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Ionic Equilibrium Short Notes for JEE, Schemes and Mind Maps of Chemistry

Chemistry

Ionic Equilibrium Short Notes. Helpfull for revising all The pH calculations

Typology: Schemes and Mind Maps

2025/2026

Uploaded on 06/27/2026

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himanshu-yadav-33 🇮🇳

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Ionic Equilibrium — pH Notes
IONIC EQUILIBRIUM
pH CALCULATIONS — COMPLETE NOTES
Every type of pH problem asked at JEE Main + Advanced level
These notes are written the way I'd want a sharp dropper to read them — every formula JEE actually
tests, the assumption that lets you use it, and the trap that catches 90% of students. Don't just
memorize the formulas; understand WHY each one applies only under specific conditions, because
JEE loves to break those conditions in disguise.
1. Foundation — What You Must Never Forget
Quantity Definition / Relation
pH −log [H ]₁₀ ⁺
pOH −log [OH ]₁₀ ⁻
pKw pH + pOH (at given T)
Kw at 25°C 1 × 10 ¹⁴ ⇒ pKw = 14⁻
pKa, pKb −log Ka , −log Kb
Relation pKa + pKb = pKw (for a conjugate acid–base pair)
JEE Tip: ⚠Kw is NOT always 10 ¹⁴. JEE often changes temperature (e.g. Kw = 10 ¹² at some T) —⁻ ⁻
then neutral pH ≠ 7. Always recompute pH(neutral) = ½ pKw for that temperature.
2. Strong Acids and Strong Bases
Strong acids/bases are assumed to dissociate 100%. Direct concentration = ion concentration —
EXCEPT at very low concentration, where water's own H can no longer be ignored.⁺
(a) Normal concentration (C ≥ 10 ⁶ M)⁻
Strong monoprotic acid: [H ] = C ⇒ pH = −log C⁺
Strong monoprotic base: [OH ] = C ⇒ pOH = −log C , pH = 14 − pOH⁻
• For diprotic strong species (e.g. dilute H SO treated as fully ionised), [H ] = 2C.₂ ₄ ⁺
• Ca(OH) , Ba(OH) fully dissociated: [OH ] = 2C.₂ ₂ ⁻
(b) Very dilute strong acid/base (C ≤ 10 ⁶ to 10 ⁸ M) — favourite JEE trap⁻ ⁻
When C is comparable to 10 ⁷ M, water's self-ionisation contributes significantly. You CANNOT say pH ⁻
= −log C; in fact at extreme dilution pH → 7, never beyond 7 for an acid.
[H ] − C[H ] − Kw = 0 (quadratic in [H ])⁺ ² ⁺ ⁺
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IONIC EQUILIBRIUM

pH CALCULATIONS — COMPLETE NOTES

Every type of pH problem asked at JEE Main + Advanced level These notes are written the way I'd want a sharp dropper to read them — every formula JEE actually tests, the assumption that lets you use it, and the trap that catches 90% of students. Don't just memorize the formulas; understand WHY each one applies only under specific conditions, because JEE loves to break those conditions in disguise.

1. Foundation — What You Must Never Forget

Quantity Definition / Relation pH −log ₁₀[H ] ⁺ pOH −log ₁₀[OH ] ⁻ pKw pH + pOH (at given T) Kw at 25°C 1 × 10 ¹⁴ ⇒ pKw = 14⁻ pKa, pKb −log Ka , −log Kb Relation pKa + pKb = pKw (for a conjugate acid–base pair) ⚠ JEE Tip: Kw is NOT always 10 ¹⁴. JEE often changes temperature (e.g. Kw = 10 ¹² at some T) —⁻ ⁻ then neutral pH ≠ 7. Always recompute pH(neutral) = ½ pKw for that temperature.

2. Strong Acids and Strong Bases

Strong acids/bases are assumed to dissociate 100%. Direct concentration = ion concentration — EXCEPT at very low concentration, where water's own H ⁺can no longer be ignored.

(a) Normal concentration (C ≥ 10 ⁶ M)⁻

Strong monoprotic acid: [H ] = C ⁺ ⇒ pH = −log C Strong monoprotic base: [OH ] = C ⁻ ⇒ pOH = −log C , pH = 14 − pOH

  • For diprotic strong species (e.g. dilute H SO₂ ₄ treated as fully ionised), [H ] = 2C.⁺
  • Ca(OH) , Ba(OH)₂ ₂ fully dissociated: [OH ] = 2C.⁻

(b) Very dilute strong acid/base (C ≤ 10 ⁶ to 10 ⁸ M) — favourite JEE trap⁻ ⁻

When C is comparable to 10 ⁷ M, water's self-ionisation contributes significantly. You CANNOT say pH⁻ = −log C; in fact at extreme dilution pH → 7, never beyond 7 for an acid. [H ] ⁺ ² − C[H ] − Kw = 0⁺ (quadratic in [H ])⁺

Solve: [H ] = { C + √(C² + 4Kw) } / 2⁺ ⚠ JEE Tip: If C < 10 ⁶ M roughly, you MUST use this quadratic. A C = 10 ⁸ M HCl solution does⁻ ⁻ NOT have pH = 8 — it stays slightly acidic, pH just below 7 (~6.98), because acid can never make pOH < pH cross to basic.

