FAD1018 W2-W3 — Ionic Equilibria
Weeks 2–3 lectures covering ionic equilibria, acids and bases. Source files: W2 (1).pdf (33 slides), W3 (1).pdf (34 slides, Dr. Fauzani Md. Salleh) from lecture notes folders.
Lecture Metadata
- Course: FAD 1018 Basic Chemistry 2 (2024/2025)
- Lecturer: Dr. Fauzani Md. Salleh, Chemistry Division, Centre for Foundation Studies in Science
- Topic: Ionic Equilibria (Part 1)
- Slides: 33 pages
Learning Objectives
- Define the basic terms of acid and base.
- Describe the Arrhenius, Lewis and Brønsted-Lowry theories of acids and bases.
- Identify the conjugate pairs of acid and base.
- Identify the strength of acid & base, and the use of $K_a$ & $K_b$.
- Calculate the pH and read the pH scale.
- Explain the acid-base properties of water and concept of auto-ionisation.
- Describe the degree of dissociation.
- Explain the relationship between $K_a$ & $K_b$.
1. Introduction to Ionic Equilibria
Ionic equilibria refer to equilibrium reactions in aqueous solution or molten state that show the presence of ions.
- Reaction contains positively charged ions (cations) and negatively charged ions (anions).
- Strong acids/bases: fully ionised / dissociate
- Weak acids/bases: partially dissociate
Cl
2. Applications of Ionic Equilibria
Electrolytes
Ionic equilibria are fundamental to electrolysis and electrochemistry:
- NaCl(s) —(Δ)→ Na⁺(ℓ) + Cl⁻(ℓ) (molten)
- NaCl(aq) —(H₂O)→ Na⁺(aq) + Cl⁻(aq) (aqueous)
Applications include electrolytic cells and Daniell cells (galvanic cells).
3. Acid-Base Concepts
Neutralization
Acid + Base → Salt + Water
$$\text{HCl} + \text{NaOH} \rightarrow \text{H}_2\text{O} + \text{NaCl}$$
Key ions:
- Cation (+ve): H₃O⁺ (hydronium) or H⁺
- Anion (-ve): OH⁻ (hydroxide)
Differences Between Acid and Base
| Property | Acids | Bases |
|---|---|---|
| Taste | Sour (lemons – citric acid) | Bitter |
| Common examples | Vinegar (ethanoic/acetic acid) | Bleach, cleaning products |
| Litmus test | Turns blue litmus to red | Turns red litmus to blue |
| pH | < 7 | > 7 |
| Neutralization | Neutralize bases | Neutralize acids |
| Other | Dissolve many metals | Feel soapy or greasy |
Common Acids
| Name | Formula | Occurrence/Uses |
|---|---|---|
| Hydrochloric acid | HCl | Metal cleaning; food preparation; ore refining; main component of stomach acid |
| Sulfuric acid | H₂SO₄ | Fertilizer and explosives manufacturing; dye and glue production; automobile batteries; electroplating of copper |
| Nitric acid | HNO₃ | Fertilizer and explosives manufacturing; dye and glue production |
| Acetic acid | CH₃COOH | Plastic and rubber manufacturing; food preservative; active component of vinegar |
| Citric acid | C₆H₈O₇ | Present in citrus fruits such as lemons and limes; used to adjust pH in foods and beverages |
CC(=O)O
O=C(O)CC(O)(CC(=O)O)C(=O)O
OS(=O)(=O)O
O[N+](=O)[O-]
Common Bases
| Name | Formula | Occurrence/Uses |
|---|---|---|
| Sodium hydroxide | NaOH | Petroleum processing; soap and plastic manufacturing |
| Potassium hydroxide | KOH | Cotton processing; electroplating; soap production; batteries |
| Sodium bicarbonate | NaHCO₃ | Antacid; ingredient of baking soda; source of CO₂ |
| Sodium carbonate | Na₂CO₃ | Manufacture of glass and soap; general cleanser; water softener |
| Ammonia | NH₃ | Detergent; fertilizer and explosives manufacturing; synthetic fiber production |
[Na+].[OH-]
[K+].[OH-]
[Na+].OC(=O)[O-]
[Na+].[Na+].[O-]C([O-])=O
4. Theories of Acid and Base
There are three main theories relating to acids and bases:
- Arrhenius Theory (A)
- Brønsted-Lowry Theory (B-L)
- Lewis Theory (L)
4.1 Arrhenius Theory (1884)
Introduced by Svante Arrhenius (1884).
