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Buffers

How does a buffer solution work?

AdvancedAcids & BasesLast reviewed 3 October 2026
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What is it?

A buffer is a solution that resists changes in pH when small amounts of acid or base are added.

A buffer contains both members of a conjugate pair in similar amounts:

  • a weak acid and its conjugate base, such as acetic acid (CHX3COOH\ce{CH3COOH}) with acetate ions (CHX3COOX−\ce{CH3COO-}, from sodium acetate); or
  • a weak base and its conjugate acid, such as ammonia (NHX3\ce{NH3}) with ammonium ions (NHX4X+\ce{NH4+}, from ammonium chloride).

Key idea

A buffer contains a “base part” that neutralizes added acid and an “acid part” that neutralizes added base, so the pH changes only a little.

Why does it matter?

  • Your blood is a buffer. Carbonic acid and hydrogen carbonate (HX2COX3\ce{H2CO3}/HCOX3X−\ce{HCO3-}) help keep blood pH between about 7.35 and 7.45.
  • Biology and medicine depend on buffers. Enzymes work only within narrow pH ranges, so cells, lab experiments and many medicines are buffered.
  • Chemistry and industry use them to keep reactions, food products, cosmetics and analytical tests at a steady pH.

How does it work?

1. How a buffer neutralizes added acid or base

Using HA for the weak acid and A⁻ for its conjugate base:

  • Acid added: AX−+HX3OX+→HA+HX2O\ce{A- + H3O+ -> HA + H2O}. The added HX3OX+\ce{H3O+} is used up.
  • Base added: HA+OHX−→AX−+HX2O\ce{HA + OH- -> A- + H2O}. The added OHX−\ce{OH-} is used up.

Each time, a strong acid or base is replaced by a weak one, so the pH barely moves.

2. Calculating buffer pH: the Henderson–Hasselbalch equation

pH=pKa+log⁡[AX−][HA]\text{pH} = \text{p}K_\text{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}
  • If [AX−]=[HA][\ce{A-}] = [\ce{HA}], then log⁡1=0\log 1 = 0 and pH = pKa.
  • More base than acid gives a pH above the pKa; more acid gives a pH below it.

3. Choosing a buffer

Choose a weak acid whose pKa is close to the pH you want (within about ±1). Acetic acid/acetate (pKa 4.74) buffers around pH 4–6; ammonium/ammonia (pKa 9.25) buffers around pH 8–10.

4. Buffer capacity

A buffer works only while it still has plenty of both HA and A⁻. More concentrated buffers can absorb more acid or base before the pH changes sharply. This is called buffer capacity.

Think of it like this

A buffer is like a goalkeeper and a defender on the same team. Whichever way the “attack” comes (extra acid or extra base), one of them blocks it. But if too many shots come at once, the defence is overwhelmed: that’s exceeding the buffer capacity.

More precisely

The Henderson–Hasselbalch equation is an approximation. It works best when both [HA] and [A⁻] are much larger than [H₃O⁺] and their ratio is between about 0.1 and 10. For calculations after adding acid or base, it is fine to use moles instead of concentrations in the ratio, because both are in the same volume.

Visualise it

How a buffer works. A buffer contains a weak acid HA and its conjugate base A minus. If acid is added, the base part removes it: A minus plus H3O plus gives HA plus water. If base is added, the acid part removes it: HA plus OH minus gives A minus plus water. Either way the pH changes only slightly, because pH depends on the ratio of A minus to HA.
Each part of the buffer neutralizes one kind of addition.

Worked example

Worked example: pH of a buffer

Question: A buffer contains 0.20 M sodium acetate and 0.10 M acetic acid. What is its pH? (pKa = 4.74)

pH=4.74+log⁡0.200.10=4.74+0.30=\text{pH} = 4.74 + \log\dfrac{0.20}{0.10} = 4.74 + 0.30 = 5.04

Worked example: Adding acid to a buffer

Question: 1.00 L of buffer contains 0.10 mol acetic acid and 0.10 mol sodium acetate (pH 4.74). What is the pH after adding 0.010 mol HCl? Compare with adding the same HCl to 1.00 L of pure water.

  1. The acetate removes the added acid: CHX3COOX−+HX3OX+→CHX3COOH+HX2O\ce{CH3COO- + H3O+ -> CH3COOH + H2O}
  2. New amounts: acetate 0.10 − 0.010 = 0.09 mol; acetic acid 0.10 + 0.010 = 0.11 mol
  3. pH=4.74+log⁡0.090.11=4.74−0.087=\text{pH} = 4.74 + \log\dfrac{0.09}{0.11} = 4.74 - 0.087 = 4.65

Compare: the buffer’s pH falls by only 0.09. In pure water, 0.010 M HCl would drop the pH from 7.00 to 2.00, a change of 5 units.

Worked example: A basic buffer

Question: What is the pH of a buffer containing 0.20 M ammonia and 0.10 M ammonium chloride? (pKa of NHX4X+\ce{NH4+} = 9.25)

Here the weak acid is NHX4X+\ce{NH4+} and its conjugate base is NHX3\ce{NH3}:

pH=9.25+log⁡0.200.10=\text{pH} = 9.25 + \log\dfrac{0.20}{0.10} = 9.55

Common mistake

Common mistake: Thinking any acid and base make a buffer

A strong acid with its salt (for example HCl with NaCl) is not a buffer: ClX−\ce{Cl-} is too weak a base to remove added acid. A buffer needs a weak acid with its conjugate base (or a weak base with its conjugate acid).

