Equilibrium

COMPREHENSIVE CHAPTER NOTES: EQUILIBRIUM (Unit 6)

INTRODUCTION TO EQUILIBRIUM

**Equilibrium** is a dynamic state where forward and reverse reactions (or processes) occur simultaneously at equal rates, resulting in constant concentrations of reactants and products over time. This is NOT a static state—continuous molecular activity occurs even though macroscopic properties remain unchanged.

**Key Concept**: Equilibrium is possible only in **closed systems** at constant temperature. In open systems, equilibrium cannot be established because products/reactants escape or are continuously added.

**Two Types of Equilibrium**:

**Physical Equilibrium**: Between two physical states (e.g., ice ⇌ water)

**Chemical Equilibrium**: Between reactants and products in chemical reactions

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6.1 EQUILIBRIUM IN PHYSICAL PROCESSES

Physical equilibrium involves phase transformations where opposing processes occur at equal rates.

6.1.1 SOLID-LIQUID EQUILIBRIUM (Melting/Freezing)

**Process**: H₂O(s) ⇌ H₂O(l) at 273 K and 1 atm

**Example**: Ice and water in an insulated thermos flask at 0°C (273 K)

**Observations**:

Mass of ice and water remains constant

Temperature remains constant (no heat exchange with surroundings)

Equilibrium is **dynamic**: molecules from liquid strike ice surface and adhere; ice molecules escape into liquid

Rate of freezing = Rate of melting (no net change)

**Key Characteristics**:

Both opposing processes occur simultaneously

Opposing processes occur at equal rates

At atmospheric pressure (1.013 bar), equilibrium occurs at **normal melting point** (0°C for water)

Only at this specific temperature can both phases coexist in equilibrium

**Exam Tip**: The equilibrium is dynamic because continuous molecular exchange happens; it's NOT that no change occurs—rather, changes in both directions balance each other.

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6.1.2 LIQUID-VAPOUR EQUILIBRIUM (Evaporation/Condensation)

**Process**: H₂O(l) ⇌ H₂O(vap) at constant temperature

**Experimental Setup**:

Closed transparent box with manometer (U-tube with mercury)

Drying agent (CaCl₂ or P₂O₅) initially removes water vapour

Watch glass with liquid water placed inside; observe mercury level rise

**Observations**:

Initially: evaporation is rapid; rate is constant

Mercury level (pressure) rises over time

After some time: pressure becomes constant (equilibrium attained)

Rate of evaporation = Rate of condensation at equilibrium

**Equilibrium Vapour Pressure**: The constant pressure exerted by vapour molecules at a given temperature when liquid and vapour are in equilibrium.

**Important Points**:

Vapour pressure **increases with temperature** (more kinetic energy → more evaporation)

Different liquids have different vapour pressures at same temperature

Example at 25°C: Acetone > Ethanol > Water (in terms of vapour pressure)

Liquids with **higher vapour pressure** are **more volatile** and have **lower boiling point**

In **open systems**, equilibrium cannot be reached because molecules disperse; evaporation continues until liquid disappears

**Normal Boiling Point**: At atmospheric pressure (1.013 bar), the temperature at which liquid vapour pressure equals atmospheric pressure. Water boils at 100°C at sea level (1.013 bar), but at lower pressure (high altitude), boiling point decreases.

**Henry's Law** (for gases dissolved in liquids):

The mass of gas dissolved in a liquid at constant temperature is **proportional to the pressure of gas above the liquid**.

**Representation**:

CO₂(g) ⇌ CO₂(aq)

At higher pressure (soda bottle sealed): more CO₂ dissolves

At lower pressure (bottle opened): CO₂ escapes; solution becomes "flat"

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6.1.3 SOLID-VAPOUR EQUILIBRIUM (Sublimation/Deposition)

**Process**: I₂(s) ⇌ I₂(vap)

**Observations**:

Solid iodine in closed vessel produces violet vapour

Intensity of colour increases initially, then becomes constant

At equilibrium: rate of sublimation = rate of condensation (deposition)

**Other Examples**:

Camphor(s) ⇌ Camphor(vap)

NH₄Cl(s) ⇌ NH₄Cl(vap)

**Key Point**: These substances sublime because vapour pressure exceeds atmospheric pressure below their melting point.

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6.1.4 EQUILIBRIUM INVOLVING DISSOLUTION

#### Solids Dissolving in Liquids

**Saturated Solution**: A solution in which no more solute can dissolve at a given temperature; a dynamic equilibrium exists between solid and dissolved solute.

