Ionic Equilibrium is the chapter where Chemical Equilibrium meets real solutions — acids, bases, buffers, salts, and the pH scale. Every concept here is a direct application of the equilibrium framework you already know. Once you see that Ka, Kb, Kw, and Ksp are just specific cases of the equilibrium constant Kc, the chapter stops being a collection of separate formulas and becomes one coherent system.
This guide covers the complete Ionic Equilibrium syllabus for JEE and NEET: the pH scale, strong and weak acid-base calculations, Ka and Kb, the relationship between Ka, Kb and Kw, buffer solutions and the Henderson-Hasselbalch equation, hydrolysis of salts, and solubility product. The final section lists the 8 traps that cost students marks every year even when they have studied the chapter.
Ionic Equilibrium contributes 3–5 questions in JEE Mains and 4–6 questions in NEET — making it one of the highest-scoring single chapters in Physical Chemistry. pH calculations, buffer questions, and salt hydrolysis are practically guaranteed in every NEET paper.
The pH Scale — Foundation of Everything
pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:
Key reference points: pure water at 25°C has pH = 7 (neutral). Acidic solutions have pH < 7 ([H⁺] > 10⁻⁷ M). Basic solutions have pH > 7. A change of 1 pH unit represents a 10-fold change in [H⁺] — so pH 3 is 100 times more acidic than pH 5. This logarithmic relationship is the source of many calculation errors.
One crucial point NEET asks about every year: Kw changes with temperature. At temperatures above 25°C, Kw > 10⁻¹⁴ and the neutral pH is less than 7. At 37°C (body temperature), Kw ≈ 2.4 × 10⁻¹⁴ and neutral pH ≈ 6.8. A solution can be neutral (equal [H⁺] and [OH⁻]) and still have pH ≠ 7 if the temperature is not 25°C.
Strong Acids and Strong Bases — Direct Calculation
Strong acids (HCl, H₂SO₄, HNO₃, HBr, HI, HClO₄) and strong bases (NaOH, KOH, Ca(OH)₂, Ba(OH)₂) dissociate completely. The pH calculation is direct — no equilibrium constant needed.
The only subtlety: for very dilute strong acids (concentration ≤ 10⁻⁶ M), the contribution of H⁺ from water autoionisation becomes significant and cannot be ignored. For a 10⁻⁸ M HCl solution, the pH is NOT 8 — it is slightly less than 7 (still acidic). JEE occasionally tests this edge case.
Weak Acids and Bases — Ka and Kb
Weak acids partially dissociate. For a weak acid HA:
Similarly for a weak base B: Kb = [BH⁺][OH⁻] / [B], and [OH⁻] = √(Kb·C).
The Ka–Kb–Kw Relationship
For a conjugate acid-base pair (HA and A⁻):
A stronger acid has a larger Ka and a weaker conjugate base (smaller Kb). This inverse relationship is tested directly in NEET in comparative questions — "which conjugate base is strongest?" always corresponds to the weakest acid in the list.
Buffer Solutions — The Henderson-Hasselbalch Equation
A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable concentrations.
Three things the Henderson-Hasselbalch equation tells you immediately:
- When [A⁻] = [HA] (equal concentrations), log(1) = 0, so pH = pKa. This is the half-equivalence point in a titration — a favourite JEE question type.
- The pH of a buffer depends on the ratio of conjugate base to acid, not their absolute concentrations. Diluting a buffer does not change its pH (as long as both components are present in significant amounts).
- Buffer capacity is maximum at pH = pKa and decreases as you move away from it.
Example: To prepare an acetate buffer at pH 5.0 using acetic acid (pKa = 4.74): log([CH₃COO⁻]/[CH₃COOH]) = 5.0 − 4.74 = 0.26, so [CH₃COO⁻]/[CH₃COOH] = 10^0.26 ≈ 1.82. You need approximately 1.82 moles of sodium acetate for every 1 mole of acetic acid.
pH Calculations Tripping You Up?
One-to-one sessions with PK Sir means you drill exactly the calculation types that appear in your target exam. Book a free demo and bring any question from this chapter.
Book Free DemoHydrolysis of Salts — Which Way Does pH Go?
When a salt dissolves in water, the ions may react with water (hydrolyse) to produce acidic or basic solutions. The direction depends on which parent acid and base formed the salt:
- Salt of strong acid + strong base (e.g. NaCl, KNO₃): No hydrolysis. pH = 7. Neither Na⁺ nor Cl⁻ reacts with water.
- Salt of weak acid + strong base (e.g. CH₃COONa, Na₂CO₃): Anion hydrolyses. CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Solution is basic (pH > 7).
- Salt of strong acid + weak base (e.g. NH₄Cl, FeCl₃): Cation hydrolyses. NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. Solution is acidic (pH < 7).
- Salt of weak acid + weak base (e.g. CH₃COONH₄): Both ions hydrolyse. pH depends on relative Ka and Kb values. If Ka = Kb, pH = 7. If Ka > Kb, solution is acidic. If Kb > Ka, solution is basic.
pH of a Salt of Weak Acid + Strong Base
pH of a Salt of Strong Acid + Weak Base
Solubility Product (Ksp) — Precipitation and Dissolution
Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For a salt MₓAy dissolving in water:
Common Ion Effect
Adding a common ion to a saturated solution suppresses the solubility of the sparingly soluble salt. For example, adding NaCl to a saturated AgCl solution increases [Cl⁻], which pushes the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left — AgCl precipitates further and [Ag⁺] decreases. This is Le Chatelier's principle applied to Ksp.
