NEET (UG) · Chemistry, Class 11
Equilibrium: NEET PYQ, 2018–2026
Equilibrium (Chemistry, Class 11) has been asked 14 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 2 questions a paper, 5% of all Chemistry questions. It is the 10th most-asked of 21 Chemistry chapters. NEET 2026 had 1 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 3 |
| 2020 | 2 |
| 2021 | 1 |
| 2023 | 1 |
| 2024 | 3 |
| 2025 | 3 |
| 2026 | 1 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Chemical equilibrium4
- Solubility product4
- Ionic equilibrium3
- Le Chatelier's principle2
- pH and buffers1
Try it
10 Equilibrium questions to practise.
Our own questions, written in the NEET 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
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.
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 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.
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
The solubility product of Mg(OH)₂ at 25 °C is 3.2 × 10⁻¹¹ mol³ L⁻³. Neglecting the OH⁻ contributed by water, what is the pH of its saturated aqueous solution? (Take log 2 = 0.30.)
Answer: (A)
For Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻, Ksp = s(2s)² = 4s³. So 4s³ = 3.2 × 10⁻¹¹, which gives s³ = 8 × 10⁻¹² and s = 2 × 10⁻⁴ M. Then [OH⁻] = 2s = 4 × 10⁻⁴ M. pOH = 4 − log 4 = 4 − 0.60 = 3.40, so pH = 14 − 3.40 = 10.60.
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 14 real NEET 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 NEET.
How many questions come from Equilibrium in NEET?
Equilibrium (Chemistry, Class 11) has been asked 14 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 2 questions a paper, 5% of all Chemistry questions. It is the 10th most-asked of 21 Chemistry chapters. NEET 2026 had 1 from it.
Which topics of Equilibrium are asked most in NEET?
Chemical equilibrium (4), Solubility product (4), Ionic equilibrium (3), Le Chatelier's principle (2), counted over the papers from 2018–2026.
Is Equilibrium important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 10 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Equilibrium NEET previous year questions?
On Nikaalle, free: sign up and practise every past NEET 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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