Equilibrium — NEET Chemistry Notes & MCQs

1. Chapter at a Glance

Many reactions do not go to completion. Instead, the forward and backward reactions continue at equal rates, and the concentrations stop changing: the system is at equilibrium. This chapter covers chemical equilibrium (Kc, Kp, Le Chatelier's principle) and ionic equilibrium (acids and bases, pH, buffers, salt hydrolysis and solubility product). NEET asks about two questions every year, many of them numerical.

What NEET asks most

Writing Kc and Kp; relation Kp = Kc(RT)^Δn.
Effect of concentration, pressure, temperature and catalyst (Le Chatelier).
pH of strong and weak acids and bases; degree of ionisation.
Buffer pH (Henderson equation) and salt hydrolysis.
Solubility product, solubility and the common-ion effect.

2. Dynamic Equilibrium

At equilibrium, the rate of the forward reaction equals the rate of the backward reaction. Both reactions still happen; only the net change is zero.
Physical equilibria: ice ⇌ water at 0 °C; liquid ⇌ vapour in a closed container; a saturated solution ⇌ undissolved solid.
Equilibrium can be reached from either side and needs a closed system.

Concentrations change quickly at first and then level off. After equilibrium they stay constant, although both reactions keep running at equal rates.

3. Law of Mass Action, Kc and Kp

aA + bB ⇌ cC + dD Kc = [C]^c[D]^d/([A]^a[B]^b) Kp = Kc(RT)^Δn

Pure solids and pure liquids do not appear in K expressions.
If a reaction is reversed, K becomes 1/K. If it is multiplied by n, K becomes Kⁿ. If two reactions are added, their K values are multiplied.
Δn = moles of gaseous products − moles of gaseous reactants. Use R = 0.0831 L bar K⁻¹ mol⁻¹ with pressures in bar.
Reaction quotient Q has the same form as K but uses the current concentrations. If Q < K the reaction moves forward; if Q > K it moves backward; if Q = K it is at equilibrium.
A large K (> 10³) means products dominate; a small K (< 10⁻³) means reactants dominate.
K depends only on temperature. It is not changed by concentration, pressure or a catalyst.

4. Le Chatelier's Principle

If a system at equilibrium is disturbed, it shifts in the direction that reduces the disturbance.

Change	Shift	Effect on K	NEET Note
Add reactant / remove product	Forward	None	Q becomes less than K
Increase pressure (decrease volume)	Towards fewer gas moles	None	No effect if Δn = 0
Increase temperature	Endothermic direction	Changes	K rises for endothermic reactions
Add catalyst	No shift	None	Equilibrium reached faster
Add inert gas at constant volume	No shift	None	Partial pressures unchanged
Add inert gas at constant pressure	Towards more gas moles	None	Like increasing volume

Example: in the Haber process, N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0), high pressure and a moderate temperature (about 700 K) with an iron catalyst favour ammonia.

5. Acids, Bases and pH

Arrhenius: acids give H⁺, bases give OH⁻ in water. Brønsted–Lowry: acids donate protons, bases accept them; conjugate pairs differ by one H⁺. Lewis: acids accept electron pairs (BF₃, AlCl₃, H⁺), bases donate them (NH₃, OH⁻).
Ionic product of water: K_w = [H⁺][OH⁻] = 1 × 10⁻¹⁴ at 298 K. K_w increases with temperature.
pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = 14 at 298 K.
Weak acid: Ka = cα²/(1 − α) ≈ cα². So α = √(Ka/c) and [H⁺] = √(Ka c) (Ostwald's dilution law: α increases as the solution is diluted).
For a conjugate acid–base pair: Ka × Kb = K_w. A stronger acid has a weaker conjugate base.

6. Buffers and Salt Hydrolysis

acidic buffer: pH = pKa + log([salt]/[acid]) basic buffer: pOH = pKb + log([salt]/[base])

A buffer resists changes in pH when small amounts of acid or base are added. Examples: CH₃COOH + CH₃COONa (acidic) and NH₄OH + NH₄Cl (basic). Blood is buffered near pH 7.4.
A buffer works best when [salt] = [acid], so that pH = pKa.

