JEE Main PYQ — The Buffer, Ksp and Inert-Gas Traps That Decide This Chapter (2015-2026)
Equilibrium JEE Main PYQ analysis (2015-2026). 227 questions mapped: Kp-Kc, degree of dissociation, pH, buffers, Ksp, 10 traps, and 12 exam-level questions.
Two to three questions a shift, and you can no longer skip the long ones.
Equilibrium is one of the heaviest chapters in JEE Main Chemistry. We counted 227 questions from 2015 to 2026. That is 101 from Chemical Equilibrium and 126 from Ionic Equilibrium. Together they give two to three questions a shift, about 6 to 8 percent of the Chemistry marks.
The chapter is also getting bigger. In the last five years, Chemical Equilibrium averaged 11 questions a year and Ionic Equilibrium 13. The 2026 sessions gave 29 between them, the highest two-session total so far.
The 2025 change hit this chapter hardest. Until 2024, students could skip a long pH or buffer numerical by picking an easier NVQ. Now all five are compulsory. Roughly half of recent Equilibrium questions sit in Section B. A buffer ratio written upside down, or a mixture you forgot to dilute, now costs marks with no options to warn you.
We analysed every Equilibrium question across JEE Main sessions from 2015 to 2026, including January and April 2026. This is part of Logic Bloom's JEE Main analysis series, after Chemical Kinetics, Thermodynamics, and Electrochemistry.
| 🎯 We mapped every JEE Main Equilibrium question of the decade. The app has them all, ready to play and practice. | |
|---|---|
| This chapter is won by running the same few setups until they are automatic. Logic Bloom's Playground turns it into practice: add helium to a flask and watch whether the equilibrium moves, mix an acid with a base and see the buffer form, and push a salt past its Ksp to watch it precipitate. Then drill every PYQ, including the compulsory NVQ type. When an upside-down buffer ratio or a missed dilution catches you, TarQ teaches the fix, and your Mistake Book logs it. | Get the app → Free to start. |
The decade at a glance: two halves, two formats
| Chemical Equilibrium | Ionic Equilibrium | |
|---|---|---|
| Questions, 2015 to 2026 | 101 | 126 |
| Average per year, last five years | 11.2 | 13.4 |
| 2026 (January + April) | 15 | 14 |
| Share that are NVQs (full bank) | 36% | 29% |
The NVQ share is climbing. NVQs only began in 2020, and from 2022 to 2026 close to half of all Equilibrium questions were NVQs. The split by sub-topic follows a simple rule. If the answer is a number, it can go in Section B. If the answer is a direction or an expression, it stays in Section A.
| Sub-topic | How often | Format lean |
|---|---|---|
| Ksp and the common-ion effect | 3 to 5 shifts every year; the numerical anchor of Ionic Equilibrium | NVQ and MCQ |
| Degree of dissociation, total-pressure setups | The most repeated derivation in Chemical Equilibrium | MCQ when the answer is an expression, NVQ when it is a number |
| Kp, Kc and K manipulation | Almost every session | NVQ-leaning |
| Buffers | One of the highest-converting NVQ topics | NVQ |
| ΔG° and K, van't Hoff | A steady Section B source | NVQ |
| Le Chatelier's principle | Rotates through every session | MCQ |
| pH of acids, bases and mixtures | Regular, often with dilution steps | NVQ and MCQ |
| Salt hydrolysis | Heavy in 2022 to 2024 | MCQ |
| Acid-base theories, indicators | 1 to 2 a year | MCQ, often Statement I and II |
One more pattern matters. NTA does not invent new systems each session. It rotates a small set of templates across shifts, changing only the numbers. A setup from one 2022 shift came back word for word in a 2024 shift. Learn the templates and you have seen most of the paper before you sit it.