**Process**: Solute(s) ⇌ Solute(aq)

**Example**: Sugar(s) ⇌ Sugar(aq)

**Observations**:

Rate of dissolution = Rate of crystallisation

Concentration of solute in solution is constant at given temperature

**Radioactive tracer experiment** (proof of dynamic equilibrium):

Add radioactive sugar to saturated solution of non-radioactive sugar

Over time: radioactivity appears in both solid and solution

Shows continuous exchange between phases despite constant concentration

Ratio of radioactive to non-radioactive molecules becomes constant

**Key Point**: Solubility (concentration of saturated solution) increases with temperature for most solids.

#### Gases Dissolving in Liquids

**Process**: Gas(g) ⇌ Gas(aq)

**Example**: CO₂(g) ⇌ CO₂(aq) in soda water

**Henry's Law**: At constant temperature, the **concentration (or mass) of gas dissolved in liquid is proportional to the pressure of gas above the liquid**.

**Mathematical Form**:

C(gas) ∝ P(gas)

Or: P(gas) = K_H × C(gas), where K_H is Henry's law constant

**Real-World Examples**:

**Soda water bottle** sealed under high CO₂ pressure; dissolved CO₂ concentration is high

When bottle is opened, pressure decreases to atmospheric CO₂ partial pressure; excess CO₂ escapes to reach new equilibrium

Solution becomes "flat" if left open because equilibrium shifts to gas phase

**Oxygen in blood**: Hemoglobin picks up O₂ in lungs (high pO₂); releases O₂ in muscles (lower pO₂) through equilibrium:

Hb + O₂ ⇌ HbO₂

**Carbon monoxide toxicity**: CO binds hemoglobin more strongly than O₂, preventing O₂ transport by competing for same equilibrium sites

**Temperature Effect**: Gas solubility **decreases with increase in temperature** (reverse of solids).

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6.1.5 GENERAL CHARACTERISTICS OF EQUILIBRIA IN PHYSICAL PROCESSES

All physical equilibria share these characteristics:

**(i) Closed System Requirement**: Equilibrium is possible only in closed systems at a given temperature. In open systems, one phase escapes or is added continuously.

**(ii) Dynamic Nature**:

Both opposing processes occur simultaneously

Both occur at equal rates

No net change in amount of phases/dissolved substance

**(iii) Constant Measurable Properties**:

All macroscopic properties (temperature, pressure, concentration, colour, etc.) remain constant

This constancy indicates equilibrium is attained

**(iv) Characteristic Parameter at Given Temperature**:

**Liquid ⇌ Vapour**: Vapour pressure is constant

**Solid ⇌ Liquid**: Temperature (melting point) is fixed at given pressure

**Solid ⇌ Vapour**: Sublimation pressure is constant

**Solute(s) ⇌ Solute(aq)**: Solute concentration (solubility) is constant

**Gas(g) ⇌ Gas(aq)**: Ratio [gas(aq)]/P(gas) is constant (Henry's law)

**(v) Extent Indicator**: The magnitude of characteristic parameter indicates how far the physical process has proceeded before equilibrium.

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6.2 EQUILIBRIUM IN CHEMICAL PROCESSES – DYNAMIC EQUILIBRIUM

**Chemical Equilibrium**: State attained when rates of forward and reverse reactions become equal, resulting in constant concentrations of all reactants and products.

GENERAL REVERSIBLE REACTION

**Representation**: A + B ⇌ C + D

**Graphical Changes** (Fig. 6.2):

**Initially**: High concentration of A and B; low/zero concentration of C and D

**Forward reaction rate**: High initially (high [A] and [B]); decreases as reactants consumed

**Reverse reaction rate**: Zero initially; increases as C and D accumulate

**At equilibrium**: Forward rate = Reverse rate; all concentrations become constant

**Key Insight**: Equilibrium can be attained from **either direction**—starting from pure reactants (A + B) or pure products (C + D) leads to same equilibrium composition if total atoms of each element are identical.

