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Class 11 Chemistry Notes

Equilibrium Class 11 Notes

Complete, exam-ready notes on equilibrium: the law of mass action, equilibrium constants Kc and Kp, Le Chatelier's principle, ionic equilibrium, pH and pOH, buffer solutions and the solubility product — written for CBSE, JEE and NEET revision.

Class11SubjectChemistryCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is equilibrium in chemistry in one line?

A reversible reaction is at equilibrium when the rates of the forward and reverse reactions are equal, so the concentrations of reactants and products stop changing.

Reversible Reactions and Equilibrium State

Dynamic equilibrium

In a reversible reaction both the forward and reverse reactions continue at equal rates, so the concentrations of all species become constant. The equilibrium is dynamic — reactions keep occurring, they just balance out.

Law of mass action

aA+bB⇌cC+dD,Kc=[C]c[D]d[A]a[B]baA + bB \rightleftharpoons cC + dD,\qquad K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

At a given temperature, the product of the concentrations of the products (raised to their stoichiometric coefficients) divided by the product of the concentrations of the reactants gives a constant, Kc. Only aqueous and gaseous species appear in the expression; pure solids and liquids are omitted.

Equilibrium Constants Kc and Kp

Kc and Kp

Kp=Kc (RT)ΔnK_p = K_c\,(RT)^{\Delta n}

Kc uses molar concentrations; Kp uses partial pressures of gases. They are related by Δn, the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants).

  • K >> 1: the reaction goes nearly to completion (products dominate).
  • K << 1: the reaction hardly proceeds (reactants dominate).
  • K ~ 1: both reactants and products are present in comparable amounts.
  • For a reaction written in reverse, K' = 1/K; doubling the equation squares K.

Le Chatelier's Principle

Le Chatelier's principle

If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that partly opposes the change.

  • Adding a reactant or removing a product shifts the equilibrium forward.
  • Increasing pressure favours the side with fewer moles of gas; decreasing pressure favours more gas moles.
  • For an exothermic reaction, heating shifts equilibrium backward; cooling shifts it forward (and the reverse for endothermic).
  • Adding an inert gas at constant volume does not change the equilibrium.

Ionic Equilibrium — Acids and Bases

Ionic product of water

Kw=[H+][OH−]=1×10−14  (25∘C)K_w = [\text{H}^+][\text{OH}^-] = 1\times10^{-14}\; (25^\circ\text{C})

Water ionises slightly: H₂O ⇌ H⁺ + OH⁻. The ionic product Kw is always 1×10⁻¹⁴ at 25 °C whether the solution is acidic, basic or neutral. Aqueous solutions therefore obey [H⁺][OH⁻] = Kw.

pH=−log⁡[H+],pOH=−log⁡[OH−],pH+pOH=14\text{pH} = -\log[\text{H}^+],\qquad \text{pOH} = -\log[\text{OH}^-],\qquad \text{pH} + \text{pOH} = 14
pH and pOH

Ionisation Constants — Ka, Kb and Degree of Dissociation

Weak acid and base constants

Ka=[H+][A−][HA],Ka×Kb=KwK_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]},\qquad K_a \times K_b = K_w

For a weak acid HA ⇌ H⁺ + A⁻, the acid dissociation constant Ka measures acid strength. For a conjugate acid–base pair, Ka × Kb = Kw, so a strong acid has a weak conjugate base and vice versa.

degree of dissociation α≈KaC,[H+]=Cα\text{degree of dissociation } \alpha \approx \sqrt{\frac{K_a}{C}},\qquad [\text{H}^+] = C\alpha
Ostwald's dilution law (weak acid, degree α)

Buffer Solutions and Salt Hydrolysis

Buffer solution

pH=pKa+log⁡[A−][HA]\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}

A buffer resists pH change on adding small amounts of acid or base. An acidic buffer is a weak acid + its salt (Henderson–Hasselbalch equation above); a basic buffer is a weak base + its salt, with pH = 14 − pKb − log[base+]/[base].

