Acid Strength: Determined by the extent of proton donation in water.
Ka (Acid Dissociation Constant): Quantitative measure of acid strength.
Base Strength: Determined by proton acceptance ability.
Kb (Base Dissociation Constant): Quantitative measure of base strength.
Definition: Solutions that resist pH changes when small amounts of acid/base are added.
Types:
Buffer Action: Based on common ion effect and Le Chatelier’s principle.
Henderson-Hasselbalch Equation:
pH = pKa + log([A⁻]/[HA])
Solubility Product (Ksp):
For AmBn ⇌ mA⁺ + nB⁻
Ksp = [A⁺]m[B⁻]n
Applications:
Definition: Suppression of ionization/solubility by adding a common ion.
Mechanism: Based on Le Chatelier’s principle.
Examples:
Definition: Ratio of solute concentrations in two immiscible solvents at equilibrium.
KPC = [X]solvent1 / [X]solvent2
Factors Affecting KPC:
Blood Buffer System:
CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻
Normal Blood pH: 7.35-7.45
Enzyme: Carbonic anhydrase catalyzes the reaction
Clinical Importance: pH < 6.8 or > 7.8 can be fatal
The strength of an acid is determined by its ability to donate protons (H⁺) in aqueous solution.
Acid Dissociation: HX + H₂O ⇌ H₃O⁺ + X⁻
Acid Dissociation Constant: Ka = [H₃O⁺][X⁻] / [HX]
pKa: pKa = -log Ka
| Name of Acid | Formula | Ka | pKa |
|---|---|---|---|
| Perchloric acid | HClO₄ | 1.0 × 10¹⁰ | -10.0 |
| Hydroiodic acid | HI | 1.0 × 10¹⁰ | -10.0 |
| Hydrobromic acid | HBr | 1.0 × 10⁹ | -9.0 |
| Hydrochloric acid | HCl | 1.0 × 10⁶ | -6.0 |
| Hydrofluoric acid | HF | 7.2 × 10⁻⁴ | +3.1 |
| Acetic acid | CH₃COOH | 1.8 × 10⁻⁵ | +4.7 |
The strength of a base is determined by its ability to accept protons (H⁺) in aqueous solution.
Base Dissociation: B + H₂O ⇌ BH⁺ + OH⁻
Base Dissociation Constant: Kb = [BH⁺][OH⁻] / [B]
pKb: pKb = -log Kb
Calculate the concentration of H⁺ ions in a 1.0M HF solution (Ka = 7.2 × 10⁻⁴)
Solution:
HF ⇌ H⁺ + F⁻
Initial: [HF] = 1.0M, [H⁺] = 0, [F⁻] = 0
Equilibrium: [HF] = 1.0 – x, [H⁺] = x, [F⁻] = x
Ka = [H⁺][F⁻] / [HF] = x² / (1.0 – x) ≈ x² / 1.0
7.2 × 10⁻⁴ = x²
x = √(7.2 × 10⁻⁴) = 0.0268M
∴ [H⁺] = 0.0268M
Buffer solutions resist changes in pH when small amounts of acid or base are added.
For Acetic Acid-Acetate Buffer:
CH₃COOH ⇌ CH₃COO⁻ + H⁺
CH₃COONa → CH₃COO⁻ + Na⁺
When acid (H⁺) is added:
CH₃COO⁻ + H⁺ → CH₃COOH (consumes added H⁺)
When base (OH⁻) is added:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O (consumes added OH⁻)
Calculate pH of acetic acid-sodium acetate buffer with 1.0M each component (Ka = 1.8 × 10⁻⁵)
Solution:
Ka = [CH₃COO⁻][H⁺] / [CH₃COOH]
1.8 × 10⁻⁵ = (1.0)[H⁺] / (1.0)
[H⁺] = 1.8 × 10⁻⁵ M
pH = -log(1.8 × 10⁻⁵) = 4.745
For slightly soluble ionic compounds in equilibrium with their ions:
General Equation: AmBn(s) ⇌ mA⁺(aq) + nB⁻(aq)
Solubility Product: Ksp = [A⁺]m[B⁻]n
CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)
Ksp = [Ca²⁺][F⁻]²
Calculate concentration of AgCl in aqueous solution (Ksp = 1.8 × 10⁻¹⁰)
Solution:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Let solubility = x mol/dm³
Ksp = [Ag⁺][Cl⁻] = x × x = x²
1.8 × 10⁻¹⁰ = x²
x = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/dm³
The suppression of ionization or solubility by adding a common ion.
