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Overview

🌳 Concept tree

Lesson 2 Ionic Equilibrium, pH Scale, Hydrolysis and Solubility Product (Ksp)
├── Strong and weak electrolytes
├── Hydronium ion
├── Definitions
├── Ostwald's dilution law
├── Hydronium ion concentration of a weak acid
├── Hydroxide ion concentration of a weak base
├── Ionization of water and the ionic product Kw
├── pH and pOH
├── Solubility and solubility product
└── Summary of key relations

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Strong and weak electrolytes

Comparison aspect	Main idea	Keywords
Ionic compounds	(e.g. NaCl) are completely ionized in solution and their solutions conduct electricity their positive and negative ions are held together by electrostatic attraction	NaClcompletelyionizedsolution
Polar covalent compounds	(HCl gas pure acetic acid) ionize when dissolved in water. HCl is ionized about 100 % whereas the ionization of acetic acid is very limited. This can be shown by experiment	pureaceticacidionize
•	Experiments	Experiments

Test the conductivity of pure (glacial) acetic acid and of HCl gas dissolved in benzene: the lamp does not light in either case — no ions.
Dissolve 0.1 mol of HCl gas in 1 L of water and 0.1 mol of pure acetic acid in 1 L of water, two solutions of equal concentration: the lamp lights strongly for HCl and faintly for acetic acid. HCl gives many more ions.
Dilute the solutions (0.01 → 0.001 mol/L): the lighting of the lamp does not change for hydrochloric acid (already completely ionized) but it increases on dilution for acetic acid (ionization increases).

Conclusion.→Dry hydrogen chloride and pure acetic acid ionize in water. HCl ionizes completely (strong electrolyte): HCl → H⁺ + Cl⁻. Acetic acid ionizes to a very limited extent (weak electrolyte), and its ionization increases with dilution, which shows that unionized molecules are present: CH₃COOH ⇌ CH₃COO⁻ + H⁺.

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Hydronium ion

There is no free proton in aqueous solution it is attracted to the lone pair of electrons on the oxygen of a water molecule and joins it by a coordinate bond, forming the hydronium (hydroxonium) ion H₃O⁺ (a hydrated proton): HCl + H₂O → H₃O⁺ + Cl⁻.
Hydronium ions are also produced when weak acids ionise.
In a weak acid only a small part of the molecules break up in water.
For example, when acetic acid CH3COOH dissolves, a molecule passes a proton to a water molecule, giving an acetate ion CH3COO– and a hydronium ion H3O+.
The reverse reaction also takes place: an acetate ion and a hydronium ion combine again to form acetic acid molecules and water.
Both forms therefore exist together in the solution.

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Definitions

Comparison aspect	Main idea	Keywords
Ionization	conversion of unionized molecules to ions	conversionunionizedmoleculesions
Complete ionization	all molecules become ions (strong electrolytes)	moleculesbecomeionsstrong
Incomplete (weak) ionization	only a small part forms ions two opposite processes always occur in the solution dissociation of molecules into ions and recombination of ions into molecules so a dynamic equilibrium exists between the ions and the unionized molecules: AB ⇌ A⁺ + B⁻	smallpartformsions
Ionic equilibrium	the equilibrium between the molecules of a weak electrolyte and the ions produced from it. The law of mass action is not applied to strong electrolytes (they are completely ionized)	betweenmoleculesweakelectrolyte

Ionization constants of weak acids (illustration, Ka) H₂SO₃ 1.7 × 10⁻²; HF 6.7 × 10⁻⁴; HNO₂ 5.1 × 10⁻⁴; CH₃COOH 1.8 × 10⁻⁵; H₂CO₃ 4.4 × 10⁻⁷; H₃BO₃ 5.8 × 10⁻¹⁰.
A larger K_a means a stronger weak acid, so H₂SO₃ is the strongest of this list and H₃BO₃ the weakest.

