Buffer solutions and the Henderson equation

What is a buffer solution?

In plain words: a buffer holds a weak acid and its conjugate base side by side. Add a little strong acid and the conjugate base mops up the H⁺; add a little alkali and the weak acid mops up the OH⁻. Either way the ratio of the two changes only slightly, so the pH hardly moves. Blood pH is maintained in a narrow range by buffers together with lungs and kidneys; urine pH varies with physiological conditions, and many medical and cosmetic preparations must be kept at a particular pH (NCERT page 202).

The calculations below use the dilute ideal-solution approximation at 298 K unless stated otherwise. Exact pH uses hydrogen-ion activity; concentration ratios approximate activity ratios.

Designing an acidic buffer

For a weak acid HA + H₂O ⇌ H₃O⁺ + A⁻ (NCERT Section 6.12.1, page 203), K_a = [H₃O⁺][A⁻]/[HA], so [H₃O⁺] = K_a[HA]/[A⁻]. Taking −log of both sides gives pH = pK_a − log([HA]/[A⁻]); turning the ratio over changes the sign of its log:

In the usual buffer approximation, the weak acid is hardly ionised, so [HA] is practically the concentration of acid taken; and nearly all of the A⁻ comes from the fully ionised salt, so [A⁻] is practically the concentration of salt (page 203).

• If [salt] = [acid], log 1 = 0 and pH = pK_a. To make a buffer of a given pH, choose an acid whose pK_a is close to it, then adjust the ratio. A buffer works well only while the ratio stays between about 0.1 and 10, that is, pH = pK_a ± 1.
• For acetic acid, pK_a = 4.76, so equal concentrations of acetic acid and sodium acetate give pH about 4.76.

Dilution. For either kind of buffer, diluting changes [salt] and [acid] in the same proportion, so the prepared concentration ratio, and approximately the pH, stays the same at fixed temperature (page 204), as long as the solution is not so dilute that these approximations fail. Dilution reduces buffer capacity even when the pH hardly changes; changing ionic strength can also change activity coefficients.

Designing a basic buffer

For a weak base B and its conjugate acid BH⁺ (page 203):

At 298 K, since pH + pOH = 14 and pK_a + pK_b = 14, with K_a now the constant of the conjugate acid BH⁺:

This has the same form as the acidic case: base over acid. NCERT’s Eq. 6.42 (page 203) prints the ratio upside down, as log([BH⁺]/[B]); its own previous line (the first line above) gives the minus sign, which is the equation above. The same slip appears in its ammonia equation on page 204. Two more wording slips: the 9.25 is the pK_a of the ammonium ion, NH₄⁺, not “for ammonia” or “for the base” (ammonia’s pK_b is 4.75), and it is misprinted “s9.25”.

With equal concentrations of ammonia and ammonium chloride the log term is zero and pH = 9.25.

How a buffer resists added acid or alkali

Take an acetic acid–sodium acetate buffer (beyond the NCERT text, but the reason buffers work):

• added H⁺ is taken up by acetate: CH₃COO⁻ + H⁺ → CH₃COOH;
• added OH⁻ is taken up by acetic acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.

The strong acid or alkali is replaced by a weak one, and only the ratio [salt]/[acid] shifts a little.

A buffer can also be made by part-neutralising a weak acid with a strong base: 0.10 mol CH₃COOH + 0.05 mol NaOH leaves 0.05 mol of each, pH 4.76. With 0.10 mol NaOH all the acid is used up; sodium acetate alone is not a buffer.

Solved examples

Common mistakes to avoid

Quick revision summary

• Buffer: resists pH change on dilution or on adding small amounts of acid or alkali.
• Acidic buffer: weak acid + its salt; pH = pK_a + log([salt]/[acid]); acetate buffer about 4.76.
• Basic buffer: weak base + its salt; pOH = pK_b + log([salt]/[base]); NH₃/NH₄Cl about 9.25.
• Equal concentrations give pH = pK_a; choose an acid with pK_a near the pH wanted.
• Dilution approximately preserves pH at fixed temperature while reducing buffer capacity.

