Buffer Ratio Design

Introduction

Once a conjugate pair has been chosen, the next decision is how much of each form to use. Two separate numbers must be fixed: the ratio of base to acid, which sets the pH, and the total concentration, which sets how much acid or base the buffer can absorb. This page focuses on turning a target pH into a precise ratio and then into amounts of each component for a given total concentration.

Core explanation

From target pH to ratio. Rearranging Henderson–Hasselbalch,

log([A⁻]/[HA]) = pH − pKa, so r = [A⁻]/[HA] = 10^(pH − pKa)

If the target equals pKa, r = 1. If it is 0.30 above pKa, r = 2. If it is 0.30 below, r = 0.5. A positive difference means more base than acid.

Splitting the total. Buffers are usually specified by a total concentration C, such as "0.100 mol dm⁻³ phosphate buffer, pH 7.40". With C = [HA] + [A⁻] and [A⁻] = r[HA]:

[HA] = C ÷ (1 + r) and [A⁻] = rC ÷ (1 + r)

The base fraction r/(1 + r) is the same as the distribution fraction α(A⁻) at the target pH, which links buffer design to speciation diagrams.

Choosing the total concentration. The ratio fixes pH but not capacity. A total concentration of 0.01–0.2 mol dm⁻³ is typical: high enough to absorb expected additions, low enough to avoid large ionic-strength effects or interference. The capacity required should be estimated from the amount of acid or base the system will generate.

Ratio sensitivity. Small errors in weighing or measuring change r slightly. Because d(pH) = 0.434 × dr/r, a 5% error in the ratio gives a pH error of only about 0.02. This forgiving behaviour is one reason buffers are practical: pH is much less sensitive to proportions than to the choice of pKa.

Polyprotic ratio design. For a pair from a polyprotic acid, use the pKa of that particular step and remember that small amounts of the neighbouring species may be present. For phosphate at pH 7.40, pKa₂ applies and H₃PO₄ and PO₄³⁻ are negligible, so the two-species treatment is accurate.

Real preparation routes. The required ratio may be achieved by mixing the two components directly, or by partially neutralising the weak acid (or weak base) with strong base (or strong acid). The same ratio target applies in both cases; only the stoichiometry of getting there differs.

Formulae

r = 10^(pH − pKa); [HA] = C/(1 + r); [A⁻] = rC/(1 + r); moles = concentration × volume.

Step-by-step reasoning

To design a buffer of given pH, total concentration and volume:

1. Find pH − pKa for the chosen pair. 2. Calculate r = 10^(pH − pKa). 3. Find [HA] and [A⁻] from C and r. 4. Multiply by the volume to get moles of each component. 5. Convert moles to masses or volumes of stock solutions, then check the pH experimentally and adjust.

Visual explanation

Picture a bar of fixed length C divided into two segments, HA and A⁻. Sliding the divider changes the pH: at the centre pH = pKa; moving it so the A⁻ segment is twice the HA segment raises pH by 0.30. Changing the bar length changes capacity but not pH.

Real-world analogy

Buffer design is like mixing a drink from concentrate. The ratio of concentrate to water sets the flavour strength (the pH), while the total volume made determines how many people can be served (the capacity). You can make a small or large jug with exactly the same taste.

Real-world example

Biochemistry laboratories keep tables of volumes of NaH₂PO₄ and Na₂HPO₄ stock solutions that give each pH from about 5.8 to 8.0. These tables are calculated from exactly this ratio method, then corrected empirically for activity effects.

Why?

Why does the ratio, not the absolute amounts, fix pH? In the Ka expression, [H₃O⁺] = Ka[HA]/[A⁻]. Doubling both [HA] and [A⁻] leaves their quotient unchanged, so [H₃O⁺] is unchanged.

Common misconception

"To raise the pH of a buffer by one unit, double the amount of base." Doubling the ratio raises pH by only log 2 = 0.30. A one-unit rise requires a tenfold increase in the ratio.

Worked example

Question: Design 1.00 dm³ of an ethanoate buffer with pH 5.00 and total concentration 0.200 mol dm⁻³ (pKa = 4.76).

Reasoning: pH − pKa = 0.24, so r = 10^0.24 = 1.74. [HA] = 0.200 ÷ 2.74 = 0.0730 mol dm⁻³. [A⁻] = 1.74 × 0.0730 = 0.127 mol dm⁻³. Check: 0.0730 + 0.127 = 0.200.

Answer: 0.073 mol ethanoic acid and 0.127 mol sodium ethanoate in 1.00 dm³.

Quick check

1. What conjugate ratio [A⁻]/[HA] is needed for a buffer whose pH is exactly 1.00 unit below its pKa? Answer: 0.10, which is one part base to ten parts acid.

Exam focus

Exam questions often give pKa, target pH and total concentration or volume. Show the rearranged equation, calculate r, split the total and give moles or masses. Keep enough significant figures in r to avoid rounding errors in the split.

Advanced insight

For accurate work, design using the apparent pKa at the intended ionic strength. Because the ionic strength itself depends on the ratio (especially for charged pairs such as H₂PO₄⁻/HPO₄²⁻), the design becomes iterative: estimate the ratio, calculate the ionic strength, correct pKa, and recalculate until the values converge.

Summary

The target pH fixes the conjugate ratio through r = 10^(pH − pKa). The chosen total concentration C is then split into [HA] = C/(1 + r) and [A⁻] = rC/(1 + r). Ratio controls pH; total concentration controls capacity. pH is insensitive to small ratio errors, but a one-unit shift needs a tenfold change in ratio.

Practice questions

1. Calculate the ratio [NH₃]/[NH₄⁺] needed for pH 9.50 (pKa = 9.25). Answer: 10^0.25 = 1.78. 2. Design 500 cm³ of 0.100 mol dm⁻³ phosphate buffer at pH 7.20 (pKa₂ = 7.20). Give the moles of each component. Answer: r = 1, so each component is 0.0500 mol dm⁻³; moles = 0.0500 × 0.500 = 0.0250 mol each of H₂PO₄⁻ and HPO₄²⁻. 3. A buffer designed for r = 1.50 is made with r = 1.60 by mistake. Estimate the pH error. Answer: log(1.60/1.50) = 0.028, so the pH is about 0.03 too high. 4. What base fraction of a 0.200 mol dm⁻³ ethanoate buffer is present as ethanoate at pH 4.26? Answer: r = 10^(−0.50) = 0.316; fraction = 0.316/1.316 = 0.24, so 0.048 mol dm⁻³ ethanoate.