Salt type	Example	Solution	pH formula
Strong acid + strong base	NaCl	Neutral	pH = 7
Weak acid + strong base	CH₃COONa	Basic	pH = 7 + ½pKa + ½log c
Strong acid + weak base	NH₄Cl	Acidic	pH = 7 − ½pKb − ½log c
Weak acid + weak base	CH₃COONH₄	Depends on Ka, Kb	pH = 7 + ½(pKa − pKb)

7. Solubility Product and Common-ion Effect

AₓBᵧ(s) ⇌ xA^y+ + yB^x− Ksp = [A]ˣ[B]ʸ = xˣyʸ s^x+y

AB type (AgCl): Ksp = s². AB₂ type (CaF₂): Ksp = 4s³. A₂B type (Ag₂CrO₄): Ksp = 4s³. AB₃ type: Ksp = 27s⁴.
Precipitation occurs when the ionic product exceeds Ksp.
Common-ion effect: adding an ion already present in the equilibrium reduces the solubility of a salt or the ionisation of a weak electrolyte. It is used in salt analysis (e.g. NH₄Cl suppresses the ionisation of NH₄OH in group III).

8. Formula Master Table

Quantity	Formula
Kp–Kc relation	Kp = Kc(RT)^Δn
Reversed / multiplied reaction	K′ = 1/K ; K′ = Kⁿ
ΔG° and K	ΔG° = −2.303RT log K
pH	pH = −log[H⁺]; pH + pOH = 14
Weak acid	α = √(Ka/c); [H⁺] = √(Ka c)
Conjugate pair	Ka × Kb = K_w
Acid buffer	pH = pKa + log([salt]/[acid])
Hydrolysis constant	K_h = K_w/Ka or K_w/Kb
Solubility product	Ksp = xˣyʸ s^x+y
Precipitation	Ionic product > Ksp

9. Master Summary Table

Concept	Key idea	NEET Note
Equilibrium	Forward rate = backward rate	Dynamic, not static
K	Temperature only	Catalyst does not change K
Q vs K	Direction of change	Q < K ⇒ forward
Le Chatelier	Opposes the change	Pressure affects only if Δn ≠ 0
Weak acids	α ∝ 1/√c	Dilution increases α
Buffer	Henderson equation	Best at pH = pKa
Hydrolysis	Salt of weak partner	CH₃COONa basic, NH₄Cl acidic
Ksp	Saturated solution	Common ion lowers solubility

10. Mind Map

Equilibrium

Chemical

Kc, Kp
Q vs K
Le Chatelier

Acids and bases

Three theories
pH, K_w
Ka, Kb, α

Buffers and salts

Henderson
Hydrolysis

Solubility

Ksp
Precipitation
Common ion

11. Tips and Tricks

Shortcuts

pH of 10⁻ⁿ M strong acid = n (valid while n < 6). For very dilute acids (10⁻⁸ M), include water: pH is just below 7, not 8.
Mixing equal volumes of two strong acids with pH a and b (differing by ≥ 2) gives pH ≈ (lower pH) + 0.3.
For a buffer with salt/acid ratio 10, pH = pKa + 1; ratio 0.1, pH = pKa − 1.
Kp = Kc when Δn = 0.

Common NEET traps

Including solids or pure liquids in K.
Saying a catalyst shifts the equilibrium.
Writing Ksp = s² for every salt (use the stoichiometry).
Forgetting that Ba(OH)₂ gives two OH⁻ per formula unit.

12. Important One-liners

At equilibrium, ΔG = 0 and Q = K.
The value of K changes only with temperature.
BF₃ is a Lewis acid; NH₃ is a Lewis base.
The conjugate base of a strong acid is very weak.
Pure water is neutral at all temperatures, even though its pH changes with temperature.
A solution of sodium acetate is basic because acetate ions hydrolyse.
The common-ion effect decreases the solubility of a sparingly soluble salt.
Human blood is maintained at a pH of about 7.4 by buffer systems.

13. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take K_w = 1 × 10⁻¹⁴ at 298 K and log 2 = 0.301. Click to reveal each answer and working.

Average level

N1Average

For N₂O₄(g) ⇌ 2NO₂(g), the equilibrium concentrations are [N₂O₄] = 0.04 M and [NO₂] = 0.02 M. Find Kc.

Show answer & solution

Concept: Kc = [NO₂]²/[N₂O₄].