PROOF OF DYNAMIC NATURE: HABER PROCESS (Ammonia Synthesis)

**Reaction**: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

**Haber's Experiments**:

1. Started with known amounts of N₂ and H₂ at high temperature and pressure

2. At regular intervals, measured amounts of N₂, H₂, and NH₃

3. After certain time: composition became constant (equilibrium reached), yet some reactants remained unconsumed

**Proof of Dynamic Nature**:

Repeated experiment using D₂ (deuterium) instead of H₂, maintaining same P, T conditions

Both H₂-mixture and D₂-mixture reached equilibrium with same composition (except H₂ replaced by D₂, NH₃ replaced by ND₃)

When equilibrium H₂-mixture and equilibrium D₂-mixture were mixed and analysed:

Total ammonia concentration remained same

**Mass spectrometry revealed all forms present**: NH₃, NH₂D, NHD₂, and ND₃

This shows continuous molecular exchange: N₂ + 3H₂ ⇌ 2NH₃ and N₂ + 3D₂ ⇌ 2ND₃ occurring simultaneously, with H and D being exchanged continuously

**Conclusion**: Even at constant concentrations, forward and reverse reactions continue—this is dynamic equilibrium

HYDROGEN-IODINE REACTION

**Reaction**: H₂(g) + I₂(g) ⇌ 2HI(g)

**Starting from Reactants**:

Initial: [H₂] = [I₂] = high; [HI] = 0

As reaction proceeds: [H₂] and [I₂] decrease; [HI] increases

Eventually: concentrations become constant → equilibrium attained

**Starting from Product**:

Initial: [HI] = high; [H₂] = [I₂] = 0

Reverse reaction dominates: [HI] decreases; [H₂] and [I₂] increase

Eventually: same equilibrium concentrations attained

**Key Finding**: Regardless of starting composition, if **total number of H and I atoms is identical**, the **same equilibrium mixture composition** is obtained. This demonstrates that equilibrium is independent of the **direction of approach**.

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6.3 LAW OF EQUILIBRIUM (EQUILIBRIUM CONSTANT)

STATEMENT

**For a reversible reaction at constant temperature**:

**aA + bB ⇌ cC + dD**

**The ratio of product of equilibrium concentrations of products to the product of equilibrium concentrations of reactants, each raised to the power of its stoichiometric coefficient, is a constant called the equilibrium constant (K_c).**

EQUILIBRIUM CONSTANT EXPRESSION (K_c)

**Definition**:

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

Where:

[A], [B], [C], [D] are equilibrium molar concentrations (mol/L)

a, b, c, d are stoichiometric coefficients

**K_c is independent of initial concentrations** and volume (for reactions with Δn = 0)

K_c **depends only on temperature**

K_c is **dimensionless** if Δn = 0 (change in number of moles of gas)

PHYSICAL SIGNIFICANCE OF K_c

**(i) Magnitude of K_c**:

**K_c >> 1** (much larger than 1): Reaction proceeds nearly to completion; products predominate; equilibrium lies to the right

**K_c << 1** (much smaller than 1): Reaction produces very small amount of products; equilibrium lies to the left; reactants predominate

**K_c ≈ 1**: Concentrations of reactants and products are comparable; equilibrium is in middle

**(ii) Independent of Path**: Same K_c is obtained whether reaction starts from reactants or products at same temperature.

**(iii) Not Independent of Temperature**: K_c changes with temperature:

For **endothermic reactions** (ΔH > 0): K_c increases with increasing temperature

For **exothermic reactions** (ΔH < 0): K_c decreases with increasing temperature

CALCULATION OF EQUILIBRIUM CONSTANT

**Example Problem 1**:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

At equilibrium, [N₂] = 0.5 M, [H₂] = 1.5 M, [NH₃] = 2.0 M

**Solution**:

$$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(2.0)^2}{(0.5)(1.5)^3} = \frac{4.0}{0.5 × 3.375} = \frac{4.0}{1.6875} = 2.37 \text{ M}^{-2}$$

Note: Exponent of [H₂] is 3 (stoichiometric coefficient); denominator has Δn = (1 + 3) – 2 = 2, so dimension is M^{-2}.

**Example Problem 2**:

H₂(g) + I₂(g) ⇌ 2HI(g)

Initial: [H₂]₀ = 1.0 M, [I₂]₀ = 1.0 M, [HI]₀ = 0

At equilibrium: [HI] = 1.6 M

Find K_c.