Common mistake: Putting the ratio upside down

In Henderson–Hasselbalch, the base (A⁻) goes on top: log([A⁻]/[HA]). More base should raise the pH above the pKa.

Common mistake: Thinking buffers keep pH perfectly constant

Buffers reduce pH changes; they don’t prevent them. Add enough acid or base and the buffer is used up, after which the pH changes sharply.

Notation note

  • Henderson–Hasselbalch is sometimes written as pH = pKa + log([base]/[acid]) or pH = pKa + log([salt]/[acid]).
  • For basic buffers, some courses use pOH = pKb + log([BH⁺]/[B]); it gives the same answer.

Remember this

Remember this

  • Buffer = weak acid + its conjugate base (or weak base + conjugate acid).
  • A⁻ removes added acid; HA removes added base.
  • pH = pKa + log([A⁻]/[HA]); equal amounts give pH = pKa.
  • Choose a buffer with pKa close to the pH you need.

Test yourself

Check your understanding before moving on.

Flashcards

9 cards

  1. Question
    What is a buffer?
    Answer

    A solution that resists changes in pH when small amounts of acid or base are added.

  2. Question
    What does a buffer contain?
    Answer

    A weak acid and its conjugate base (or a weak base and its conjugate acid), in similar amounts.

  3. Question
    How does a buffer remove added acid?
    Answer

    The conjugate base reacts with it: AX−+HX3OX+→HA+HX2O\ce{A- + H3O+ -> HA + H2O}

  4. Question
    How does a buffer remove added base?
    Answer

    The weak acid reacts with it: HA+OHX−→AX−+HX2O\ce{HA + OH- -> A- + H2O}

  5. Question
    Write the Henderson–Hasselbalch equation.
    Answer

    pH=pKa+log⁡[AX−][HA]\text{pH} = \text{p}K_\text{a} + \log\dfrac{[\ce{A-}]}{[\ce{HA}]}

  6. Question
    What is the pH of a buffer with equal concentrations of HA and A⁻?
    Answer

    pH = pKa (because log 1 = 0).

  7. Question
    How do you choose a weak acid for a buffer at a particular pH?
    Answer

    Pick one whose pKa is close to (within about 1 unit of) the target pH.

  8. Question
    Is HCl + NaCl a buffer?
    Answer

    No. Cl⁻ is too weak a base to remove added acid. A buffer needs a weak acid and its conjugate base.

  9. Question
    What is buffer capacity?
    Answer

    How much acid or base a buffer can absorb before its pH changes sharply. Higher concentrations give a larger capacity.

Quiz

7 questions

  1. Question 1EasyWhich pair of substances forms a buffer?
    Show answer

    Answer: Acetic acid and sodium acetate

    A buffer needs a weak acid and its conjugate base. Acetic acid is weak and acetate is its conjugate base. HCl/NaCl does not buffer because Cl⁻ is too weak a base.

  2. Question 2EasyA buffer contains equal concentrations of acetic acid and acetate (pKa = 4.74). What is its pH?
    Show answer

    Answer: 4.74

    With [A⁻] = [HA], log(1) = 0, so pH = pKa = 4.74.

  3. Question 3MediumWhat is the pH of a buffer that is 0.10 M acetic acid and 0.20 M sodium acetate? (pKa = 4.74)
    Show answer

    Answer: 5.04

    pH = 4.74 + log(0.20/0.10) = 4.74 + 0.30 = 5.04. The answer 4.44 puts the ratio upside down.

  4. Question 4MediumWhat is the pH of a buffer that is 0.20 M acetic acid and 0.10 M sodium acetate? (pKa = 4.74)
    Show answer

    Answer: 4.44

    pH = 4.74 + log(0.10/0.20) = 4.74 − 0.30 = 4.44. More acid than base gives a pH below the pKa.

  5. Question 5HardA buffer contains 0.10 mol of acetic acid and 0.10 mol of acetate. 0.010 mol of NaOH is added. What is the new pH? (pKa = 4.74)
    Show answer

    Answer: 4.83

    OH⁻ reacts with acetic acid: HA = 0.09 mol, A⁻ = 0.11 mol. pH = 4.74 + log(0.11/0.09) = 4.83. The pH rises only slightly.

  6. Question 6MediumYou need a buffer at pH 9.5. Which conjugate pair is the best choice?
    Show answer

    Answer: Ammonium/ammonia (pKa 9.25)

    Choose a pair whose pKa is within about 1 unit of the target pH. 9.25 is close to 9.5.

  7. Question 7EasyIn a buffer, which species removes added HX3OX+\ce{H3O+}?
    Show answer

    Answer: The conjugate base, A⁻

    AX−+HX3OX+→HA+HX2O\ce{A- + H3O+ -> HA + H2O}. The base part of the buffer neutralizes added acid; the acid part neutralizes added base.

Notes and downloads

  • Worksheet

    Buffers Worksheet

    8 questions on how buffers work, Henderson–Hasselbalch calculations and adding acid to a buffer. Answer key included.

    AdvancedFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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