Buffers and Henderson–Hasselbalch

A buffer solution is a mixture of a weak acid and its conjugate base, such as acetic acid and sodium acetate, that holds its pH nearly constant when acid or base is added. Added hydrogen ions are taken up by the base (A⁻ + H⁺ → HA); added hydroxide is neutralized by the acid (HA + OH⁻ → A⁻ + H₂O). The pH is set by the ratio of the two, through the Henderson–Hasselbalch equation, pH = pK_a + log₁₀([A⁻]/[HA]), and the ratio changes only slowly as the pair absorbs what you add.

Buffers are why blood stays near pH 7.4, why biochemists can study enzymes at a fixed pH, and why the middle of a weak-acid titration curve is so flat. This chapter shows a buffer at work, takes the Henderson–Hasselbalch equation apart, and maps the conditions where it gives the wrong answer.

Water against a buffer§

The two curves are the same chemistry seen at different scales. Pure water has nothing to absorb protons except its own trace of OH⁻, so the first milliliter of acid sends it from 7 to 2. The buffer holds 5 mmol of acetate ready to take protons. Each millimole of HCl you add turns one millimole of acetate into acetic acid, and the pH follows the ratio: 4.58 after 1 mL, 4.39 after 2, and so on. Near 5 mL the acetate runs out, and the buffer curve drops to meet the water curve. That point is the buffer’s limit, and it scales with its concentration: the 0.01 M buffer is overwhelmed by a single milliliter (pH 2.30).

Notice that this is the middle section of the acetic acid titration curve from Chapter 3, run in both directions from its center. A buffer is a titration paused halfway. You can make that literal: stop the titration bench at its half-equivalence point and use the link under it to carry that very flask into the figure above.

The Henderson–Hasselbalch equation§

Take the logarithm of the acid dissociation constant, K_a = [H⁺][A⁻]/[HA], and rearrange:

So far this is exact, written in activities: it is the equilibrium condition, nothing more. The approximation comes in the next step, when you substitute the amounts you mixed for the amounts present at equilibrium. For 0.05 M acetic acid with 0.05 M sodium acetate, equation (4.1) then says pH = 4.756 + log 1 = 4.756, and the exact answer is 4.756 to three decimals. To make a buffer of pH 5.00 from the same pair, you need [A⁻]/[HA] = 10^{5.00 − 4.756} = 1.75.

Henderson’s paper in the American Journal of Physiology^[1] and the history of its logarithmic form are reviewed by Po and Senozan, who also catalogue its limitations^[2][3]. documented Henderson’s question, why blood is so well buffered, is the subject of Chapter 5.

Where the approximation fails§

The substitution “as mixed = at equilibrium” is safe only if the acid and its base barely react with water, so that the protons they exchange with water are negligible compared with the amounts you put in^[3]. That fails in three situations:

• The buffer is dilute. At 10⁻⁴ M of each, acetic acid dissociates appreciably on its own, adding acetate. Henderson–Hasselbalch says 4.756; the true pH is 4.873.
• The ratio is extreme. Mix 0.1 M acetic acid with only 10⁻⁴ M acetate and the equation predicts pH 1.76. But acetic acid alone at 0.1 M already has pH 2.88; the tiny amount of acetate is swamped by what the acid makes itself. The error is more than a full unit.
• The pKa is near the ends of the scale. For phosphoric acid (pK_a 2.15) at 0.01 M of each form, the hydrogen ions are comparable to the buffer components, and the error is 0.31. A base pair with pK_a 11 fails the same way, through hydroxide.

For acetic acid, the pale region where the error is below 0.01 covers 1:1 mixtures above about 3 mM and base-rich mixtures above about 1 mM. Acid-rich mixtures need more: the acid’s own dissociation adds acetate, so a 1:10 mixture is off by 0.07 even at 0.01 M total. For strong-ish or very weak acids the safe region shrinks to a narrow band. The exact charge balance has no such limits, and it is what every figure on this site solves.

To see that gap rather than read about it, go back to the figure at the top of the page, choose the phosphate buffer and tick “Real solution”. At 0.1 M the dashed Davies curve starts 0.38 below the ideal one (6.82 against 7.20), and about 0.09 below for acetate; the shape, and so the buffer’s resistance, hardly changes. Only the level does, which is why a meter reads a real buffer lower than the equation predicts.