**Solution**:

Using ICE (Initial-Change-Equilibrium) table:

| | H₂ | I₂ | HI |

|---|---|---|---|

| Initial | 1.0 | 1.0 | 0 |

| Change | –x | –x | +2x |

| Equilibrium | 1.0–x | 1.0–x | 2x |

Given [HI] at equilibrium = 1.6 M, so 2x = 1.6, thus x = 0.8

[H₂] = 1.0 – 0.8 = 0.2 M

[I₂] = 1.0 – 0.8 = 0.2 M

$$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(1.6)^2}{(0.2)(0.2)} = \frac{2.56}{0.04} = 64 \text{ (dimensionless)}$$

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6.4 RELATIONSHIP BETWEEN K_p AND K_c

For gaseous reactions, equilibrium constant can be expressed in terms of **partial pressures** (K_p) or **molar concentrations** (K_c).

PARTIAL PRESSURE

For gas: **P_i = n_i(RT/V)** or **P_i = C_i(RT)**

Where:

P_i = partial pressure of gas i

C_i = molar concentration of gas i (mol/L)

R = 0.0821 L·bar·K⁻¹·mol⁻¹ or 8.314 J·K⁻¹·mol⁻¹

T = absolute temperature (K)

RELATIONSHIP: K_p = K_c(RT)^{Δn}

**For reaction**: aA(g) + bB(g) ⇌ cC(g) + dD(g)

$$K_p = \frac{P_C^c × P_D^d}{P_A^a × P_B^b}$$

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

**Derivation**:

P_C = [C]RT, P_D = [D]RT, etc.

Substituting in K_p expression:

$$K_p = \frac{([C]RT)^c([D]RT)^d}{([A]RT)^a([B]RT)^b} = \frac{[C]^c[D]^d}{[A]^a[B]^b} × \frac{(RT)^{c+d}}{(RT)^{a+b}}$$

$$K_p = K_c(RT)^{(c+d)-(a+b)} = K_c(RT)^{Δn}$$

Where **Δn = (sum of molar coefficients of gaseous products) – (sum of molar coefficients of gaseous reactants)**

KEY POINTS

**If Δn = 0**: K_p = K_c (no change in moles of gas)

Example: H₂(g) + I₂(g) ⇌ 2HI(g); Δn = 2 – 2 = 0; so K_p = K_c

**If Δn > 0**: K_p > K_c (pressure favours forward reaction)

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); Δn = 2 – (1+3) = –2; so K_p = K_c(RT)^{-2}; thus K_p < K_c

**If Δn < 0**: K_p < K_c (pressure favours reverse reaction)

Example: 2HCl(g) ⇌ H₂(g) + Cl₂(g); Δn = 2 – 2 = 0; K_p = K_c

UNITS OF K_p

K_p uses **bar** or **atm** for partial pressures

**Dimensionless** if Δn = 0

**Unit**: (bar)^{Δn} or (atm)^{Δn} if Δn ≠ 0

CALCULATION EXAMPLE

**Example**: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 400 K

Given: K_c = 0.005 M⁻²

Find K_p (use R = 0.0821 L·bar·K⁻¹·mol⁻¹)

**Solution**:

Δn = 2 – (1 + 3) = –2

$$K_p = K_c(RT)^{Δn} = 0.005 × (0.0821 × 400)^{-2}$$

$$= 0.005 × (32.84)^{-2} = 0.005 × 0.000927 = 4.64 × 10^{-6} \text{ bar}^{-2}$$

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6.5 LE CHATELIER'S PRINCIPLE AND FACTORS AFFECTING EQUILIBRIUM

**Le Chatelier's Principle**: When a system at equilibrium is subjected to a change in any of the factors affecting equilibrium (concentration, pressure, temperature, etc.), the system shifts in a direction that tends to counteract the effect of the change.

EFFECT OF CONCENTRATION CHANGES

**Case 1: Increasing concentration of reactant**

For reaction: A + B ⇌ C + D

**Change**: Increase [A]

**Direction of Shift**: Forward (right) to consume excess A

**Effect**: [C] and [D] increase; [B] decreases

**Result**: K_c remains unchanged; system reaches new equilibrium with higher [C] and [D]

**Example**: Haber Process N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Increase pressure (increases concentrations of all gases): equilibrium shifts right (favours fewer moles = products)

**Case 2: Increasing concentration of product**

**Change**: Increase [C]

**Direction of Shift**: Reverse (left) to consume excess C

**Effect**: [A] and [B] increase; [D] decreases

**Result**: K_c remains constant; new equilibrium established

**Industrial Application**: In ammonia synthesis, unreacted N₂ and H₂ are recycled; as products (NH₃) are continuously removed (liquefied), equilibrium shifts right, increasing overall yield.