Acid: A substance that dissociates in water to produce an excess of hydrogen ions, H⁺(aq) [or hydronium H₃O⁺] in aqueous solution.
Compounds (acids) dissolve in water in two ways:
-
Dissociates without reacting with water molecules: $$\text{HCl(aq)} \xrightarrow{\text{H}_2\text{O}} \text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$$
-
Dissociates when reacted with water molecules: $$\text{HCl(aq)} + \text{H}_2\text{O}(\ell) \rightarrow \text{H}_3\text{O}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$$
Base: A substance that dissociates in water to produce an excess of hydroxide ions, OH⁻(aq) in aqueous solution.
$$\text{NaOH(aq)} \xrightarrow{\text{H}_2\text{O}} \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)}$$
Limitation: Applicable only to aqueous solution for compounds containing H⁺ and OH⁻.
4.2 Brønsted-Lowry Theory (1923)
Proposed by Johannes Brønsted and Thomas M. Lowry (1923).
Acid:
- Are proton (H⁺) donor
- Becomes its conjugate base when it donates a proton
Base:
- Are proton (H⁺) acceptor
- Becomes its conjugate acid when it accepts a proton
Example 1: $$\text{HCl} + \text{H}_2\text{O} \rightarrow \text{Cl}^- + \text{H}_3\text{O}^+$$
- HCl / Cl⁻ is called a conjugate pair
- H₂O / H₃O⁺ is also called a conjugate pair
Example 2: $$\text{H}_2\text{O} + \text{NH}_3 \rightleftharpoons \text{OH}^- + \text{NH}_4^+$$
- H₂O acts as Brønsted-Lowry Acid (donates proton)
- NH₃ acts as Brønsted-Lowry Base (accepts proton)
Example 3: $$\text{H}_2\text{SO}_4 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^- + \text{H}_3\text{O}^+$$
- H₂SO₄ is a strong acid, diprotic acid (dissociates 2 H⁺)
Water as Amphoteric Solvent
- H₂O is an amphoteric solvent
- Acting both as a base and an acid
- It undergoes auto-ionisation (self-ionisation):
$$\text{H}_2\text{O} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$$
Worked Examples (Conjugate Pairs)
Identify Arrhenius acid, Arrhenius base, Brønsted-Lowry acid, Brønsted-Lowry base, conjugate acid or conjugate base:
(a) H₂CO₃(aq) + H₂O(aq) ⇌ HCO₃⁻(aq) + H₃O⁺(aq)
- H₂CO₃ is Arrhenius acid and Brønsted-Lowry acid
- H₂O is Brønsted-Lowry base
- HCO₃⁻ is conjugate base of H₂CO₃
- H₃O⁺ is conjugate acid of H₂O
(b) NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)
- NH₄⁺ is Arrhenius acid and Brønsted-Lowry acid
- H₂O is Brønsted-Lowry base
- NH₃ is conjugate base of NH₄⁺
- H₃O⁺ is conjugate acid of H₂O
(c) NH₃(aq) + H₃O⁺(aq) ⇌ NH₄⁺(aq) + H₂O(l)
(d) CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq)
Exercises
Exercise 1: Select (i) the Brønsted-Lowry acid (ii) the conjugate acid for the following reversible reactions:
(a) HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq) → B-L Acid: HSO₄⁻; Conjugate acid: H₃O⁺ (b) NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) → B-L Acid: NH₄⁺; Conjugate acid: H₃O⁺ (c) N₂H₄(aq) + H₂O(l) ⇌ N₂H₅⁺(aq) + OH⁻(aq) → B-L Acid: H₂O; Conjugate acid: N₂H₅⁺ (d) Fe[(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ Fe[(H₂O)₅OH]²⁺(aq) + H₃O⁺(aq) → B-L Acid: Fe[(H₂O)₆]³⁺; Conjugate acid: H₃O⁺
Exercise 2: What is the conjugate base for each of the following acids:
| Acid | Conjugate Base |
|---|---|
| HClO₄ | ClO₄⁻ |
| H₂S | HS⁻ |
| PH₄⁺ | PH₃ |
| HCO₃⁻ | CO₃²⁻ |
Exercise 3: What is the conjugate acid for each of the following bases:
| Base | Conjugate Acid |
|---|---|
| CN⁻ | HCN |
| SO₄²⁻ | HSO₄⁻ |
| H₂O | H₃O⁺ |
| HCO₃⁻ | H₂CO₃ |
4.3 Lewis Theory (1938)
Proposed by G. N. Lewis (1938).
