Unit 2: Ionic Equilibrium

Introduction to Acids and Bases

Acids and bases are fundamental concepts in chemistry that describe substances which donate or accept protons (H⁺) or electron pairs. Over time, three major theories have been developed to explain their behaviour:

Arrhenius Concept

Arrhenius acid: a substance that dissociates in aqueous solution to produce hydrogen ions (H⁺).
Arrhenius base: a substance that dissociates in aqueous solution to produce hydroxide ions (OH⁻).

Examples: HCl → H⁺ + Cl⁻; NaOH → Na⁺ + OH⁻.

Limitation: The Arrhenius definition fails to explain the basicity of ammonia (NH₃) in water and does not apply to reactions occurring in non‑aqueous solvents.

Bronsted‑Lowry Concept

Bronsted‑Lowry acid: a proton (H⁺) donor.
Bronsted‑Lowry base: a proton (H⁺) acceptor.

This theory extends the acid‑base concept to any solvent and explains why NH₃ acts as a base: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ (NH₃ accepts a proton from water).

Lewis Concept

Lewis acid: an electron‑pair acceptor.
Lewis base: an electron‑pair donor.

Examples of Lewis acids: BF₃, AlCl₃ (they accept a lone pair).
Examples of Lewis bases: NH₃, H₂O, OH⁻ (they donate a lone pair).

The Lewis theory is the most general and includes the Bronsted‑Lowry and Arrhenius definitions as special cases.

Relative Strength of Acids and Bases

The strength of an acid or base is quantified by its tendency to ionise in solution.

Strong Acids and Bases

• Strong acids are completely ionised in dilute aqueous solution.
• Common strong acids: HCl, HBr, HI, HNO₃, H₂SO₄ (first proton), HClO₄.
• Strong bases are completely dissociated to give OH⁻ ions.
• Common strong bases: NaOH, KOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂.

Weak Acids and Bases

• Weak acids only partially ionise; an equilibrium exists between the undissociated acid and its ions.
• Examples: acetic acid (CH₃COOH), hydrocyanic acid (HCN), carbonic acid (H₂CO₃).
• Weak bases similarly partially accept protons.
• Examples: ammonia (NH₃), amines, carbonate ion (CO₃²⁻).

Acid Dissociation Constant (Ka)

For a weak acid HA:

HA ⇌ H⁺ + A⁻

The acid dissociation constant is:

Ka = [H⁺][A⁻] / [HA]

A larger Ka indicates a stronger acid. The constant is temperature‑dependent.

Acid	Ka (approx.)	pKa = –log Ka
HCl (strong)	~10⁷	–7
CH₃COOH	1.8 × 10⁻⁵	4.74
HCN	4.9 × 10⁻¹⁰	9.31
H₂CO₃ (first)	4.3 × 10⁻⁷	6.37

Conjugate Acid‑Base Pairs

According to Bronsted‑Lowry theory, when an acid donates a proton it forms its conjugate base; when a base accepts a proton it forms its conjugate acid.

• Conjugate acid: species formed after a base gains H⁺.
• Conjugate base: species formed after an acid loses H⁺.

Examples:

• CH₃COOH (acid) ⇌ CH₃COO⁻ (conjugate base) + H⁺
• NH₃ (base) + H⁺ ⇌ NH₄⁺ (conjugate acid)
• H₂O (acid) ⇌ OH⁻ (conjugate base) + H⁺
• H₂O (base) + H⁺ ⇌ H₃O⁺ (conjugate acid)

For any conjugate pair, the product of their dissociation constants equals the ionic product of water:

Ka (acid) × Kb (conjugate base) = Kw

Ostwald's Dilution Law

Ostwald’s law relates the dissociation constant of a weak electrolyte to its degree of ionisation (α) and concentration (C).

For a weak binary electrolyte AB ⇌ A⁺ + B⁻:

Ka = C α² / (1 – α)

Where:

• α = degree of ionisation = fraction of total molecules that dissociate.
• C = initial concentration of the electrolyte (mol L⁻¹).

At very low concentrations (infinite dilution), the denominator (1 – α) approaches 1 because α → 1, and the law simplifies to Ka ≈ C α². This shows that dilution increases the degree of ionisation of a weak electrolyte.

