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Acid-Base Theories and pH Scale | Study Mondo
Topics / Acids and Bases / Acid-Base Theories and pH Scale Acid-Base Theories and pH Scale Understand Arrhenius, Brønsted-Lowry, and Lewis theories; master pH, pOH, and the pH scale.
BC Written and reviewed by Brendan Cusack , Study Mondo Education Team • Last updated April 7, 2026
🎯 ⭐ INTERACTIVE LESSON
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Start Interactive Lesson → Acid-Base Theories and pH Scale
Arrhenius Theory
💡 Study Tips✓ Work through examples step-by-step ✓ Practice with flashcards daily ✓ Review common mistakes Simplest definition (aqueous solutions):
Acid: Produces H⁺ in water
Base: Produces OH⁻ in water
Limitation: Only for aqueous solutions
Brønsted-Lowry Theory Acid: Proton (H⁺) donor
Base: Proton (H⁺) acceptor
H C l + H 2 O → H 3 O + + C l − HCl + H2O \rightarrow H3O^+ + Cl^- H Cl + H 2 O → H 3 O + + C l −
HCl: acid (donates H⁺)
H₂O: base (accepts H⁺)
Conjugate acid-base pairs:
H A + B ⇌ A − + H B + HA + B \rightleftharpoons A^- + HB^+ H A + B ⇌ A − + H B +
HA/A⁻: conjugate pair (differ by H⁺)
B/HB⁺: conjugate pair
Example: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
NH₃/NH₄⁺: conjugate pair
H₂O/OH⁻: conjugate pair
Lewis Theory Acid: Electron pair acceptor
Base: Electron pair donor
Example: BF₃ + NH₃ → F₃B-NH₃
BF₃: Lewis acid (accepts electron pair)
NH₃: Lewis base (donates electron pair)
All Brønsted acids/bases are Lewis, but not vice versa
Autoionization of Water 2 H 2 O ⇌ H 3 O + + O H − 2H2O \rightleftharpoons H3O^+ + OH^- 2 H 2 O ⇌ H 3 O + + O H −
Simplified: H₂O ⇌ H⁺ + OH⁻
Ion product constant (K_w):
K w = [ H + ] [ O H − ] = 1.0 × 10 − 14 at 25°C K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ at 25°C} K w = [ H + ] [ O H − ] = 1.0 × 1 0 − 14 at 25°C
Pure water: [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M
Acidic: [H⁺] > [OH⁻]
Basic: [H⁺] < [OH⁻]
Always: [H⁺][OH⁻] = 1.0 × 10⁻¹⁴
pH Scale pH = − log [ H + ] \text{pH} = -\log[H^+] pH = − log [ H + ]
pOH = − log [ O H − ] \text{pOH} = -\log[OH^-] pOH = − log [ O H − ]
pH + pOH = 14.00 at 25°C \text{pH} + \text{pOH} = 14.00 \text{ at 25°C} pH + pOH = 14.00 at 25°C
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
pH < 7: Acidic
pH = 7: Neutral
pH > 7: Basic
Calculating pH pH = − log [ H + ] \text{pH} = -\log[H^+] pH = − log [ H + ]
Example: [H⁺] = 1.0 × 10⁻³ M
pH = -log(1.0 × 10⁻³) = 3.00
Calculate pOH = -log[OH⁻]
pH = 14.00 - pOH
Example: [OH⁻] = 1.0 × 10⁻⁴ M
pOH = -log(1.0 × 10⁻⁴) = 4.00
pH = 14.00 - 4.00 = 10.00
Calculating [H⁺] from pH [ H + ] = 10 − pH [H^+] = 10^{-\text{pH}} [ H + ] = 1 0 − pH
[H⁺] = 10⁻⁵·⁰⁰ = 1.0 × 10⁻⁵ M
Strong Acids and Bases Strong acids (completely ionize):
HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄
For strong monoprotic acid:
[H⁺] = 0.010 M
pH = -log(0.010) = 2.00
Strong bases (completely dissociate):
Group 1A hydroxides: LiOH, NaOH, KOH
Group 2A hydroxides: Ca(OH)₂, Ba(OH)₂, Sr(OH)₂
[OH⁻] = [base] × (number of OH⁻)
[OH⁻] = 0.010 M
pOH = 2.00
pH = 12.00
[OH⁻] = 2 × 0.010 = 0.020 M
pOH = -log(0.020) = 1.70
pH = 14.00 - 1.70 = 12.30
Significant Figures in pH pH has decimal places = sig figs in [H⁺]
Example: [H⁺] = 2.5 × 10⁻³ M (2 sig figs)
pH = 2.60 (2 decimal places)
The digits before decimal point come from exponent
📚 Practice Problems
1 Problem 1easy ❓ Question:Identify the conjugate acid-base pairs in: HF(aq) + NH₃(aq) ⇌ F⁻(aq) + NH₄⁺(aq)
