JEE (Main) · Chemistry, Class 11
Equilibrium: JEE Main PYQ, 2020–2025
Equilibrium (Chemistry, Class 11) has been asked 117 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.1 questions a paper, 4% of all Chemistry questions. It is the 12th most-asked of 21 Chemistry chapters. The 2025 shifts had 20 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 17in 14 papers |
| 2021 | 27in 26 papers |
| 2022 | 24in 24 papers |
| 2023 | 12in 12 papers |
| 2024 | 17in 18 papers |
| 2025 | 20in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Chemical equilibrium42
- pH and buffers30
- Solubility product27
- Ionic equilibrium13
- Le Chatelier's principle5
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10 Equilibrium questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Equilibrium
What is the pH at 298 K of a 0.01 M aqueous solution of sodium hydroxide, a strong base that dissociates completely? (Kw = 1.0 × 10^-14 at 298 K)
Answer: (C)
For a strong base, [OH-] equals the base concentration, here 0.01 M = 10^-2 M. Then pOH = -log(10^-2) = 2. Since pH + pOH = 14 at 298 K (from Kw = 10^-14), pH = 12. Always compute pOH from [OH-] first and then convert, rather than taking -log of [OH-] and calling it pH.
Q2 · Chemistry · Equilibrium
The solubility of Ag₂CrO₄ in water at a certain temperature is 1 × 10⁻⁴ mol/L. What is its solubility product?
Answer: (B)
Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻, so [Ag⁺] = 2s and [CrO₄²⁻] = s. Ksp = (2s)² × s = 4s³ = 4 × (10⁻⁴)³ = 4 × 10⁻¹².
Q3 · Chemistry · Equilibrium
Which of the following mixtures, each made up to one litre with water, acts as a buffer solution?
Answer: (C)
A buffer needs a weak acid together with its conjugate base (or a weak base with its conjugate acid) in comparable amounts. 0.1 mol of NaOH converts 0.1 mol of the acetic acid into sodium acetate and leaves 0.1 mol of acetic acid, so the solution holds CH₃COOH and CH₃COO⁻ in equal amounts: an acidic buffer with pH = pK_a.
Q4 · Chemistry · Equilibrium
The solubility product of magnesium hydroxide, Mg(OH)₂, at 25 °C is 5.6 × 10⁻¹² (mol L⁻¹)³. What is its molar solubility in pure water at this temperature?
Answer: (A)
Principle: for a salt that gives n ions in total, Ksp is the product of the ion concentrations, each raised to its stoichiometric power. Application: Mg(OH)₂ gives one Mg²⁺ and two OH⁻ per formula unit. If the molar solubility is s, then [Mg²⁺] = s and [OH⁻] = 2s, so Ksp = s·(2s)² = 4s³. Hence s = (Ksp/4)^(1/3) = (1.4 × 10⁻¹²)^(1/3) ≈ 1.1 × 10⁻⁴ mol L⁻¹.
Q5 · Chemistry · Equilibrium
At a certain temperature, N₂(g) + O₂(g) ⇌ 2NO(g) has K₁ = 2.0 × 10⁻³ and 2NO(g) + O₂(g) ⇌ 2NO₂(g) has K₂ = 4.0 × 10². What is the equilibrium constant of N₂(g) + 2O₂(g) ⇌ 2NO₂(g) at the same temperature?
Answer: (B)
Adding the two equations gives the required one. K₁ × K₂ = ([NO]²/[N₂][O₂]) × ([NO₂]²/[NO]²[O₂]) = [NO₂]²/[N₂][O₂]², so K = K₁K₂ = 2.0 × 10⁻³ × 4.0 × 10² = 0.80.
Q6 · Chemistry · Equilibrium
A solution contains 0.010 M Mg²⁺. Its pH is raised to 11.0 using a buffer, which keeps the pH constant and does not change the volume. Excess Mg(OH)₂ precipitates. Ksp of Mg(OH)₂ at 25 °C is 4.0 × 10⁻¹² mol³ L⁻³. What is [Mg²⁺] in solution at equilibrium?