Ionic Product of Water (Kw)

Water undergoes self‑ionisation:

2 H₂O ⇌ H₃O⁺ + OH⁻ (often written as H₂O ⇌ H⁺ + OH⁻)

The equilibrium constant for this process is the ionic product of water:

Kw = [H⁺][OH⁻]

At 25 °C, Kw = 1.0 × 10⁻¹⁴ (mol² L⁻²). Consequently, in pure water:

[H⁺] = [OH⁻] = √Kw = 1.0 × 10⁻⁷ M

Taking negative logarithms gives the relationship:

pH + pOH = pKw = 14 (at 25 °C)

Dissociation Constants (Ka and Kb)

For a weak acid HA:

Ka = [H⁺][A⁻] / [HA]

For a weak base B:

B + H₂O ⇌ BH⁺ + OH⁻

Kb = [BH⁺][OH⁻] / [B]

For a conjugate acid‑base pair:

Ka × Kb = Kw

Thus, knowing Ka of an acid allows calculation of Kb of its conjugate base and vice‑versa.

pH and pOH Calculations

The pH of a solution is defined as:

pH = –log₁₀[H⁺]

Similarly, pOH = –log₁₀[OH⁻].

Strong Acid

For a strong acid HA at concentration C (assuming complete dissociation):

[H⁺] = C → pH = –log C

Weak Acid (approximation)

When α is small (α ≪ 1), Ostwald’s law gives:

Ka ≈ C α² → α ≈ √(Ka / C)

Hence:

[H⁺] = C α ≈ √(Ka × C)

pH ≈ –log √(Ka × C) = ½(pKa – log C)

Strong Base

For a strong base BOH at concentration C:

[OH⁻] = C → pOH = –log C → pH = 14 – pOH

Weak Base (approximation)

Analogous to weak acid:

[OH⁻] ≈ √(Kb × C)

pOH ≈ –log √(Kb × C) → pH = 14 – pOH

Numerical Problems

1. Problem 1: Calculate the pH of 0.01 M HCl solution.
2. Solution: HCl is a strong acid → [H⁺] = 0.01 M → pH = –log(0.01) = 2.
3. Problem 2: Find the pH of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵).
4. Solution: Using weak‑acid approximation: [H⁺] ≈ √(Ka·C) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ M. pH = –log(1.34×10⁻³) ≈ 2.87.
5. Problem 3: Determine the pH of 0.05 M NaOH solution.
6. Solution: NaOH is a strong base → [OH⁻] = 0.05 M → pOH = –log(0.05) ≈ 1.30 → pH = 14 – 1.30 = 12.70.
7. Problem 4: Calculate the pH of 0.2 M ammonia solution (Kb = 1.8 × 10⁻⁵).
8. Solution: [OH⁻] ≈ √(Kb·C) = √(1.8×10⁻⁵ × 0.2) = √(3.6×10⁻⁶) ≈ 1.90×10⁻³ M. pOH = –log(1.90×10⁻³) ≈ 2.72 → pH = 14 – 2.72 = 11.28.

Solubility Product (Ksp)

For a sparingly soluble salt AₓBᵧ that dissociates as:

AₓBᵧ (s) ⇌ x Aⁿ⁺ (aq) + y Bᵐ⁻ (aq)

The solubility product constant is:

Ksp = [Aⁿ⁺]^x [Bᵐ⁻]^y

If the molar solubility of the salt is s mol L⁻¹, then:

[Aⁿ⁺] = x·s and [Bᵐ⁻] = y·s

Substituting gives:

Ksp = (x·s)^x (y·s)^y = x^x y^y s^{x+y}

Hence, the solubility can be expressed as:

s = ( Ksp / (x^x y^y) )^{1/(x+y)}

For salts where x = y = 1 (e.g., AgCl, BaSO₄), this simplifies to:

s = √Ksp

Ion Product vs. Ksp

• Ion product (Q) = [Aⁿ⁺]^x [Bᵐ⁻]^y calculated from the actual concentrations in solution.
• If Q < Ksp → solution is unsaturated; more solid can dissolve.
• If Q = Ksp → solution is saturated; equilibrium exists.
• If Q > Ksp → solution is supersaturated; precipitation occurs until Q = Ksp.