Part 1: Chemical equilibrium
K and how to manipulate it
Most Chemical Equilibrium marks start with writing K correctly. Three rules cover almost every manipulation.
| 🧠 The K rules | |
|---|---|
| Reverse the reaction | K becomes 1/K |
| Multiply the equation by n | K becomes Kⁿ (halving means √K) |
| Add two reactions | Multiply their K values |
| Kp and Kc | Kp = Kc(RT)^Δn, where Δn = gas moles of products − gas moles of reactants |
| Solids and pure liquids | Leave them out of K completely |
Degree of dissociation and total pressure
This is the most repeated setup in the chapter. Start with 1 mol of reactant, let a fraction α break up, and count moles. Each partial pressure is its mole fraction times the total pressure P.
| Reaction | Total moles | Kp in terms of α and P |
|---|---|---|
| PCl₅ ⇌ PCl₃ + Cl₂ | 1 + α | α²P / (1 − α²) |
| N₂O₄ ⇌ 2NO₂ | 1 + α | 4α²P / (1 − α²) |
A quick check saves time on the MCQ version. Both reactions make more gas than they use, so Kp must grow with P in the expression. Options with P in the denominator can go before you write a single line.
From vapour density. If the theoretical vapour density is D and the measured one is d, then α = (D − d) / [(n − 1)d]. Here n is the number of gas particles one molecule breaks into. For N₂O₄, n = 2, so α = (D − d)/d.
A worked JEE Main question
Q. (NVQ) At 300 K and 1 atm, ozone is 50% dissociated by 2O₃(g) ⇌ 3O₂(g). Find the standard free energy change, in J mol⁻¹, as a magnitude. Round off to the nearest integer. (Given R = 8.3 J K⁻¹ mol⁻¹, ln 1.35 = 0.3)
Solution. Start with 1 mol O₃. Half breaks up, leaving 0.5 mol O₃. Each 2 mol of O₃ gives 3 mol O₂, so 0.5 mol gives 0.75 mol O₂. Total moles = 1.25. The mole fractions are 0.4 for O₃ and 0.6 for O₂. So Kp = (0.6)³ / (0.4)² = 0.216 / 0.16 = 1.35. Then ΔG° = −RT ln K = −8.3 × 300 × 0.3 = −747 J mol⁻¹.
Answer: 747. Trap: making 0.5 mol of O₂ (forgetting the 3:2 ratio) breaks the match with ln 1.35. Multiplying by 2.303 when ln is already given ruins the integer.
K, ΔG° and temperature
Two equations link K to energy. ΔG° = −RT ln K connects K to free energy. The van't Hoff equation, log(K₂/K₁) = [ΔH° / 2.303R] × (T₂ − T₁)/(T₁T₂), gives K at a new temperature. K equals 1 exactly when ΔG° = 0, which happens at T = ΔH°/ΔS°.
Remember one fact above all. K changes only with temperature. Changing pressure, concentration, or adding a catalyst never changes K. It only changes how fast, or which way, the system moves to reach it.
Le Chatelier's principle: the inert-gas trap
A disturbed equilibrium shifts to undo the change. Concentration and temperature are routine. The question NTA keeps rotating is adding an unreactive gas like helium or xenon. The answer depends on one word in the stem.
At fixed volume, the helium adds to the total pressure but leaves every partial pressure where it was. Nothing shifts. At fixed pressure, the container must grow. Every partial pressure falls, so the system shifts toward the side with more gas moles. For N₂O₄ ⇌ 2NO₂ or PCl₅ ⇌ PCl₃ + Cl₂, that means more dissociation. If both sides have equal gas moles, nothing shifts either way.
A newer twist came in April 2026: a closed flask containing reactive gases plus helium. You must subtract helium's partial pressure from the total before putting pressures into Kp.
A worked JEE Main question
Q. (MCQ) In a blast furnace, Fe₂O₃(s) + 3CO(g) ⇌ 2Fe(l) + 3CO₂(g) is at equilibrium. Which change will NOT disturb the equilibrium?
(A) Adding Fe₂O₃ (B) Adding CO₂ (C) Removing CO (D) Removing CO₂
Solution. Solids and pure liquids stay out of K, so Kp = (p_CO₂)³ / (p_CO)³. Adding solid Fe₂O₃ changes neither pressure, so nothing moves. Adding CO₂ or removing CO raises Q above K, which pushes the reaction backward. Removing CO₂ lowers Q, which pushes it forward.
Answer: (A). Trap: treating the solid as if it had a concentration.
Part 2: Ionic equilibrium
pH of weak acids and bases
For a weak acid of concentration C, [H⁺] = √(Ka C) when it ionises only a little. That holds when the degree of ionisation, α = √(Ka/C), is small. For a weak base, use Kb the same way to get [OH⁻]. Diluting a weak acid raises α, so its pH changes more slowly than a strong acid's does.
