AP Chemistry Buffers

Which mixtures are buffers, what happens when acid or base is added, and how to make or choose one, in five worked examples.

A buffer is a solution that resists large changes in pH when small amounts of strong acid or strong base are added. The AP exam asks about buffers in words more often than in numbers: identify a buffer, write the reaction that happens when acid or base is added, and explain why the pH barely moves. The Unit 8 review summarizes the idea, and the Henderson-Hasselbalch equation page handles the pH arithmetic. This page covers the reasoning in between, with no numerical pH-change calculations, which the course framework leaves out.

What Makes a Mixture a Buffer

A buffer needs two things at once:

• A weak acid and its conjugate base (HF and F⁻) or a weak base and its conjugate acid (NH₃ and NH₄⁺).
• Substantial amounts of both. One component has to absorb added H⁺, the other added OH⁻, so neither can be nearly absent.

The conjugate base is the part that reacts with added acid, and the weak acid is the part that reacts with added base. Because each reaction swaps one component for the other, the ratio [A⁻]/[HA] changes only a little, and since pH depends on that ratio, pH changes only a little. A buffer works best when the ratio is between about 0.1 and 10, which puts the pH within one unit of the pK_a.

Worked Example 1: Deciding Which Mixtures Are Buffers

Decide whether each solution is a buffer: (a) HF and NaF, (b) HCl and NaCl, (c) NH₃ and NH₄Cl, (d) acetic acid alone, (e) NaOH and NaCl.

(a) Yes. HF is a weak acid and F⁻, from NaF, is its conjugate base.
(b) No. HCl is a strong acid and Cl⁻ is a negligible base, so there is no conjugate pair to absorb added acid.
(c) Yes. NH₃ is a weak base and NH₄⁺, from NH₄Cl, is its conjugate acid.
(d) No. Acetic acid alone has almost no acetate ion, so nothing can absorb added H⁺.
(e) No. NaOH is a strong base and Na⁺ and Cl⁻ do not react with added acid or base.

The test is the same each time: is there a weak acid or base, with a meaningful amount of its conjugate? Strong acids and bases and their spectator ions fail it. Sorting acids by strength is covered in strong acids and bases.

Worked Example 2: Writing the Reactions When Acid or Base Is Added

A buffer contains HF and NaF. Write the net ionic equation for the reaction when (a) HCl is added and (b) NaOH is added.

(a) Added acid reacts with the conjugate base: F⁻(aq) + H⁺(aq) → HF(aq).
(b) Added base reacts with the weak acid: HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l).

Both reactions have large equilibrium constants. For (a), K = 1/K_a = 1/(6.8 × 10⁻⁴) ≈ 1.5 × 10³, and for (b), K = K_a/K_w ≈ 6.8 × 10¹⁰, so the added strong acid or base is consumed almost completely. The same pattern holds for NH₃/NH₄⁺: NH₃ + H⁺ → NH₄⁺ and NH₄⁺ + OH⁻ → NH₃ + H₂O. More on writing these is in net ionic equations.

Worked Example 3: Why the pH Barely Moves and When It Finally Does

A 1.00 L buffer contains 0.30 M weak acid HA and 0.45 M conjugate base A⁻. Compare what happens when 0.020 mol of strong acid is added to it and to 1.00 L of pure water, then to a buffer ten times as concentrated.

Buffer, strong acid added: 0.020 mol H⁺ consumes 0.020 mol A⁻ and makes 0.020 mol HA, so A⁻ goes from 0.45 to 0.43 mol and HA from 0.30 to 0.32 mol. The ratio [A⁻]/[HA] falls only from 1.50 to 1.34, so the pH shifts only slightly.
Pure water: nothing absorbs the acid, so [H⁺] = 0.020 M and pH = 1.70, a drop of more than five units.
Ten times more concentrated buffer (3.0 M HA, 4.5 M A⁻): the same 0.020 mol moves the ratio only from 1.50 to 1.48.

The comparison shows two ideas. A buffer resists change because the ratio moves a little, not because nothing happens, and a larger amount of each component, which means greater buffer capacity, resists even more. Capacity runs out when one component is used up: adding 0.50 mol of strong acid to the original buffer would consume all 0.45 mol of A⁻, leave excess H⁺, and the solution would behave like a strong acid solution again. The exam asks you to explain this behavior, not to compute the exact pH; the calculation method is in the Henderson-Hasselbalch guide.

Worked Example 4: Making a Buffer by Partial Neutralization

A student starts with 0.100 mol HF. Which of these additions of NaOH gives a buffer: (a) 0.060 mol, (b) 0.100 mol, (c) 0.140 mol?