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Ostwald’s dilution law

Ostwald discovered the relation between the degree of ionization α and the concentration C (mol/L) of a weak electrolyte.
For a weak monoprotic acid HA ⇌ H⁺ + A⁻ the law of mass action gives Ka = [H⁺][A⁻]/[HA], where Ka is the equilibrium constant of ionization.
Let 1 mole of HA be dissolved in V litres, with degree of ionization α (= number of ionized moles / total number of moles before ionization).
At equilibrium: unionized moles = (1 − α), moles of H⁺ and of A⁻ = α each, so [HA] = (1 − α)/V, [H⁺] = [A⁻] = α/V.

Then

K_a = α² / ((1 − α) V) = α² C / (1 − α) (C = 1/V mol/L)
This relation is Ostwald’s dilution law. For a weak electrolyte α is small, so 1 − α ≈ 1:

K_a = α² C → α = √(K_a / C)

Comparison aspect	Main idea	Keywords
•	At constant temperature K_a is constant, so α increases on dilution (C decreases) and decreases when the solution is more concentrated.	constanttemperatureincreasesdilution
•	To double α the concentration must be reduced to a quarter	doubleconcentrationmustreduced
Example 9 (new numbers).	Degree of ionization of 0.05 M HCN (K_a = 4.9 × 10⁻¹⁰): α = √(4.9 × 10⁻¹⁰/0.05) = √(9.8 × 10⁻⁹) ≈ 9.9 × 10⁻⁵.	Degreeionization
Example 10.	Degree of ionization of 0.02 M acetic acid (K_a = 1.8 × 10⁻⁵): α = √(1.8 × 10⁻⁵/0.02) = √(9 × 10⁻⁴) = 0.03, i.e. 3 %. If the solution is diluted to 0.005 M (a quarter of the concentration): α = √(1.8 × 10⁻⁵/0.005) = √(3.6 × 10⁻³) = 0.06, i.e. twice as large.	Degreeionizationaceticacid
Example 11 (finding Ka).	A 0.1 M solution of a weak monoprotic acid is 1 % ionized. K_a = α² C = (0.01)² × 0.1 = 1 × 10⁻⁵.	solutionweakmonoproticacid

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Hydronium ion concentration of a weak acid

For acetic acid CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, the ions are produced in equal amounts, so [H₃O⁺] = [CH₃COO⁻]; Ka = [CH₃COO⁻][H₃O⁺]/[CH₃COOH].
Because the acid is weak, the amount ionized is very small and the equilibrium concentration of the acid ≈ its original concentration Ca.

Then Ka = [H₃O⁺]²/Ca, giving

[H₃O⁺] = √(Ka × Ca)

Example 12.→The hydronium concentration in 0.05 M nitrous acid HNO₂ (K_a = 5.1 × 10⁻⁴): [H₃O⁺] = √(5.1 × 10⁻⁴ × 0.05) = √(2.55 × 10⁻⁵) = 5.05 × 10⁻³ M; pH = −log(5.05 × 10⁻³) = 2.30.

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Hydroxide ion concentration of a weak base

Weak bases are partially ionized in water.
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻; Kb = [NH₄⁺][OH⁻]/[NH₃] (Kb = 1.8 × 10⁻⁵ for ammonia).
Since [NH₄⁺] = [OH⁻] and ionization is small, [NH₃] ≈ Cb:

[OH⁻] = √(Kb × Cb)

Example 13.→For 0.04 M ammonia: [OH⁻] = √(1.8 × 10⁻⁵ × 0.04) = √(7.2 × 10⁻⁷) = 8.5 × 10⁻⁴ M; pOH = 3.07, so pH = 14 − 3.07 = 10.93.

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Ionization of water and the ionic product Kw

Pure water is a very poor conductor of electricity but not a non-conductor, so a little ionization must occur: 2H₂O ⇌ H₃O⁺ + OH⁻, simplified to H₂O ⇌ H⁺ + OH⁻.
The equilibrium constant is K = [H⁺][OH⁻]/[H₂O].
Very few water molecules ionize, so the number of unionized molecules is practically constant, [H₂O] is included in the constant and
Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴ at 25 °C (the ionic product of water).
Since water is neutral to litmus, the concentration of H⁺ (responsible for acidic properties) equals that of OH⁻ (responsible for basic properties): [H⁺] = [OH⁻] = 10⁻⁷ mol/L.
If [H⁺] increases (by adding acid), [OH⁻] decreases so that the product stays 10⁻¹⁴ — so knowing one ion concentration gives the other.
For example, if [H⁺] = 10⁻² then [OH⁻] = 10⁻¹² mol/L.

