Equilibrium Notes for NEET: Class 11 NCERT Chemistry

Q: What is the Henderson-Hasselbalch equation?

pH = pKa + log([A⁻]/[HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the weak acid in the buffer. When [A⁻] = [HA], pH = pKa. Buffers work best when pH is within 1 unit of pKa.

Q: When does precipitation occur in a Ksp problem?

A precipitate forms when the ionic product (IP) of the ions in solution exceeds Ksp. If IP Ksp, precipitation occurs. The common ion effect lowers solubility by increasing one of the ions already present.

Introduction to Equilibrium

Equilibrium is one of the most important and high-frequency topics in NEET Chemistry. It appears almost every year and can involve anything from calculating an equilibrium constant to predicting how a system responds to a change in conditions.

At its core, equilibrium means that the rate of the forward reaction equals the rate of the reverse reaction. The concentrations of reactants and products are not necessarily equal, but they are constant at equilibrium.

This chapter covers two closely related topics: chemical equilibrium (for reactions in which molecules rearrange) and ionic equilibrium (for reactions involving acids, bases, and sparingly soluble salts in water).

Equilibrium in Physical Processes

Before looking at chemical reactions, you should understand equilibrium in physical processes. These are simpler to visualise.

Liquid-Vapour Equilibrium

Place water in a sealed container. Water molecules evaporate from the surface and vapour molecules condense back into liquid. At equilibrium, the rate of evaporation equals the rate of condensation. The pressure exerted by the vapour at this point is called the vapour pressure. It depends only on temperature, not on the amount of liquid present (as long as some liquid remains).

Solid-Liquid Equilibrium

Ice and water at 0 °C and 1 atm are in equilibrium. The rate of melting equals the rate of freezing. This is why the temperature stays at 0 °C as long as both phases are present.

Gas in Liquid (Henry's Law)

Carbon dioxide in a sealed soft-drink bottle is in equilibrium with CO₂ dissolved in the liquid. Henry's law states that the concentration of a dissolved gas is proportional to the partial pressure of that gas above the solution.

In all physical equilibria, the concentrations (or pressures) of the phases involved are constant at a given temperature. This is the same idea that applies to chemical equilibrium.

Chemical Equilibrium

Consider the reaction for synthesis of ammonia:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)

At the start, only N₂ and H₂ are present. The forward reaction begins. As NH₃ builds up, the reverse reaction (decomposition of NH₃) also begins. Over time, the rates of the forward and reverse reactions become equal. At this point, the concentrations of N₂, H₂, and NH₃ stop changing. This is the state of chemical equilibrium.

Important features of chemical equilibrium:

It is dynamic, meaning both forward and reverse reactions continue at equal rates.
It can be reached from either direction (start with reactants or with products and you end up at the same equilibrium state at a given temperature).
The concentrations of reactants and products are constant but not necessarily equal.
It requires a closed system.

Law of Mass Action and Kc

For a general reaction at equilibrium:

a A + b B ⇌ c C + d D

The equilibrium constant Kc is defined as:

K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

The square brackets [ ] denote molar concentration (mol/L) at equilibrium. The exponents are the stoichiometric coefficients from the balanced equation.

Important Rules for Writing Kc

Pure solids and pure liquids are not included in the expression because their concentrations are constant (their activity = 1).
For water as a solvent in dilute solution, [H₂O] is not included.
If you reverse a reaction, the new K is the reciprocal: K' = 1/K.
If you multiply a reaction by a factor n, the new K is K^n.
If two reactions are added together, their equilibrium constants are multiplied.

Example for a heterogeneous equilibrium:

CaCO_{3} (s) ⇌ CaO (s) + CO_{2} (g)

K_{c} = [CO_{2}]

Only CO₂ appears in Kc because CaCO₃ and CaO are pure solids.

Kp and the Kp-Kc Relationship

For reactions involving gases, you can write the equilibrium constant in terms of partial pressures. For the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g):

K_{p} = \frac{( P _{C} ) ^{c} ( P _{D} ) ^{d}}{( P _{A} ) ^{a} ( P _{B} ) ^{b}}

The relationship between Kp and Kc is:

K_{p} = K_{c} (R T)^{Δ n_{g}}

where $Δ n_{g}$ = (moles of gaseous products) – (moles of gaseous reactants), R = 0.0821 L·atm/(mol·K), and T is in Kelvin.