NaOH converts HF into F⁻ one-for-one: HF + OH⁻ → F⁻ + H₂O.
(a) 0.060 mol NaOH leaves 0.040 mol HF and makes 0.060 mol F⁻. Both are present, so it is a buffer. It has more base than acid, so it absorbs added acid better than added base.
(b) 0.100 mol NaOH converts all the HF to 0.100 mol F⁻. No HF is left, so it is not a buffer; it is a weak base solution at the equivalence point.
(c) 0.140 mol NaOH uses up all the HF and leaves 0.040 mol excess OH⁻. This is not a buffer; the strong base controls the pH.

A weak acid becomes a buffer only when the strong base added is more than zero and less than the moles of acid. When exactly half the acid has been neutralized (0.050 mol NaOH here), HF and F⁻ are equal and pH = pK_a, the half-equivalence point on a titration curve. The titration view is in titration curve pH calculations.

Worked Example 5: Choosing the Right Buffer for a Target pH

A buffer is needed at pH 7.40. Which conjugate pair is best: acetic acid/acetate (K_a = 1.8 × 10⁻⁵), HOCl/OCl⁻ (K_a = 3.0 × 10⁻⁸), NH₄⁺/NH₃ (K_a = 5.6 × 10⁻¹⁰), or HF/F⁻ (K_a = 6.8 × 10⁻⁴)?

Convert each K_a to pK_a: acetic acid 4.74, HOCl 7.52, NH₄⁺ 9.25, HF 3.17.
Pick the pK_a closest to the target pH: HOCl, at 7.52, is only 0.12 from 7.40, and the others are 2 to 4 units away.
Check the ratio: [OCl⁻]/[HOCl] = 10^{(7.40 − 7.52)} ≈ 0.76, comfortably between 0.1 and 10. Acetic acid would need 10^{(7.40 − 4.74)} ≈ 457, almost no acid left, so it would barely resist added base.

Choosing by pK_a is the standard design rule: use the pair whose pK_a is within one unit of the target, then set the ratio with the Henderson-Hasselbalch equation. The acid-base vocabulary behind these pairs is in Brønsted-Lowry acids and bases.

Five Buffer Practice Questions With Answers

1. Which of these mixtures can act as a buffer: HCl and NaCl, CH₃COOH and CH₃COONa, or NaOH and NaCl?
   Show answer

   CH₃COOH and CH₃COONa, a weak acid with its conjugate base. The other two contain a strong acid or base and spectator ions only.
2. A buffer contains HCOOH and HCOONa. Write the net ionic equation for the reaction with added NaOH.
   Show answer

   HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(l).
3. Which component of an NH₃/NH₄Cl buffer reacts with added HCl?
   Show answer

   NH₃, the weak base: NH₃ + H⁺ → NH₄⁺.
4. Buffer X is 0.10 M HA and 0.10 M A⁻; buffer Y is 1.0 M HA and 1.0 M A⁻. Both have the same pH. Which resists added acid better, and why?
   Show answer

   Buffer Y. It has the same pH but ten times as many moles of A⁻ to react with added H⁺, so its capacity is greater and the ratio changes less.
5. 0.20 mol of acetic acid reacts with 0.10 mol of NaOH. Is the result a buffer, and how does its pH compare with the pK_a?
   Show answer

   Yes. It leaves 0.10 mol CH₃COOH and makes 0.10 mol CH₃COO⁻. The two are equal, so the ratio is 1 and pH = pK_a = 4.74.

Common Buffer Mistakes

• Counting a strong acid and its salt as a buffer. HCl with NaCl has no weak conjugate pair, so it cannot absorb added acid or base.
• Saying a buffer keeps the pH constant. It limits the change; the ratio, and so the pH, still shifts a little.
• Reacting added acid with the wrong component. Added H⁺ reacts with the conjugate base; added OH⁻ reacts with the weak acid.
• Requiring equal amounts. Any ratio from about 0.1 to 10 gives a working buffer; equal amounts are only the special case where pH = pK_a.
• Forgetting that capacity is limited. Once a component is used up, the buffer fails and the leftover strong acid or base sets the pH.
• Calling the equivalence point a buffer. At equivalence only the conjugate remains, so one half of the pair is missing.

A solid grip on Unit 8 helps on both the multiple-choice and free-response sections. To see how it affects your overall estimate, the AP Chem Score Calculator turns section scores into a 1–5 range.

Related Resources

• AP Chem Score Calculator
• Unit 8 Review: Acids and Bases
• AP Chemistry Henderson-Hasselbalch Equation
• AP Chemistry Titration Curve pH Calculations
• AP Chemistry Strong Acids and Bases
• AP Chemistry Bronsted-Lowry Acids and Bases
• How to Write Net Ionic Equations
• AP Chemistry Study Guide

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