Ionic equilibrium for JEE Main

Ionic equilibrium is the most calculation-heavy part of the equilibrium unit, and JEE Main uses it for both multiple-choice and numerical-value questions. The good news is that almost every problem is one of six types: a strong acid or base, a weak acid or base, a mixture, a salt, a buffer, or a sparingly soluble salt. Identify the type first, then use its formula. Most errors come from using the right formula on the wrong type.

The official 2026 syllabus lists weak and strong electrolytes, the Arrhenius, Brønsted–Lowry and Lewis concepts, acid–base equilibria including multistage ionisation, ionisation constants, the ionisation of water, the pH scale, the common ion effect, hydrolysis of salts and the pH of their solutions, solubility and solubility products, and buffer solutions. The practical chemistry unit adds acid–base titrations and indicators.

Acids and bases: three definitions

In the Brønsted picture, every acid has a conjugate base that differs by one H⁺: HSO₄⁻ has conjugate base SO₄²⁻ and conjugate acid H₂SO₄. A strong acid has a weak conjugate base, and for any conjugate pair Ka × Kb = Kw.

Water and the pH scale

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K, so pH + pOH = 14 at that temperature. Kw rises with temperature. Pure water at a higher temperature therefore has a pH below 7 but is still neutral, because [H⁺] = [OH⁻].

pH = −log[H⁺]. Strong acids and bases ionise completely:

• 0.01 M HCl: pH = 2
• 0.001 M NaOH: pOH = 3, pH = 11
• mixing a strong acid and a strong base: find the excess mmol of H⁺ or OH⁻ and divide by the total volume

Weak acids and bases

For a weak acid HA of concentration C and degree of ionisation α:

• Ka = Cα²/(1 − α) ≈ Cα² (Ostwald's dilution law, for small α)
• α = √(Ka/C), so dilution increases α
• [H⁺] = √(KaC) and pH = ½(pKa − log C)

The weak-base forms are the same with Kb and [OH⁻]. The approximation holds when α is below about 5%; otherwise solve the quadratic.

Polyprotic acids such as H₂S, H₂CO₃ and H₃PO₄ ionise in steps with Ka₁ much larger than Ka₂. The pH comes almost entirely from the first step.

Very dilute strong acids

For 10⁻⁸ M HCl, pH is not 8: an acid cannot give a basic solution. Water's own ionisation matters now. Let [H⁺] = h, so [OH⁻] = h − 10⁻⁸ (the H⁺ beyond the acid's share comes from water).

h(h − 10⁻⁸) = 10⁻¹⁴ gives h ≈ 1.05 × 10⁻⁷ M, so pH ≈ 6.98.

Salt hydrolysis

A salt of a weak acid or weak base reacts with water. At 298 K:

The hydrolysis constant is Kh = Kw/Ka (or Kw/Kb), and the degree of hydrolysis is h = √(Kh/C). For the weak acid–weak base salt, pH does not depend on concentration.

Buffers

A buffer resists changes in pH when a little acid or base is added. It is a weak acid with its salt (CH₃COOH + CH₃COONa) or a weak base with its salt (NH₄OH + NH₄Cl).

• Acidic buffer: pH = pKa + log([salt]/[acid])
• Basic buffer: pOH = pKb + log([salt]/[base])

Buffer capacity is greatest when [salt] = [acid], where pH = pKa. Partially neutralising a weak acid with a strong base creates a buffer; work in mmol.

Indicators change colour over a range: methyl orange about 3.1–4.4, phenolphthalein about 8.2–10. Choose one whose range falls on the steep part of the titration curve: phenolphthalein for a weak acid against a strong base, methyl orange for a strong acid against a weak base.

Solubility product

For a sparingly soluble salt with molar solubility s (mol/L of salt that dissolves), write Ksp from the ions it gives:

• Precipitation happens when the ionic product exceeds Ksp.
• Common ion effect: adding an ion already present lowers solubility. For AgCl in c M NaCl, s ≈ Ksp/c.
• Compare solubilities through s, not Ksp, when the salt types differ.

Worked problems

Problem 1: weak acid. Find the pH and degree of ionisation of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵).

[H⁺] = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M.
pH = 3 − log 1.34 = 2.87. α = 1.34 × 10⁻³/0.1 = 1.34%, so the approximation holds.

Problem 2: buffer by partial neutralisation. 100 mL of 0.1 M CH₃COOH is mixed with 20 mL of 0.1 M NaOH. Find the pH (pKa = 4.74).

Acid = 10 mmol, NaOH = 2 mmol. After reaction: acid 8 mmol, salt 2 mmol.
pH = 4.74 + log(2/8) = 4.74 − 0.60 = 4.14.

Problem 3: common ion effect. Ksp of AgCl is 1.8 × 10⁻¹⁰. Find its solubility in pure water and in 0.1 M NaCl.

Pure water: s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M.
In 0.1 M NaCl: s × 0.1 = 1.8 × 10⁻¹⁰, so s = 1.8 × 10⁻⁹ M, more than 7,000 times lower.

Problem 4 (numerical answer): mixing a strong acid and base. 100 mL of 0.1 M HCl is mixed with 100 mL of 0.08 M NaOH. Find the pH.

H⁺ = 10 mmol, OH⁻ = 8 mmol. Excess H⁺ = 2 mmol in 200 mL = 0.01 M.
pH = 2.

Traps to watch

• pH of very dilute acids. Include water's ionisation below about 10⁻⁶ M.
• Mixing without dividing by the total volume. Always use the combined volume.
• Using the weak-acid formula for a buffer or the buffer formula for a pure salt. Identify the type first.
• Assuming pH 7 is always neutral. That is true only at 298 K.
• Comparing Ksp values of different salt types to rank solubility.

Practice set

1. Find the pH of 10⁻⁸ M HCl.
2. The solubility of AgCl in water at some temperature is 10⁻⁵ M. Find its Ksp.
3. Find the pH of 0.1 M CH₃COONa (pKa of acetic acid = 4.74).
4. Find the pH of 0.005 M H₂SO₄, assuming complete ionisation.
5. Ksp of CaF₂ is 3.2 × 10⁻¹¹. Find its molar solubility.
6. Find the pH of 0.01 M NH₃ (Kb = 1.8 × 10⁻⁵).
7. Name the conjugate base of HSO₄⁻, and state whether BF₃ is a Lewis acid or base.

Answers

1. About 6.98, not 8.
2. Ksp = s² = 10⁻¹⁰.
3. pH = 7 + ½(4.74 − 1) = 8.87.
4. [H⁺] = 2 × 0.005 = 0.01 M, so pH = 2.
5. 4s³ = 3.2 × 10⁻¹¹, so s³ = 8 × 10⁻¹² and s = 2 × 10⁻⁴ M.
6. [OH⁻] = √(1.8 × 10⁻⁷) = 4.24 × 10⁻⁴ M, pOH = 3.37, so pH = 10.63.
7. SO₄²⁻; BF₃ is a Lewis acid (it accepts an electron pair).

What to do next

• Sort twenty past pH questions into the six types before solving any of them.
• Learn the four salt-hydrolysis formulas and derive one of them once.
• Revise chemical equilibrium if ICE tables still feel slow.
• Practise titration and indicator questions with the practical chemistry guide, then move on to redox and electrochemistry.