Common Ksp Values (25 °C)

Salt	Ksp
AgCl	1.8 × 10⁻¹⁰
BaSO₄	1.1 × 10⁻¹⁰
PbI₂	7.1 × 10⁻⁹
CaF₂	3.9 × 10⁻¹¹
Mg(OH)₂	5.6 × 10⁻¹²

Numerical Problems

1. Problem 1: Calculate the solubility of AgCl in pure water.
2. Solution: AgCl ⇌ Ag⁺ + Cl⁻; Ksp = [Ag⁺][Cl⁻] = s² → s = √Ksp = √(1.8×10⁻¹⁰) ≈ 1.34×10⁻⁵ M.
3. Problem 2: Determine the solubility of PbI₂ in water.
4. Solution: PbI₂ ⇌ Pb²⁺ + 2 I⁻; Ksp = [Pb²⁺][I⁻]² = s·(2s)² = 4s³ → s = (Ksp/4)^{1/3} = (7.1×10⁻⁹ / 4)^{1/3} ≈ (1.775×10⁻⁹)^{1/3} ≈ 1.22×10⁻³ M.
5. Problem 3: Will BaSO₄ precipitate when 0.001 M BaCl₂ is mixed with 0.001 M Na₂SO₄ (equal volumes)?
6. Solution: After mixing, concentrations halve: [Ba²⁺] = 0.0005 M, [SO₄²⁻] = 0.0005 M. Q = [Ba²⁺][SO₄²⁻] = (5×10⁻⁴)² = 2.5×10⁻⁷. Since Q (2.5×10⁻⁷) > Ksp (1.1×10⁻¹⁰), precipitation will occur.

Common Ion Effect

The solubility of a sparingly soluble salt decreases when a soluble salt sharing a common ion is added to the solution. This is a direct consequence of Le Chatelier’s principle.

Example: The solubility of AgCl in pure water is s ≈ 1.34×10⁻⁵ M. Adding NaCl (which provides Cl⁻ ions) shifts the equilibrium:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Increased [Cl⁻] drives the reaction left, reducing [Ag⁺] and thus the solubility of AgCl.

Numerical Problem

1. Problem: Calculate the solubility of AgCl in 0.10 M NaCl solution.
2. Solution: Let s' be the solubility of AgCl in presence of 0.10 M Cl⁻ from NaCl. Then [Ag⁺] = s', [Cl⁻] ≈ 0.10 + s' ≈ 0.10 (since s' ≪ 0.10). Ksp = [Ag⁺][Cl⁻] = s' × 0.10 → s' = Ksp / 0.10 = (1.8×10⁻¹⁰) / 0.10 = 1.8×10⁻⁹ M. The solubility drops from 1.34×10⁻⁵ M to 1.8×10⁻⁹ M.

Buffer Solutions

A buffer solution resists changes in pH upon addition of small amounts of acid or base. It consists of a weak acid and its conjugate base (acidic buffer) or a weak base and its conjugate acid (basic buffer).

Acidic Buffer

Example: acetic acid (CH₃COOH) + sodium acetate (CH₃COONa).

Basic Buffer

Example: ammonia (NH₃) + ammonium chloride (NH₄Cl).

Henderson‑Hasselbalch Equation

For an acidic buffer:

pH = pKa + log₁₀([salt]/[acid])

For a basic buffer (using pKb):

pOH = pKb + log₁₀([salt]/[base]) → pH = 14 – pOH

The buffer capacity is greatest when [salt] ≈ [acid] (or [base]), giving pH ≈ pKa (or pOH ≈ pKb).

Numerical Problems

1. Problem 1: Calculate the pH of a buffer containing 0.20 M CH₃COOH and 0.30 M CH₃COONa (Ka = 1.8×10⁻⁵).
2. Solution: pKa = –log(1.8×10⁻⁵) = 4.74. Using Henderson‑Hasselbalch: pH = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.176 = 4.92.
3. Problem 2: A basic buffer is prepared with 0.15 M NH₃ and 0.10 M NH₄Cl (Kb = 1.8×10⁻⁵). Find its pH.
4. Solution: pKb = –log(1.8×10⁻⁵) = 4.74. pOH = pKb + log([salt]/[base]) = 4.74 + log(0.10/0.15) = 4.74 + log(0.6667) = 4.74 – 0.176 = 4.56. pH = 14 – 4.56 = 9.44.

Indicators and Selection

Indicators are weak organic acids or bases that exhibit different colours in their protonated and deprotonated forms, thus signalling the pH of a solution.

Common Indicators:

• Phenolphthalein: colourless (pH < 8.2) → pink (pH 8.2–10.0).
• Methyl orange: red (pH < 3.1) → yellow (pH > 4.4).
• Methyl red: red (pH < 4.4) → yellow (pH > 6.2).
• Bromothymol blue: yellow (pH < 6.0) → blue (pH > 7.6).
• Thymol blue: yellow (pH < 1.2) → blue (pH 8.0–9.6).