Acid (Lewis acid):
- Is an atom, ion or molecule that accepts a pair of electrons to form a coordinate covalent bond
- Electrophile (usually +ve charge)
- Examples: BF₃, AlCl₃ (incomplete octet)
Cl[Al](Cl)Cl
Base (Lewis base):
- Is an atom, ion or molecule that donates a pair of electrons to form a coordinate covalent bond
- Nucleophile (usually -ve charge)
Examples:
$$\text{NH}_3 + \text{BF}_3 \rightarrow \text{adduct}$$
FB(F)F
N
$$\text{H}_3\text{C-NH}_2 + \text{BH}_3 \rightarrow \text{H}_3\text{C-NH}_2-BH}_3$$
$$\text{H}_3\text{C-NH}_2 + \text{HCl} \rightarrow \text{H}_3\text{C-NH}_3\text{Cl}$$
$$\text{F}^- + \text{BF}_3 \rightarrow [\text{F-BF}_3]^-$$
$$2\text{NH}_3 + \text{Ag}^+ \rightarrow [\text{H}_3\text{N-Ag-NH}_3]^+$$
The products of the reactions are called adducts.
Lewis Base vs Lewis Acid Comparison
| Category | Lewis Base (e⁻ donor) | Lewis Acid (e⁻ acceptor) |
|---|---|---|
| General | Something that complexes with a Lewis Acid | Something that complexes with a Lewis Base |
| Lone-Pair | Lone-Pair donors: :NH₃, H₂O: | Lone-Pair acceptors: BF₃, AlCl₃ |
| Brønsted relation | Brønsted bases: ⁻OH, ⁻CH₃ | Metal cations: Al³⁺, Fe³⁺ |
| Nucleophile/Electrophile | Nucleophiles: CH₃-S⁻ | Electrophiles: CH₃CO⁺ |
| Ligands | [Fe(H₂O)₆]³⁺ | The proton: H⁺ |
| Counter ions | Anionic counter ions: SO₄²⁻, NO₃⁻ | Cationic spectator ions: K⁺ |
| π systems | Electron-rich π systems (benzene) | Electron-poor π systems: [CH₂-CH=CH₂]⁺ |
Important: Not all Lewis Bases are Brønsted-Lowry Bases because the definition of a Lewis Base is broader than that of a Brønsted-Lowry Base.
Lewis Acid-Base Exercise
In the following reactions, identify the Lewis acid and the Lewis base:
(a) NH₃(aq) + H⁺(aq) ⇌ NH₄⁺(aq) → Lewis acid: H⁺; Lewis base: NH₃ (b) Cu²⁺(aq) + 4 NH₃(aq) ⇌ Cu(NH₃)₄²⁺(aq) → Lewis acid: Cu²⁺; Lewis base: NH₃
5. Amphoteric Species
An amphoteric species is a substance that has the ability to act either as an acid or a base depending on what other substances it is reacting with.
- Acids donate protons (or accept electron pairs) when reacting with basic substances.
- Bases accept protons when reacting with acidic substances.
Examples of amphoteric species: water, aluminium hydroxide, bicarbonate ion.
Other amphoteric substances can be formed from the oxides and hydroxides of metals like beryllium, zinc, and lead.
Amino acids, which are the building blocks of proteins, are also amphoteric.
5.1 Water as Amphoteric Species
- Normally considered a neutral substance.
- If water reacts with a base like ammonia, it acts as an acid by donating a proton:
$$\text{H}_2\text{O}(l) + \text{NH}_3\text{(aq)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$$
- If water comes in contact with an acid like HCl, it acts as a base by receiving a proton:
$$\text{HCl(aq)} + \text{H}_2\text{O}(l) \rightarrow \text{Cl}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$$
H₂O is an amphoteric species.