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pH and pOH

Comparison aspect	Main idea	Keywords
pH = −log[H⁺]	(the negative logarithm base 10 of the hydrogen-ion concentration) pOH = −log[OH⁻] the symbol p means −log. This is a convenient way of expressing the acidity or alkalinity of a solution	negativelogarithmbasehydrogen
Taking −log of Kw	pH + pOH = 14 (pKw = 14)

Comparison aspect	Main idea	Example	Keywords
Neutral	pH = 7 = pOH	—
Acidic	pH < 7 (pOH > 7)	pH 5 → pOH 9
Basic	pH > 7 (pOH < 7)	pH 8 → pOH 6
Example 14.	Find the pH of (a) 0.001 M HCl (b) 0.01 M NaOH (c) a solution with [H⁺] = 2 × 10⁻⁴ M	—	FindNaOHsolution
•	(a) [H⁺] = 10⁻³ → pH = 3. (b) [OH⁻] = 10⁻² → pOH = 2 → pH = 12. (c) pH = −log(2 × 10⁻⁴) = 4 − 0.30 = 3.70	—
Example 15.	A solution has pH 9.4. Then pOH = 4.6 [OH⁻] = 10⁻⁴·⁶ = 2.5 × 10⁻⁵ M and [H⁺] = 10⁻⁹·⁴ = 4.0 × 10⁻¹⁰ M. (Check: product 1.0 × 10⁻¹⁴.)	—	solutionThenCheckproduct
Example 16 (dilution).	10 mL of 0.1 M HCl are diluted with water to 1 L: [H⁺] = 0.1 × 10/1000 = 10⁻³ M pH = 3. Diluting an acid 100 times raises its pH by 2 units	—	dilutedwater1000Diluting

Typical pH values (illustration)→1 M HCl 0; 0.1 M HCl 1; gastric juice 1.6–1.8; vinegar 2.9; orange juice 3.5; tomato juice 4.2; coffee 5; milk 6.3–6.6; pure water 7; blood 7.35–7.45; sea water 8.4; magnesia emulsion 10.5; 0.1 M ammonia 11; washing soda 12.9; 0.1 M NaOH 13; 1 M NaOH 14.

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Solubility and solubility product

At a given temperature a solid salt has a limit of solubility in water.
When this limit is reached, a dynamic equilibrium exists between the solid (solute) and the dissolved ions, and the solution is saturated.
Solubilities differ widely KNO₃ about 31.6 g/100 mL at 20 °C but AgCl only 0.0016 g/100 mL.
The degree of solubility is the concentration of the saturated solution of a sparingly soluble salt at a given temperature.
For lead(II) bromide PbBr₂(s) ⇌ Pb²⁺ + 2Br⁻. The law of mass action applies; the concentration of the solid is constant and is included in the constant:
Ksp = [Pb²⁺][Br⁻]² (solubility product)
Solubility product (Ksp) the product of the concentrations (mol/L) of the ions of a sparingly soluble salt in its saturated solution, each raised to a power equal to the number of such ions in the formula, at a given temperature.