Δn_g	Relationship	Example
0	Kp = Kc	H₂ + I₂ ⇌ 2HI
+1	Kp = Kc × RT	PCl₅ ⇌ PCl₃ + Cl₂
-2	Kp = Kc / (RT)²	N₂ + 3H₂ ⇌ 2NH₃ (Δn = -2)

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Significance of the Equilibrium Constant

The value of K tells you the relative amounts of products and reactants at equilibrium.

K value	Meaning
K >> 1	Reaction goes almost to completion; products favoured
K << 1	Very little product forms; reactants favoured
K ≈ 1	Appreciable amounts of both reactants and products at equilibrium

K does NOT tell you anything about the rate of the reaction. A reaction with a large K can still be very slow.

Reaction Quotient Q

The reaction quotient Q is calculated using the same expression as Kc but with the current concentrations (not necessarily equilibrium concentrations):

Q_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} (at any time)

Comparing Q with K tells you which direction the reaction will move:

Q < K: The reaction moves forward (more products will form) to reach equilibrium.
Q > K: The reaction moves backward (products decompose to reactants).
Q = K: The system is at equilibrium already.

Le Chatelier's Principle

Le Chatelier's Principle: when a system at equilibrium is disturbed, it adjusts to oppose the disturbance.

N₂ + 3H₂ ⇌ 2NH₃

H₂ + I₂ ⇌ 2HI

N₂O₄ ⇌ 2NO₂

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

ΔH = -92 kJ/mol (exothermic) | Δn(gas) = -2

Apply a stress

Increase T

Decrease T

Increase P

Decrease P

Add reactant

Remove reactant

Add product

Remove product

Add catalyst

Select a stress above to see the equilibrium shift.

Le Chatelier's principle: If an equilibrium system is disturbed, it will shift in the direction that partially reduces the disturbance.

Effect of Changing Concentration

Adding a reactant: equilibrium shifts toward products (forward direction) to consume the added reactant.
Removing a product: equilibrium shifts forward (toward products) to replace what was removed.
Removing a reactant or adding a product: equilibrium shifts backward.

Note: Adding a pure solid or changing the amount of a pure solid has no effect because it does not appear in the equilibrium expression.

Effect of Pressure (for Gas-Phase Reactions)

Increasing pressure (by reducing volume): equilibrium shifts toward the side with fewer moles of gas.
Decreasing pressure: shifts toward the side with more moles of gas.
Adding an inert gas at constant volume: no effect (partial pressures of reacting gases do not change).
Adding an inert gas at constant pressure (allowing volume to expand): partial pressures of reacting gases decrease, so equilibrium shifts toward more moles of gas.
If Δn_g = 0, pressure change has no effect on the position of equilibrium.

Effect of Temperature

Temperature change actually changes the value of K (unlike concentration or pressure changes, which shift equilibrium without changing K).

Increasing temperature: favours the endothermic direction (the reaction that absorbs heat). K increases if the forward reaction is endothermic; K decreases if the forward reaction is exothermic.
Decreasing temperature: favours the exothermic direction.

Effect of a Catalyst

A catalyst speeds up both the forward and reverse reactions equally. It helps the system reach equilibrium faster but does not change the position of equilibrium or the value of K.

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Ionic Equilibrium

ICE tables help find equilibrium concentrations from initial concentrations and the equilibrium constant Kc.

Problem 1

Problem 2

Problem 3

A(aq) + B(aq) ⇌ C(aq)

Kc = 4 | [A]₀ = 1 M, [B]₀ = 1 M, [C]₀ = 0 M

Species

Initial (M)

Change

Equilibrium (M)

When an acid or base dissolves in water, it establishes an equilibrium between its ionised and un-ionised forms. This is ionic equilibrium.

Strong Acids and Bases

Strong acids (HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄) and strong bases (NaOH, KOH, Ca(OH)₂) are completely dissociated in dilute aqueous solution. There is no equilibrium to set up; [H⁺] or [OH⁻] equals the initial concentration directly.

Weak Acids: Ka and Degree of Ionisation

A weak acid HA ionises partially:

HA ⇌ H^{+} (a q) + A^{-} (a q)

K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}

Ka is the acid dissociation constant. A larger Ka means a stronger weak acid.

The degree of ionisation α is the fraction of acid molecules that have ionised:

α = \frac{x}{C}

where x = [H⁺] and C is the initial acid concentration. For a weak acid with small Ka:

K_{a} \approx C α^{2} \Rightarrow α = \frac{K _{a}}{C}

This is called Ostwald's dilution law. As you dilute the acid (decrease C), α increases. At infinite dilution, α → 1 (complete ionisation).

