Buffers: How a Weak Acid and Its Conjugate Base Hold pH Steady

Two partners that each neutralize one threat

Take a mixture of acetic acid, CH₃COOH, and sodium acetate, which supplies CH₃COO⁻. If strong acid is added, the acetate ion takes up the extra protons: CH₃COO⁻ + H₃O⁺ → CH₃COOH + H₂O. If strong base is added, the acetic acid neutralizes it: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.

Both reactions go essentially to completion, so the added strong acid or base is converted into more of the weak partner. The pH still moves, because the ratio of acid to base changes, but only slightly. A solution of a weak acid alone cannot do this, because it has almost no conjugate base to absorb added acid.

Common buffer pairs include acetic acid with acetate, ammonium with ammonia, dihydrogen phosphate with hydrogen phosphate, and carbonic acid with hydrogen carbonate. The last pair helps hold human blood near pH 7.4, as OpenStax section 14.6 describes.

Deriving Henderson–Hasselbalch from the Ka expression

Start from Ka = [H₃O⁺][A⁻]/[HA] and solve for hydronium: [H₃O⁺] = Ka × [HA]/[A⁻]. Take −log of both sides to get pH = pKa + log([A⁻]/[HA]). This is the Henderson–Hasselbalch equation.

In a buffer, both HA and A⁻ are present in substantial amounts, and each suppresses the other's reaction with water. Their equilibrium concentrations are therefore almost equal to the amounts you mixed. That is why you can put the prepared concentrations straight into the equation without an ICE table.

Because both partners sit in the same volume, the ratio of concentrations equals the ratio of moles. You can use moles directly, which saves a step after acid or base is added.

• When [A⁻] = [HA], the log term is zero and pH = pKa.
• Each tenfold increase in the base-to-acid ratio raises the pH by one unit.
• More conjugate base than acid gives a pH above pKa; more acid gives a pH below it.

Worked example: the pH of an acetate buffer

A buffer contains 0.300 M CH₃COOH and 0.250 M CH₃COONa. Appendix H gives Ka = 1.8 × 10⁻⁵, so pKa = 4.745. Then pH = 4.745 + log(0.250/0.300) = 4.745 − 0.079 = 4.67.

Carry the unrounded pKa until the end. Rounding it first to 4.74 gives 4.66. The difference is within the uncertainty of a two-significant-figure Ka, but rounding once at the end keeps your answers consistent with an answer key. The pH is slightly below pKa, as expected, because there is a little more acid than base.

Worked example: adding strong acid, then strong base

Add 0.0200 mol HCl to 1.00 L of the buffer above. Do the neutralization first, as a stoichiometry step: the H₃O⁺ converts 0.0200 mol acetate into acetic acid. Acetic acid rises from 0.300 to 0.320 mol and acetate falls from 0.250 to 0.230 mol. Then pH = 4.745 + log(0.230/0.320) = 4.60. The pH fell from 4.67 to 4.60, less than a tenth of a unit.

Adding the same 0.0200 mol HCl to 1.00 L of pure water would give [H₃O⁺] = 0.0200 M and a pH of 1.70, a fall of more than five units from neutral.

Now add 0.0500 mol NaOH to a fresh 1.00 L of the original buffer. The hydroxide converts 0.0500 mol acetic acid into acetate, leaving 0.250 mol acid and 0.300 mol base. pH = 4.745 + log(0.300/0.250) = 4.82, a rise from 4.67 of less than 0.2. In pure water the same NaOH would give pH 12.70.

The same logic lets you make a buffer by partial neutralization. Mixing 0.0400 mol acetic acid with 0.0150 mol NaOH leaves 0.0250 mol acid and 0.0150 mol acetate, so pH = 4.745 + log(0.0150/0.0250) = 4.52, whatever the final volume.

Designing a buffer for a target pH

First choose a conjugate pair whose pKa is within one unit of the pH you want. Then use Henderson–Hasselbalch backward to find the ratio, and stoichiometry to turn the ratio into amounts.

Worked example. Prepare a pH 9.50 buffer from 500.0 mL of 0.200 M ammonia by adding solid ammonium chloride. The acid partner is NH₄⁺. From Appendix I, Kb(NH₃) = 1.8 × 10⁻⁵, so pKb = 4.745 and pKa(NH₄⁺) = 14.00 − 4.745 = 9.255. The required ratio is [NH₃]/[NH₄⁺] = 10 raised to (9.50 − 9.255) = 1.76.

The flask holds 0.5000 L × 0.200 mol/L = 0.100 mol NH₃, so it needs 0.100 / 1.76 = 0.0569 mol NH₄⁺. With M(NH₄Cl) = 53.49 g/mol, that is about 3.0 g of ammonium chloride, reported to two significant figures because Kb has two. In practice you would then check the pH with a calibrated meter and adjust.

Buffer capacity and the useful pH range

Buffer range is the pH span over which a buffer works well. A ratio of 10:1 or 1:10 corresponds to pH = pKa ± 1, and outside that span one partner is too scarce to absorb much more acid or base. Acetate buffers are useful from about pH 3.7 to 5.7, and ammonia buffers from about 8.3 to 10.3.

Buffer capacity is the amount of acid or base a buffer can absorb before its pH changes greatly. It depends on how much of each partner is present, not just their ratio. Compare two acetate buffers, each 1.00 L with equal acid and base. Both start at pH = pKa = 4.74. With 0.100 M of each, adding 0.00500 mol HCl lowers the pH by 0.04, to 4.70. With 0.0100 M of each, the same addition lowers it by nearly half a unit, to 4.27. The ratio was the same, but the dilute buffer had a tenth of the capacity.

A buffer is exhausted when the added acid or base consumes all of one partner. After that, the pH is set by the excess strong acid or base, and it changes sharply.

Mistakes that make a buffer answer wrong

Buffer problems mix two kinds of reasoning: stoichiometry for the neutralization and equilibrium for the pH. Keep them in that order.

• Writing the ratio upside down. It is base over acid: log([A⁻]/[HA]).
• Putting the added HCl or NaOH into the Henderson–Hasselbalch equation before doing the neutralization step.
• Using pKb in place of pKa for an ammonia buffer. Convert with pKa + pKb = 14.00 at 25 °C.
• Mixing moles for one partner with concentration for the other when they are in different volumes.
• Assuming a buffer still works after the added acid or base exceeds the amount of one partner.
• Treating a weak acid alone, or a weak acid with a strong acid, as a buffer.