EFFECT OF PRESSURE AND VOLUME CHANGES

**General Rule**: Pressure increase favours the direction with **fewer moles of gas** (Δn < 0).

**Case 1: Increasing pressure (decreasing volume)**

For: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); Δn = 2 – 4 = –2

**Change**: Increase pressure

**Direction of Shift**: Forward (right) — shifts towards fewer moles (4 → 2)

**Effect**: [NH₃] increases; [N₂] and [H₂] decrease

**Result**: Increases ammonia yield

**Mathematical Basis**:

Increased pressure → increased concentration of all species

Forward reaction: consumes 4 moles → produces 2 moles (net reduction beneficial)

System shifts right to relieve pressure by reducing total moles

**Case 2: For H₂(g) + I₂(g) ⇌ 2HI(g); Δn = 0**

**Change**: Increase pressure

**Direction of Shift**: No shift (equal moles on both sides)

**Effect**: All concentrations increase equally; K_c unchanged

**Result**: Equilibrium position unaffected

**Case 3: Decreasing pressure (increasing volume)**

For: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

**Change**: Decrease pressure

**Direction of Shift**: Reverse (left) — towards more moles (2 → 4)

**Effect**: [NH₃] decreases; [N₂] and [H₂] increase

**Industrial Importance**:

High pressure (200-300 atm) used in ammonia synthesis to shift equilibrium right

Lower pressures favour dissociation of compounds (e.g., PCl₅ ⇌ PCl₃ + Cl₂)

EFFECT OF TEMPERATURE CHANGES

**Critical Point**: Temperature is the ONLY factor that changes K_c value.

**For Endothermic Reactions** (ΔH > 0):

**Change**: Increase temperature

**Direction of Shift**: Forward (right) — absorbs added heat

**Effect**: All equilibrium concentrations change; K_c increases

**Example**: N₂O₄ ⇌ 2NO₂ (ΔH = +58 kJ/mol)

Increase T: shifts right; brown colour (NO₂) intensifies; K_c increases

**For Exothermic Reactions** (ΔH < 0):

**Change**: Increase temperature

**Direction of Shift**: Reverse (left) — minimizes increase in heat

**Effect**: K_c decreases; equilibrium shifts left

**Example**: 2NO₂ ⇌ N₂O₄ (ΔH = –58 kJ/mol)

Increase T: shifts left; brown colour decreases; K_c decreases

Decrease T: shifts right; colourless N₂O₄ increases; K_c increases

**Mathematical Relationship (van't Hoff Equation)**:

$$\ln\left(\frac{K_2}{K_1}\right) = \frac{-ΔH°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

Where K₁, K₂ are equilibrium constants at T₁, T₂ respectively.

**Haber Process Optimization**:

Reaction N₂ + 3H₂ ⇌ 2NH₃ is highly exothermic (ΔH = –92.4 kJ/mol)

High K_c desired → low temperature favoured

BUT: Low temperature → very slow reaction rate

**Industrial compromise**: 400-500°C temperature (higher than optimal K_c, but reasonable reaction rate); 200-300 atm pressure (drives equilibrium right)

EFFECT OF INERT GAS (UNREACTIVE GAS)

**At Constant Volume**:

Adding inert gas at constant volume does NOT change partial pressures of reactants/products

No shift in equilibrium position

**At Constant Pressure**:

Adding inert gas at constant pressure increases total volume

Partial pressures of all reactants/products decrease

**Effect**: Same as decreasing pressure

Example: For N₂ + 3H₂ ⇌ 2NH₃, equilibrium shifts left

EFFECT OF CATALYST

**Important**: Catalyst does NOT affect equilibrium position or K_c.

**Effect**:

**Lowers activation energy** for both forward and reverse reactions equally

**Increases reaction rate** in both directions equally

**Reaches equilibrium faster** but at same composition

**No change in K_c or equilibrium concentrations**

**Haber Process**: Iron catalyst used to speed up reaction; does NOT increase ammonia yield, only reduces time to reach equilibrium.