The very dilute acid trap. For 10⁻⁸ M HCl, students write pH = 8, which would make an acid basic. Water already supplies 10⁻⁷ M H⁺, which is bigger than the acid's share. Solving properly gives pH ≈ 6.98, just below 7.
Salt hydrolysis: three formulas
| Salt of | Example | pH |
|---|---|---|
| Weak acid + strong base | CH₃COONa | 7 + ½pKa + ½ log C (above 7) |
| Strong acid + weak base | NH₄Cl | 7 − ½pKb − ½ log C (below 7) |
| Weak acid + weak base | CH₃COONH₄ | 7 + ½(pKa − pKb), with no C at all |
The third row is a favourite Statement I and II trap. The pH of a weak acid and weak base salt does not depend on its concentration. Often the salt forms by mixing an acid and a base. Then find C from the total volume of the mixture, not one solution's volume.
Buffers and the titration curve
A buffer holds a weak acid and its salt, or a weak base and its salt. Its pH comes from the Henderson equation: pH = pKa + log([salt]/[acid]). The curve below shows where a buffer lives inside a titration.
Two facts come straight from the curve. Halfway to the equivalence point, [salt] = [acid], so pH = pKa exactly. And a weak acid titrated with a strong base reaches equivalence above pH 7. That is why phenolphthalein works and methyl orange does not.
The ratio check. If the target pH is above pKa, the salt must be in excess. If it is below pKa, the acid must be. Check this before entering any buffer NVQ. It catches the upside-down ratio every time.
A worked JEE Main question
Q. (NVQ) A buffer is made from 1 M benzoic acid (pKa = 4.2) and 1 M sodium benzoate. Its pH is 4.5. What volume of benzoic acid solution is in 300 mL of this buffer, in mL? Round off to the nearest integer. (Given log 2 = 0.3)
Solution. Both solutions are 1 M, so the mole ratio equals the volume ratio. Henderson gives 4.5 = 4.2 + log(V_salt / V_acid), so log(V_salt / V_acid) = 0.3 and V_salt = 2 V_acid. With V_salt + V_acid = 300 mL, V_acid = 100 mL.
Answer: 100. Trap: solving for the salt volume, or flipping the ratio, gives 200. The check: pH 4.5 is above pKa 4.2, so salt must be in excess.
The predicted 2027 twist is a buffer made by partial neutralisation. Here a weak acid is mixed with less strong base than it needs. Subtract the base from the acid first. What is left of the acid, plus the salt just formed, goes into Henderson.
Solubility product, the numerical anchor
Ksp shows up in three to five shifts every year. Every version starts by writing Ksp in terms of the molar solubility s. The power and the number in front depend on the salt's formula.
| Salt type | Examples | Ksp in terms of s |
|---|---|---|
| AB | AgCl, BaSO₄ | s² |
| AB₂ or A₂B | Mg(OH)₂, CaF₂, Ag₂CrO₄ | 4s³ |
| AB₃ or A₃B | Fe(OH)₃, Ag₃PO₄ | 27s⁴ |
| A₃B₂ | Ca₃(PO₄)₂ | 108s⁵ |
Common ion. If the salt dissolves in a solution that already holds one of its ions, solubility falls sharply. Put the added ion's concentration in place of the tiny amount from the salt. For AB₂ in c M of B⁻, s = Ksp / c².
Precipitation. A solid starts to form when the ionic product reaches Ksp. For a hydroxide, that gives a minimum [OH⁻], so a pH at which precipitation begins.
Ion concentrations given directly. Sometimes the stem gives the ion concentrations, not s. Then just multiply them, raised to their powers. The 2 in front of an ion like Ag⁺ is already inside the given number. Do not double it again.
A worked JEE Main question
Q. (MCQ) The solubility of calcium phosphate (molar mass M) in water is W g per 100 mL at 25°C. Its solubility product is about:
(A) 108(W/M)⁵ (B) 108(10W/M)⁵ (C) 54(10W/M)⁵ (D) 27(10W/M)⁴
Solution. W g per 100 mL is 10W g per litre, so s = 10W/M mol L⁻¹. Ca₃(PO₄)₂ gives 3Ca²⁺ and 2PO₄³⁻. So Ksp = (3s)³(2s)² = 27 × 4 × s⁵ = 108(10W/M)⁵.