Comparison aspect	Main idea	Keywords
Example 17 (Ksp from solubility).	The solubility of lead iodide PbI₂ is 1.2 × 10⁻³ mol/L. PbI₂ ⇌ Pb²⁺ + 2I⁻: [Pb²⁺] = 1.2 × 10⁻³, [I⁻] = 2.4 × 10⁻³. K_sp = (1.2 × 10⁻³)(2.4 × 10⁻³)² = 6.9 × 10⁻⁹.	leadiodidePbI₂
Example 18 (solubility from Ksp).	K_sp of calcium fluoride CaF₂ is 3.2 × 10⁻¹¹. CaF₂ ⇌ Ca²⁺ + 2F⁻; with solubility s, [Ca²⁺] = s and [F⁻] = 2s, so K_sp = s(2s)² = 4s³. s = (3.2 × 10⁻¹¹/4)¹/3) = (8 × 10⁻¹²)¹/3) = 2.0 × 10⁻⁴ mol/L.	calciumfluorideCaF₂
Example 19 (a 1:1 salt).	K_sp of AgBr is 5.0 × 10⁻¹³: K_sp = s² so s = √(5.0 × 10⁻¹³) = 7.1 × 10⁻⁷ mol/L.	AgBr
General relation	for AB (1:1) Ksp = s² for AB₂ or A₂B (3 ions) Ksp = 4s³	ions
•	Application — will a precipitate form?	Applicationwillprecipitateform

In a saturated solution the ionic product equals K_sp.
If the product of the ion concentrations (each raised to its power) exceeds K_sp, a precipitate forms; if it is smaller, the solution is unsaturated and no precipitate forms.
This is used for precipitating sparingly soluble salts in qualitative and gravimetric analysis.
Example 20. Equal volumes of 2 × 10⁻⁴ M AgNO₃ and 2 × 10⁻⁴ M NaCl are mixed. Each ion is diluted to 1 × 10⁻⁴ M, so the ionic product = 10⁻⁴ × 10⁻⁴ = 10⁻⁸, which is greater than K_sp(AgCl) = 1.7 × 10⁻¹⁰: AgCl precipitates. If the stock solutions were 2 × 10⁻⁶ M, the ionic product would be (10⁻⁶)² = 10⁻¹², less than K_sp, so no precipitate forms.

Example 21 (extra: common ion).→Adding NaCl to a saturated AgCl solution so that [Cl⁻] = 0.01 M lowers [Ag⁺] to K_sp/[Cl⁻] = 1.7 × 10⁻¹⁰/0.01 = 1.7 × 10⁻⁸ M, so AgCl becomes less soluble.

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Summary of key relations

Quantity	Expression
Equilibrium constant	K_c = [C]^c[D]^d / ([A]^a[B]^b)
Pressure form	K_p from partial pressures
Ostwald’s law (weak)	K_a = α²C ; α = √(K_a/C)
Weak acid	[H₃O⁺] = √(K_a C_a)
Weak base	[OH⁻] = √(K_b C_b)
Ionic product of water	K_w = [H⁺][OH⁻] = 10⁻¹⁴
pH	pH = −log[H⁺]; pH + pOH = 14
Solubility product	K_sp = product of ion concentrations (powers)

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Part 12

📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Chemistry Thanaweya Amma (2026/2027)

Lesson 2 – Ionic Equilibrium, pH Scale, Hydrolysis and Solubility Product (Ksp)
Detailed Academic Lesson Overview:
Ionic equilibrium applies to weak electrolytes where un-ionized molecules dynamically equilibrate with their ions.
We analyze Ostwald's dilution law, acid dissociation constant Ka, the ion product of water Kw, pH and pOH calculations, salt hydrolysis, and solubility product constant Ksp.

🌳 Hierarchy Classification & Conceptual Roadmap

Ionic Equilibrium in Aqueous Solutions

├── 1. Electrolytes & Ostwald's Dilution Law

│ ├── Strong ElectrolytesCompletely ionized (HCl, NaOH, NaCl); no ionic equilibrium

