A buffer keeps its pH almost constant when small amounts of acid or alkali are added. The trick is two large reservoirs: a weak acid to mop up OH− and its conjugate base to mop up H+.
A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid and its conjugate base, for example CH3COOH with CH3COONa: added H+ is removed by CH3COO− + H+ → CH3COOH, and added OH− is removed by CH3COOH + OH− → CH3COO− + H2O. An alkaline buffer such as NH3 with NH4Cl works the same way using NH3 and NH4+. Because both reservoirs are large, their ratio hardly changes, so the pH, given by [H+] = Ka × [HA] ÷ [A−], hardly changes. Blood is buffered by H2CO3 / HCO3−.
A buffer needs a large amount of both partners. Compare each mixture below: only those with a weak acid and a large amount of its conjugate base (or a weak base and its conjugate acid) can buffer.
| Buffer | Removes added H+ | Removes added OH− |
|---|---|---|
| CH3COOH + CH3COONa (acidic, pH about 4 to 6) | CH3COO− + H+ → CH3COOH | CH3COOH + OH− → CH3COO− + H2O |
| NH3 + NH4Cl (alkaline, pH about 9) | NH3 + H+ → NH4+ | NH4+ + OH− → NH3 + H2O |
| H2CO3 + HCO3− (blood, pH 7.35 to 7.45) | HCO3− + H+ → H2CO3 | H2CO3 + OH− → HCO3− + H2O |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What is a buffer solution?
Which mixture forms a buffer?
What is the pH of a buffer containing 0.100 mol dm−3 ethanoic acid and 0.100 mol dm−3 sodium ethanoate (Ka = 1.74 × 10−5 mol dm−3)?
Which indicator should be used to titrate ethanoic acid with sodium hydroxide?
Write Ka for the weak acid, rearrange for [H+] and use the concentrations of acid and salt directly. Ethanoic acid: Ka = 1.74 × 10−5 mol dm−3.
| Buffer | [HA] : [A−] | [H+] / mol dm−3 | pH |
|---|---|---|---|
| 0.100 mol dm−3 acid + 0.100 mol dm−3 salt | 1 : 1 | 1.74 × 10−5 | 4.76 (= pKa) |
| 0.200 mol dm−3 acid + 0.100 mol dm−3 salt | 2 : 1 | 3.48 × 10−5 | 4.46 |
| 50.0 cm3 of 0.200 mol dm−3 acid + 25.0 cm3 of 0.100 mol dm−3 NaOH | 0.00750 : 0.00250 mol = 3 : 1 | 5.22 × 10−5 | 4.28 |
| 1.00 dm3 with 0.0100 mol of each, then 0.00100 mol HCl added | 0.0110 : 0.00900 | 2.13 × 10−5 | 4.67 |
Run each acid and base pairing in the simulator. Notice the buffer region on the weak acid curve, where pH = pKa at half-neutralisation, and check which indicator changes colour inside the steep section.
| Titration | pH at equivalence | Suitable indicator |
|---|---|---|
| strong acid + strong base | 7 | methyl orange (3.1 to 4.4) or phenolphthalein (8.3 to 10.0) |
| weak acid + strong base | about 9 | phenolphthalein |
| strong acid + weak base | about 5 | methyl orange |
| weak acid + weak base | no steep section | none suitable |
A buffer is not a solution whose pH never changes: it resists change from small additions, and fails once one reservoir is used up. In buffer calculations use the salt concentration for [A−], not [H+]; the √(Ka c) shortcut is for a weak acid on its own. When acid and alkali are mixed to make a buffer, work in moles after reaction before taking the ratio.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
Calculate the pH of a buffer made from 0.200 mol dm−3 ethanoic acid and 0.100 mol dm−3 sodium ethanoate (Ka = 1.74 × 10−5 mol dm−3).
50.0 cm3 of 0.200 mol dm−3 ethanoic acid is mixed with 25.0 cm3 of 0.100 mol dm−3 NaOH. What is the pH (Ka = 1.74 × 10−5 mol dm−3)?
1.00 dm3 of buffer contains 0.0100 mol ethanoic acid and 0.0100 mol sodium ethanoate. 0.00100 mol HCl is added. What is the new pH?
Which equation shows how an ammonia and ammonium chloride buffer removes added acid?
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