Weak Bases: Kb

A weak base B ionises as:

B + H_{2} O ⇌ BH^{+} + OH^{-}

K_{b} = \frac{[ BH ^{+} ] [ OH ^{-} ]}{[ B ]}

Similarly, [OH⁻] = √(Kb × C) for a weak base.

Relationship Between Ka and Kb

For a conjugate acid-base pair (e.g., NH₄⁺ and NH₃):

K_{a} \times K_{b} = K_{w} = 1.0 \times 1 0^{- 14} at 25 °C

where Kw is the ionic product of water.

Acids, Bases, and pH

Select a solution to calculate its pH with full step-by-step working.

HCl (0.1 M)

H₂SO₄ (0.05 M)

NaOH (0.1 M)

Ca(OH)₂ (0.01 M)

CH₃COOH (0.1 M)

HCOOH (0.1 M)

NH₃ (0.1 M)

C₅H₅N (0.1 M)

1.00

Nature

Acidic

Type

strong acid

Strong acid: fully ionizes in water.

[H⁺] = C = 1.00e-1 M

pH = −log₁₀[H⁺] = −log₁₀(1.00e-1) = 1.00

pH scale (0–14)

0 (very acidic)

7 (neutral)

14 (very basic)

Arrhenius Concept

An Arrhenius acid gives H⁺ in water. An Arrhenius base gives OH⁻ in water. This concept is limited to aqueous solutions.

Bronsted-Lowry Concept

A Bronsted-Lowry acid is a proton (H⁺) donor. A Bronsted-Lowry base is a proton acceptor. Every acid has a conjugate base (formed by losing a proton) and every base has a conjugate acid (formed by gaining a proton).

HCl + H_{2} O \to H_{3} O^{+} + Cl^{-}

Here HCl is the acid, H₂O is the base, H₃O⁺ is the conjugate acid of water, and Cl⁻ is the conjugate base of HCl.

Lewis Concept

A Lewis acid accepts an electron pair. A Lewis base donates an electron pair. This is the broadest definition and covers reactions where no proton is involved (e.g., BF₃ + NH₃ → F₃B-NH₃, where BF₃ is the Lewis acid).

The pH Scale

pH is defined as:

pH = - lo g_{10} [H^{+}]

Similarly:

pOH = - lo g_{10} [OH^{-}]

pH + pOH = 14 (at 25 °C)

Key values to remember:

Solution	[H⁺] (mol/L)	pH
Pure water at 25 °C	10⁻⁷	7
Acidic solution	> 10⁻⁷	< 7
Basic solution	< 10⁻⁷	> 7
0.1 M HCl	0.1 = 10⁻¹	1
0.01 M NaOH	10⁻¹² (from Kw/[OH⁻])	12

For weak acids, [H⁺] = √(Ka × C), so:

pH = \frac{1}{2} (pK_{a} - lo g C)

Buffer Solutions

A buffer is a solution that resists a change in pH when a small amount of acid or base is added. A typical buffer contains a weak acid and its conjugate base (or a weak base and its conjugate acid).

Example: CH₃COOH (acetic acid) + CH₃COONa (sodium acetate) is an acidic buffer.

The pH of a buffer is given by the Henderson-Hasselbalch equation:

pH = pK_{a} + lo g \frac{[ A ^{-} ]}{[ HA ]}

When [A⁻] = [HA] (equal concentrations of acid and conjugate base), pH = pKa. This is the half-neutralisation point.

How a buffer works: if you add a small amount of HCl (strong acid), the conjugate base A⁻ absorbs the extra H⁺ to form HA, so pH barely changes. If you add NaOH, HA donates a proton to neutralise the OH⁻, again minimising pH change.

Buffers have a limited capacity. If you add too much acid or base, the buffer is overwhelmed and pH changes sharply.

Solubility Product Ksp

For a sparingly soluble salt, the equilibrium between the solid and its ions in a saturated solution is:

AgCl (s) ⇌ Ag^{+} (a q) + Cl^{-} (a q)

K_{s p} = [Ag^{+}] [Cl^{-}]

The solid AgCl is not included because it is a pure solid. Ksp is the solubility product constant.

For a salt MₘXₙ dissolving as MₘXₙ ⇌ mM(n+) + nX(m-):

K_{s p} = [M^{n +}]^{m} [X^{m -}]^{n}

Relating Ksp to Molar Solubility

Let s = molar solubility (mol/L) of the salt. Then [M^n+] = ms and [X^m-] = ns. Substituting:

K_{s p} = (m s)^{m} (n s)^{n} = m^{m} n^{n} \cdot s^{m + n}

Solve for s.