Kc = (0.02)²/0.04 = 4 × 10⁻⁴/0.04.
Kc = 0.01 mol L⁻¹.

Answer: 0.01

N2Average

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K, Kc = 0.061 (concentrations in mol L⁻¹). Find Kp in bar units (R = 0.0831 L bar K⁻¹ mol⁻¹).

Show answer & solution

Concept: Kp = Kc(RT)^Δn.

Δn = 2 − 4 = −2; RT = 0.0831 × 500 = 41.55.
Kp = 0.061/(41.55)² = 0.061/1726.4.
Kp ≈ 3.5 × 10⁻⁵.

Answer: about 3.5 × 10⁻⁵

N3Average

Find the pH of (a) 0.001 M HCl and (b) 0.005 M Ba(OH)₂.

Show answer & solution

Concept: strong electrolytes ionise completely.

(a) [H⁺] = 10⁻³ ⇒ pH = 3.
(b) [OH⁻] = 2 × 0.005 = 0.01 ⇒ pOH = 2 ⇒ pH = 12.

Answer: (a) 3; (b) 12

N4Average

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

Show answer & solution

Concept: [H⁺] = √(Ka c); α = [H⁺]/c.

[H⁺] = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³ M.
pH = −log(1.34 × 10⁻³) ≈ 2.87.
α = 1.34 × 10⁻³/0.1 ≈ 0.0134 (1.34%).

Answer: pH about 2.87; α about 1.34%

N5Average

The solubility of AgCl in water at 298 K is 1.3 × 10⁻⁵ mol L⁻¹. Find its Ksp.

Show answer & solution

Concept: AB type: Ksp = s².

Ksp = (1.3 × 10⁻⁵)².
Ksp = 1.69 × 10⁻¹⁰.

Answer: 1.69 × 10⁻¹⁰

Advanced level

N6Advanced

A buffer contains 0.2 M acetic acid and 0.1 M sodium acetate (pKa = 4.74). Find its pH.

Show answer & solution

Concept: Henderson equation.

pH = 4.74 + log(0.1/0.2) = 4.74 + log 0.5.
log 0.5 = −0.301.
pH = 4.44.

Answer: about 4.44

N7Advanced

1 mol of PCl₅ is heated in a 1 L vessel: PCl₅ ⇌ PCl₃ + Cl₂, Kc = 0.5 mol L⁻¹. Find the degree of dissociation and the equilibrium concentrations.

Show answer & solution

Concept: ICE table: (1 − x), x, x.

Kc = x²/(1 − x) = 0.5 ⇒ x² + 0.5x − 0.5 = 0.
x = [−0.5 + √(0.25 + 2)]/2 = (−0.5 + 1.5)/2 = 0.5.
α = 0.5 (50%); [PCl₅] = [PCl₃] = [Cl₂] = 0.5 M.

Answer: 50%; each 0.5 M

N8Advanced

Ksp of CaF₂ is 4 × 10⁻¹¹. Find its solubility (a) in pure water and (b) in 0.1 M NaF.

Show answer & solution

Concept: AB₂ type: Ksp = 4s³; common ion F⁻ in (b).

(a) 4s³ = 4 × 10⁻¹¹ ⇒ s³ = 10⁻¹¹ ⇒ s ≈ 2.15 × 10⁻⁴ M.
(b) [F⁻] ≈ 0.1 M: Ksp = s(0.1)² ⇒ s = 4 × 10⁻¹¹/0.01 = 4 × 10⁻⁹ M.
The common ion lowers the solubility about 50 000 times.

Answer: (a) about 2.2 × 10⁻⁴ M; (b) 4 × 10⁻⁹ M

N9Advanced

Find the pH of a 0.1 M NH₄Cl solution (Kb of NH₄OH = 1.8 × 10⁻⁵).

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Concept: NH₄⁺ hydrolyses: K_h = K_w/Kb; [H⁺] = √(K_h c).

K_h = 10⁻¹⁴/1.8 × 10⁻⁵ ≈ 5.56 × 10⁻¹⁰.
[H⁺] = √(5.56 × 10⁻¹⁰ × 0.1) ≈ 7.45 × 10⁻⁶ M.
pH ≈ 5.13 (acidic, as expected for a salt of a strong acid and weak base).