Answer: (B). Trap: (A) forgets to convert 100 mL to a litre; (C) does not square the 2; (D) uses the A₃B pattern.
| ⚡ The 10 traps, one line each | |
|---|---|
| Δn in Kp = Kc(RT)^Δn | Products minus reactants, gas moles only. |
| Solids and liquids | Leave them out of K. |
| Inert gas | Fixed volume: no shift. Fixed pressure: shifts to more gas moles. |
| What changes K | Only temperature. Not pressure, not concentration, not a catalyst. |
| Very dilute acid | Add water's 10⁻⁷ M H⁺. 10⁻⁸ M HCl has pH ≈ 6.98. |
| Ksp power | AB₂ is 4s³, A₃B is 27s⁴, A₃B₂ is 108s⁵. |
| Ion concentrations given | Plug them in directly. Do not double again. |
| Mixing dilutes | Use millimoles over the total volume. |
| Buffer ratio | pH above pKa means salt in excess. |
| Weak acid + weak base salt | Its pH does not depend on concentration. |
The formula sheet you need cold
| Quantity | Formula |
|---|---|
| Kp and Kc | Kp = Kc(RT)^Δn |
| Free energy and K | ΔG° = −RT ln K = −2.303 RT log K |
| van't Hoff | log(K₂/K₁) = [ΔH° / 2.303R] × (T₂ − T₁)/(T₁T₂) |
| From vapour density | α = (D − d) / [(n − 1)d] |
| Weak acid | [H⁺] = √(Ka C), α = √(Ka/C) |
| Water | Kw = 10⁻¹⁴ at 25°C, pH + pOH = 14, Ka × Kb = Kw |
| Acidic buffer | pH = pKa + log([salt]/[acid]) |
| Basic buffer | pOH = pKb + log([salt]/[base]) |
| Ksp of AB₂ in pure water | Ksp = 4s³ |
| Indicators | Methyl orange about 3.1 to 4.4; phenolphthalein about 8.3 to 10 |
| Handy values | log 2 = 0.30, log 3 = 0.48, log 5 = 0.70, log 7 = 0.85 |
| 🎯 Drill the whole decade of Equilibrium PYQs, sorted by sub-topic and format. | |
|---|---|
| Knowing the ratio check is not the same as using it with no options and a −1 waiting. Logic Bloom's Playground has every JEE Main Equilibrium PYQ from 2015 to 2026, tagged by sub-topic, format, and the exact trap it hides. Miss a dilution step or a Ksp power and TarQ walks you through the fix, then your Mistake Book brings it back until it sticks. | Get the app → Free to start. |
12 must-attempt JEE Main questions, built on real PYQ patterns
These twelve follow the templates JEE Main rotates most. Seven are MCQs and five are NVQs, close to the real mix. Each needs at least two steps or turns on a trap. Try each before reading the answer.
1. (MCQ) At 300 K, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has Kc = 0.25 mol⁻² L². What is Kp for NH₃(g) ⇌ ½N₂(g) + 3/2 H₂(g) at the same temperature? (Take RT = 24.6 L atm mol⁻¹)
(A) 0.081 atm (B) 98.4 atm (C) 2.0 atm (D) 49.2 atm
Work. Reversing gives 1/0.25 = 4. Halving the equation takes the square root, so Kc = √4 = 2. For the new reaction, Δn = (½ + 3/2) − 1 = +1. So Kp = 2 × (24.6)¹ = 49.2 atm.
Answer: (D). Trap: (A) uses Δn = −1; (B) reverses but forgets the square root; (C) stops at Kc.
2. (MCQ) N₂O₄ (molar mass 92) partly dissociates into NO₂. The measured vapour density of the equilibrium mixture is 36.8. The degree of dissociation is:
(A) 20% (B) 25% (C) 75% (D) 50%
Work. Theoretical vapour density D = 92/2 = 46. Each N₂O₄ gives 2 particles, so n − 1 = 1. α = (D − d)/d = (46 − 36.8)/36.8 = 9.2/36.8 = 0.25.
Answer: (B). Trap: (A) divides by D instead of d; (C) gives the part that did not dissociate; (D) doubles α for the two NO₂.