Answer: (A)
Principle: when solid Mg(OH)₂ is in equilibrium with solution, Ksp = [Mg²⁺][OH⁻]² holds. The buffer fixes [OH⁻], so [Mg²⁺] follows directly. pH = 11 gives pOH = 14 − 11 = 3, so [OH⁻] = 1.0 × 10⁻³ M. Then [Mg²⁺] = 4.0 × 10⁻¹²/(1.0 × 10⁻³)² = 4.0 × 10⁻⁶ M. Only 0.04% of the original 0.010 M Mg²⁺ remains dissolved.
Q7 · Chemistry · Equilibrium
A buffer is prepared by mixing acetic acid and sodium acetate so that [CH₃COO⁻] = 0.10 M and [CH₃COOH] = 0.010 M at 25 °C. The pKa of acetic acid is 4.76. What is the pH of this buffer?
Answer: (C)
For an acidic buffer made of a weak acid HA and its salt with conjugate base A⁻, use the Henderson equation pH = pKa + log([A⁻]/[HA]). Here [A⁻]/[HA] = 0.10/0.010 = 10, and log 10 = 1. So pH = 4.76 + 1 = 5.76. Note that the ratio is salt over acid, so a larger salt concentration raises the pH.
Q8 · Chemistry · Equilibrium
At 298 K, 60.0 mL of 0.10 M HCl is mixed with 40.0 mL of 0.20 M NaOH. Assuming the volumes are additive, what is the pH of the resulting solution? (log 2 = 0.30; Kw = 1.0 × 10⁻¹⁴)
Answer: (A)
Strong acid and strong base react completely, so find which one is in excess. HCl = 0.10 × 60 = 6.0 mmol. NaOH = 0.20 × 40 = 8.0 mmol. Excess OH⁻ = 2.0 mmol in a total volume of 100 mL, so [OH⁻] = 0.020 M. pOH = −log(2 × 10⁻²) = 2 − 0.30 = 1.70. pH = 14 − 1.70 = 12.30.
Q9 · Chemistry · Equilibrium
2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH < 0, is at equilibrium at a fixed temperature in a cylinder with a movable piston that keeps the total pressure constant. A quantity of argon, which takes no part in the reaction, is now added. Which statement is correct?
Answer: (C)
Write Kp = p(SO₃)²/[p(SO₂)²·p(O₂)] = n(SO₃)²·n(total)/[n(SO₂)²·n(O₂)·P]. At constant total pressure P, adding argon increases n(total), so Q becomes greater than Kp. The reaction therefore shifts backward, toward SO₂ and O₂, which have more moles of gas, until Q = Kp. Kp itself is unchanged because the temperature is constant. At constant volume, argon would not change the partial pressures, and there would be no shift.
Q10 · Chemistry · Equilibrium
A solution has pH 2.0 at 25 °C. It is diluted with water exactly ten times in volume. What is the pH of the diluted solution? Assume the acid is strong and the solution remains dilute.
Answer: (C)
pH = −log[H⁺]. Initially [H⁺] = 10⁻² M. Diluting tenfold divides [H⁺] by 10, giving 10⁻³ M. So pH = 3.0. A tenfold dilution of a strong acid raises the pH by exactly one unit, because pH is a logarithm of concentration.
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All 117 real JEE Main questions from Equilibrium.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Equilibrium in JEE Main.
How many questions come from Equilibrium in JEE Main?
Equilibrium (Chemistry, Class 11) has been asked 117 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.1 questions a paper, 4% of all Chemistry questions. It is the 12th most-asked of 21 Chemistry chapters. The 2025 shifts had 20 from it in all.
Which topics of Equilibrium are asked most in JEE Main?
Chemical equilibrium (42), pH and buffers (30), Solubility product (27), Ionic equilibrium (13), counted over the papers from 2020–2025.
Is Equilibrium important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 12 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Equilibrium JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Equilibrium with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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