Questions students ask

What are some examples of buffer solutions in daily life?

Blood is normally regulated near pH 7.35 to 7.45 by buffers, respiration and kidney function. A major buffer is the carbonic acid–hydrogencarbonate buffer, H₂CO₃ ⇌ H⁺ + HCO₃⁻: at body temperature its pK_a is about 6.1 (counting dissolved CO₂ as carbonic acid), and blood holds about 20 times as much HCO₃⁻ as H₂CO₃, so pH = 6.1 + log 20 ≈ 7.4. Inside cells the H₂PO₄⁻/HPO₄²⁻ pair, pK_a 7.21, buffers near pH 7, as phosphate buffers do in the laboratory. Citric acid with sodium citrate buffers many soft drinks, jams and medicines.

What is buffer capacity, and why is it largest when pH = pK_a?

Buffer capacity is the local differential amount of strong base added per litre per unit rise in pH (or acid per unit fall), evaluated for small additions at a specified pH. For a weak-acid/conjugate-base pair at fixed total concentration, neglecting water’s contribution and activity effects, its contribution is maximal near equal concentrations, pH = pK_a. This is not an exact finite one-pH-unit dose. Adding 0.010 mol L^⁻¹ of H⁺ to an acetate buffer 0.050 M in each form lowers pH by 0.18; for 0.080 M acid and 0.020 M acetate the fall is 0.35.

How is a buffer solution related to the common ion effect?

A buffer of a weak acid and its salt relies on the common ion effect: the acetate ions from sodium acetate push the ionisation CH₃COOH ⇌ H⁺ + CH₃COO⁻ back, so almost all the acid stays un-ionised and [HA] and [A⁻] are simply the concentrations of acid and salt taken, as the prepared-concentration approximation to Henderson–Hasselbalch assumes. The common ion effect on its own only lowers the ionisation of the weak acid; the buffer action comes from having large amounts of both partners, one to take up added H⁺ and the other added OH⁻.

When can the Henderson–Hasselbalch equation be used?

Henderson–Hasselbalch written in equilibrium activities is an exact rearrangement of K_a: pH = pK_a + log(a_A−/a_HA). Using prepared acid/salt concentrations instead requires their ionisation/hydrolysis changes to be small and activity coefficients close to 1. For 1.0 × 10^⁻⁴ M each of acetic acid and sodium acetate, solving the ideal mass balance gives pH about 4.88; inserting the prepared equal concentrations gives 4.76. The approximation fails, not the activity equation. Ratios about 0.1–10 describe a useful buffering range, not a mathematical restriction on the identity.

Are buffer solutions in the JEE Main 2027 and NEET (UG) 2026 syllabuses?

Yes, buffer solutions are listed under ionic equilibrium in Unit 6, Equilibrium, of both the JEE Main 2027 syllabus (NTA) and the NEET (UG) 2026 syllabus (NMC). Buffer calculations find the pH of an acidic or basic buffer from the Henderson–Hasselbalch equation, the ratio of salt to acid needed for a given pH, or the small change in pH when a little strong acid or alkali is added.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the pH of a solution 0.20 M in formic acid and 0.10 M in sodium formate (K_a = 1.8 × 10⁻⁴, NCERT Table 6.6).

   Show answerHide answer

   pK_a = −log(1.8 × 10⁻⁴) = 3.74. pH = 3.74 + log(0.10/0.20) = 3.74 − 0.30 = 3.44.

2. Q2. The buffer of Example 1 is diluted ten times with water. What is its new pH?

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   Both concentrations become 0.010 M; the ratio is still 1, so the approximate pH remains 4.76; the buffer capacity is reduced.

3. Q3. Which pair would you use for a buffer of pH about 9: acetic acid/sodium acetate or ammonia/ammonium chloride? Why?

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   Ammonia/ammonium chloride: the pK_a of NH₄⁺, 9.25, is close to 9, so a ratio near 1 (0.56 for [NH₃]/[NH₄⁺]) gives pH 9. The acetate pair buffers around 4.76.