│ └── Weak ElectrolytesPartially ionized (CH3COOH, NH4OH, HCN); Ka = α² · C

├── 2. Ionization of Water & pH Scale

│ ├── Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C

│ ├── pH = -log[H⁺] and pOH = -log[OH⁻]; pH + pOH = 14
│ └── For weak acid [H⁺] = √(Ka · Ca) and pH = -log√(Ka · Ca)
├── 3. Salt Hydrolysis (Acidic, Basic, or Neutral)
│ ├── Strong Acid + Weak Base (NH₄Cl) → Acidic solution (pH < 7)
│ ├── Weak Acid + Strong Base (CH₃COONa) → Basic solution (pH > 7)
│ └── Strong Acid + Strong Base (NaCl) → Neutral solution (pH = 7)
└── 4. Solubility Product (Ksp) of Sparingly Soluble Salts
├── Degree of solubility (x) Concentration of saturated solution
└── For salt AxBy ⇌ x A^y+ + y B^x−, Ksp = [A^y+]^x · [B^x−]^y
📋 Master Teacher's Blackboard Rules & Exam Secrets
RULE 1 Ostwald's Law: For weak electrolytes, as dilution increases (concentration C decreases), the degree of ionization α increases to keep Ka constant.
RULE 2 Kw = 10⁻¹⁴ only at 25°C. Since water auto-ionization is endothermic, increasing temperature increases Kw and decreases the neutral pH value below 7.
RULE 3 A salt of a strong acid and a strong base (like NaCl, KNO3) does not undergo hydrolysis because neither Na⁺ nor Cl⁻ can combine with OH⁻ or H⁺ to form weak molecules.
📊 Detailed Comparison Matrix Hydrolysis Behavior of Common Salts Matrix

Salt Formula	Parent Acid / Strength	Parent Base / Strength	Aqueous Solution pH
Sodium Chloride (NaCl)	HCl (Strong acid)	NaOH (Strong base)	pH = 7 (Neutral)
Ammonium Chloride (NH₄Cl)	HCl (Strong acid)	NH₄OH (Weak base)	pH < 7 (Acidic)
Sodium Acetate (CH₃COONa)	CH₃COOH (Weak acid)	NaOH (Strong base)	pH > 7 (Basic)
Ammonium Acetate (CH₃COONH₄)	CH₃COOH (Weak acid)	NH₄OH (Weak base)	pH ≈ 7 (Nearly Neutral)

Lesson 2: Ionic Equilibrium, pH Scale, Hydrolysis and Solubility Product (Ksp) · 12 / 12

Curriculum

Chemistry

Chapter 1 - Transition Elements and Iron Metallurgy

Chapter 2 - Chemical Analysis

Chapter 3 - Chemical Equilibrium

Chapter 4 - Electrochemistry

Chapter 5 - Organic Chemistry

More lessons

Lesson 2: Ionic Equilibrium, pH Scale, Hydrolysis and Solubility Product (Ksp)

Lesson 2: Ionic Equilibrium, pH Scale, Hydrolysis and Solubility Product (Ksp)

Overview

🌳 Concept tree

Strong and weak electrolytes

Hydronium ion

Definitions

Ostwald’s dilution law

Then

Hydronium ion concentration of a weak acid

Then Ka = [H₃O⁺]²/Ca, giving

[H₃O⁺] = √(Ka × Ca)

Hydroxide ion concentration of a weak base

[OH⁻] = √(Kb × Cb)

Ionization of water and the ionic product Kw

pH and pOH

Solubility and solubility product

Summary of key relations

Part 12

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Chemistry Thanaweya Amma (2026/2027)

🌳 Hierarchy Classification & Conceptual Roadmap

Ionic Equilibrium in Aqueous Solutions

├── 2. Ionization of Water & pH Scale

│ ├── Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C

Well done! You finished this lesson

Curriculum

Chemistry

Chapter 1 - Transition Elements and Iron Metallurgy

Chapter 2 - Chemical Analysis

Chapter 3 - Chemical Equilibrium

Chapter 4 - Electrochemistry

Chapter 5 - Organic Chemistry

More lessons

Lesson 2: Ionic Equilibrium, pH Scale, Hydrolysis and Solubility Product (Ksp)

🌳 Concept tree

Then

Then Ka = [H₃O+]2/Ca, giving

[H₃O+] = √(Ka × Ca)

[OH−] = √(Kb × Cb)

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Chemistry Thanaweya Amma (2026/2027)

🌳 Hierarchy Classification & Conceptual Roadmap

Ionic Equilibrium in Aqueous Solutions

├── 2. Ionization of Water & pH Scale

│ ├── Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

Well done! You finished this lesson

Then Ka = [H₃O⁺]²/Ca, giving

[H₃O⁺] = √(Ka × Ca)

[OH⁻] = √(Kb × Cb)

│ ├── Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C