Mixing Two Buffer Solutions

Introduction

What pH results when two buffers are poured together? A tempting guess is the average of their pH values, but pH is logarithmic and the mixtures contain different amounts of each species, so averaging usually gives the wrong answer. The reliable route is to count moles of every species, allow any proton transfers to run to completion, and then find the single pH at which all the conjugate pairs present are simultaneously at equilibrium.

Core explanation

Same conjugate pair. If both buffers are made from the same weak acid, for example two acetate buffers of different ratios, there is no new chemistry on mixing. The total moles of HA and of A⁻ are simply added:

n(HA)total = n(HA)₁ + n(HA)₂ and n(A⁻)total = n(A⁻)₁ + n(A⁻)₂

The pH then follows from pH = pKa + log(n(A⁻)total/n(HA)total). The answer lies between the two original pH values, but it is weighted by amounts, not by volume or by pH.

Different conjugate pairs. When buffers built from different weak acids are mixed, the stronger acid of one pair may react with the stronger base of the other. The procedure is:

1. List every acid and every base present, with their pKa values. 2. Identify the strongest acid (lowest pKa) and the strongest base (whose conjugate acid has the highest pKa). 3. If the acid's pKa is lower than the base's conjugate-acid pKa, a proton transfer occurs with K = 10^(pKa(base's conjugate acid) − pKa(acid)). 4. When K is large, let the reaction run to completion, limited by whichever reactant runs out. 5. Repeat if another favourable acid–base pair remains. 6. Use the pair that still contains substantial amounts of both members to calculate the pH.

One solution, one pH. Every conjugate pair in the final solution must satisfy its own Ka at the same [H₃O⁺]. Once the pH is known from the dominant pair, the ratio of every other pair follows, which acts as a useful check. A pair whose pKa is several units away from the final pH will be almost entirely in one form.

Mass balance still applies. The total amount of each acid framework (HA + A⁻ for acetate, NH₄⁺ + NH₃ for ammonia) is conserved; only the distribution of protons among them changes. This is the same mass-balance principle used for single solutions, now applied to a mixture.

Formulae

Same pair: pH = pKa + log[(n(A⁻)₁ + n(A⁻)₂)/(n(HA)₁ + n(HA)₂)]. Proton transfer between HA and B: K = Ka(HA)/Ka(BH⁺) = 10^(pKa(BH⁺) − pKa(HA)).

Step-by-step reasoning

1. Convert all concentrations and volumes into moles. 2. Decide whether the buffers share the same conjugate pair. 3. If not, find and complete any strongly favourable proton transfer. 4. Recalculate the amounts of each species. 5. Calculate the pH from the pair containing both members in substantial amounts. 6. Check the other pairs against that pH.

Visual explanation

Draw a pKa ladder with acetic acid (4.76) low and ammonium (9.25) high. Protons flow downhill from the lower-pKa acid to the base on the higher rung. When the flow stops, the pH sits on the rung of the pair that still holds both members.

Real-world analogy

Mixing buffers is like merging two bank accounts held in two currencies. You first convert what can be exchanged at a favourable rate, then count the balance in the currency that remains in both forms; averaging the two original statements tells you nothing.

Real-world example

In biochemistry, a sample stored in Tris buffer is often added to an assay mixture buffered with phosphate. Analysts estimate the final pH by counting moles of each buffer species and checking for proton transfer between Tris and phosphate, rather than assuming the assay buffer wins automatically.

Why?

Why can pH values not be averaged? pH is a logarithm, so averaging pH is equivalent to taking the geometric mean of [H₃O⁺], which has no physical basis. Only conserved quantities, moles, can be added on mixing.

Common misconception

"Mixing equal volumes of pH 4 and pH 6 buffers gives pH 5." This is only true by coincidence. The result depends on the moles of each species and whether the pairs react, so the true pH may lie well away from the arithmetic mean.

Worked example

Question: 0.050 mol CH₃COOH and 0.050 mol CH₃COO⁻ are mixed with 0.020 mol NH₃ and 0.020 mol NH₄⁺. Find the pH (pKa acetic acid 4.76; pKa NH₄⁺ 9.25).

Reasoning: The strongest acid is CH₃COOH (4.76); the strongest base is NH₃ (conjugate acid 9.25). K = 10^(9.25 − 4.76) = 10^4.49 ≈ 3 × 10⁴, so CH₃COOH + NH₃ → CH₃COO⁻ + NH₄⁺ goes to completion. NH₃ is limiting (0.020 mol). After: CH₃COOH 0.030 mol, CH₃COO⁻ 0.070 mol, NH₄⁺ 0.040 mol, NH₃ ≈ 0. pH = 4.76 + log(0.070/0.030) = 4.76 + 0.37 = 5.13.

Answer: pH ≈ 5.13. Check: at this pH, [NH₃]/[NH₄⁺] = 10^(5.13 − 9.25) ≈ 8 × 10⁻⁵, confirming NH₃ is negligible.

Quick check

1. Two acetate buffers are mixed. Which quantities must be added together to find the final pH? Answer: The total moles of CH₃COOH and the total moles of CH₃COO⁻ from both buffers.

Exam focus

Examiners test whether you add moles rather than pH values. For mixtures of different pairs, a clearly written proton-transfer equation, a statement of which reactant is limiting, and a final check of the minor pair usually secure full marks.

Advanced insight

When the proton-transfer constant is modest, for example between pairs whose pKa values differ by only one or two units, the reaction does not go to completion. The composition must then be found by solving the mass balances for both frameworks together with the single-pH condition, a genuinely coupled equilibrium problem.

Summary

Buffers of the same conjugate pair are mixed by adding moles of HA and A⁻ separately and recalculating the ratio. Buffers of different pairs may undergo proton transfer, which runs to completion when the pKa difference is large. The final pH is set by the pair still containing both members, and every other pair must agree with that single pH.

Practice questions

1. 100 cm³ of buffer (0.10 M HA, 0.20 M A⁻) is mixed with 200 cm³ of buffer (0.20 M HA, 0.10 M A⁻), pKa 4.76. Find the pH. Answer: HA = 0.010 + 0.040 = 0.050 mol; A⁻ = 0.020 + 0.020 = 0.040 mol; pH = 4.76 + log(0.80) = 4.66. 2. Why is the answer to question 1 not the mean of the original pH values (5.06 and 4.46)? Answer: The mean, 4.76, ignores the different amounts of each species; pH depends on the total mole ratio, not on averaged logarithms. 3. When an acetate buffer is mixed with an ammonia buffer, write the proton-transfer reaction that occurs. Answer: CH₃COOH + NH₃ → CH₃COO⁻ + NH₄⁺, with K ≈ 3 × 10⁴. 4. After mixing, a solution contains 0.040 mol NH₄⁺, 0.030 mol NH₃ and 0.060 mol CH₃COO⁻. Which pair sets the pH, and what is it? Answer: The ammonium pair, since it has both members; pH = 9.25 + log(0.030/0.040) = 9.25 − 0.12 = 9.13.