3. Weak Acids and Weak Bases (Ostwald's Dilution Law) Weak electrolytes partially dissociate. Let C = initial concentration, α = degree of dissociation. HA ⇌ H ⁺ + A⁻ Ka = Cα ²/ (1−α) - If α is small (α << 1, valid when Ka/C ≤ 10 ² roughly, i.e. C/Ka ≥ 100): Ka ≈ Cα²⁻ α = √(Ka/C) [H ] = Cα = √(Ka·C) ⁺ pH = ½(pKa − log C) Similarly for a weak base BOH: [OH ] = √(Kb·C) ⁻ pOH = ½(pKb − log C) pH = 14 − pOH ⚠ JEE Tip: Always check the validity condition (C/Ka ≥ 100, i.e. α < 0.05–0.1) before using the square-root shortcut. If Ka is comparably large or C is small, you MUST solve the full quadratic Ka = Cα²/(1−α) — a classic JEE Advanced twist where the 'easy' formula deliberately gives a wrong answer.

Effect of dilution on α and pH

  • On dilution, α increases (Ostwald: α ∝ 1/√C), but [H ] still decreases overall — pH increases⁺ (becomes less acidic) on diluting a weak acid, same direction as strong acid, just a gentler change. 4. Polyprotic (Diprotic/Triprotic) Weak Acids For acids like H S, H CO , H PO₂ ₂ ₃ ₃ ₄with Ka1 >> Ka2 (>> Ka3), the second and further ionisations are negligible for [H ] determination.⁺ [H ] ≈ √(Ka1·C) ⁺ (treat as if only 1st ionisation occurs) ⚠ JEE Tip: To find [A² ] (conjugate base of 2nd ionisation) in a diprotic acid solution (no other⁻ source of A² ), a beautiful shortcut: [A² ] ≈ Ka2 itself (independent of C), since [HA ] ≈ [H ] from step⁻ ⁻ ⁻ ⁺ 1, cancelling out in the Ka2 expression. This appears repeatedly in H S problems with sulfide₂ precipitation. 5. Salt Hydrolysis — 4 Cases A salt formed from acid + base hydrolyses only if at least one parent is weak. Let C = salt concentration, h = degree of hydrolysis.

Case 1: Salt of Strong Acid + Strong Base

  • e.g. NaCl, KNO ₃— No hydrolysis. Solution is neutral, pH = 7 (at 25°C).

Case 2: Salt of Weak Acid + Strong Base (anion hydrolysis)

e.g. CH COONa, Na CO. The anion A₃ ₂ ₃ ⁻ acts as a weak base: A ⁻ + H O₂ ⇌ HA + OH⁻ Kh = Kw/Ka h = √(Kw / (Ka·C)) [OH ] = √(Kw·C / Ka) ⁻

(b) Two weak acids together

[H ] = √(Ka1C1 + Ka2C2) ⁺ (when both are weak, comparable, and assumptions α<<1 hold for each)

(c) Weak acid + its salt (common ion suppression, non-buffer framing)

Ka = [H ][A ]/[HA] ⁺ ⁻ → [H ] ≈ Ka × (C_acid / C_salt)⁺ This is identical algebraically to Henderson's equation — same physics, just without explicitly calling it a 'buffer'.

8. Amphiprotic / Amphoteric Salts (e.g. NaHCO , NaHC O , NaH PO )₃ ₂ ₄ ₂ ₄ These salts contain an ion (like HCO ₃⁻) that can BOTH donate and accept a proton, governed by two successive Ka values of the parent polyprotic acid (Ka1, Ka2). [H ] = √(Ka1 · Ka2) ⁺ pH = ½(pKa1 + pKa2) ⚠ JEE Tip: This pH is independent of the salt's concentration C — a frequently tested conceptual MCQ. Don't try to apply hydrolysis or Henderson formulas here; amphiprotic species need this special geometric-mean relation. 9. Quick Decision Table — “Which Formula Do I Use?” Given species Formula to apply Strong acid/base, normal C [H ] or [OH ] = C directly⁺ ⁻ Strong acid/base, C ≤ 10 ⁶ M⁻ Quadratic with Kw: [H ]² − C[H ] − Kw = 0⁺ ⁺ Single weak acid/base [H ]=√(KaC) or [OH ]=√(KbC), check C/Ka ≥ 100⁺ ⁻ Polyprotic weak acid (Ka1>>Ka2) Use only Ka1: [H ]=√(Ka1C)⁺ Salt of WA + SB pH = 7 + ½pKa + ½log C Salt of SA + WB pH = 7 − ½pKb − ½log C Salt of WA + WB pH = 7 + ½pKa − ½pKb (independent of C) Buffer (acid + salt) Henderson: pH = pKa + log(salt/acid) Amphiprotic salt (e.g. NaHCO )₃ pH = ½(pKa1 + pKa2) (independent of C) Strong acid + weak acid mix [H ] ≈ strong acid's own C (common ion effect)⁺ 10. High-Yield Traps JEE Repeats Every Year - Forgetting the validity check (C/Ka ≥ 100) before using α = √(Ka/C). - Using pH = −log C for very dilute strong acids/bases instead of the Kw quadratic. - Mixing up which hydrolysis formula gives pH > 7 vs pH < 7 — anion hydrolysis (conjugate base) → basic; cation hydrolysis (conjugate acid) → acidic. - Forgetting factor of 2 for diprotic strong species like H SO₂ ₄ or Ba(OH) ₂when fully dissociated.

  • Plugging raw added moles of strong acid/base directly into Henderson equation without first neutralising stoichiometrically against existing buffer components.
  • Assuming amphiprotic salt pH depends on concentration — it does NOT (to first approximation).
  • Forgetting that at non-25°C temperature, neutral pH ≠ 7; recompute from given Kw. 11. One-Look Revision Card Type Key formula 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) 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 salt pH = ½(pKa1+pKa2) Very dilute strong acid [H ] = {C+√(C²+4Kw)}/2⁺ Work through one solved numeric per row of the revision card on your own — the formula sticks only once your hand has solved it, not just your eye read it. All the best for Mains and Advanced.