HF(aq) ⇌ H⁺(aq) + F⁻(aq) (weak electrolyte, slight dissociation)
NaF(s) → Na⁺(aq) + F⁻(aq) (strong electrolyte, complete dissociation)
Added F⁻ from NaF shifts equilibrium left, decreasing HF dissociation
NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq)
HCl(aq) → H⁺(aq) + Cl⁻(aq)
Added Cl⁻ from HCl shifts equilibrium left, causing NaCl precipitation
The partition coefficient (KPC) describes how a solute distributes between two immiscible solvents.
Definition: KPC = [X]solvent2 / [X]solvent1
Where [X] is the concentration of solute X in each solvent at equilibrium
100 cm³ of 0.150M aqueous methylamine shaken with 75 cm³ organic solvent
After equilibrium, 25 cm³ aqueous layer titrated with 0.225M HCl: used 7.05 cm³
Solution:
Total moles CH₃NH₂ = 0.100 × 0.150 = 0.015 mol
Moles in 25 cm³ aqueous layer = 0.225 × (7.05/1000) = 0.001586 mol
Moles in 100 cm³ aqueous layer = 0.001586 × (100/25) = 0.00634 mol
Moles in organic layer = 0.015 – 0.00634 = 0.00866 mol
[CH₃NH₂]aq = 0.00634 / 0.100 = 0.0634 M
[CH₃NH₂]org = 0.00866 / 0.075 = 0.115 M
KPC = 0.115 / 0.0634 = 1.81
Strong Acids: “So I Brought No Clean Clothes” – H₂SO₄, HI, HBr, HNO₃, HCl, HClO₄
Strong Bases: Group 1 hydroxides (LiOH, NaOH, KOH, etc.) and heavy Group 2 hydroxides (Ca(OH)₂, Sr(OH)₂, Ba(OH)₂)
For conjugate acid-base pairs: Ka × Kb = Kw = 1.0 × 10⁻¹⁴
pKa + pKb = 14 (at 25°C)
Stronger acid = Weaker conjugate base
“Weak acid with its salt, or weak base with its salt”
Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA])
When [A⁻] = [HA], pH = pKa
Most salts of Na⁺, K⁺, NH₄⁺ are soluble
Most nitrates, acetates, chlorates are soluble
Most chlorides are soluble except AgCl, PbCl₂, Hg₂Cl₂
Most sulfates are soluble except BaSO₄, PbSO₄, CaSO₄
Imagine equilibrium as a seesaw – adding weight (common ion) to one side shifts the balance
More common ions = Less dissociation/solubility
“Like dissolves like” – Polar solutes prefer polar solvents
KPC > 1: Solute prefers solvent 2
KPC < 1: Solute prefers solvent 1
i. Which phenomenon describes a shift in equilibrium due to the addition of an ion already involved in the equilibrium?
Answer: b) Common Ion Effect
Explanation: The common ion effect specifically refers to the suppression of ionization or solubility when a common ion is added to an equilibrium system.
ii. What property is used to distinguish between strong and weak acids?
Answer: c) Extent of ionization
Explanation: Strong acids ionize completely in water, while weak acids only partially ionize. The extent of ionization is quantified by the acid dissociation constant Ka.
iii. Which factor is considered when distinguishing between strong and weak bases?
Answer: c) Extent of ionization
Explanation: Similar to acids, strong bases ionize completely in water, while weak bases only partially ionize. The extent of ionization is quantified by the base dissociation constant Kb.
iv. What defines a buffer solution?
Answer: b) Presence of a weak acid and its conjugate base (or a weak base and its conjugate acid)
Explanation: Buffer solutions contain a weak acid-base conjugate pair that can neutralize added acids or bases, maintaining relatively constant pH.
v. How can a buffer solution be made?
Answer: b) Mixing a weak acid and its conjugate base (or a weak base and its conjugate acid)
Explanation: Buffer solutions are specifically prepared by combining a weak acid with its salt (conjugate base) or a weak base with its salt (conjugate acid).
vi. What role does HCO₃⁻ play in controlling pH in blood?
