Worked example 1
A buffer contains 0.40 mol CH₃COOH and 0.25 mol CH₃COO⁻ in 1.00 L. Ka = 1.8 × 10⁻⁵. Find the pH.
Try it first: Convert Ka to pKa before anything else.
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Acids, Bases & Aqueous EquilibriaAcids and BasesContent level: Core 26 min
Buffers and the Henderson-Hasselbalch Equation
What you'll be able to do: Explain how a buffer resists pH change and calculate buffer pH before and after adding acid or base.
Introduction
Blood holds its pH within about 0.05 units of 7.4 despite everything you eat. The chemistry that makes that possible is a buffer, and it is the most practically important idea in this unit.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Recognize what makes a mixture a buffer
- Apply the Henderson-Hasselbalch equation
- Calculate the new pH after adding strong acid or strong base to a buffer
- Distinguish buffer range from buffer capacity
Lesson
What a buffer is
A buffer is a solution containing significant amounts of a conjugate pair, such as CH₃COOH with CH₃COO⁻, or NH₄⁺ with NH₃. Added OH⁻ is consumed by the acid component and added H₃O⁺ is consumed by the base component, so neither accumulates.
You can make a buffer by mixing a weak acid with its salt, or by partially neutralizing a weak acid with a limited amount of strong base.
The Henderson-Hasselbalch equation
Rearranging Ka = [H₃O⁺][A⁻]/[HA] and taking negative logs gives the working equation. Because it involves a ratio, moles can be used instead of concentrations whenever both species share the same volume.
pH = pKa + log([A⁻]/[HA])
Ratio, not amount
Doubling the volume halves both concentrations and leaves the ratio, and hence the pH, unchanged. This is why diluting a buffer does not move its pH appreciably, unlike diluting a strong acid.
Buffer pH depends on the ratio; buffer capacity depends on the absolute amounts.
Choosing a buffer
Equal amounts of the two components gives log 1 = 0, so pH = pKa exactly. Beyond a 10:1 or 1:10 ratio the buffer is nearly exhausted on one side, which sets the practical range of pKa plus or minus 1. To buffer at pH 7.4 you choose an acid with pKa near 7.4, which is why the dihydrogen phosphate system (pKa2 = 7.21) is used in the lab.
Adding strong acid or base
Treat the addition as a reaction that runs to completion: added OH⁻ converts HA into A⁻, and added H₃O⁺ converts A⁻ into HA. Update the moles, then use Henderson-Hasselbalch again. If the addition exceeds the buffer component available, the buffer is destroyed and you have a simple excess strong acid or base problem.
n(HA) - n(added OH⁻) and n(A⁻) + n(added OH⁻)
Buffers in the body
Blood uses the carbonic acid / bicarbonate pair. Its pKa1 of 6.35 is not close to 7.4, but the lungs continuously exhale CO₂, holding the carbonic acid side low and keeping the ratio around 20:1. An open system beats the usual range rule.
Same pH can mean different capacity
Henderson-Hasselbalch depends on the ratio [A⁻]/[HA], so two buffers with the same ratio have about the same initial pH. Capacity depends on the absolute amounts available to consume added acid or base. A ten-times more concentrated matched pair resists a given addition better even though its initial pH is the same.
Ratio sets initial pH; total moles set resistance to disturbance.
Key ideas
Definition
Buffer
A solution containing appreciable amounts of both members of a conjugate pair.
Rule
Henderson-Hasselbalch
pH = pKa + log([A⁻]/[HA]); moles may replace concentrations.
Rule
Buffer range
Approximately pKa plus or minus 1 pH unit.
Rule
Buffer capacity
Set by absolute concentrations, not the ratio.
Rule
Additions
Strong acid or base reacts completely, converting one component into the other.
Equation
Henderson-Hasselbalch
pH = pKa + log([A⁻] / [HA])
- [A⁻] = conjugate base amount
- [HA] = weak acid amount
Equation
Equal components
pH = pKa when [A⁻] = [HA]
- pKa = -log Ka of the weak acid
Worked examples
Worked example 2
To 1.00 L of that buffer, 0.10 mol of NaOH is added. Find the new pH.
Try it first: Decide which component the hydroxide attacks.
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Worked example 3
You need a buffer at pH 9.00. Choose between acetic acid (pKa 4.74) and ammonium (pKa 9.25), and give the required ratio.
Try it first: Match the pKa to the target pH first.
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Worked example 4
Buffer A has 0.10 mol HA and 0.10 mol A⁻. Buffer B has 1.0 mol of each in the same volume. Compare initial pH and capacity.
Try it first: Each buffer has base-to-acid ratio 1.
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Common mistakes
Calling a mixture of a strong acid and its salt a buffer.
Why it's wrong: Cl⁻ has no basicity, so nothing can neutralize added acid.
Check instead: A buffer needs a weak conjugate pair.
Thinking dilution changes buffer pH substantially.
Why it's wrong: Both components dilute equally and the ratio is unchanged.
Check instead: Only capacity falls.
Inverting the ratio in Henderson-Hasselbalch.
Why it's wrong: Base goes on top; inverting flips the sign of the correction.
Check instead: More base than acid must give a pH above pKa.
Using an ICE table for the strong base addition.
Why it's wrong: The neutralization goes to completion, so it is a stoichiometry step.
Check instead: Subtract moles first, then apply the buffer equation.
Applying Henderson-Hasselbalch after the buffer is exhausted.
Why it's wrong: With one component gone the ratio is meaningless.
Check instead: Check that both components survive the addition.
After the lesson · Question bank
Independent practice
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What you should now know
A buffer contains appreciable amounts of both a weak acid and its conjugate base, so it can neutralize added base with the acid component and added acid with the base component. Its pH follows the Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), which shows that pH depends on the ratio of the two components rather than their absolute amounts: diluting a buffer barely changes its pH. A buffer works best when the ratio is near 1, that is when pH is near pKa, and the useful range is roughly pKa plus or minus 1. Buffer capacity, the amount of acid or base that can be absorbed, depends on the absolute concentrations. To handle an addition, work in moles: apply the strong-acid-strong-base reaction to convert one component into the other, then put the new mole amounts back into the equation, since the volume cancels in the ratio.
- A buffer holds appreciable amounts of both members of a conjugate pair
- pH = pKa + log([A⁻]/[HA]), so the ratio sets the pH
- Buffers work best within about one unit of the pKa
- Dilution changes capacity, not pH
- Handle additions as complete reactions in moles, then reapply the equation
Sources and further reading
This lesson is original Chem Help content. No external sources were adapted.