JEE Main Chemistry Equilibrium 2027: Kc, Kp, pH, Buffers & Solubility Product

If you had to pick one Chemistry chapter to master for JEE Main, Equilibrium would be a strong candidate. It spans physical equilibrium, chemical equilibrium and ionic equilibrium, and it feeds directly into Electrochemistry and Thermodynamics.

At equilibrium, the reaction has not stopped. Forward and reverse rates are equal.

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Chapter at a Glance

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

• Dynamic equilibrium, equilibria involving physical processes, Henry's law
• Law of chemical equilibrium, Kp and Kc, significance of ΔG and ΔG°, factors affecting equilibrium, Le Chatelier's principle
• Ionic equilibrium: weak and strong electrolytes, acids and bases (Arrhenius, Bronsted-Lowry, Lewis), multistage ionisation, ionisation of water, pH scale
• Common ion effect, hydrolysis of salts, solubility products and buffer solutions

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Master These Topics

1. Kc, Kp and Le Chatelier

For aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]^c [D]^d / [A]^a [B]^b. In terms of pressures, Kp = Kc (RT)^Δn, where Δn is moles of gaseous products minus reactants.

The reaction quotient Q has the same form as K but uses current concentrations. If Q < K the reaction goes forward, if Q > K it goes backward.

Le Chatelier's principle: a system disturbed from equilibrium shifts to oppose the change.

• Adding reactant shifts the reaction forward.
• Raising pressure favours the side with fewer gas moles.
• Raising temperature favours the endothermic direction.
• A catalyst speeds up both directions equally and does not change K.

Trap: Only temperature changes the value of K. Concentration, pressure and catalysts change only how fast or in which direction the system moves.

2. pH of Acids and Salts

pH = −log[H⁺]. For a weak acid of concentration C with small dissociation, [H⁺] = √(Ka C).

Worked example: 0.1 M acetic acid with Ka = 1.8 × 10⁻⁵ has [H⁺] = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³ M, so pH ≈ 2.87.

Very dilute acid trap: For 10⁻⁸ M HCl, do not answer pH 8. Water contributes 10⁻⁷ M of H⁺, so [H⁺] ≈ 1.1 × 10⁻⁷ M and pH ≈ 6.96.

Salt hydrolysis: a salt of a weak acid and strong base (like sodium acetate) is basic, and pH = 7 + ½(pKa + log C).

3. Buffer Solutions and the Henderson Equation

A buffer resists pH change. For an acidic buffer, pH = pKa + log([salt]/[acid]).

Worked example: An acetic acid and sodium acetate buffer with pKa = 4.74 has pH 4.74 when both are 0.1 M. If the salt concentration is doubled to 0.2 M, pH = 4.74 + log 2 ≈ 5.04.

Buffer capacity is maximum when [salt] = [acid], so pH = pKa.

4. Solubility Product and the Common Ion Effect

For AgCl ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻]. If Ksp = 1.8 × 10⁻¹⁰, the molar solubility is s = √Ksp ≈ 1.34 × 10⁻⁵ M. For a salt of type AB₂ or A₂B, Ksp = 4s³.

Precipitation occurs when the ionic product exceeds Ksp. Adding a common ion lowers solubility, because the equilibrium shifts towards the solid.

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Common Traps to Avoid

• Thinking a catalyst or pressure change alters the equilibrium constant. Only temperature does.
• Reporting pH 8 for 10⁻⁸ M HCl instead of about 6.96.
• Using Ksp = s² for salts that dissociate into three ions.
• Forgetting Δn in the relation Kp = Kc (RT)^Δn.

60-Second Revision Sheet

• Kp = Kc (RT)^Δn; compare Q with K for direction
• Weak acid [H⁺] = √(Ka C)
• Buffer pH = pKa + log(salt/acid)
• AB salt Ksp = s²; AB₂ or A₂B salt Ksp = 4s³

Your Study Plan

1. Day 1: Kc, Kp, Q and Le Chatelier with ICE tables.
2. Day 2: acids, bases, Ka, Kb and the pH of weak acids and bases.
3. Day 3: hydrolysis of salts and buffer calculations.
4. Day 4: Ksp, common ion effect, precipitation conditions and a timed set.

Practice Equilibrium Questions Free → (opens in a new tab)

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Continue Your Chemistry Journey

• Previous chapter: Solutions
• Next chapter: Redox Reactions & Electrochemistry
• All JEE Main Chemistry chapters
• Complete JEE Main Syllabus 2027 guide

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Frequently Asked Questions