Answer: about 5.13

N10Advanced

For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 50 at a certain temperature. A mixture has [H₂] = [I₂] = 0.1 M and [HI] = 0.5 M. In which direction will it move? Find the equilibrium concentrations.

Show answer & solution

Concept: compare Q with K, then use an ICE table.

Q = (0.5)²/(0.1 × 0.1) = 25 < 50 ⇒ moves forward.
Let x react: (0.5 + 2x)²/(0.1 − x)² = 50 ⇒ (0.5 + 2x)/(0.1 − x) = 7.07.
0.5 + 2x = 0.707 − 7.07x ⇒ x ≈ 0.0228.
[H₂] = [I₂] ≈ 0.077 M; [HI] ≈ 0.546 M.

Answer: forward; [H₂] = [I₂] ≈ 0.077 M, [HI] ≈ 0.546 M

14. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

A catalyst added to a reaction at equilibrium:

Increases K
Decreases K
Helps reach equilibrium faster without changing K
Shifts the equilibrium forward

Show solution

A catalyst speeds up the forward and backward reactions equally, so the system reaches equilibrium sooner. The position of equilibrium and the value of K do not change.

Answer: (C)

Q2Easy

Which is a Lewis acid?

NH₃
BF₃
H₂O
OH⁻

Show solution

BF₃ has an incomplete octet, so it can accept an electron pair, which makes it a Lewis acid. NH₃, H₂O and OH⁻ all have lone pairs to donate, so they are Lewis bases.

Answer: (B)

Q3Medium

For which reaction does increasing the pressure have no effect on the equilibrium?

N₂ + 3H₂ ⇌ 2NH₃
PCl₅ ⇌ PCl₃ + Cl₂
N₂ + O₂ ⇌ 2NO
2SO₂ + O₂ ⇌ 2SO₃

Show solution

Pressure matters only if Δn ≠ 0. For N₂ + O₂ ⇌ 2NO, Δn = 2 − 2 = 0, so pressure has no effect. The other three reactions change the number of gas moles.

Answer: (C)

Q4Medium

The aqueous solution of which salt is basic?

NaCl
NH₄Cl
CH₃COONa
KNO₃

Show solution

CH₃COONa is the salt of a weak acid and a strong base. Acetate hydrolyses to give OH⁻, so the solution is basic. NaCl and KNO₃ solutions are neutral, and NH₄Cl is acidic.

Answer: (C)

Q5Medium

If the solubility of Ag₂CrO₄ is s mol L⁻¹, its Ksp is:

s²
2s³
4s³
27s⁴

Show solution

Ag₂CrO₄ gives 2Ag⁺ + CrO₄²⁻, so Ksp = (2s)²(s) = 4s³. s² is for AB salts, and 27s⁴ is for AB₃ salts.

Answer: (C)

Q6Medium

For an exothermic reaction at equilibrium, increasing the temperature will:

Increase K and favour products
Decrease K and favour reactants
Not change K
Double K

Show solution

Adding heat favours the endothermic (backward) direction. K falls and the reactants are favoured. Temperature is the only factor that changes K.

Answer: (B)

Q7Hard

The pH of a 10⁻⁸ M HCl solution is:

8
Slightly less than 7
Exactly 7
6

Show solution

An acid cannot have a pH above 7. Water itself gives 10⁻⁷ M H⁺, which is much more than 10⁻⁸ M. The total [H⁺] ≈ 1.05 × 10⁻⁷ M, so the pH is about 6.98. (A) ignores water completely.

Answer: (B)

Q8Hard

The equilibrium constant for A ⇌ B is 4. What is the equilibrium constant for 2B ⇌ 2A?

1/16
1/4
16
1/8

Show solution

Reversing the reaction gives 1/4. Multiplying it by 2 squares this value: (1/4)² = 1/16. (B) forgets the multiplication, and (C) forgets the reversal.

Answer: (A)

15. Quick Revision Summary

Remember before the exam

Kc from mass action; Kp = Kc(RT)^Δn; K depends only on T.
Q < K forward; Le Chatelier for concentration, pressure, temperature.
pH = −log[H⁺]; weak acid [H⁺] = √(Ka c); Ka Kb = K_w.
Buffer pH = pKa + log(salt/acid); hydrolysis depends on the weak partner.
Ksp = xˣyʸs^x+y; common ion lowers solubility.

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