💡 Show Solution Reaction: HF(aq) + NH₃(aq) ⇌ F⁻(aq) + NH₄⁺(aq)
Identify acids and bases:
Left side (reactants):
HF: Has H⁺ to donate → acid NH₃: Can accept H⁺ → base
Right side (products):
F⁻: Accepted H⁺ to become HF → base NH₄⁺: Donated H⁺ to become NH₃ → acid
Conjugate pairs differ by one H⁺:
Pair 1: HF and F⁻
HF → F⁻ + H⁺
HF is acid, F⁻ is its conjugate base
HF/F⁻ conjugate acid-base pair
Pair 2: NH₃ and NH₄⁺
NH₃ + H⁺ → NH₄⁺
NH₃ is base, NH₄⁺ is its conjugate acid
NH₃/NH₄⁺ conjugate acid-base pair
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:
Pair 1: HF/F⁻Pair 2: NH₃/NH₄⁺
Pattern: Conjugate pairs always differ by exactly one proton (H⁺)
2 Problem 2easy ❓ Question:(a) Calculate the pH of a solution with [H⁺] = 2.5 × 10⁻⁴ M. (b) Calculate the pOH. (c) Is this solution acidic, basic, or neutral?
💡 Show Solution Solution:
(a) Calculate pH:
pH = -log[H⁺]
pH = -log(2.5 × 10⁻⁴)
pH = 3.60
(b) Calculate pOH:
pH + pOH = 14.00
pOH = 14.00 - 3.60 = 10.40
(c) Acidic, basic, or neutral?
pH < 7, so the solution is acidic
3 Problem 3medium ❓ Question:Calculate the pH and pOH of a solution with [OH⁻] = 3.5 × 10⁻⁴ M at 25°C.
💡 Show Solution Given:
[OH⁻] = 3.5 × 10⁻⁴ M
T = 25°C (K_w = 1.0 × 10⁻¹⁴)
Calculate pOH:
pOH = − log [ O H − ] \text{pOH} = -\log[OH^-] pOH = −
4 Problem 4medium ❓ Question:Calculate the pH of a 0.025 M HCl solution. Assume HCl is a strong acid that completely dissociates.
💡 Show Solution Solution:
HCl dissociation:
HCl → H⁺ + Cl⁻
Since HCl is a strong acid, it completely dissociates:
[H⁺] = 0.025 M
Calculate pH:
pH = -log[H⁺]
pH = -log(0.025)
pH = -log(2.5 × 10⁻²)
pH = 1.60
Note: For strong acids (HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄), pH calculation is straightforward because [H⁺] = initial acid concentration.
5 Problem 5hard ❓ Question:A solution of Ba(OH)₂ has pH = 12.60 at 25°C. Calculate the concentration of Ba(OH)₂.
💡 Show Solution Given:
Compound: Ba(OH)₂ (strong base)
pH = 12.60
T = 25°C
Note: Ba(OH)₂ → Ba²⁺ + 2OH⁻ (2 OH⁻ per formula unit)
Calculate pOH:
pH + pOH = 14.00 \text{pH} + \text{pOH} = 14.00 pH +
Explain using: 📝 Simple words 🔗 Analogy 🎨 Visual desc. 📐 Example 💡 Explain
📋 AP Chemistry — Exam Format Guide⏱ 3 hours 15 minutes 📝 67 questions 📊 3 sections
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% ✅
💡 Key Test-Day Tips✓ Memorize common polyatomic ions✓ Practice dimensional analysis✓ Know your gas laws⚠️ Common Mistakes: Acid-Base Theories and pH ScaleAvoid these 3 frequent errors
1 Not balancing equations before doing stoichiometry
▾ 2 Confusing molarity (M) with molality (m)
▾ 3 Forgetting to convert temperature to Kelvin for gas law problems
▾ 🌍 Real-World Applications: Acid-Base Theories and pH ScaleSee how this math is used in the real world
🌍 Water Purification
Environment
▾ 💻 Battery Technology
Technology
▾
📝 Worked Example: Stoichiometry — Limiting ReagentProblem: 2 2 2 H 2 H_2 H 2 1 1 1 O 2 O_2 O 2 2 H 2 + O 2 → 2 H 2 O 2H_2 + O_2 \to 2H_2O 2 H 2 + O 2 → 2 H 2 O
1 Write the balanced equation Click to reveal →
2 Determine the limiting reagent
3 Calculate moles of product
🧪 Practice Lab Interactive practice problems for Acid-Base Theories and pH Scale
▾ 📌 Related Topics in Acids and Bases❓ Frequently Asked QuestionsWhat is Acid-Base Theories and pH Scale?▾ Understand Arrhenius, Brønsted-Lowry, and Lewis theories; master pH, pOH, and the pH scale.
