Acid-Base Theories and pH Scale - Complete Interactive Lesson
Part 1: Arrhenius & Brønsted-Lowry
🧪 Arrhenius Acids and Bases
Part 1 of 7 — The First Modern Definition
Three Acid-Base Theories — Where We're Headed
Theory
Acid Is...
Base Is...
This Part
Arrhenius
Produces H+ in water
Produces OH− in water
✅ Part 1
Brønsted-Lowry
Proton donor
Proton acceptor
Part 2
Lewis
Electron pair acceptor
Electron pair donor
Part 3
🔑 Why this matters: The Arrhenius model is the foundation — every acid-base theory that follows builds on these ideas.
What You'll Master in Part 1
Defining Arrhenius acids and bases by what they produce in water
Identifying limitations of the Arrhenius model
Recognizing strong acids and the hydronium ion concept
📖 The Arrhenius Definition
In 1884, Svante Arrhenius proposed a simple classification:
Type
Definition
Example
Arrhenius Acid
Produces H+ ions in aqueous solution
HCl(aq
🧪 Common Arrhenius Acids
Strong Acids (Complete Dissociation)
Formula
Name
Dissociation
HCl
Hydrochloric acid
HCl→H++
⚛️ The Hydronium Ion
In reality, free H+ ions (bare protons) don't exist in water. Instead, they bond to water molecules:
H+(aq)+
Arrhenius Concept Check 🎯
📌 Arrhenius Neutralization
When an Arrhenius acid reacts with an Arrhenius base, they undergo neutralization:
Acid+Base→Salt+Water
Examples
Arrhenius Classification 🔍
Exit Quiz — Arrhenius Acids & Bases ✅
Part 2: Conjugate Acid-Base Pairs
🔄 Brønsted-Lowry Acids and Bases
Part 2 of 7 — Proton Donors and Acceptors
Arrhenius vs. Brønsted-Lowry
Feature
Arrhenius
Brønsted-Lowry
Acid definition
Produces H+ in water
Donates a proton (H+)
Base definition
Produces in water
Part 3: The pH Scale
🔬 Lewis Acids and Bases
Part 3 of 7 — Electron Pair Donors and Acceptors
How the Three Theories Compare
Theory
Key Question
Broadest?
Arrhenius
Does it produce H+ or OH−?
Narrowest
Brønsted-Lowry
Part 4: Strong Acids & Bases
📊 The pH Scale
Part 4 of 7 — Measuring Acidity and Basicity
The pH Scale at a Glance
pH
[H+] (M)
Character
Example
0
100
Strongly acidic
Part 5: Calculating pH & pOH
💪 Strong Acids and Bases — pH Calculations
Part 5 of 7 — Complete Dissociation Means Easy Math
Strong = Complete Dissociation
Type
Example
Key Calculation
Strong monoprotic acid
0.025 M HCl
[H+]=0.025 M, pH = 1.60
Strong diprotic acid
0.010 M
Part 6: Problem-Solving Workshop
🛠️ Problem-Solving Workshop
Part 6 of 7 — Acid-Base Theories and pH
Problem Types You'll Practice
Problem Type
Skills Combined
Theory identification
Arrhenius vs. Brønsted-Lowry vs. Lewis
Multi-step pH
Dilution → dissociation → −log
Conceptual reasoning
Very dilute acid pH limits
Conjugate pair analysis
Identifying donors/acceptors
🔑 Why this matters: AP Chemistry free-response questions often combine acid-base theory with pH calculations — exactly the type of multi-step problems in this workshop.
What You'll Master in Part 6
Solving multi-step pH problems with dilution
Identifying acid-base behavior across all three theories
Reasoning about edge cases like very dilute strong acids
🔑 A conjugate pair always differs by exactly one proton (H+).