What dilution does to a buffer§

Dilution leaves the ratio alone, so Henderson–Hasselbalch predicts no change at all, and the exact answer agrees to 0.003. Dilute another tenfold, to 0.0005 M of each, and the pH has moved by 0.03; at 5 × 10⁻⁵ M, by 0.2. A buffer keeps its pH under dilution far better than an acid does, but what it loses is its ability to absorb acid or base.

Buffer capacity§

Buffer capacity is how much strong acid or base a buffer can absorb for a given change in its pH. Two buffers can share a pH and differ a hundredfold in capacity, because the pH depends on the ratio of the pair and the capacity on how much of the pair there is. In 1922 Donald Van Slyke defined a buffer’s strength as a number: the buffer value β, the amount of strong base per liter needed to raise the pH by one unit^[6]. established For a weak acid of total concentration C it is

The first two terms are water’s own buffering, by its H⁺ and OH⁻; they matter only below pH 3 or above pH 11. The third is the buffer pair’s, and it peaks exactly at pH = pK_a, where it equals ln 10 · C/4 ≈ 0.58 C. One unit either side it has fallen to about a third. That is the origin of the rule that a buffer works within about one unit of its pK_a. For acids with more than one proton, each pK_a contributes its own peak; the figure computes the general case, in which each acid’s share is the variance of the number of protons it holds.

For 0.1 M acetate at pH 4.76, β = 0.058 mol/L per pH unit: adding 1 mmol of strong acid per liter moves the pH by about 0.017. At pH 7, pure water’s β is 5 × 10⁻⁷, a hundred thousand times smaller, which is why one drop of acid moved the beaker in Chapter 1 by three units. And notice the curious fact the figure shows at the extremes: at pH 2 or 12, water itself is a good buffer, because there is so much H⁺ or OH⁻ already that adding a little more barely changes the logarithm.

How much acid can a buffer take?§

β is a slope, the capacity at one pH, and it changes as the buffer is used. To ask how much acid a real buffer can take you have to add up the slope over the whole change. Take the buffer in the figure at the top of the page: 100 mL of acetate buffer, 0.05 M acid and 0.05 M acetate, pH 4.76. How much 1 M HCl does it take to push the pH down a full unit, to 3.76? The exact answer, with dilution included, is 4.1 mL. illustrative You can check it by hand: at pH 3.76 the ratio [A⁻]/[HA] has to be 1:10, so of the 10 mmol of the pair only 0.91 mmol can still be acetate, and the other 4.1 mmol of acetate must have taken a proton.

Spread over that one unit, 4.1 mmol in 0.1 L is an average capacity of 0.041 mol/L per pH, less than the 0.058 at the start. The buffer is strongest at its pK_a and gets weaker as it is pushed off it, which is why a buffer that has absorbed half its capacity behaves noticeably worse than a fresh one. Make the same buffer ten times weaker, 0.005 M of each, and 0.43 mL does the same damage. Base works symmetrically: 4.1 mL of 1 M NaOH lifts the 0.1 M buffer a unit, to 5.76.

Buffer capacity on a titration curve§

Buffer capacity is the same thing you see in a weak-acid titration, read sideways. The flat middle of the acetic acid curve in Chapter 3 is the stretch where the flask holds a buffer: every milliliter of NaOH turns some acetic acid into acetate, and the pH barely moves. Apart from the effect of dilution, the slope of a titration curve is inversely proportional to β, so the curve is flattest at the half-equivalence point, where pH = pK_a and β peaks, and steepest at the equivalence point, where the acid is gone and the flask has almost no buffer capacity left. The jump that makes titration precise and the plateau that makes a buffer useful are one fact seen from two sides.

Choosing and making a buffer§

Three decisions make a buffer. First, choose a pair whose pK_a is within one unit of the pH you want, so that both forms are present in comparable amounts. Second, choose the total concentration for the capacity you need: enough to absorb the acid or base your experiment will produce, but no more than the system tolerates. Third, set the ratio from equation (4.1), and finish by adjusting with a calibrated pH meter, because of the activity corrections the equation ignores.

Temperature is a fourth decision that is easy to forget. The pK_a of acetic acid hardly changes with temperature, because its ionization releases almost no heat. Tris and ammonium, whose ionization absorbs about 47 and 52 kJ/mol, lose about 0.03 pH units for every degree they warm^[7]. A 1:1 Tris buffer at pH 8.07 at 25 °C is near pH 8.7 in a cold room at 4 °C. Goldberg and co-workers tabulate the enthalpies needed to make this correction for every common buffer.