Common Ion Effect

If one of the ions is already present in solution (e.g., you add AgNO₃ to a solution that already contains Cl⁻), the solubility of AgCl decreases. This is the common ion effect. It follows Le Chatelier's principle: the extra Cl⁻ shifts the equilibrium backward, reducing dissolution.

Ionic Product and Precipitation

IP < Ksp: Solution is unsaturated; more salt can dissolve.
IP = Ksp: Saturated solution; equilibrium.
IP > Ksp: Solution is supersaturated; precipitation occurs.

Worked NEET Problems

NEET-style problem · Kp and Kc

Question

For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 1.8 × 10⁻⁷ at 300 K. Find Kp. (R = 0.0821 L·atm/mol·K)

Solution

Step 1: Find Δn_g. Moles of gaseous products = 1 (PCl₃) + 1 (Cl₂) = 2. Moles of gaseous reactants = 1 (PCl₅). So Δn_g = 2 − 1 = 1.

Step 2: Apply Kp = Kc(RT)^Δn_g.

K_{p} = 1.8 \times 1 0^{- 7} \times (0.0821 \times 300)^{1}

K_{p} = 1.8 \times 1 0^{- 7} \times 24.63 = 4.43 \times 1 0^{- 6}

NEET-style problem · Reaction Quotient Q

Question

For N₂ + 3H₂ ⇌ 2NH₃, Kc = 2.0 × 10⁻⁵ at a certain temperature. At a particular moment, [N₂] = 2.0 mol/L, [H₂] = 0.5 mol/L, [NH₃] = 0.1 mol/L. In which direction will the reaction proceed?

Solution

Step 1: Calculate Q.

Q_{c} = \frac{[ NH _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}} = \frac{( 0.1 ) ^{2}}{( 2.0 ) ( 0.5 ) ^{3}} = \frac{0.01}{2.0 \times 0.125} = \frac{0.01}{0.25} = 0.04

Step 2: Compare Q with Kc. Q = 0.04 and Kc = 2.0 × 10⁻⁵. Since Q >> Kc, the reaction will shift in the reverse direction (toward reactants) to reduce [NH₃] and increase [N₂] and [H₂].

NEET-style problem · pH of a Weak Acid

Question

Calculate the pH of a 0.1 M solution of acetic acid. Ka = 1.8 × 10⁻⁵.

Solution

Step 1: Set up the ICE table. CH₃COOH ⇌ H⁺ + CH₃COO⁻. Initial: [CH₃COOH] = 0.1, [H⁺] = 0, [CH₃COO⁻] = 0. At equilibrium: [CH₃COOH] = 0.1 − x, [H⁺] = x, [CH₃COO⁻] = x.

K_{a} = \frac{x ^{2}}{0.1 - x} \approx \frac{x ^{2}}{0.1} = 1.8 \times 1 0^{- 5}

x^{2} = 1.8 \times 1 0^{- 6} \Rightarrow x = 1.34 \times 1 0^{- 3} mol/L

Check: x/C = 1.34 × 10⁻³/0.1 = 1.34% < 5%, so the approximation is valid.

pH = - lo g (1.34 \times 1 0^{- 3}) = 3 - lo g (1.34) = 3 - 0.127 \approx 2.87

NEET-style problem · Solubility Product

Question

The molar solubility of BaSO₄ is 1.05 × 10⁻⁵ mol/L at 25 °C. Calculate Ksp.

Solution

BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. If molar solubility = s = 1.05 × 10⁻⁵ mol/L, then [Ba²⁺] = [SO₄²⁻] = 1.05 × 10⁻⁵ mol/L.

K_{s p} = [Ba^{2 +}] [SO_{4}^{2 -}] = (1.05 \times 1 0^{- 5})^{2} = 1.10 \times 1 0^{- 10}

NEET-style problem · Buffer pH (Henderson-Hasselbalch)

Question

Calculate the pH of a buffer containing 0.2 M CH₃COOH and 0.3 M CH₃COONa. pKa of acetic acid = 4.74.

