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9476 / 2027
Chemistry of Aqueous Solutions

Lesson 4 of 6

Buffers in calculations and oceans

Follow consumption of a small acid or base addition before using equilibrium.

A-Level 9476 (2026-2027)

  • I can explain buffer action and applications.Syllabus

    Syllabus 9476, 10.1(f). Explain buffer action and applications.

  • I can calculate buffer pH.Syllabus

    Syllabus 9476, 10.1(g). Calculate buffer pH.

A buffer contains a reservoir for both added acid and added base

A weak acid/conjugate-base pair or weak base/conjugate-acid pair limits small pH changes.

Make a guess. You are not marked.

A few drops of dilute HCl are added to a buffer of ethanoic acid and sodium ethanoate. What happens to the pH?

  1. It falls only very slightly
  2. It rises, because the buffer makes more base
  3. It falls sharply, as it would in water
Show the answer

It falls only very slightly

The added H+ ions react with the large reservoir of ethanoate ions to make ethanoic acid, so few free H+ ions remain.

An acidic buffer contains a good amount of a weak acid, HA, and of its conjugate base, A-. You can make one by mixing the acid with its soluble salt, or by partly neutralising the acid. Added H+ reacts with A-. Added OH- reacts with HA. An alkaline buffer works the same way with a weak base and its conjugate acid, such as NH3/NH4+.

Follow the added species, not just an equilibrium arrow
AdditionAcidic buffer reactionAlkaline ammonia buffer reaction
A small amount of acidA- + H+ → HANH3 + H+ → NH4+
A small amount of baseHA + OH- → A- + H2ONH4+ + OH- → NH3 + H2O

These reactions stop the added strong acid or base from staying free in the solution. If the addition is small compared with both components, their ratio changes only a little, so the pH changes only a little. A buffer does not keep the pH perfectly fixed. It stops working once one component is nearly used up.

Buffers keep a suitable pH for enzyme reactions, chemical measurements and calibrating pH meters. Diluting both components by the same factor leaves their ratio about the same, so the pH changes little. But there is less of each component in every cm3, so the buffer can absorb less acid or base. At extreme dilution the simple formula no longer works.

Exam-style question

A solution contains similar, large amounts of NH3 and NH4Cl. Explain, with equations, how this solution keeps its pH almost constant when a small amount of acid or a small amount of alkali is added. [3]

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Neutralise the added reagent first, then calculate the new ratio

The equilibrium formula uses the remaining buffer components.

Rearranging Ka = [H+][A-]/[HA] gives pH = pKa + log10([A-]/[HA]). Both components are in the same volume, so their mole ratio equals their concentration ratio. Use the amounts left after any added strong acid or base has reacted. This works when the further ionisation of HA is small compared with those amounts.

Worked exampleA buffer before and after added acid

100.0 cm3 of buffer contains 0.0200 mol HA and 0.0300 mol A-. Ka = 1.80 × 10-5 mol dm-3. Find its initial pH and its pH after adding 0.00500 mol HCl, with volume effects common to both components.

Try it first. Then show the worked stepsHide the worked steps
  1. pKa = -log10(1.80 × 10-5) = 4.7447. Take the log of the number only; pKa has no units.
  2. Initial pH = 4.7447 + log10(0.0300/0.0200) = 4.92.
  3. Added H+ reacts with A-: new A- amount = 0.0250 mol and new HA amount = 0.0250 mol.
  4. The new ratio is 1, so pH = pKa.
Answer

pH changes from 4.92 to 4.74. Substituting the added HCl concentration directly into -log[H+] would ignore its reaction with the buffer.

Try it step by step

Work it out in steps: pKa = -log Ka, then Ratio [A-]/[HA] after the added acid or base reacts, then pH = pKa + log([A-]/[HA]).

  1. Problem 1. A buffer contains 0.0400 mol HA and 0.0400 mol A-, with Ka = 1.80 x 10-5 mol dm-3. 0.00500 mol NaOH is added. Find the new pH to 2 decimal places, and the ratio to 3 significant figures.
  2. Problem 2. A buffer contains 0.0300 mol HA and 0.0200 mol A-, with Ka = 6.30 x 10-5 mol dm-3. 0.00500 mol HCl is added. Find the new pH to 2 decimal places, and the ratio to 3 significant figures.
  3. Problem 3. A buffer contains 0.0500 mol HA and 0.0250 mol A-, with Ka = 1.00 x 10-5 mol dm-3. 0.00500 mol NaOH is added. Give pKa and pH to 2 decimal places, and the ratio to 3 significant figures.
Show the answer

Problem 1.

pKa = -log Ka: pKa = -log(1.80 x 10-5) = 4.74 (keep 4.7447 in the calculator).

