pH and Buffer Solutions
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Direct answer
Add a drop of strong acid to pure water and the pH crashes; add the same drop to a mixture of acetic acid and sodium acetate and almost nothing happens. That mixture is a buffer — a weak acid with its conjugate base (CH3COOH/CH3COONa), or a weak base with its conjugate acid (NH4OH/NH4Cl) — and its pH is set by the Henderson-Hasselbalch equation: pH = pKa + log([salt]/[acid]). At 25°C the supporting facts are pH + pOH = 14 and Kw = 1.0 × 10^-14, while a lone weak acid obeys [H+] = sqrt(Ka × C), giving pH = 1/2(pKa − log C). A buffer works because added H+ is eaten by the acetate ion and added OH- by the acetic acid, so the ratio inside the logarithm barely moves.
What you must remember
- Henderson-Hasselbalch, both forms: pH = pKa + log([salt]/[acid]) for acidic buffers; pOH = pKb + log([salt]/[base]) for basic buffers — convert to pH at the end.
- Capacity and range: capacity is maximum when acid and salt are equimolar (then pH = pKa), and the useful working range is pKa ± 1, a ten-fold ratio either way.
- Dilution immunity: diluting a buffer leaves the salt-to-acid ratio unchanged, so pH stays put — only the capacity (total reserve) falls; this is a favourite assertion-reason pair.
- Half-neutralisation identity: when a weak acid is exactly half titrated by strong base, pH = pKa — the cleanest single inference on any titration curve.
- Blood buffer: pH = 6.1 + log([HCO3-]/[H2CO3]); with the bicarbonate-to-carbonic acid ratio near 20, the pH sits at 7.4 — the physiological anchor JEE loves to quote.
- Constant-product pair: Ka × Kb = Kw, so pKa + pKb = 14 at 25°C for any conjugate pair.
- Weak-acid approximation: [H+] = sqrt(Ka C) holds when C is at least about 100 times Ka; otherwise solve the quadratic.
A buffer defending itself, with numbers
Build the classic: 0.2 M CH3COOH with 0.3 M CH3COONa, pKa 4.74. pH = 4.74 + log(0.3/0.2) = 4.74 + log 1.5 = 4.74 + 0.18 = 4.92. Now punish it with 0.01 mol of HCl per litre: the acetate consumes it, salt falls to 0.29 M and acid rises to 0.21 M, so pH = 4.74 + log(0.29/0.21) = 4.74 + 0.14 = 4.88. A shift of four-hundredths of a unit. The same 0.01 mol of HCl dropped into pure water gives [H+] = 0.01 M and pH 2 — two whole units. The entire pedagogy of buffers sits in that comparison.
For the basic buffer, 0.1 M NH4OH with 0.2 M NH4Cl: pOH = pKb + log(0.2/0.1) = 4.74 + 0.30 = 5.04, so pH = 8.96. Note the working habit — compute pOH first, then subtract from 14 — because students who force the acidic formula onto ammonia buffers get answers that are wrong by units, not decimals.
How the exam frames it
JEE Main's versions are engineered around clean logs (log 2 = 0.30, log 3 = 0.48, log 5 = 0.70), so a well-set buffer question answers itself in one line if you know the equation and fights you for four minutes if you do not. The standard traps: applying Henderson-Hasselbalch to a strong acid with its salt (no buffer exists there); using the equation on a weak acid alone (no conjugate reservoir); and forgetting that mixing a weak acid with strong base consumes the acid — only the leftover acid plus the salt formed makes a buffer, so acid must be in excess. JEE Advanced extends to selection problems: which pair among four options buffers at pH 7 (equal molar ammonium hydroxide and ammonium chloride buffers near 9.25, so the answer is usually an equimolar weak acid-salt pair whose pKa is near 7) and to titration-curve reading, where the half-equivalence point hands you pKa for free.
Frequently asked questions
When is the Henderson-Hasselbalch equation valid?
For a buffer built from a weak acid or base and its conjugate, with concentrations well above Ka, typically within the pKa ± 1 working range.
Why does dilution not change a buffer's pH?
The salt-to-acid concentration ratio is unchanged by dilution, and pH depends on that ratio alone; only the buffer capacity decreases.
At what point in a titration does pH equal pKa?
At half-neutralisation of a weak acid by a strong base, where [acid] = [salt], the log term vanishes and pH = pKa.
How does the bicarbonate buffer hold blood at pH 7.4?
Through pH = 6.1 + log([HCO3-]/[H2CO3]) with the ratio held near 20 by respiration and kidney excretion, giving 6.1 + 1.3 = 7.4.
Can a mixture of HCl and NaCl act as a buffer?
No — chloride is the conjugate of a strong acid and has no tendency to consume added base, so there is no conjugate pair reservoir.