Solution

pH = pK_{a} + lo g \frac{[ A ^{-} ]}{[ HA ]} = 4.74 + lo g \frac{0.3}{0.2} = 4.74 + lo g (1.5)

pH = 4.74 + 0.176 = 4.92

Summary Cheat Sheet

Concept	Formula / Key Point
Equilibrium constant Kc	[products]^coeff / [reactants]^coeff (at equilibrium)
Kp vs Kc	Kp = Kc(RT)^Δn_g
Reaction quotient Q	Same formula as Kc but with current concentrations
Predicting direction	Q < K → forward; Q > K → reverse; Q = K → equilibrium
Le Chatelier (concentration)	Add reactant → shift forward; add product → shift reverse
Le Chatelier (pressure)	Increase P → shift to fewer gas moles side
Le Chatelier (temperature)	Increase T → favour endothermic; changes K
Weak acid [H⁺]	[H⁺] = √(Ka × C)
Weak base [OH⁻]	[OH⁻] = √(Kb × C)
Ka × Kb relationship	Ka × Kb = Kw = 10⁻¹⁴ at 25 °C (for conjugate pair)
pH	pH = −log[H⁺]; pH + pOH = 14
Buffer pH	pH = pKa + log([A⁻]/[HA])
Ksp and solubility	Ksp = [ions]^stoich; IP > Ksp → precipitation
Common ion effect	Decreases solubility (shifts equilibrium backward)

Frequently asked questions

What is the difference between Kc and Kp?

Kc is the equilibrium constant written in terms of molar concentrations (mol/L). Kp is written in terms of partial pressures (atm or bar). They are related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). For reactions where Δn = 0, Kp = Kc.

How do I predict which direction a reaction will proceed?

Calculate the reaction quotient Q using the current concentrations (same formula as Kc but with current values, not equilibrium values). If Q < K, the reaction goes forward (towards products). If Q > K, the reaction goes backward (towards reactants). If Q = K, the system is already at equilibrium.

What does Le Chatelier's principle mean in simple terms?

It means a system at equilibrium resists change. If you disturb the equilibrium (by adding a reactant, increasing pressure, or changing temperature), the system shifts to partially undo that disturbance. Adding a reactant shifts equilibrium toward products. Increasing pressure shifts toward fewer moles of gas. Increasing temperature shifts toward the endothermic direction.

How do you calculate the pH of a weak acid?

For a weak acid HA with concentration C and dissociation constant Ka: HA ⇌ H⁺ + A⁻. At equilibrium, [H⁺] = [A⁻] = x. Ka = x²/(C − x). If Ka is small (weak acid), C − x ≈ C, so x = √(Ka × C). Then pH = −log(x). Always check the approximation: it is valid when x/C < 5%.

What is the Henderson-Hasselbalch equation?

pH = pKa + log([A⁻]/[HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the weak acid in the buffer. When [A⁻] = [HA], pH = pKa. Buffers work best when pH is within 1 unit of pKa.

When does precipitation occur in a Ksp problem?

A precipitate forms when the ionic product (IP) of the ions in solution exceeds Ksp. If IP < Ksp, the solution is unsaturated and no precipitate forms. If IP = Ksp, the solution is just saturated. If IP > Ksp, precipitation occurs. The common ion effect lowers solubility by increasing one of the ions already present.

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EquilibriumNEET Chemistry · Class 11 · NCERT Chapter 6

Introduction to Equilibrium

Equilibrium in Physical Processes

Liquid-Vapour Equilibrium

Solid-Liquid Equilibrium

Gas in Liquid (Henry's Law)

Chemical Equilibrium

Law of Mass Action and Kc

Important Rules for Writing Kc

Kp and the Kp-Kc Relationship

Significance of the Equilibrium Constant

Reaction Quotient Q

Le Chatelier's Principle

Effect of Changing Concentration

Effect of Pressure (for Gas-Phase Reactions)

Effect of Temperature

Effect of a Catalyst

Ionic Equilibrium

Strong Acids and Bases

Weak Acids: Ka and Degree of Ionisation

Weak Bases: Kb

Relationship Between Ka and Kb

Acids, Bases, and pH

Arrhenius Concept

Bronsted-Lowry Concept

Lewis Concept

The pH Scale

Buffer Solutions

Solubility Product Ksp

Relating Ksp to Molar Solubility

Common Ion Effect

Ionic Product and Precipitation

Worked NEET Problems

Summary Cheat Sheet

Frequently asked questions

What is the difference between Kc and Kp?

How do I predict which direction a reaction will proceed?

What does Le Chatelier's principle mean in simple terms?

How do you calculate the pH of a weak acid?

What is the Henderson-Hasselbalch equation?

When does precipitation occur in a Ksp problem?

Continue with the next chapter notes

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