Ratio [A-]/[HA] after the added acid or base reacts: OH- reacts with HA: HA = 0.0400 - 0.00500 = 0.0350 mol and A- = 0.0400 + 0.00500 = 0.0450 mol. Ratio = 0.0450/0.0350 = 1.29. Both are in the same volume, so amounts give the same ratio as concentrations.

pH = pKa + log([A-]/[HA]): pH = 4.7447 + log(1.29) = 4.85. A rise of only 0.11 for added strong base.

Problem 2.

pKa = -log Ka: pKa = -log(6.30 x 10-5) = 4.20.

Ratio [A-]/[HA] after the added acid or base reacts: H+ reacts with A-: A- = 0.0200 - 0.00500 = 0.0150 mol and HA = 0.0300 + 0.00500 = 0.0350 mol. Ratio = 0.0150/0.0350 = 0.429.

pH = pKa + log([A-]/[HA]): pH = 4.2007 + log(0.429) = 3.83.

Problem 3.

pKa = -log Ka: pKa = -log(1.00 x 10-5) = 5.00.

Ratio [A-]/[HA] after the added acid or base reacts: OH- reacts with HA: HA = 0.0450 mol and A- = 0.0300 mol. Ratio = 0.0300/0.0450 = 0.667.

pH = pKa + log([A-]/[HA]): pH = 5.00 + log(0.667) = 4.82.

For a weak base B and its conjugate acid BH+, you have two choices. Use Ka of BH+ in the same equation. Or use pOH = pKb + log10([BH+]/[B]), then pH = pKw - pOH. Equal amounts of NH3 and NH4+, with Kb = 1.80 × 10-5 mol dm-3, give a pH of about 9.26 at 298 K.

Check your understanding

The buffer above holds 0.0300 mol of A-. Can the buffer equation be used after adding 0.0400 mol of HCl?

  1. No. The buffer equation works for weak bases, and HA is a weak acid.
  2. Yes. Put the new amounts of HA and A- into the buffer equation as usual.
  3. No. The A- is used up, and 0.0100 mol of strong acid is left in excess.
Show the answer

No. The A- is used up, and 0.0100 mol of strong acid is left in excess.

Only 0.0300 mol A- can react, so the buffer is exhausted. Find the excess H+ concentration in the final volume and calculate pH from that.

Carbonate buffering limits acidification but cannot stop unlimited carbon dioxide input

More dissolved CO2 consumes carbonate and changes the balance of the ocean carbon system.

Follow added carbon dioxide into the carbonate system
  1. CO2 enters the water

    CO2(g) ⇌ CO2(aq). Dissolved CO2 then reacts with water. In simple form: CO2(aq) + H2O(l) ⇌ H+(aq) + HCO3-(aq).

  2. Carbonate consumes some added acid

    CO32-(aq) + H+(aq) → HCO3-(aq). This stops the free hydrogen-ion concentration rising as much.

  3. The buffer balance changes

    Carbonate is used up and hydrogencarbonate builds up, so the carbonate to hydrogencarbonate ratio falls and the buffered pH falls. Overall: CO2 + CO32- + H2O → 2HCO3-.

The pair HCO3-/CO32- is an acid and conjugate-base buffer. Carbonate accepts H+. Hydrogencarbonate can react with OH- to form carbonate and water. Seawater has other buffering species too, but this pair links rising CO2 in the air directly to how much carbonate is available.

Carbon dioxide in the air is rising fast, so surface seawater takes up more of it. The buffer resists the extra acid by using up carbonate. Its capacity is limited, so the equilibrium pH still falls. Ocean acidification means a fall in pH. Seawater does not have to drop below pH 7. With less carbonate available, it is also harder for sea creatures to form calcium carbonate shells and skeletons.

Check your understanding

The ocean contains the HCO3-/CO32- buffer. Why does ocean pH still fall slowly as atmospheric CO2 keeps rising?

  1. CO2 dissolves to form hydroxide ions, which the hydrogencarbonate buffer cannot remove.
  2. The buffer works below 25 degrees C, and warming seas have stopped it from working.
  3. Each H+ removed uses up a CO32-, so the [CO32-]/[HCO3-] ratio, and the pH, keep falling.
Show the answer

Each H+ removed uses up a CO32-, so the [CO32-]/[HCO3-] ratio, and the pH, keep falling.

CO2 + H2O <=> H+ + HCO3-, then CO32- + H+ -> HCO3-. Carbonate is used up and hydrogencarbonate builds up. A buffer resists small additions; a steady input slowly uses up its reservoir.