Predicting Precipitation
Compare the ionic product (IP) — calculated using actual current concentrations — with Ksp:
- IP > Ksp: Solution is supersaturated → precipitation occurs until equilibrium is reached.
- IP = Ksp: Solution is exactly at saturation — equilibrium.
- IP < Ksp: Solution is unsaturated → no precipitation, more solid can dissolve.
The 8 Traps Examiners Set Every Year
Saying pH of Very Dilute Strong Acid is Greater Than 7
For 10⁻⁸ M HCl, students write pH = 8. But an acid can never make a solution basic. The H⁺ from water must be included: total [H⁺] = 10⁻⁸ + 10⁻⁷ ≈ 1.1 × 10⁻⁷ M, giving pH ≈ 6.96 — slightly acidic, as it must be.
Treating pH = 7 as "Neutral" at All Temperatures
Neutral means [H⁺] = [OH⁻], not pH = 7. At temperatures above 25°C, Kw > 10⁻¹⁴ and neutral pH < 7. NEET asks this in the form: "At 60°C, pH of pure water is 6.5. Is it acidic?" Answer: No — it is still neutral because [H⁺] = [OH⁻].
Using Ka × Kb = Kw for Non-Conjugate Pairs
Ka × Kb = Kw applies ONLY to a conjugate acid-base pair — the acid HA and its conjugate base A⁻. It does NOT apply to an arbitrary acid and an arbitrary base. For example, Ka(HCN) × Kb(NH₃) ≠ Kw.
Forgetting to Check the 5% Rule Before Using the Approximation
The simplification [H⁺] = √(Ka·C) assumes α << 1. This fails when Ka/C > 0.05. If Ka = 10⁻² and C = 0.1 M, Ka/C = 0.1 — the approximation gives a 10% error. In such cases, you must solve the full quadratic Cα²/(1−α) = Ka.
Assuming Diluting a Buffer Changes Its pH
Buffer pH depends on the ratio [A⁻]/[HA], not on absolute concentrations. When you dilute a buffer, both numerator and denominator decrease proportionally — the ratio stays the same, so pH stays the same (in the ideal case). Students often calculate a new pH after dilution when no calculation is needed.
Using Ksp = s² for All Sparingly Soluble Salts
Ksp = s² only for 1:1 salts like AgCl or BaSO₄. For Ag₂CrO₄ (2:1 salt), Ksp = 4s³. For PbCl₂ (1:2 salt), Ksp = 4s³. For Al(OH)₃ (1:3 salt), Ksp = 27s⁴. Using s² for all these types is one of the most common errors in NEET numerical questions.
Ranking Salts by Ksp to Determine Solubility
A larger Ksp does not necessarily mean greater molar solubility — unless both salts have the same formula type. AgCl (Ksp = 1.8×10⁻¹⁰, type 1:1, s = √Ksp) vs. Ag₂CrO₄ (Ksp = 1.1×10⁻¹², type 2:1, s = (Ksp/4)^(1/3)). Ag₂CrO₄ has a lower Ksp but higher molar solubility. Always calculate s before comparing.
Getting the Sign of Salt Hydrolysis pH Wrong
A salt of a weak acid + strong base gives basic solution (anion hydrolyses, produces OH⁻). A salt of strong acid + weak base gives acidic solution (cation hydrolyses, produces H₃O⁺). Students frequently reverse these — a quick memory hook: the "surviving" ion wins. In CH₃COONa, NaOH "wins" (strong base) → basic solution.
A Worked Example: Buffer pH Calculation
This type appears in both JEE Mains and NEET almost every year.
Problem: Calculate the pH of a buffer made by mixing 0.3 mol of sodium acetate and 0.2 mol of acetic acid in 1 L of solution. Ka of acetic acid = 1.8 × 10⁻⁵.
Your Ionic Equilibrium Revision Checklist
- Write pH = −log[H⁺] and relate pH + pOH = 14 (at 25°C); explain why neutral pH ≠ 7 at other temperatures.
- Calculate pH of strong acids and bases directly; handle very dilute solutions by including water's H⁺ contribution.
- Set up the Ka expression for any weak acid; use [H⁺] = √(Ka·C) after verifying the 5% approximation holds.
- Apply Ka × Kb = Kw only to conjugate acid-base pairs.
- Use the Henderson-Hasselbalch equation to calculate buffer pH; identify the half-equivalence point (pH = pKa).
- Predict whether a salt solution is acidic, basic, or neutral from the nature of parent acid and base.
- Write the Ksp expression for any salt, accounting for stoichiometry; calculate molar solubility s from Ksp correctly for each salt type.
- Apply the common ion effect to predict decreased solubility; use IP vs. Ksp to predict whether precipitation occurs.
Ionic Equilibrium is the chapter that separates students who have memorised formulas from students who understand equilibrium. Once the framework is clear — every constant (Ka, Kb, Kw, Ksp) is just a Kc in disguise — the chapter becomes one of the fastest to master. Invest two focused sessions in it and it pays back in every exam you write.
This chapter connects directly to the Chemical Equilibrium guide (same Kc framework applied to ions) and to the Electrochemistry guide (where pH appears in the Nernst equation). If you want to work through specific pH, buffer, or Ksp numericals with immediate feedback, book a free 30-minute demo class. Bring the problem type you find hardest — we will crack the pattern together.