# pH and Buffer Solutions > pH and buffers for JEE Chemistry: weak acid pH, Henderson-Hasselbalch equation, buffer range and capacity, blood buffer and worked buffer questions. - Canonical URL: https://prepelephant.com/topics/jee/chemistry/ph-and-buffer-jee - Exam / course: JEE · Subject: Chemistry - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "pH and Buffer Solutions", PrepElephant, https://prepelephant.com/topics/jee/chemistry/ph-and-buffer-jee ## 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.