3. (NVQ) At 300 K and 1 atm total pressure, N₂O₄(g) ⇌ 2NO₂(g) is one-third dissociated. Find ΔG° for the reaction in J mol⁻¹. Round off to the nearest integer. (Given R = 8.3 J K⁻¹ mol⁻¹, ln 2 = 0.693)
Work. Start with 1 mol N₂O₄. At equilibrium there are 2/3 mol N₂O₄ and 2/3 mol NO₂, so 4/3 mol in all. Each mole fraction is 0.5. Kp = (0.5)² / 0.5 = 0.5. ΔG° = −RT ln 0.5 = RT ln 2 = 8.3 × 300 × 0.693 ≈ 1726 J mol⁻¹.
Answer: 1726. Trap: putting mole amounts into Kp instead of mole fractions gives Kp = 0.67 and about 1010.
4. (MCQ) N₂O₄(g) ⇌ 2NO₂(g) is at equilibrium in a cylinder with a piston. Which change makes more N₂O₄ dissociate?
(A) Adding helium at constant volume (B) Adding helium at constant pressure (C) Adding a catalyst (D) Pushing the piston in at constant temperature
Work. At constant pressure, added helium forces the volume up. The partial pressures of N₂O₄ and NO₂ fall, so the system shifts to the side with more gas moles, toward NO₂. At constant volume the partial pressures do not change, so nothing shifts.
Answer: (B). Trap: (A) assumes any added gas shifts it; (C) a catalyst never shifts equilibrium; (D) compression shifts it back toward N₂O₄.
5. (NVQ) A reaction has ΔH° = +40 kJ mol⁻¹ and ΔS° = +125 J K⁻¹ mol⁻¹. Treat both as constant. Above what temperature, in K, does its equilibrium constant exceed 1? Round off to the nearest integer.
Work. K = 1 exactly when ΔG° = −RT ln K = 0. So ΔH° − TΔS° = 0, and T = ΔH°/ΔS° = 40000/125 = 320 K. Above this, ΔG° turns negative and K rises above 1.
Answer: 320. Trap: leaving ΔH° in kJ gives 0.32.
6. (NVQ) 10 mL of 0.1 M weak acid HA (Ka = 1.0 × 10⁻⁵) is diluted with water to 1 L. Find the pH of the diluted solution. Round off to the nearest integer.
Work. Dilution by 100 times makes C = 1.0 × 10⁻³ M. For a weak acid, [H⁺] = √(Ka C) = √(10⁻⁵ × 10⁻³) = 10⁻⁴ M. So pH = 4.
Answer: 4. Trap: treating HA as a strong acid gives pH 3. Adding 2 to the old pH, as for a strong acid diluted 100 times, gives 5.
7. (NVQ) 50 mL of 0.3 M acetic acid (pKa = 4.74) is mixed with 50 mL of 0.1 M NaOH. The pH of the mixture is x × 10⁻². Find x. Round off to the nearest integer. (Given log 2 = 0.30, log 3 = 0.48)
Work. Acid = 15 mmol and base = 5 mmol. The base uses up 5 mmol of acid and makes 5 mmol of acetate. Left: 10 mmol acid and 5 mmol salt. pH = 4.74 + log(5/10) = 4.74 − 0.30 = 4.44. So x = 444.
Answer: 444. Trap: skipping the subtraction (15 acid, 5 salt) gives 426; flipping the ratio gives 504.
8. (MCQ) 20 mL of 0.1 M NH₄OH is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to: (Given Kb of NH₄OH = 1 × 10⁻⁵, log 3 = 0.48)
(A) 5.0 (B) 6.0 (C) 5.2 (D) 8.8
Work. Both are 2 mmol, so they neutralise exactly. That leaves 2 mmol NH₄Cl in 60 mL, so C = 1/30 M. For a strong acid and weak base salt, pH = 7 − ½pKb − ½ log C = 7 − 2.5 + ½(1.48) = 5.24.
Answer: (C). Trap: (A) ignores the dilution and uses C = 0.1 M; (B) forgets the ½ on log C; (D) uses the weak acid and strong base formula.
9. (MCQ) Ag₂CrO₄ has Ksp = 4 × 10⁻¹². How many times larger is its solubility in pure water than in 0.01 M AgNO₃?
