A Short Recap: The Pair and the Ratio
Topic 8.4 built the buffer: appreciable amounts of a weak acid with its conjugate base, or a weak base with its conjugate acid.
Both members, in real quantity. A buffer defends in two directions, so both members must stand by in quantity. A trace of one member, such as the trace conjugate base a weak acid makes by ionizing, defends nothing.
The ratio sets the pH. Everything here hangs off one line and one question: how much of each member is there?
pH = pKa + log([A⁻] ÷ [HA])
The Effective Range: Within One Unit of pKa
A buffer works well only while both members make up a serious share of the pair: pH within one unit of pKa.
At the edges the ratio is 10 or 0.10. pH = pKa + 1 requires [A⁻] ÷ [HA] = 10, so the acid member is only about 9% of the pair; pH = pKa − 1 leaves the base member at about 9%. Beyond the edge, a modest addition aimed at the scarce member exhausts it.
Dominance is the quick read. At pH = pKa the two forms are equal. One unit away, the dominant form is about 91% of the pair; two units away, about 99%.
Additions and Their Limits: Moles First
An addition to a buffer is a stoichiometry problem first: the added ion converts one member into the other completely, and only then is the pH recalculated.
Added strong acid attacks A⁻:
A⁻ + H₃O⁺ → HA + H₂O
Added strong base attacks HA:
HA + OH⁻ → A⁻ + H₂O
Each stock moves by the moles added, one up and one down.
Why the pH barely moves. pH answers to the logarithm of the ratio, so a small percentage change in the ratio moves it only a little. Both members must still stand at the end: a stock driven to zero means the buffer is gone and the leftover strong acid or base sets the pH alone.
Capacity is the stock, per direction. Defence against acid spends A⁻, so the base member’s stock caps that direction; defence against base spends HA. A buffer rich in A⁻ but poor in HA shields well against acid and weakly against base.
Concentration scales it. Two buffers at the same pH and ratio can differ tenfold in what they absorb: 0.50 mol L⁻¹ of each member absorbs ten times the attack of 0.050 mol L⁻¹ of each, because ten times the moles stand ready. Equal members, at pH = pKa, stock both directions evenly.
Choosing and Preparing a Buffer
Pick the weak acid whose pKa is closest to the target pH.
Target pH 4.0: CH₃COOH / CH₃COO⁻ (pKa 4.74) is the pick, needing a ratio of 10⁽⁴·⁰⁰⁻⁴·⁷⁴⁾ = 0.18. HCl / Cl⁻ is never a candidate: Cl⁻ is not a base in water at all. Target pH 9.3: NH₄⁺ / NH₃ (pKa 9.26) wins, ratio about 1.1.
The blood exception. Blood holds pH 7.4 with H₂CO₃ / HCO₃⁻ (pKa 6.37), one unit below target, ratio about 10.7. It works because the system is open: breathing regulates CO₂.
CO₂ + H₂O ⇌ H₂CO₃
Three routes to make one. Mix the pair directly, in amounts that give the target ratio. Or partly neutralize a weak acid: 0.20 mol of CH₃COOH with 0.10 mol of NaOH leaves 0.10 mol of CH₃COOH alongside 0.10 mol of CH₃COO⁻ formed in place, a buffer at pH = pKa = 4.74. Or partly neutralize a weak base with strong acid.
The stop rule. The strong reagent must run out first. Add a full equivalent or more and one member is never formed; the product is a salt solution, or excess strong acid or base.
Dilution: The Same pH behind a Thinner Shield
Dilute a buffer and both members fall by the same factor, so their ratio, and therefore the pH, stays put: a 0.20 ÷ 0.20 acetate buffer diluted tenfold to 0.020 ÷ 0.020 still reads pH 4.74 (the exact solve agrees within 0.01).
What dilution takes away is staying power. Each litre of the diluted buffer holds a tenth of the A⁻ stock, so a tenth of the added acid exhausts it. The pH reading hides the loss. Pushed far enough, the members end up so dilute that water’s own ions are no longer negligible beside them.
