# Ionic Equilibrium

> Ionic equilibrium for JEE Chemistry: pH of acids and bases, buffers, solubility product, Ostwald dilution law and the common ion effect.

- Canonical URL: https://prepelephant.com/topics/jee/chemistry/ionic-equilibrium
- 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: "Ionic Equilibrium", PrepElephant, https://prepelephant.com/topics/jee/chemistry/ionic-equilibrium

## Direct answer

Ionic equilibrium applies equilibrium ideas to weak electrolytes, which ionise only partially in water. The degree of ionisation obeys Ka = C alpha^2/(1 - alpha) (Ostwald), so dilution raises alpha; for a weak acid [H+] = sqrt(Ka × C) and pH = 1/2 (pKa - log C). Hydrolysis sets the pH of salt solutions, the Henderson equation pH = pKa + log([salt]/[acid]) governs buffers, and solubility product Ksp governs sparingly soluble salts.

## What you must remember

- pH = -log[H+]; pH + pOH = 14 at 25 °C; Kw = 1.0 × 10^-14 at 25 °C and rises with temperature because water's ionisation is endothermic; pKa + pKb = 14 for a conjugate pair.
- Weak acid [H+] = sqrt(Ka × C); weak base [OH-] = sqrt(Kb × C); for alpha far below 1, alpha = sqrt(Ka/C) — dilution raises alpha but lowers [H+].
- Salt hydrolysis: weak acid + strong base (CH3COONa) is alkaline, pH = 7 + 1/2(pKa + log C); strong acid + weak base (NH4Cl) is acidic; NaCl neutral; CH3COONH4 near neutral when Ka = Kb.
- Henderson equation: acidic buffer pH = pKa + log([salt]/[acid]); basic buffer pOH = pKb + log([salt]/[base]); capacity is maximum at equal salt and acid.
- Solubility product: AB gives Ksp = s^2; AB2 or A2B gives 4s^3; A2B3 gives 108 s^5. Precipitation begins when the ionic product exceeds Ksp.
- Common ion effect: a shared ion suppresses ionisation and solubility — the basis of group separation in salt analysis.
- Indicators: methyl orange 3.1-4.4, phenolphthalein 8.3-10.0; strong acid-weak base ends acidic (methyl orange), weak acid-strong base ends basic (phenolphthalein).

## Common confusion

The recurring error is treating every salt as neutral: NH4Cl is acidic, sodium acetate alkaline, and only strong-strong salts sit at pH 7. The second confusion is dilution — a weak acid ionises more (alpha rises) yet becomes less acidic ([H+] falls); the two quantities move differently. In Ksp problems students forget the stoichiometric multipliers — AB2 carries 2s squared times s, giving 4s^3, not s^3.

## Exam-focused takeaway

JEE Main tests computed pH values: weak acids and bases, salt solutions, buffers, Ksp-to-solubility conversions, often as numerical-value questions. JEE Advanced layers concepts — pH change on dilution or partial neutralisation, simultaneous equilibria linked by a common ion, precipitation sequencing between competing Ksp values, and indicator choice justified by the end-point pH jump. Note the temperature: every relation assumes 25 °C unless stated.

## Frequently asked questions

### Why does pure water's pH fall below 7 on heating?

Ionisation is endothermic, so Kw exceeds 10^-14; [H+] rises above 10^-7 even though the water stays neutral.

### What is the common ion effect?

Adding an ion already present suppresses further ionisation of the weak electrolyte — acetate added to acetic acid holds the acid back.

### How is buffer pH calculated?

By pH = pKa + log([salt]/[acid]); it works best when both concentrations are well above Ka.

### What relates Ksp and solubility?

Ksp is the ion product at saturation: for AB, s = sqrt(Ksp); for AB2, s = (Ksp/4)^(1/3).

### Which indicator suits a weak acid versus strong base titration?

Phenolphthalein, since the equivalence point lies alkaline within its 8.3-10.0 range.

### Why does dilution increase the degree of ionisation?

alpha = sqrt(Ka/C): lowering C raises alpha even as the absolute [H+] decreases.