(A) 2.5 × 10³ (B) 50 (C) 2.5 × 10⁵ (D) 1.0 × 10⁴
Work. In water, Ksp = 4s³, so s = (10⁻¹²)^(1/3) = 10⁻⁴ M. In 0.01 M Ag⁺, Ksp = (0.01)² s′, so s′ = 4 × 10⁻¹² / 10⁻⁴ = 4 × 10⁻⁸ M. The ratio is 10⁻⁴ / (4 × 10⁻⁸) = 2.5 × 10³.
Answer: (A). Trap: (B) uses s = √Ksp in water; (C) forgets to square [Ag⁺]; (D) doubles [Ag⁺] to 0.02 M.
10. (NVQ) A solution has [Mg²⁺] = 0.1 M. Mg(OH)₂ has Ksp = 1 × 10⁻¹¹. At what pH does Mg(OH)₂ just begin to precipitate? Round off to the nearest integer.
Work. Precipitation starts when [Mg²⁺][OH⁻]² = Ksp. So [OH⁻]² = 10⁻¹¹ / 0.1 = 10⁻¹⁰, and [OH⁻] = 10⁻⁵ M. pOH = 5, so pH = 9.
Answer: 9. Trap: entering the pOH gives 5.
11. (MCQ) In a saturated solution of the salt A₂B, [A⁺] = 2 × 10⁻⁴ M and [B²⁻] = 1 × 10⁻⁴ M. The Ksp of A₂B is:
(A) 3.2 × 10⁻¹¹ (B) 1.6 × 10⁻¹¹ (C) 4.0 × 10⁻¹² (D) 2.0 × 10⁻⁸
Work. Ksp = [A⁺]²[B²⁻]. The given [A⁺] already includes the 2 from the formula. So Ksp = (2 × 10⁻⁴)² × (1 × 10⁻⁴) = 4 × 10⁻¹².
Answer: (C). Trap: (A) uses 4s³ with s = 2 × 10⁻⁴; (B) doubles [A⁺] again; (D) forgets to square [A⁺].
12. (MCQ) Consider two statements.
Statement I: Phenolphthalein is a suitable indicator for titrating acetic acid with NaOH.
Statement II: In a weak acid and strong base titration, the equivalence point lies above pH 7, inside phenolphthalein's range of about 8.3 to 10.
Choose the correct answer.
(A) Both I and II are correct (B) Both I and II are incorrect (C) I is correct but II is incorrect (D) I is incorrect but II is correct
Work. At equivalence, the solution holds sodium acetate, which hydrolyses to give a basic pH near 8.7. That sits inside phenolphthalein's colour change, so both statements are true.
Answer: (A). Trap: (D) assumes equivalence is always at pH 7.
What to expect in JEE Main 2027
The 2027 syllabus is locked to 2026's, and the five NVQs stay compulsory. Expect one Chemical Equilibrium and one or two Ionic Equilibrium questions per shift, with at least one in Section B. Five setups look most likely.
1. Buffers made by partial neutralisation. 2024 and 2026 used ready-mixed buffers. The next step is mixing a weak acid with less strong base. Subtract the base first, then use Henderson.
2. van't Hoff and ΔG° hybrids. Find K at a second temperature, or find the temperature where K = 1 using T = ΔH°/ΔS°.
3. Selective precipitation. Two ions compete for the same reagent. Find which precipitates first, and how much of it remains when the second just starts.
4. Inert gas in a closed flask. Subtract helium's partial pressure from the total before using Kp.
5. Statement I and II pairs. Expect hydrolysis or common-ion observations paired with a reason based on Ksp, buffer action, or indicator choice.
How to split your time: 40% on Ionic numericals: pH of mixtures, buffers, and Ksp with common ion. 35% on Chemical Equilibrium numericals: K rules, Kp and Kc, degree of dissociation, and ΔG°. 25% on Le Chatelier, hydrolysis rules, and indicators, the MCQ side.
| 🚀 Turn this analysis into a real score. | |
|---|---|
| You now know the templates and the ten traps. Close the gap between knowing and scoring. Logic Bloom's Playground drills every JEE Main Equilibrium PYQ of the decade, and Battleground lets you test the chapter in 1v1 duels against other aspirants. Understand through games. Score through practice. | Get the app → Free to start. |