5.2 Bicarbonate Ion (HCO₃⁻) as Amphoteric Species
- When reacting with the acid hydronium ion, HCO₃⁻ acts as a base by accepting a proton:
$$\text{H}_3\text{O}^+\text{(aq)} + \text{HCO}_3^-\text{(aq)} \rightleftharpoons \text{H}_2\text{CO}_3\text{(aq)} + \text{H}_2\text{O}(l)$$
- When reacting with the base hydroxide ion, HCO₃⁻ acts as an acid by donating a proton:
$$\text{HCO}_3^-\text{(aq)} + \text{OH}^-\text{(aq)} \rightleftharpoons \text{CO}_3^{2-}\text{(aq)} + \text{H}_2\text{O}(l)$$
HCO₃⁻ is an amphoteric species.
5.3 Aluminum Hydroxide (Al(OH)₃) as Amphoteric Species
- If it reacts with a base like NaOH, it acts as an acid:
$$\text{Al(OH)}_3\text{(aq)} + \text{NaOH(aq)} \rightarrow \text{Na[Al(OH)}_4]\text{(aq)}$$
- With HCl, it acts as a base:
$$\text{Al(OH)}_3\text{(aq)} + 3\text{HCl(aq)} \rightleftharpoons \text{AlCl}_3\text{(aq)} + 3\text{H}_2\text{O}(l)$$
Al(OH)₃ is an amphoteric species.
5.4 Amphoteric Species Exercise
Select an amphoteric species from the following reactions:
$$\text{HSO}_4^-\text{(aq)} + \text{H}_2\text{O}(l) \rightleftharpoons \text{SO}_4^{2-}\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$$
$$\text{H}_2\text{SO}_4\text{(aq)} + \text{H}_2\text{O}(l) \rightleftharpoons \text{HSO}_4^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$$
Answer: HSO₄⁻ is the amphoteric species (can act as acid or base).
6. Strength of Acid & Base
6.1 Strong Acid
- Dissociate completely (100% ionized) in aqueous solution to form H⁺ or H₃O⁺ ions.
- Examples: HCl(aq), H₂SO₄(aq), HNO₃(aq), HBr(aq), HClO₄(aq), HI(aq)
Br
OCl(=O)(=O)=O
I
- The molarities of H⁺ or H₃O⁺ ions are high for strong acids.
- pH 1 – very acidic
- Low pH means high molarities of H⁺ or H₃O⁺ ions
- The degree of dissociation ($\alpha$) for strong acid is 1 or 100%
$$\text{HCl(aq)} \rightarrow \text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$$
6.2 Strong Base
- Dissociate completely (100% ionized) in aqueous solution to form OH⁻ ion.
- Examples: NaOH(aq), Mg(OH)₂(aq), Ca(OH)₂(aq)
[Mg+2].[OH-].[OH-]
[Ca+2].[OH-].[OH-]
- The molarities of OH⁻ are high for strong bases.
- pH 14 – very basic
- High pH means high molarities of OH⁻ ion
- The degree of dissociation ($\alpha$) for strong base is 1 or 100%
$$\text{NaOH(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)}$$
6.3 Weak Acid
- Not dissociate completely (partially ionized) in aqueous solution to form H⁺ or H₃O⁺ ions.
- Therefore the concentrations of H⁺ or H₃O⁺ ions in the solutions are always less than the concentrations of the original acids.
- Examples: CH₃COOH(aq), HCN(aq), HNO₂(aq), HF(aq)
C#N
O=NO
F
- The system exists as an equilibrium between the un-dissociated acid and the dissociated ions.
$$\text{HA}{\text{(aq)}} + \text{H}2\text{O}{(\ell)} \rightleftharpoons \text{A}^-{\text{(aq)}} + \text{H}3\text{O}^+{\text{(aq)}}$$
At equilibrium: (1-x) M HA, x M A⁻, x M H₃O⁺
$$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$$
- $K_a$ is a dissociation constant for acid, measures the acid strength.
- Values of $K_a$ are unaffected by the concentration of acids (only temperature – Le Chatelier's principle).
- Larger $K_a$ means stronger acid.
- Higher $K_a$ = lower $pK_a$
- $pK_a = -\log K_a$
6.4 Weak Base
- Not dissociate completely (partially ionized) in aqueous solution to form OH⁻ ion.
- Therefore the concentrations of OH⁻ ions in the solutions are always less than the concentrations of the original bases.
- Examples: NH₄OH(aq), CH₃NH₂(aq), C₅H₅N(aq)
[NH4+].[OH-]
CN
c1ccncc1
- The system exists as an equilibrium between the un-dissociated base and the dissociated ions.
$$\text{NH}_3\text{(aq)} + \text{H}_2\text{O}(\ell) \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$$
At equilibrium: (1-x) M NH₃, x M NH₄⁺, x M OH⁻
$$K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$$
- $K_b$ is a dissociation constant for base, measures the base strength.
