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IONIC EQUILIBRIUM — PH CALCULATIONS
(COMPLETE NOTES)
Concise, exam-focused guide covering every pH problem JEE Main & Advanced asks: core definitions, solution strategies, validity checks, common traps, and one- look revision formulas. Read as a sharp dropper would — know when a shortcut applies, the assumptions that justify it, and the traps that make 90% of students lose marks.
FOUNDATIONS (MUST NEVER FORGET)
DEFINITIONS
pH = −log10[H⁺]; pOH = −log10[OH⁻]; pKw = pH + pOH (at given T). At 25°C Kw = 1×10⁻ 14 → pKw = 14. pKa = −log Ka; pKb = −log Kb; pKa + pKb = pKw for conjugate pairs.
TEMPERATURE TRAP
Kw may change with T (e.g., 10⁻12 at some T). Neutral pH = ½ pKw — recompute when T ≠ 25°C.
STRONG ACIDS & BASES
NORMAL CONCENTRATION (C ≥ 10 ⁻ 6 M)
Assume 100% dissociation: [H⁺] = C ⇒ pH = −log C. For strong base, [OH⁻] = C ⇒ pH = 14 − (−log C). Diprotic strong species: factor of 2 (e.g. H₂SO₄ ⇒ [H⁺] = 2C).
VERY DILUTE (C ≲ 10 ⁻6)
Water auto-ionization matters. Use quadratic: [H⁺]² − C[H⁺] − Kw = 0 → [H⁺] = (C + √(C² + 4Kw))/2. Acid cannot make pH > neutral; extreme dilutions trend toward neutral pH.
WEAK ACIDS/BASES — OSTWALD'S LAW & VALIDITY
For HA ⇌ H⁺ + A⁻, Ka = Cα²/(1−α). If α ≪ 1 (C/Ka ≥ 100), approximate Ka ≈ Cα² → α = √(Ka/C), [H⁺] = √(Ka·C), pH = ½(pKa − log C). For bases: [OH⁻] = √(Kb·C), pOH = ½(pKb − log C), pH = 14 − pOH.
Always check C/Ka ≥ 100 before using √ shortcut; otherwise solve the full quadratic — classic Advanced twist.
POLYPROTIC WEAK ACIDS
When Ka1 ≫ Ka2 (≫ Ka3), second ionisations are negligible for [H⁺]: treat as monoprotic with Ka1 → [H⁺] ≈ √(Ka1·C). Shortcut: for [A2−] in diprotic acid (no other sources), [A2−] ≈ Ka2 (independent of C) — appears often in precipitation problems.
SALT HYDROLYSIS — FOUR CASES
CASE 1: STRONG ACID + STRONG BASE
No hydrolysis (NaCl, KNO₃). Solution neutral: pH = 7 at 25°C (recompute if Kw differs).
CASE 2: WEAK ACID + STRONG BASE (A ⁻
HYDROLYSIS)
Kh = Kw/Ka. [OH⁻] = √(Kw·C / Ka). pH = 7 + ½ pKa + ½ log C (basic).
CASE 3: STRONG ACID + WEAK BASE (B ⁺
HYDROLYSIS)
Kh = Kw/Kb. [H⁺] = √(Kw·C / Kb). pH = 7 − ½ pKb − ½ log C (acidic).
CASE 4: WEAK ACID + WEAK BASE
Kh = Kw/(Ka·Kb). [H⁺] = √(Kw·Ka / Kb). pH = 7 + ½ pKa − ½ pKb (nearly C-independent). If Ka = Kb → pH = 7.
Formulas assume h ≪ 1. If h > 0.1, solve the exact quadratic Kh = Ch²/(1−h).
BUFFERS — HENDERSON–HASSELBALCH & PRACTICAL STEPS
Acidic buffer: pH = pKa + log([salt]/[acid]). Basic buffer: pOH = pKb + log([salt]/[base]) → pH = 14 − pOH. Max buffer capacity when [salt] = [acid] (pH = pKa). When adding strong acid/base, neutralize stoichiometrically first, then apply Henderson. On dilution, ideal buffer pH ≈ unchanged because ratio is constant.
MIXTURES & COMMON-ION EFFECTS
SALT WA+WB
pH = 7 + ½ pKa − ½ pKb
BUFFER
pH = pKa + log(salt/acid)
AMPHIPROTIC
pH = ½(pKa1 + pKa2)
HIGH-YIELD JEE TRAPS (ALWAYS CHECK)
Validate C/Ka ≥ 100 before √ shortcut. Use Kw quadratic for very dilute strong acids/bases — never pH = −log C when C ≲ 10 ⁻6 M. Distinguish anion hydrolysis (basic) vs cation hydrolysis (acidic). Remember factor of 2 for fully dissociated diprotic species (H₂SO₄, Ca(OH)₂). Neutralize added strong acid/base to buffers before applying Henderson. Amphiprotic pH ≠ dependent on concentration (first approx.). Recompute neutral pH from Kw if T ≠ 25°C.
ONE-LOOK REVISION CARD
CORE FORMULAS
Strong acid: pH = −log C. Strong base: pH = 14 + log C. Weak acid: pH = ½(pKa − log C). Weak base: pOH = ½(pKb − log C). Very dilute strong: [H⁺] = (C + √(C² + 4Kw))/2.
SALTS, BUFFERS, AMPHIPROTIC
Salt WA+SB: pH = 7 + ½pKa + ½log C. Salt SA+WB: pH = 7 − ½pKb − ½log C. Salt WA+WB: pH = 7 + ½pKa − ½pKb. Buffer: pH = pKa + log(salt/acid). Amphiprotic: pH = ½(pKa1 + pKa2).
Work through a solved numeric for each row — formulas stick only after you hand-solve them. Best wishes for Mains and Advanced.
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