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SUMMARY TABLE: EFFECT OF FACTORS ON EQUILIBRIUM

| Factor | Change | Effect on Equilibrium | Effect on K_c | Shifts |

|---|---|---|---|---|

| **Concentration (Reactant)** | Increase | Increases concentration of products | No change | Right |

| **Concentration (Product)** | Increase | Increases concentration of reactants | No change | Left |

| **Pressure (Δn < 0)** | Increase | Shifts to fewer moles | No change | Right |

| **Pressure (Δn > 0)** | Increase | Shifts to fewer moles | No change | Left |

| **Pressure (Δn = 0)** | Increase | No shift in position | No change | None |

| **Temperature (Endothermic)** | Increase | Shifts right; K_c increases | **Increases** | Right |

| **Temperature (Exothermic)** | Increase | Shifts left; K_c decreases | **Decreases** | Left |

| **Catalyst** | Added | No effect on equilibrium | No change | None |

| **Inert gas (constant V)** | Added | No effect | No change | None |

| **Inert gas (constant P)** | Added | Acts like pressure decrease | No change | Depends on Δn |

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IONIZATION CONCEPTS AND ACIDS-BASES

ARRHENIUS CONCEPT (LIMITED)

**Acid**: Substance that increases [H⁺] in aqueous solution

Example: HCl → H⁺ + Cl⁻

**Base**: Substance that increases [OH⁻] in aqueous solution

Example: NaOH → Na⁺ + OH⁻

**Limitation**: Applies only to aqueous solutions; doesn't explain basic nature of ammonia or role of solvents.

BRØNSTED-LOWRY CONCEPT (MODERN, PREFERRED)

**Acid**: Proton (H⁺) donor

**Base**: Proton (H⁺) acceptor

**Advantage**: Applies to any solvent, not just water.

**Examples**:

**HCl (acid)**: HCl → H⁺ + Cl⁻ (donates proton)

**NH₃ (base)**: NH₃ + H⁺ → NH₄⁺ (accepts proton)

**H₂O (amphoteric)**: Can act as acid or base

As acid: H₂O → H⁺ + OH⁻

As base: H₂O + H⁺ → H₃O⁺

**Conjugate Acid-Base Pairs**:

When acid donates proton: A – H → A⁻ + H⁺; A⁻ is conjugate base of A–H

When base accepts proton: B + H⁺ → BH⁺; BH⁺ is conjugate acid of B

**Example**: HCl ⇌ H⁺ + Cl⁻

HCl is acid; Cl⁻ is its conjugate base

H⁺ is acid; base is whatever accepts it

**Reaction**: HCl + NH₃ ⇌ NH₄⁺ + Cl⁻

HCl (acid, donates H⁺) + NH₃ (base, accepts H⁺) → products

NH₄⁺ (conjugate acid of NH₃) + Cl⁻ (conjugate base of HCl)

LEWIS CONCEPT (MOST GENERAL)

**Acid**: Electron pair acceptor (Lewis acid)

**Base**: Electron pair donor (Lewis base)

**Advantage**: Explains reactions not involving proton transfer; covers all solvents.

**Examples**:

**BF₃ + NH₃ → BF₃·NH₃**

BF₃ is acid (accepts electron pair from N)

NH₃ is base (donates lone pair)

No proton transfer, yet acid-base reaction occurs

**Al³⁺ + 6H₂O ⇌ [Al(H₂O)₆]³⁺**

Al³⁺ is acid (accepts electron pairs from O)

H₂O is base (donates lone pairs)

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IONIZATION OF ELECTROLYTES IN AQUEOUS SOLUTION

STRONG ELECTROLYTES

**Definition**: Electrolytes that ionize **completely** in aqueous solution.

**Examples**:

**Strong Acids**: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄

**Strong Bases**: Group 1 hydroxides (NaOH, KOH), Group 2 hydroxides [Ca(OH)₂, Ba(OH)₂]

**Salts**: Most soluble salts (NaCl, KNO₃, etc.)

**Ionization**:

HCl → H⁺(aq) + Cl⁻(aq) [complete, 100%]

NaOH → Na⁺(aq) + OH⁻(aq) [complete, 100%]

**Equilibrium Constant Expression**: Not meaningful; equilibrium lies almost entirely to the right.

WEAK ELECTROLYTES

**Definition**: Electrolytes that ionize **partially** in aqueous solution; significant undissociated molecules remain.

**Examples**:

**Weak Acids**: CH₃COOH, HF, HNO₂, H₂CO₃, H₂S

**Weak Bases**: NH₃, hydrazine (N₂H₄), methylamine (CH₃NH₂)

**Ionization Process** (Reversible Equilibrium):

**Weak Acid**: HA ⇌ H⁺ + A⁻

Example: CH₃COOH ⇌ H⁺ + CH₃COO⁻

Only small fraction ionizes; mostly exists as molecules

**Weak Base**: B + H₂O ⇌ BH⁺ + OH⁻

Example: NH₃ + H