Worked Examples
Example 1: an acetate buffer under acid attack. One litre holds 0.30 mol of CH₃COOH and 0.30 mol of CH₃COO⁻, Ka = 1.8 × 10⁻⁵. Equal members, so pH = pKa = 4.74. Add 0.020 mol of HCl: its H₃O⁺ converts CH₃COO⁻ into CH₃COOH mole for mole, leaving 0.32 mol of CH₃COOH and 0.28 mol of CH₃COO⁻. The ratio route gives pH = 4.74 + log(0.28 ÷ 0.32) = 4.69, a fall of 0.06. The full equilibrium returns [H₃O⁺] = 2.1 × 10⁻⁵ mol L⁻¹ and pH 4.69.
Example 2: an ammonia buffer under base attack. One litre holds 0.20 mol of NH₃ and 0.20 mol of NH₄Cl, Kb of NH₃ = 1.8 × 10⁻⁵. Equal members, so pH 9.26. Add 0.020 mol of NaOH: its OH⁻ converts NH₄⁺ into NH₃, leaving 0.22 mol of NH₃ and 0.18 mol of NH₄⁺. Then [OH⁻] = Kb × [NH₃] ÷ [NH₄⁺] = 1.8 × 10⁻⁵ × 0.22 ÷ 0.18 = 2.2 × 10⁻⁵ mol L⁻¹, so pOH = 4.66 and pH = 9.34. The exact solve agrees.
Example 3: an HF buffer under base attack. One litre holds 0.30 mol of HF and 0.10 mol of F⁻, pKa = 3.17. The starting pH = 3.17 + log(0.10 ÷ 0.30) = 3.17 − 0.48 = 2.69. Add 0.050 mol of NaOH: its OH⁻ converts HF into F⁻, leaving 0.25 mol of HF and 0.15 mol of F⁻. Then pH = 3.17 + log(0.15 ÷ 0.25) = 3.17 − 0.22 = 2.95, a rise of 0.26. The exact solve agrees.
Example 4: dilution counts moles, not litres. A 250 mL acetate buffer at 0.20 mol L⁻¹ of each member holds 0.050 mol of each and absorbs 0.050 mol of HCl before its CH₃COO⁻ runs out. Diluted to 1.0 L, it still holds 0.050 mol of each member, still reads pH 4.74, and still absorbs the same 0.050 mol. What fell is the capacity per litre, from 0.20 mol to 0.050 mol.
Traps That Cost Marks in Topic 8.8
“Any conjugate pair in solution is a working buffer.” A pair with one member at trace level, or both extremely dilute, has the right names and no defence.
“HCl / Cl⁻ buffers because they are conjugates.” Cl⁻ never accepts a proton in water.
“Diluting a buffer changes its pH.” Dilution divides both members by the same factor; ratio and pH stand still.
“A buffer at pH = pKa + 1.5 is fine because its pH is right.” At that distance the ratio is about 32 and one member is about 3% of the pair; the first push in its direction ends it.
“More A⁻ always means a better buffer.” Only against acid: defence against base spends HA, and a lopsided buffer can be formidable one way and defenceless the other. The ratio sets the pH; amounts set the absorbable attack.
Key Relationships
- A working buffer needs both members of a weak conjugate pair, in appreciable and comparable amounts
- Effective range: pH within about 1 unit of pKa, where [A⁻] ÷ [HA] stays between 0.10 and 10 and neither member falls below about 9% of the pair
- At pH = pKa the members are equal and the buffer is most balanced; one unit away the dominant form is about 91% of the pair
- Added H₃O⁺ converts A⁻ into HA; added OH⁻ converts HA into A⁻; recount the moles first, then use the new ratio
- Defence against acid is capped by the A⁻ stock, against base by the HA stock; once a member is exhausted, the excess sets the pH
- Capacity grows with the absolute amounts of the members, not with the ratio; the ratio sets the pH
- Choose the pair whose pKa lies closest to the target pH; prepare by mixing the pair, or by partial neutralization
- Dilution leaves ratio and pH unchanged but shrinks both stocks
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