Strength Relationship
⚠️ Strong acid → very weak conjugate base (and vice versa)
HCl is strong → Cl− is a negligible base (does not accept protons)
CH3COO is weak → is a moderate conjugate base
Brønsted-Lowry Concept Check 🎯
⚗️ Identifying Conjugate Pairs in Reactions
For any Brønsted-Lowry reaction, there are always two conjugate pairs:
acid1HF+base2Hconj. base1Fconj. acid2H
Pair 1:HF/F−
Pair 2:H2O/H3O+
Steps to Identify
Find the species that lost a proton → that's the acid; its product is the conjugate base
Find the species that gained a proton → that's the base; its product is the conjugate acid
Each acid is paired with its conjugate base (they differ by one H+)
Conjugate Pair Identification 🔍
For the reaction: NH3+H2O⇌NH4++OH−
Conjugate Pair Practice 🧮
Identify the conjugate partners:
1) What is the conjugate base of H2CO3? (Enter the chemical formula, e.g. Cl-)
2) What is the conjugate acid of PO43−? (Enter the chemical formula, e.g. H2SO4)
3) What is the conjugate base of H2O? (Enter the chemical formula, e.g. F-)
Exit Quiz — Brønsted-Lowry Theory ✅
Does it donate or accept H+?
Middle
Lewis
Does it donate or accept electron pairs?
Broadest
The Lewis model captures reactions that have nothing to do with protons!
🔑 Why this matters: Lewis acid-base theory explains coordination chemistry, organic reactions, and metal complex formation — all tested on the AP exam.
What You'll Master in Part 3
Defining Lewis acids (electron pair acceptors) and bases (electron pair donors)
Identifying Lewis acids: metal cations, incomplete octets, H+
Comparing all three acid-base theories on the AP exam
📖 The Lewis Definition
Type
Definition
Key Feature
Lewis Acid
Electron pair acceptor
Has an empty orbital or can make room for electrons
Lewis Base
Electron pair donor
Has a lone pair of electrons to share
Comparison of All Three Theories
Theory
Acid
Base
Arrhenius
Produces H+ in water
Produces OH− in water
Brønsted-Lowry
Proton donor
Proton acceptor
Lewis
Electron pair acceptor
Electron pair donor
Key Insight
Every Arrhenius acid is a Brønsted-Lowry acid, and every Brønsted-Lowry acid involves a Lewis acid interaction. The Lewis definition is the most inclusive.
Arrhenius⊂Brønsted-Lowry⊂Lewis
🔑 If you can’t explain a reaction with Arrhenius or Brønsted-Lowry, try Lewis — it covers everything.
🧪 Common Lewis Acids
1. Metal Cations
Metal ions have empty orbitals and accept electron pairs from ligands:
Cu2++4NH3→[Cu(NH3)4]2+
Cu2+: Lewis acid (accepts electron pairs)
NH3: Lewis base (donates lone pair)
2. Molecules with Incomplete Octets
BF3+NH3→F
BF3: Lewis acid (boron has only 6 electrons, empty p orbital)
NH3: Lewis base (nitrogen has a lone pair)
💡 Molecules with incomplete octets (like BF3 and AlCl3) are classic Lewis acids.
3. Protons (H+)
The proton itself is a Lewis acid — it accepts an electron pair:
H++OH−→H2O
This shows how the Lewis definition encompasses the Brønsted-Lowry definition.
✏️ Common Lewis Bases
🔑 Any species with a lone pair can be a Lewis base:
NH3, H2O, OH, ,
Lewis Acid-Base Concept Check 🎯
🔗 Coordinate Covalent Bonds
When a Lewis base donates an electron pair to a Lewis acid, the resulting bond is called a coordinate covalent bond (or dative bond).
F3B+:NH3→F3B←NH3
The arrow ← shows that both electrons in the bond came from the nitrogen of NH3.
💡 A coordinate covalent bond (dative bond) is formed whenever a Lewis base donates a lone pair to a Lewis acid.