Answer: b) Buffering against changes in pH
Explanation: HCO₃⁻ is part of the bicarbonate buffer system in blood, which maintains pH within the narrow range of 7.35-7.45 by neutralizing both acids and bases.
vii. How is the concentration of a slightly soluble salt calculated?
Answer: a) Using the solubility product constant (Ksp)
Explanation: The solubility of slightly soluble salts is calculated from their Ksp values using equilibrium expressions.
viii. Which term is used to describe the strength of an acid in terms of its ionization in water?
Answer: b) Acid Dissociation Constant (Ka)
Explanation: Ka quantitatively measures acid strength by representing the equilibrium constant for acid dissociation in water.
ix. Which type of solvent would favour the partitioning of a polar solute?
Answer: c) Polar solvent
Explanation: The principle “like dissolves like” means polar solutes have higher affinity for polar solvents, resulting in higher partition coefficients in polar solvents.
x. What is the partition coefficient defined as?
Answer: a) Ratio of solute concentration in one solvent to the other
Explanation: The partition coefficient KPC is defined as the ratio of solute concentrations in two immiscible solvents at equilibrium.
i. Explain the common ion effect with a suitable example.
Answer: The common ion effect is the suppression of ionization or solubility of an electrolyte when a highly soluble electrolyte containing a common ion is added to the solution.
Example: In a solution of HF (weak acid), adding NaF (strong electrolyte) provides additional F⁻ ions. According to Le Chatelier’s principle, the equilibrium HF ⇌ H⁺ + F⁻ shifts left, decreasing HF dissociation.
HF(aq) ⇌ H⁺(aq) + F⁻(aq)
NaF(s) → Na⁺(aq) + F⁻(aq)
The added F⁻ from NaF suppresses HF dissociation.
ii. Differentiate between strong and weak acids using the extent of ionization and Ka.
Answer:
Strong Acids:
Weak Acids:
iii. Differentiate between strong and weak bases using the extent of ionization and Kb.
Answer:
Strong Bases:
Weak Bases:
iv. Define a buffer solution and provide an example of how it can be made.
Answer: A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added.
Example preparation: Acetic acid-acetate buffer can be made by mixing:
This creates the conjugate pair: CH₃COOH/CH₃COO⁻ that can neutralize added H⁺ or OH⁻ ions.
v. How does a buffer solution control pH? Include chemical equations in your explanation.
Answer: Buffer solutions control pH through the common ion effect and Le Chatelier’s principle.
For acetic acid-acetate buffer:
CH₃COOH ⇌ CH₃COO⁻ + H⁺
CH₃COONa → CH₃COO⁻ + Na⁺
When acid (H⁺) is added:
CH₃COO⁻ + H⁺ → CH₃COOH (consumes added H⁺)
When base (OH⁻) is added:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O (consumes added OH⁻)
These reactions minimize pH changes by converting strong acids/bases into weak acids/bases.
i. Discuss the principles behind the common ion effect, providing two examples to illustrate its application.
Answer: The common ion effect is based on Le Chatelier’s principle, which states that when a system at equilibrium is disturbed, it will adjust to minimize that disturbance.
Principles:
Example 1: Suppression of Weak Acid Dissociation
In HF solution: HF ⇌ H⁺ + F⁻ (Ka = 7.2 × 10⁻⁴)
When NaF is added: NaF → Na⁺ + F⁻
The added F⁻ ions shift the equilibrium left, decreasing HF dissociation and [H⁺].
Example 2: Decreased Solubility
In saturated NaCl solution: NaCl(s) ⇌ Na⁺ + Cl⁻
When HCl is added: HCl → H⁺ + Cl⁻
The added Cl⁻ ions shift the equilibrium left, causing NaCl to precipitate.
Applications:
ii. Compare and contrast strong and weak acids based on their extent of ionization and the value of Ka.
Answer:
Comparison of Strong and Weak Acids:
Extent of Ionization:
Acid Dissociation Constant (Ka):
pKa Values:
pH of Solutions:
Conductivity:
Reaction Rates:
Examples:
Both strong and weak acids follow the same general dissociation equilibrium: HA ⇌ H⁺ + A⁻, but the position of this equilibrium differs significantly between strong and weak acids.