Buffer Solutions & pH Calculations: JEE Main Guide

Buffer solutions and pH calculations sit within the ionic equilibrium chapter, one of the most formula-rich in JEE Main Chemistry. These topics contribute two to three questions per session in Physical Chemistry and require systematic application of equilibrium principles to acid-base systems. This guide covers buffer theory, Henderson-Hasselbalch equation, salt hydrolysis, and the pH calculation chain that JEE Main uses most frequently.

pH and pOH: Basics

pH = −log[H⁺]. pOH = −log[OH⁻]. At 25°C: pH + pOH = 14 (from K_w = [H⁺][OH⁻] = 10⁻¹⁴). Strong acid: pH = −log C_acid. Strong base: pOH = −log C_base, then pH = 14 − pOH. Weak acid HA ⇌ H⁺ + A⁻: [H⁺] = √(K_a × C), so pH = ½(pK_a − log C). Degree of dissociation α = √(K_a/C). For K_a << C, this approximation is valid; check that α < 0.05 to justify ignoring α² vs. C. For the general ionic equilibrium framework, see our Ionic Equilibrium Guide.

Buffer Solutions: Mechanism

A buffer resists pH change on addition of small amounts of acid or base. An acidic buffer contains a weak acid and its conjugate base (salt): e.g., CH₃COOH + CH₃COONa. If strong acid is added, the conjugate base neutralises it: CH₃COO⁻ + H⁺ → CH₃COOH. If strong base is added, the weak acid neutralises it: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. A basic buffer contains a weak base and its conjugate acid (salt): e.g., NH₃ + NH₄Cl.

Henderson-Hasselbalch equation for acidic buffer: pH = pK_a + log([A⁻]/[HA]) = pK_a + log([salt]/[acid]). Maximum buffer capacity occurs at pH = pK_a (equal concentrations of acid and salt). For basic buffer: pOH = pK_b + log([BH⁺]/[B]) = pK_b + log([salt]/[base]). JEE Main uses these equations to: (1) find pH given acid, salt concentrations and pK_a; (2) find the ratio [salt]/[acid] needed to achieve a target pH; (3) find pH after adding strong acid or base to a buffer. Take a free mock test on ionic equilibrium and buffer problems to practise these calculations.

Salt Hydrolysis and pH of Salt Solutions

Salt of strong acid and strong base (NaCl): does not hydrolyse, pH = 7. Salt of weak acid and strong base (CH₃COONa): anion hydrolyses — CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. pH > 7. pH = 7 + ½(pK_a − log C) approximately = 7 + ½pK_a + ½log C. Salt of strong acid and weak base (NH₄Cl): cation hydrolyses — NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. pH < 7. pH = 7 − ½(pK_b − log C). Salt of weak acid and weak base (CH₃COONH₄): pH = 7 + ½(pK_a − pK_b). JEE Main tests all four cases — identify which type of salt, then apply the formula.

Degree of Hydrolysis

Degree of hydrolysis h = √(K_h/C), where K_h = K_w/K_a (for anion hydrolysis) or K_h = K_w/K_b (for cation hydrolysis). For NH₄⁺ hydrolysis: h = √(K_w/(K_b × C)). JEE Main asks: "What fraction of NH₄⁺ is hydrolysed at concentration C?" — apply the formula and express h as percentage. Alternatively, given that 1% is hydrolysed (h = 0.01), find K_h = h²C.

Common Ion Effect on pH

Adding a common ion to a weak acid solution suppresses its dissociation (Le Chatelier's principle), raising pH. Example: adding 0.1 M NaA to 0.1 M HA solution — the [A⁻] from the salt dominates; [H⁺] = K_a × [HA]/[A⁻] ≈ K_a × 0.1/0.1 = K_a, so pH = pK_a. This is exactly the Henderson-Hasselbalch equation at equal concentrations. The common ion effect is the buffer mechanism — the common ion (A⁻) suppresses [H⁺] changes.

Exam Strategy

Buffer and pH problems follow a fixed algorithm: (1) identify acid-base type; (2) choose the correct formula (weak acid approximation, Henderson-Hasselbalch, salt hydrolysis); (3) substitute and compute. Common errors: forgetting to take log base 10 (not natural log), using moles instead of concentrations, or confusing pK_a and pK_b. For the solubility equilibrium that connects to ionic equilibrium, see our Solubility Product Guide. For the broader equilibrium context, see our Chemical Equilibrium Guide. Upgrade for ₹149/month for 200+ ionic equilibrium problems including buffers, hydrolysis, and Ksp.