In Coordination Chemistry
Metal ions form coordination compounds with Lewis bases (called ligands):
Fe3++6CN−→[Fe(CN)
Lewis Acid
Lewis Base (Ligand)
Product
Fe3+
CN−
Lewis Acid-Base Classification 🔍
Theory Comparison 🧮
For each species, identify which acid-base theory can explain its behavior as an acid or base:
1)NaOH acting as a base — which is the simplest theory that explains this? (Enter: Arrhenius, Bronsted-Lowry, or Lewis)
2)NH3 acting as a base (no OH− in its formula) — simplest theory? (Enter: Arrhenius, Bronsted-Lowry, or Lewis)
3)BF3 acting as an acid (no H+ to donate) — simplest theory? (Enter: Arrhenius, Bronsted-Lowry, or Lewis)
Exit Quiz — Lewis Acids & Bases ✅
Battery acid
3
10−3
Acidic
Vinegar
7
10−7
Neutral
Pure water
11
10−11
Basic
Ammonia
14
10−14
Strongly basic
Drain cleaner
Each pH unit = a 10-fold change in [H+].
🔑 Why this matters: pH calculations are on virtually every AP Chemistry exam — mastering the logarithmic relationship between [H+] and pH is essential.
What You'll Master in Part 4
Converting between [H+], pH, [OH−], and pOH
Understanding the pH + pOH = 14 relationship
Interpreting what pH values mean for acidity and basicity
🔗 pH, pOH, and Their Relationship
pH Definition
pH=−log[H+]
pOH Definition
pOH=−log[OH−]
The Key Relationship
At 25°C:
pH+pOH=14
This comes from Kw:
[H+][OH−]=1.0×10−14
Taking −log of both sides:
−log[H+]+(−log[OH−])
pH+pOH=14
Interpreting pH
pH Range
Solution Type
[H+] vs [OH−]
🔢 pH Calculations
From [H+] to pH
Problem:[H+]=3.2×10−4 M. Find pH.
Solution:
pH=−log(3.2×10−4)=−(−3.49)=3.49
From pH to [H+]
Problem:pH=5.60. Find [H+].
Solution:
[H+]=10−pH=10
From [OH−] to pH
Problem:[OH−]=4.0×10−3 M. Find pH.
Solution:
Step 1: pOH=−log(4.0×10−3)=2.40
Step 2: pH=14−pOH=14−2.40=11.60
The "p" Notation
🔑 The prefix "p" always means −log:
pX=−logX
So pKa=−logKa, ,
pH Concept Check 🎯
📌 Significant Figures in pH
An important AP Chemistry rule:
🔑 The number of decimal places in the pH equals the number of significant figures in [H+].
Examples
[H+]
Sig Figs
pH
Decimal Places
1.0×10−4
2
4.00
2
2.5×10−6
⚠️ The digits before the decimal in pH only indicate the order of magnitude — they don't count as sig figs!
pH Calculation Drill 🧮
1) What is the pH of a solution with [H+]=5.0×10−9 M? (2 decimal places)
2) What is the [H+] in a solution with pH=4.30? (Enter in scientific notation, e.g. 2.5e-3)
3) What is the pH of a solution with [OH−]=2.0×10−4 M? (2 decimal places)
pH Scale Understanding 🔍
Exit Quiz — pH Scale ✅
H2SO4
[H+]≈0.020 M
Strong base (Group 1)
0.010 M NaOH
[OH−]=0.010 M, pOH = 2.00
Strong base (Group 2)
0.005 M Ba(OH)2
[OH−]=0.010 M
🔑 Why this matters: Strong acid/base pH problems are the foundation for all later calculations — titrations, buffers, and equilibrium all build from here.
What You'll Master in Part 5
Calculating pH of strong monoprotic and diprotic acids
Calculating pOH and pH of strong bases (Group 1 and Group 2)
Handling dilution before calculating pH
🧪 pH of Strong Acids
For a strong acid HA at concentration C:
HA→H++A−
Since dissociation is 100% complete: [H+]=C
pH=−logC
Example 1
Problem: What is the pH of 0.025 M HCl?
Solution:
[H+]=0.025 MpH=−log(0.025)=1.60
Example 2
Problem: What is the pH of 0.0040 M HNO3?