- Larger $K_b$ means stronger base.
- Higher $K_b$ = lower $pK_b$
- $pK_b = -\log K_b$
- pH between 8 to 10
6.5 Ka and Kb Tables
Table 3: Values of $K_a$ for some common acids
| Name of Acid | Formula | $K_a$ |
|---|---|---|
| Hydrochloric acid | HCl | ∞ |
| Nitric acid | HNO₃ | ∞ |
| Hydrofluoric acid | HF | 7.1 × 10⁻⁴ |
| Nitrous acid | HNO₂ | 4.5 × 10⁻⁴ |
| Formic acid | HCOOH | 1.7 × 10⁻⁴ |
| Acetic acid | CH₃COOH | 1.8 × 10⁻⁵ |
| Hydrocyanic acid | HCN | 4.9 × 10⁻¹⁰ |
| Aspirin | C₉H₈O₄ | 3.0 × 10⁻⁴ |
| Ascorbic acid (Vitamin C) | C₆H₈O₆ | 8.0 × 10⁻⁵ |
| Benzoic acid | C₆H₅COOH | 6.5 × 10⁻⁵ |
| Phenol | C₆H₅OH | 1.3 × 10⁻¹⁰ |
| Hypochlorous acid | HOCl | 3.0 × 10⁻⁸ |
O=CO
CC(=O)Oc1ccccc1C(=O)O
OC1=C(O)C(=O)OC1C(O)CO
O=C(O)c1ccccc1
Oc1ccccc1
OCl
Table 4: Values of $K_b$ for some common bases
| Name of Base | Formula | $K_b$ |
|---|---|---|
| Ethylamine | CH₃CH₂NH₂ | 5.6 × 10⁻⁴ |
| Methylamine | CH₃NH₂ | 4.4 × 10⁻⁴ |
| Ammonia | NH₃ | 1.8 × 10⁻⁵ |
| Aniline | C₆H₅NH₂ | 3.8 × 10⁻¹⁰ |
| Pyridine | C₅H₅N | 1.7 × 10⁻⁹ |
| Caffeine | C₈H₁₀N₄O₂ | 4.1 × 10⁻⁴ |
| Urea | N₂H₄CO | 1.5 × 10⁻¹⁴ |
| Carbonate ion | CO₃²⁻ | 1.8 × 10⁻⁴ |
CCN
Nc1ccccc1
CN1C=NC2=C1C(=O)N(C(=O)N2C)C
NC(=O)N
[O-]C([O-])=O
6.6 Acid and Base Strength Exercises
Exercise 1: Arrange the following acids in the order of increasing acid strength: Aspirin, formic acid, phenol, vitamin C and nitrous acid.
Answer: Phenol < Vitamin C < Formic acid < Aspirin < Nitrous acid
(Using Ka values: Phenol 1.3×10⁻¹⁰, Vitamin C 8.0×10⁻⁵, Formic acid 1.7×10⁻⁴, Aspirin 3.0×10⁻⁴, Nitrous acid 4.5×10⁻⁴)
Exercise 2: Arrange the following base in the order of decreasing base strength: Aniline, Ammonia, Ethylamine, Urea.
Answer: Ethylamine > Ammonia > Aniline > Urea
(Using Kb values: Ethylamine 5.6×10⁻⁴, Ammonia 1.8×10⁻⁵, Aniline 3.8×10⁻¹⁰, Urea 1.5×10⁻¹⁴)
Exercise 3: Identify the weakest base from the information of Kb value in Table 4 above.
Answer: Urea (Kb = 1.5 × 10⁻¹⁴)
Key Equations
$$pH = -\log[H^+]$$
$$pOH = -\log[OH^-]$$
$$pH + pOH = 14 \text{ (at 25°C)}$$
$$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$
$$K_b = \frac{[BH^+][OH^-]}{[B]}$$
$$pK_a = -\log K_a$$
$$pK_b = -\log K_b$$
$$K_a \times K_b = K_w = 1.0 \times 10^{-14} \text{ (at 25°C)}$$
W3(1) — Ionic Equilibria (Part 3): Salt Hydrolysis
Lecturer: Dr. Fauzani Md. Salleh
Slides: 34 pages
Focus: Acid-base properties of salts, hydrolysis mechanisms, and quantitative pH calculations.
