Question
A buffer solution is prepared by mixing 0.1 mol of acetic acid (CH₃COOH) and 0.15 mol of sodium acetate (CH₃COONa) in 1 litre of water. Calculate the pH of this buffer. Given: Ka of acetic acid = 1.8 × 10⁻⁵.
Solution — Step by Step
We need three things for Henderson-Hasselbalch: pKa, moles of acid, moles of conjugate base.
- Acid (CH₃COOH): 0.1 mol
- Conjugate base (CH₃COO⁻ from sodium acetate): 0.15 mol
- Ka = 1.8 × 10⁻⁵
We round to pKa ≈ 4.74. Memorise this — acetic acid’s pKa is one of the most frequently tested values in board and JEE alike.
Since both species are in the same 1 L solution, the ratio of moles equals the ratio of concentrations:
This log calculation is worth practising — JEE often gives values where you need log 2 (0.301) and log 3 (0.477) from memory.
pH of the buffer = 4.92
Why This Works
The Henderson-Hasselbalch equation comes directly from the equilibrium expression for a weak acid. We take the Ka expression, apply -log to both sides, and the [H⁺] term becomes pH while -log(Ka) becomes pKa. The magic is that [HA] and [A⁻] don’t change significantly in a buffer — the conjugate pair resists pH change by neutralising any added acid or base.
Notice the ratio in the log term: when [A⁻] > [HA], the log is positive and pH > pKa. When [A⁻] < [HA], log is negative and pH < pKa. At equal concentrations, log(1) = 0 and pH = pKa exactly. This is why buffers work best near the pKa of the acid used.
The volume cancels in the ratio — 0.15 mol / 0.10 mol gives the same number whether they’re in 1 L or 2 L. This is a key shortcut when volume isn’t given in the problem.
Alternative Method
We can solve this from first principles using the Ka expression directly, without memorising the H-H equation.
Same answer. This method is more reliable in exams if you forget the H-H form, and it also makes the assumption explicit: we treat the equilibrium concentrations as approximately equal to the initial concentrations (valid because the acid is weak and the buffer salt suppresses dissociation).
Common Mistake
Putting acid in numerator and base in denominator.
The Henderson-Hasselbalch equation is pH = pKa + log([A⁻]/[HA]) — conjugate base on top, acid on bottom. Students who flip this get pH = 4.74 + log(0.10/0.15) = 4.74 − 0.176 = 4.56, which is wrong. A quick sanity check: since we have more base than acid (0.15 > 0.10), the pH must be above pKa (4.74). Our answer 4.92 > 4.74 ✓. If you’d got 4.56, that’s below pKa — should have been a red flag.
PYQ pattern: JEE Main 2023 asked this with slightly different concentrations but the same acid. NEET has repeatedly tested buffer pH with NH₃/NH₄Cl systems. The method is identical — just use pKb and the pOH form, then convert. Same equation, same ratio logic.