Solution:
[H+]=0.0040 MpH=−log(0.0040)=2.40
Diprotic Strong Acid (H2SO4)
For the first dissociation (strong): H2SO4→H++
For dilute solutions, each mole of H2SO4 produces approximately 2 moles of H+:
[H+]≈2C (for dilute solutions)
⚠️ The second dissociation of H2SO4 (HSO) is weak (), so at higher concentrations the approximation may not hold exactly.
📌 pH of Strong Bases
For a strong base like NaOH at concentration C:
NaOH→Na++OH−
[OH−]=C, then:
pOH=−logC⇒pH=14−pOH
Example 1
Problem: What is the pH of 0.010 M NaOH?
Solution:
[OH−]=0.010 MpOH=−log(0.010)=
Group 2 Hydroxides
For Ba(OH)2 or Ca(OH)2:
Ba(OH)2→Ba2++2OH
[OH−]=2C
Example 2
Problem: What is the pH of 0.0050 M Ba(OH)2?
Solution:
[OH−]=2(0.0050)=0.010 MpOH=
Strong Acid/Base pH Check 🎯
📌 Mixing and Dilution
Diluting a Strong Acid
🔑 Use M1V1=M2V2 for dilution problems:
Example: 25.0 mL of 0.10 M HCl is diluted to 100.0 mL. What is the new pH?
M2=V2
pH=−log(0.025)=1.60
Mixing Strong Acid and Strong Base
Example: 50.0 mL of 0.10 M HCl + 30.0 mL of 0.10 M NaOH
Moles H+ = 0.050×0.10=0.0050 mol
Moles OH− = 0.030×0.10=0.0030 mol
Excess H+ = 0.0050−0.0030=0.0020 mol
Total volume = 80.0 mL = 0.0800 L
[H+]=0.08000.0020=0.025 M
pH=−log(0.025)=1.60
Strong Acid/Base Calculation Drill 🧮
1) What is the pH of 0.0020 M HClO4? (2 decimal places)
2) What is the pH of 0.050 M Ba(OH)2? (2 decimal places)
3) 40.0 mL of 0.15 M HNO3 is mixed with 20.0 mL of 0.15 M NaOH. What is the pH? (2 decimal places)
Strong Acid/Base Reasoning 🔍
Exit Quiz — Strong Acid/Base pH ✅
Reaction A:
HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)
Reaction B:BF3+F−→BF4−
Reaction C:NaOH(aq)→Na+(aq)+OH−(aq)
For each reaction, the acid-base theory required is:
Reaction A: Brønsted-Lowry (proton transfer from HF to H2O)
Reaction B: Lewis (BF3 accepts electron pair from F−)
Reaction C: Arrhenius (NaOH produces OH− in water)
🔑 When classifying, always use the simplest theory that explains the observation.
Problem 1 Practice 🎯
Reaction A:HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)
Reaction B:BF3+F−→BF4
🔢 Problem 2: Multi-Step pH Calculation
A chemist prepares the following solutions:
Solution A: 0.035 M HCl
Solution B: 0.035 M NaOH
Solution C: 50.0 mL of Solution A mixed with 30.0 mL of Solution B
Solution A pH
[H+]=0.035 M → pH=−log(0.035)=1.46
Solution B pH
[OH−]=0.035 M → pOH=−log(0.035)= →
Solution C pH
Moles H+ = (0.050)(0.035)=1.75×10−3 mol
Moles OH− = (0.030)(0.035)=1.05×10−3 mol
Excess H+ = 1.75×10−3−1.05× mol
[H+]=0.0807.0×10 M
pH=−log(8.75×10−3)=2.06
Problem 2 Practice 🧮
A student mixes 25.0 mL of 0.080 M HNO3 with 15.0 mL of 0.080 M KOH.
1) How many moles of excess H+ remain? (Enter in scientific notation, e.g. 3.5e-3)
2) What is the total volume in liters? (3 decimal places)
3) What is the pH of the resulting solution? (2 decimal places)
📌 Problem 3: Conceptual Reasoning
The pH of Very Dilute Strong Acids
When a strong acid is extremely dilute (e.g., 10−8 M HCl), you cannot simply say pH=8.
⚠️ An acid solution can never have pH>7!
