one proton
Henderson–Hasselbalch equation
pH = pKa + log([A−]/[HA])
buffer capacity
Amount of strong acid/base a buffer can absorb before pH changes much
buffer range
pH window ≈ pKa ± 1 over which the buffer works well
pKₐ
-logKa, the pH where [HA] = [A−]

Sources & references

  1. openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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