Loading…
Understand Arrhenius, Brønsted-Lowry, and Lewis theories; master pH, pOH, and the pH scale.
Learn step-by-step with practice exercises built right in.
Simplest definition (aqueous solutions):
Acid: Produces H⁺ in water
Base: Produces OH⁻ in water
Limitation: Only for aqueous solutions
More general definition:
Acid: Proton (H⁺) donor Base: Proton (H⁺) acceptor
Example:
Identify the conjugate acid-base pairs in: HF(aq) + NH₃(aq) ⇌ F⁻(aq) + NH₄⁺(aq)
Reaction: HF(aq) + NH₃(aq) ⇌ F⁻(aq) + NH₄⁺(aq)
Identify acids and bases:
Left side (reactants):
Right side (products):
| Section | Format | Questions | Time | Weight | Calculator |
|---|---|---|---|---|---|
| Multiple Choice | MCQ | 60 | 90 min | 50% | ✅ |
| Free Response (Long) | FRQ | 3 | 69 min | 30% | ✅ |
| Free Response (Short) | FRQ | 4 | 36 min | 20% | ✅ |
Avoid these 3 frequent errors
See how this math is used in the real world
mol of reacts with mol of . How many grams of water are produced? Which is the limiting reagent? ()
Review key concepts with our flashcard system
Explore more AP Chemistry topics
Conjugate acid-base pairs:
Example: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Most general:
Acid: Electron pair acceptor Base: Electron pair donor
Example: BF₃ + NH₃ → F₃B-NH₃
All Brønsted acids/bases are Lewis, but not vice versa
Water self-ionizes:
Simplified: H₂O ⇌ H⁺ + OH⁻
Ion product constant (K_w):
Key relationships:
pH definition:
pOH definition:
Relationship:
pH Scale interpretation:
| pH | [H⁺] (M) | Type |
|---|---|---|
| 0 | 1 | Very acidic |
| 1 | 0.1 | Strong acid |
| 7 | 1.0 × 10⁻⁷ | Neutral |
| 13 | 1.0 × 10⁻¹³ | Strong base |
| 14 | 1.0 × 10⁻¹⁴ | Very basic |
Ranges:
From [H⁺]:
Example: [H⁺] = 1.0 × 10⁻³ M
pH = -log(1.0 × 10⁻³) = 3.00
From [OH⁻]:
Example: [OH⁻] = 1.0 × 10⁻⁴ M
pOH = -log(1.0 × 10⁻⁴) = 4.00 pH = 14.00 - 4.00 = 10.00
Inverse relationship:
Example: pH = 5.00
[H⁺] = 10⁻⁵·⁰⁰ = 1.0 × 10⁻⁵ M
Strong acids (completely ionize):
For strong monoprotic acid:
Example: 0.010 M HCl
Strong bases (completely dissociate):
For strong base:
Example: 0.010 M NaOH
Example: 0.010 M Ca(OH)₂
pH has decimal places = sig figs in [H⁺]
Example: [H⁺] = 2.5 × 10⁻³ M (2 sig figs)
The digits before decimal point come from exponent
Conjugate pairs differ by one H⁺:
Pair 1: HF and F⁻
Pair 2: NH₃ and NH₄⁺
Summary:
| Species | Role | Conjugate |
|---|---|---|
| HF | Acid | F⁻ (conjugate base) |
| NH₃ | Base | NH₄⁺ (conjugate acid) |
| F⁻ | Conjugate base | of HF |
| NH₄⁺ | Conjugate acid | of NH₃ |
Answers:
Pattern: Conjugate pairs always differ by exactly one proton (H⁺)
Calculate the pH and pOH of a solution with [OH⁻] = 3.5 × 10⁻⁴ M at 25°C.
Given:
Calculate pOH:
Using calculator:
Round to 2 decimal places (2 sig figs in 3.5):
pOH = 3.46
Calculate pH:
Use relationship:
Verify using [H⁺]:
From K_w:
pH:
✓
Answers:
Interpretation: pH > 7, solution is basic
A solution of Ba(OH)₂ has pH = 12.60 at 25°C. Calculate the concentration of Ba(OH)₂.
Given:
Note: Ba(OH)₂ → Ba²⁺ + 2OH⁻ (2 OH⁻ per formula unit)
Calculate pOH:
Calculate [OH⁻]:
Calculate [Ba(OH)₂]:
Stoichiometry: Ba(OH)₂ → Ba²⁺ + 2OH⁻
Each Ba(OH)₂ produces 2 OH⁻:
Answer: [Ba(OH)₂] = 0.020 M or 2.0 × 10⁻² M
Verify:
If [Ba(OH)₂] = 0.020 M:
Key point: Remember to account for stoichiometry!
Ba(OH)₂ produces 2 moles OH⁻ per mole compound.