How can I study Acid-Base Theories and pH Scale effectively?▾ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Acid-Base Theories and pH Scale study guide free?▾ Yes — all study notes, flashcards, and practice problems for Acid-Base Theories and pH Scale on Study Mondo are free to access. No account is needed.
What course covers Acid-Base Theories and pH Scale?▾ Acid-Base Theories and pH Scale is part of the AP Chemistry course on Study Mondo, specifically in the Acids and Bases section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Acid-Base Theories and pH Scale?▾ Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
log
[
O
H −
]
pOH = − log ( 3.5 × 10 − 4 ) \text{pOH} = -\log(3.5 \times 10^{-4}) pOH = − log ( 3.5 × 1 0 − 4 )
pOH = − ( log 3.5 + log 10 − 4 ) \text{pOH} = -(\log 3.5 + \log 10^{-4}) pOH = − ( log 3.5 + log 1 0 − 4 )
pOH = − ( 0.544 − 4 ) \text{pOH} = -(0.544 - 4) pOH = − ( 0.544 − 4 )
pOH = − ( − 3.456 ) \text{pOH} = -(-3.456) pOH = − ( − 3.456 )
pOH = 3.456 \text{pOH} = 3.456 pOH = 3.456
Round to 2 decimal places (2 sig figs in 3.5):
pH + pOH = 14.00 \text{pH} + \text{pOH} = 14.00 pH + pOH = 14.00
pH = 14.00 − pOH \text{pH} = 14.00 - \text{pOH} pH = 14.00 − pOH
pH = 14.00 − 3.46 \text{pH} = 14.00 - 3.46 pH = 14.00 − 3.46
pH = 10.54 \text{pH} = 10.54 pH = 10.54
[ H + ] = K w [ O H − ] = 1.0 × 10 − 14 3.5 × 10 − 4 [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}} [ H + ] = [ O H − ] K w = 3.5 × 1 0 − 4 1.0 × 1 0 − 14
[ H + ] = 2.86 × 10 − 11 M [H^+] = 2.86 \times 10^{-11} \text{ M} [ H + ] = 2.86 × 1 0 − 11 M
pH = − log ( 2.86 × 10 − 11 ) = 10.54 \text{pH} = -\log(2.86 \times 10^{-11}) = 10.54 pH = − log ( 2.86 × 1 0 − 11 ) = 10.54
Interpretation: pH > 7, solution is basic
pOH =
14.00
pOH = 14.00 − 12.60 = 1.40 \text{pOH} = 14.00 - 12.60 = 1.40 pOH = 14.00 − 12.60 = 1.40
[ O H − ] = 10 − pOH [OH^-] = 10^{-\text{pOH}} [ O H − ] = 1 0 − pOH
[ O H − ] = 10 − 1.40 [OH^-] = 10^{-1.40} [ O H − ] = 1 0 − 1.40
[ O H − ] = 3.98 × 10 − 2 M [OH^-] = 3.98 \times 10^{-2} \text{ M} [ O H − ] = 3.98 × 1 0 − 2 M
[ O H − ] = 0.0398 M [OH^-] = 0.0398 \text{ M} [ O H − ] = 0.0398 M
Stoichiometry: Ba(OH)₂ → Ba²⁺ + 2OH⁻
Each Ba(OH)₂ produces 2 OH⁻:
[ B a ( O H ) 2 ] = [ O H − ] 2 [Ba(OH)_2] = \frac{[OH^-]}{2} [ B a ( O H ) 2 ] = 2 [ O H − ]
[ B a ( O H ) 2 ] = 0.0398 2 [Ba(OH)_2] = \frac{0.0398}{2} [ B a ( O H ) 2 ] = 2 0.0398
[ B a ( O H ) 2 ] = 0.0199 M [Ba(OH)_2] = 0.0199 \text{ M} [ B a ( O H ) 2 ] = 0.0199 M
Answer: [Ba(OH)₂] = 0.020 M or 2.0 × 10⁻² M
[OH⁻] = 2 × 0.020 = 0.040 M
pOH = -log(0.040) = 1.40
pH = 14.00 - 1.40 = 12.60 ✓
Key point: Remember to account for stoichiometry!
Ba(OH)₂ produces 2 moles OH⁻ per mole compound.