Learning Objectives
- Identify acid-base properties of salts
- Distinguish salts of strong acid + strong base (SA-SB), strong acid + weak base (SA-WB), and strong base + weak acid (SB-WA)
- Apply qualitative and quantitative analysis to salt solutions
Hydrolysis
Hydrolysis is the reaction of a cation or an anion (or both) with water. The reaction is reversible.
- Cation hydrolysis (produces H₃O⁺): $$\text{X}^+_{(aq)} + \text{H}2\text{O}{(l)} \rightleftharpoons \text{XOH} + \text{H}3\text{O}^+{(aq)}$$
- Anion hydrolysis (produces OH⁻): $$\text{Y}^-{(aq)} + \text{H}2\text{O}{(l)} \rightleftharpoons \text{HY} + \text{OH}^-{(aq)}$$
Relationship between Ka and Kb
For a conjugate acid-base pair:
$$K_a \times K_b = K_w = 1.0 \times 10^{-14} \text{ (at 25°C)}$$
$$pK_a + pK_b = pK_w = 14$$
Derivation:
- Acid dissociation: $\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CH}_3\text{COO}^-$ $$K_a = \frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$$
- Conjugate base hydrolysis: $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{OH}^- + \text{CH}_3\text{COOH}$ $$K_b = \frac{[\text{OH}^-][\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]}$$
Multiplying the two expressions: $K_a K_b = [\text{H}_3\text{O}^+][\text{OH}^-] = K_w$.
Neutral Salts (SA-SB)
Formed from strong acid + strong base. pH = 7.
- Neither cation nor anion reacts with water → no hydrolysis
- Both ions are spectator ions (low charge density / low proton affinity)
$$\text{HCl}{(aq)} + \text{NaOH}{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}2\text{O}{(l)}$$
$$\text{NaCl}{(aq)} \rightarrow \text{Na}^+{(aq)} + \text{Cl}^-_{(aq)}$$
[Na+].[Cl-]
[K+].[Br-]
[K+].[I-]
[Rb+].[Br-]
[Ba+2].[Cl-].[Cl-]
From ionic theory, cancelling common ions leaves $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$, so $[\text{H}^+] = [\text{OH}^-]$.
Acidic Salts (SA-WB)
Formed from strong acid + weak base. pH < 7.
- Cation undergoes hydrolysis (donates proton to water)
- Examples: NH₄Cl, NH₄NO₃
$$\text{HCl}{(aq)} + \text{NH}{3(aq)} \rightarrow \text{NH}4\text{Cl}{(aq)}$$
Dissociation: $$\text{NH}4\text{Cl}{(aq)} \rightarrow \text{NH}4^+{(aq)} + \text{Cl}^-_{(aq)}$$
Hydrolysis: $$\text{NH}4^+{(aq)} + \text{H}2\text{O}{(l)} \rightleftharpoons \text{NH}_{3(aq)} + \text{H}3\text{O}^+{(aq)}$$
[NH4+].[Cl-]
[NH4+].[O-][N+](=O)[O-]
Worked Example: pH of 0.20 M NH₄Cl
Given $K_b$ for NH₃ = $1.8 \times 10^{-5}$:
- $K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$
- ICE table for NH₄⁺ hydrolysis:
- Initial: [NH₄⁺] = 0.20 M, [NH₃] = 0, [H₃O⁺] = 0
- Change: –x, +x, +x
- Equilibrium: (0.20 – x), x, x
- $K_a = \frac{x^2}{0.20-x} \approx \frac{x^2}{0.20}$ (assuming x is small)
- $x = [\text{H}_3\text{O}^+] = 1.05 \times 10^{-5}$ mol dm⁻³
- $pH = -\log(1.05 \times 10^{-5}) = \mathbf{4.98}$
Practice Problems (Acidic Salts)
- Calculate pH of 0.10 M N₂H₅Cl ($K_b$ for N₂H₄ = $1.7 \times 10^{-7}$)
- Calculate pH of 0.42 M NH₄NO₃ ($K_b$ for NH₃ = $1.8 \times 10^{-5}$)
- Calculate pH of 0.25 M CH₃NH₃NO₃ ($K_b$ for CH₃NH₂ = $4.4 \times 10^{-4}$)
[NH3+]N.[Cl-]
C[NH3+].[O-][N+](=O)[O-]
Basic Salts (SB-WA)
Formed from weak acid + strong base. pH > 7.