Selection Principle: Choose an indicator whose pH range brackets the equivalence point pH of the titration. For a strong acid–strong base titration, equivalence point pH ≈ 7, so indicators like bromothymol blue or phenolphthalein (if the titration is slightly basic) are suitable. For a weak acid–strong base titration, equivalence point pH > 7, thus phenolphthalein is appropriate. For a strong acid–weak base titration, equivalence point pH < 7, thus methyl orange or methyl red are used.

Types of Salts

Salts can be classified based on the strength of the parent acid and base:

Category	Parent Acid	Parent Base	Example	Expected pH of Aqueous Solution
Acidic salt	Strong acid	Weak base	NH₄Cl, NaHSO₄	pH < 7 (hydrolysis of cation)
Basic salt	Weak acid	Strong base	CH₃COONa, Na₂CO₃	pH > 7 (hydrolysis of anion)
Neutral (simple) salt	Strong acid	Strong base	NaCl, KNO₃	pH ≈ 7 (no hydrolysis)
Complex salt	—	—	K₄[Fe(CN)₆], [Cu(NH₃)₄]SO₄	Depends on ligand; often neutral or slightly acidic/basic

Hydrolysis of Salts

Hydrolysis is the reaction of a salt’s cation or anion with water, producing H⁺ or OH⁻ and thus affecting pH.

Classification and Expected pH

• Strong acid + Strong base (e.g., NaCl): Neither ion hydrolyses → solution neutral, pH = 7.
• Weak acid + Strong base (e.g., CH₃COONa): Anion (CH₃COO⁻) hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ → solution basic, pH > 7.
• Strong acid + Weak base (e.g., NH₄Cl): Cation (NH₄⁺) hydrolyses: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ → solution acidic, pH < 7.
• Weak acid + Weak base (e.g., NH₄CH₃COO): Both ions hydrolyse; pH depends on relative Ka and Kb.

pH Calculation for Hydrolysed Salts

For a salt of a weak acid and strong base (e.g., NaA):

The hydrolysis constant for the anion:

Kh = Kw / Ka

Assuming hydrolysis h ≪ 1:

[OH⁻] ≈ √(Kh × C) = √( (Kw / Ka) × C )

Then pOH = –log[OH⁻] and pH = 14 – pOH.

For a salt of a strong acid and weak base (e.g., BH⁺X⁻):

Kh = Kw / Kb

[H⁺] ≈ √(Kh × C) = √( (Kw / Kb) × C )

pH = –log[H⁺].

Numerical Problems

1. Problem 1: Find the pH of 0.10 M sodium acetate solution (Ka of acetic acid = 1.8×10⁻⁵).
2. Solution: Kh = Kw / Ka = (1.0×10⁻¹⁴) / (1.8×10⁻⁵) = 5.56×10⁻¹⁰. [OH⁻] ≈ √(Kh·C) = √(5.56×10⁻¹⁰ × 0.10) = √(5.56×10⁻¹¹) ≈ 7.45×10⁻⁶ M. pOH = –log(7.45×10⁻⁶) ≈ 5.13. pH = 14 – 5.13 = 8.87.
3. Problem 2: Calculate the pH of 0.05 M ammonium chloride solution (Kb of ammonia = 1.8×10⁻⁵).
4. Solution: Kh = Kw / Kb = (1.0×10⁻¹⁴) / (1.8×10⁻⁵) = 5.56×10⁻¹⁰. [H⁺] ≈ √(Kh·C) = √(5.56×10⁻¹⁰ × 0.05) = √(2.78×10⁻¹¹) ≈ 5.27×10⁻⁶ M. pH = –log(5.27×10⁻⁶) ≈ 5.28.
5. Problem 3: Determine the pH of 0.02 M ammonium acetate (NH₄CH₃COO) solution (Ka of acetic acid = 1.8×10⁻⁵, Kb of ammonia = 1.8×10⁻⁵).
6. Solution: Since Ka = Kb, the solution is approximately neutral. More precisely, pH = 7 + ½(pKa – pKb) = 7 + ½(4.74 – 4.74) = 7.00.

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Summary: This chapter has provided a comprehensive overview of ionic equilibrium, covering acid‑base theories, quantitative measures of strength, conjugate relationships, dilution effects, water auto‑ionisation, dissociation constants, pH calculations, solubility principles, common ion effect, buffer action, indicator choice, salt classification, and hydrolysis phenomena. Each concept is supported by definitions, formulas, illustrative examples, and step‑by‑step numerical problems to reinforce understanding and problem‑solving skills.