The autoionization of water contributes [H+]=10−7 M, which is much larger than the acid's contribution.
Correct approach:
[H+]
pH=−log(1.1×10−7)=6.96
💡 This is slightly below 7, as expected for an acidic solution.
Conceptual Check 🎯
Workshop Synthesis 🔍
[
H+
]
Conjugate pairs
Acid → conjugate base + H+
Kw
[H+][OH−]=1.0×10−14 at 25°C
🔑 Why this matters: This review mirrors the AP exam format — expect questions that require you to connect theory, calculations, and conceptual reasoning in a single problem.
What You'll Master in Part 7
Tackling AP-style multiple choice across all acid-base topics
Writing free-response explanations using proper chemistry terminology
Identifying common AP traps and avoiding them
📋 Complete Summary
Three Acid-Base Theories
Theory
Acid
Base
Scope
Arrhenius
Produces H+
Produces OH−
Aqueous only
Brønsted-Lowry
Proton donor
Proton acceptor
Any solvent
Lewis
e⁻ pair acceptor
e⁻ pair donor
Broadest
Key Equations
pH=−log[H+]pOH=−log
pH+pOH=14Kw
[H+]=10−pH[OH
Strong Acid/Base Rules
Strong acids: HCl,HBr,HI,HNO3,H
🔑 Strong = complete dissociation. No Ka or Kb needed — just use the concentration directly.
AP-Style Questions — Set 1 🎯
AP Calculation Practice 🧮
1) What is the pH of a solution made by mixing 100.0 mL of 0.15 M HCl with 75.0 mL of 0.15 M NaOH? (2 decimal places)
2) A solution has a pH of 11.50. What is [H+]? (Enter in scientific notation, e.g. 4.7e-9)
3) What volume (mL) of 0.20 M NaOH is needed to exactly neutralize 50.0 mL of 0.10 M H2SO4? (Enter as whole number)
AP-Style Questions — Set 2 🎯
Comprehensive Review 🔍
Final Exit Quiz — Acid-Base Theories & pH ✅
−
2
→
Ca2++
2OH−
2
→
Ba2++
2OH−
3
O+
(
a
q
)
+
Cl−(aq)
(
l
)
⇌
NH4+(aq)+
OH−(aq)
+
+
H2O
OH−
Differs by one H+
NH4+
NH3
Differs by one H+
H2SO4
HSO4−
Differs by one H+
HSO4−
SO42−
Differs by one H+
H
CH3COO−
2
O
⇌
−
+
3
O+
3
B
-
N
H3
−
F−
CN−
Molecules with lone pairs on N, O, S, or halide ions
6
]3−
[Fe(CN)6]3−
Ag+
NH3
[Ag(NH3)2]+
Cu2+
H2O
[Cu(H2O)6]2+
=
−log(1.0×
10−14)
pH<
7
Acidic
[H+]>[OH−]
pH=7
Neutral
[H+]=[OH−]
pH>7
Basic
[H+]<[OH−]
−5.60
=
2.5×
10−6 M
p
Kb
=
−logKb
pKw=−logKw=14
2
5.60
2
3.45×10−8
3
7.462
3
H
S
O4−
4
−
Ka=0.012
[H+]=2C
2.00
pH=14−2.00=12.00
−
−log(0.010)=
2.00
pH=14−2.00=12.00
M1V1
=
100.0(0.10)(25.0)=
0.025 M
−
1.46
pH=14−1.46=12.54
10−3=
7.0×
10−4
−4
=
8.75×
10−3
total
=
[
H+
]acid
+
[
H+
]water
≈
1
0−8
+
1
0−7
=
1.1
×
1
0−7
M
[
O
H−
]
=
[
H+
]
[
O
H−
]
=
1.0
×
1
0−14
−
]
=
10−pOH
2
S
O4
,
H
Cl
O4
Strong bases: Group 1 hydroxides + Ca(OH)2,Sr(OH)2,Ba(OH)2
[H+]=Cacid for monoprotic strong acids
[OH−]=nCbase where n = number of OH− per formula unit