- Anion undergoes hydrolysis (accepts proton from water)
- Examples: CH₃COONa, KNO₂
$$\text{CH}3\text{COOH}{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{CH}3\text{COONa}{(aq)} + \text{H}2\text{O}{(l)}$$
Dissociation: $$\text{CH}3\text{COONa}{(aq)} \rightarrow \text{Na}^+_{(aq)} + \text{CH}3\text{COO}^-{(aq)}$$
Hydrolysis: $$\text{CH}3\text{COO}^-{(aq)} + \text{H}2\text{O}{(l)} \rightleftharpoons \text{CH}3\text{COOH}{(aq)} + \text{OH}^-_{(aq)}$$
CC(=O)[O-].[Na+]
[K+].[O-]N=O
Worked Example: pH of Sodium Benzoate Solution
0.05 mol C₆H₅COONa dissolved in 100 cm³ volumetric flask. $K_a$ for C₆H₅COOH = $6.5 \times 10^{-5}$.
- $[\text{C}_6\text{H}_5\text{COONa}] = \frac{0.05 \times 1000}{100} = 0.5$ mol dm⁻³
- $K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.5 \times 10^{-5}} = 1.54 \times 10^{-10}$
- ICE table for C₆H₅COO⁻ hydrolysis:
- Initial: [C₆H₅COO⁻] = 0.5, [C₆H₅COOH] = 0, [OH⁻] = 0
- Change: –x, +x, +x
- Equilibrium: (0.5 – x), x, x
- $K_b = \frac{x^2}{0.5-x} \approx \frac{x^2}{0.5}$
- $x = [\text{OH}^-] = 8.77 \times 10^{-6}$ mol dm⁻³
- $pOH = -\log(8.77 \times 10^{-6}) = 5.06$
- $pH = 14 - 5.06 = \mathbf{8.94}$ (> 7, basic salt)
O=C([O-])c1ccccc1.[Na+]
Practice Problems (Basic Salts)
- Calculate pH of 0.15 M CH₃COONa ($K_a$ for CH₃COOH = $1.8 \times 10^{-5}$)
- Calculate pH of 0.24 M HCOOK ($K_a$ for HCOOH = $1.7 \times 10^{-4}$)
- Calculate pH of 0.75 M KClO ($K_a$ for HClO = $3.0 \times 10^{-8}$)
O=C[O-].[K+]
[K+].[O-]Cl
Salts of Weak Acid & Weak Base (WA-WB)
Both cation and anion hydrolyse in water. The salt may be neutral, acidic, or basic depending on relative $K_a$ and $K_b$:
| Condition | Salt Type | Dominant Mechanism |
|---|---|---|
| $K_a \approx K_b$ | Neutral | Cation and anion hydrolysis equally extensive |
| $K_a > K_b$ | Acidic | Cation hydrolysis more extensive |
| $K_a < K_b$ | Basic | Anion hydrolysis more extensive |
Example: C₆H₅COONH₄ (ammonium benzoate) is an acidic salt because $K_a$ for C₆H₅COOH ($6.5 \times 10^{-5}$) > $K_b$ for NH₃ ($1.8 \times 10^{-5}$).
O=C([O-])c1ccccc1.[NH4+]
Example: CH₃COONH₄ (ammonium ethanoate) is a neutral salt because $K_a$ for CH₃COOH ($1.8 \times 10^{-5}$) = $K_b$ for NH₃ ($1.8 \times 10^{-5}$).
CC(=O)[O-].[NH4+]
Prediction Exercise: Predict whether solutions of the following salts are acidic, basic, or neutral: (a) NH₄I, (b) CaCl₂, (c) KCN, (d) NaClO₄, (e) CH₃NH₃Br, (f) HCOOK
[NH4+].[I-]
[Ca+2].[Cl-].[Cl-]
[K+].[C-]#N
[Na+].[O-]Cl(=O)(=O)=O
C[NH3+].[Br-]
O=C[O-].[K+]
Links
- Ionic Equilibria
- Arrhenius Theory
- Brønsted-Lowry Theory
- Lewis Theory
- Amphoteric Species
- Acid Dissociation Constant
- Base Dissociation Constant
- FAD1018 - Basic Chemistry II
- Dr Fauzani Md Salleh