  • Acidic buffer example: CH₃COOH + CH₃COONa.
  • Basic buffer example: NH₄OH + NH₄Cl.
  • Blood is a natural buffer (HCO₃⁻/H₂CO₃ system) keeping pH near 7.4.

Solubility Product

Solubility product

AxBy(s)⇌xAy++yBx−,Ksp=[Ay+]x[Bx−]yA_xB_y(s) \rightleftharpoons xA^{y+} + yB^{x-},\qquad K_{sp} = [A^{y+}]^x[B^{x-}]^y

For a sparingly soluble salt, the product of the ion concentrations at saturation is constant, Ksp. If the ionic product exceeds Ksp, precipitation occurs; if it is below, the salt dissolves.

  • For AgCl, Ksp = [Ag⁺][Cl⁻] = s² where s is the molar solubility.
  • For PbCl₂, Ksp = [Pb²⁺][Cl⁻]² = 4s³.
  • Adding a common ion suppresses solubility (common-ion effect), useful in qualitative analysis and precipitation.

Solved Examples

Example: The pH of a solution is 3. What is its [H⁺] and is the solution acidic or basic?

Solution: [H⁺] = 10⁻³ mol L⁻¹. With pH < 7 the solution is acidic, and [OH⁻] = Kw/[H⁺] = 1×10⁻¹¹ mol L⁻¹.

Example: For the reaction N₂ + 3H₂ ⇌ 2NH₃, state the direction of shift if pressure is increased.

Solution: The forward step converts 4 moles of gas (1 N₂ + 3 H₂) into 2 moles of NH₃. Increasing pressure favours the side with fewer gas moles, so the equilibrium shifts forward, producing more ammonia.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Equilibrium constant

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

Kp from Kc

Kp=Kc (RT)ΔnK_p = K_c\,(RT)^{\Delta n}

Ionic product of water

Kw=[H+][OH−]=10−14K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}

pH and pOH

pH=−log⁡[H+],  pH+pOH=14\text{pH} = -\log[\text{H}^+],\; \text{pH}+\text{pOH}=14

Acid–base relation

Ka×Kb=KwK_a \times K_b = K_w

Buffer (Henderson–Hasselbalch)

pH=pKa+log⁡[A−][HA]\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}

Solubility product

Ksp=[Ay+]x[Bx−]yK_{sp} = [A^{y+}]^x[B^{x-}]^y

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Omit pure solids and liquids from Kc and Kp expressions — only gases and aqueous species count.
  • Kp = Kc(RT)^Δn; Δn is the change in moles of gas.
  • Increase pressure → shift toward fewer gas moles; inert gas at constant volume → no shift.
  • pH = −log[H⁺]; pH + pOH = 14 at 25 °C; neutral pH = 7.
  • Ka × Kb = Kw: a strong acid pairs with a weak conjugate base.
  • Buffer resists pH change — weak acid + its salt, or weak base + its salt.
  • Precipitation occurs when the ionic product exceeds Ksp (common-ion effect suppresses solubility).

FAQ

Common questions

What is the difference between Kc and Kp?

Kc uses molar concentrations while Kp uses partial pressures of gases for the equilibrium expression. They are related by Kp = Kc(RT)^Δn, where Δn is the change in the number of moles of gaseous species.

What does Le Chatelier's principle state?

If a system at equilibrium is disturbed, it shifts in the direction that opposes the change — by adjusting concentration, pressure or temperature to partly undo the disturbance.

How do you find pH from hydrogen ion concentration?

pH = −log[H⁺]. For example, [H⁺] = 10⁻³ mol L⁻¹ gives pH 3 (acidic). At 25 °C, pH + pOH = 14 and a neutral solution has pH 7.

What is a buffer solution?

A buffer resists pH change on adding small amounts of acid or base. It contains a weak acid with its salt (e.g. CH₃COOH/CH₃COONa) or a weak base with its salt (e.g. NH₄OH/NH₄Cl).

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