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Most students can put two numbers into pH = pKa + log([salt]/[acid]). Exam questions go further: make a buffer of a given pH, find the pH after part of a weak acid is neutralised, or compare two buffers that start at the same pH. This post covers those three problem types and the limits of the equation itself.
Where does the equation come from?
The NCERT Class 11 Equilibrium chapter derives it from the ionisation constant. For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA]. Take logarithms and rearrange: pH = pKa + log([A⁻]/[HA]). NCERT then notes two consequences. With equal concentrations of acid and salt, pH = pKa, so you choose an acid whose pKa is close to the pH you want; for acetic acid that is 4.76. And because only a ratio appears, the pH of a buffer "is not affected by dilution". For a basic buffer, NCERT gives pOH = pKb + log([BH⁺]/[B]), and notes that ammonia with ammonium chloride in equal concentrations gives a buffer near pH 9.25.
The JEE (Main) 2026 and NEET (UG) 2026 syllabi both list buffer solutions under ionic equilibrium. Check your own year's syllabus. If you need the basics first, including pH of weak acids and salt hydrolysis, start with our ionic equilibrium guide.
Is the equation exact?
No. Po and Senozan traced the history of the Henderson-Hasselbalch equation and compared the approximate pH it gives with exact hydrogen ion concentrations during the titration of a weak acid with sodium hydroxide, showing the percentage error as a function of the acid's Ka and the buffer's composition. The equation assumes the acid barely ionises and the base barely hydrolyses, so the concentrations you put in stay as they are. That holds well when both forms are present at reasonable concentrations. It weakens for very dilute solutions, for fairly strong acids, and when one form is almost absent.
A common working rule follows from the log: keep the ratio between 0.1 and 10, so the pH stays within one unit of pKa.
Worked example 1: design a buffer for a target pH
Make an acetate buffer of pH 5.00 with 0.100 M acetic acid.
5.00 = 4.76 + log([acetate]/[acid]), so log(ratio) = 0.24.
Ratio = 10^0.24 = 1.74.
[acetate] = 1.74 × 0.100 = 0.174 M.
The target is within one unit of pKa, so acetic acid is a sensible choice. For pH 9, you would switch to the ammonia buffer.
Worked example 2: partial neutralisation
Mix 0.20 mol acetic acid and 0.050 mol NaOH in 1.0 L of solution. What is the pH?
The NaOH is used up: 0.050 mol acid becomes 0.050 mol acetate.
Left: 0.15 mol acid and 0.050 mol acetate.
pH = 4.76 + log(0.050 ÷ 0.15) = 4.76 − 0.48 = 4.28.
Questions rarely say "buffer". They give a weak acid and a smaller amount of strong base, and expect you to see the buffer pair after the reaction. If the base equals the acid, there is no buffer left and you need the salt hydrolysis formula instead.
Worked example 3: two buffers, same pH
Buffer A is 0.10 M acetic acid and 0.10 M acetate. Buffer B is 0.010 M of each. Both have pH 4.76. Add 0.0050 mol HCl to 1.0 L of each.
Buffer A. Acetate 0.10 − 0.005 = 0.095 mol, acid 0.105 mol. pH = 4.76 + log(0.095 ÷ 0.105) = 4.72. A change of 0.04.
Buffer B. Acetate 0.010 − 0.005 = 0.005 mol, acid 0.015 mol. pH = 4.76 + log(0.005 ÷ 0.015) = 4.28. A change of almost 0.5.
The same reasoning shows that a buffer resists change better when its ratio is near 1. Take two buffers with 0.20 M total. One is 0.10 : 0.10; the other is 0.15 M acid and 0.05 M acetate. Add 0.010 mol HCl to 1.0 L. The first moves from 4.76 to 4.67, a change of 0.09. The second moves from 4.28 to 4.16, a change of 0.12.
This resistance is called buffer capacity. Urbansky and Schock describe how to derive and compute buffer capacity, noting that the idea is used across analytical, environmental and biochemistry, physiology and medicine, but is often neglected in undergraduate teaching. For entrance exams you need only the qualitative rule: more total buffer and a ratio closer to 1 give more capacity.
Which mistakes should you avoid?
Using the equation for a weak acid alone, with no conjugate base.
Putting added strong acid or base into the log without doing the reaction in moles first.
Mixing pKa and pKb in a basic buffer. Use pOH with pKb, or pH with the pKa of the conjugate acid.
Assuming dilution never matters. The ratio is unchanged, but at very low concentrations the approximation fails.
Forgetting the 10 : 1 limit when asked which acid to choose for a target pH.
The buffer region also forms the middle of a titration curve; see our titration curves guide, where half-equivalence gives pH = pKa. If the idea of conjugate pairs is the weak link, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
A practice routine
Day 1: five design problems, each with a target pH and a choice of acid or base.
Day 2: four partial neutralisation problems, including one where the base is in excess.
Day 3: four capacity comparisons like Example 3, written as a moles table.
Day 4: a mixed set of previous-year buffer questions, timed.
Today, find the pH when 0.10 mol NH₄Cl and 0.050 mol NH₃ are dissolved in 1.0 L, using NCERT's pKa of 9.25 for the ammonium ion.
