Titration: Gradual addition of solution of known concentration (titrant) to solution being analyzed (analyte)
Purpose: Determine concentration of unknown
Equivalence point: Moles acid = moles base (stoichiometric point)
Endpoint: Where indicator changes color (ideally = equivalence point)
Types of Titrations
Four main types:
Strong acid + strong base
Weak acid + strong base
Strong acid + weak base
Weak acid + weak base (not common)
Each has distinct titration curve!
Strong Acid - Strong Base Titration
Example: HCl + NaOH
Titration curve features:
📚 Practice Problems
1Problem 1easy
❓ Question:
Calculate the pH when 25.0 mL of 0.100 M HCl is titrated with: (a) 0.0 mL NaOH, (b) 12.5 mL of 0.100 M NaOH, (c) 25.0 mL of 0.100 M NaOH, (d) 30.0 mL of 0.100 M NaOH.
💡 Show Solution
Given:
25.0 mL of 0.100 M HCl
Titrant: 0.100 M NaOH
Initial moles H⁺: 0.0250 L × 0.100 M = 0.00250 mol
(a) 0.0 mL NaOH (before titration)
Pure HCl solution:
[H⁺] = 0.100 M
pH = -log(0.100) = 1.00
pH = 1.00
Explain using:
📋 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%
✅
📊 Scoring: 1-5
5
Extremely Qualified
~12%
4
Well Qualified
~16%
3
Qualified
~24%
2
Possibly Qualified
~24%
1
No Recommendation
~24%
💡 Key Test-Day Tips
✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws
⚠️ Common Mistakes: Acid-Base Titrations and Indicators
Avoid these 3 frequent errors
🌍 Real-World Applications: Acid-Base Titrations and Indicators
See how this math is used in the real world
📝 Worked Example: Stoichiometry — Limiting Reagent
Problem:
2 mol of H2 reacts with 1 mol of O2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O)
Master titration curves, equivalence points, and choosing appropriate indicators for acid-base titrations.
How can I study Acid-Base Titrations and Indicators 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 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Acid-Base Titrations and Indicators study guide free?▾
Yes — all study notes, flashcards, and practice problems for Acid-Base Titrations and Indicators on Study Mondo are 100% free. No account is needed to access the content.
What course covers Acid-Base Titrations and Indicators?▾
Acid-Base Titrations and Indicators 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 Titrations and Indicators?
Region 1 - Before equivalence:
Excess acid present
pH low (< 7)
pH rises slowly
Equivalence point:
Moles H⁺ = moles OH⁻
Only Na⁺ and Cl⁻ (neutral)
pH = 7.00 (at 25°C)
Sharp pH jump (pH 3-11)
Region 3 - After equivalence:
Excess base present
pH high (> 7)
pH rises slowly
Buffer region: None!
Weak Acid - Strong Base Titration
Example: CH₃COOH + NaOH
Titration curve features:
Region 1 - Before adding base:
Weak acid only
pH = weak acid calculation
pH higher than strong acid
Region 2 - Buffer region:
HA + A⁻ present (buffer!)
pH changes slowly
Use Henderson-Hasselbalch
Half-equivalence point:
Moles base = ½ moles acid
[HA] = [A⁻]
pH = pK_a (important!)
Equivalence point:
All HA → A⁻
Only conjugate base present
pH > 7 (basic!)
pH jump narrower than strong/strong
Region 4 - Excess base:
pH controlled by excess OH⁻
Strong Acid - Weak Base Titration
Example: HCl + NH₃
Opposite of weak acid - strong base:
Before equivalence:
Buffer region (B + BH⁺)
pH < 7
Half-equivalence:
pH = pK_a of BH⁺ (or 14 - pK_b)
Equivalence point:
All B → BH⁺ (conjugate acid)
pH < 7 (acidic!)
After equivalence:
Excess strong acid
pH very low
Calculations During Titration
Before Equivalence (strong acid/base):
Excess reactant determines pH
Example: 25.0 mL 0.10 M HCl + 10.0 mL 0.10 M NaOH
Moles H⁺: 0.025 L × 0.10 M = 0.0025 mol
Moles OH⁻: 0.010 L × 0.10 M = 0.0010 mol
After reaction: 0.0025 - 0.0010 = 0.0015 mol H⁺ remains
Volume: 25.0 + 10.0 = 35.0 mL = 0.035 L
[H⁺]: 0.0015/0.035 = 0.043 M
pH: -log(0.043) = 1.37
At Equivalence (weak acid/strong base):
Only conjugate base present
Use K_b calculation or Henderson-Hasselbalch at equivalence
In Buffer Region (weak acid/strong base):
Use Henderson-Hasselbalch:
Moles unreacted HA and moles A⁻ formed
pH = pK_a + log(moles A⁻/moles HA)
Indicators
Indicator: Weak acid (HIn) with different colors for HIn and In⁻
HIn ⇌ H⁺ + In⁻
HIn: one color (acidic form)
In⁻: different color (basic form)
Color change: Occurs over ~2 pH units centered at pK_a(indicator)
A 25.0 mL sample of 0.150 M acetic acid (K_a = 1.8 × 10⁻⁵) is titrated with 0.150 M NaOH. Calculate the pH: (a) initially, (b) after adding 12.5 mL NaOH, (c) at equivalence point.
💡 Show Solution
Given:
25.0 mL of 0.150 M CH₃COOH
K_a = 1.8 × 10⁻⁵
Titrant: 0.150 M NaOH
Initial moles CH₃COOH: 0.0250 L × 0.150 M = 0.00375 mol
pK_a: -log(1.8 × 10⁻⁵) = 4.74
(a) Initial pH (before NaOH)
Weak acid calculation:
Ka=0.150−xx2
Approximation: x² = K_a × 0.150
x=(1.8×10−5)(0.150)
pH = -log(1.64 × 10⁻³) = 2.78
Answer (a): pH = 2.78
(b) After 12.5 mL NaOH (half-equivalence)
Moles OH⁻: 0.0125 L × 0.150 M = 0.001875 mol
Reaction: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
After reaction:
Moles CH₃COOH: 0.00375 - 0.001875 = 0.001875 mol
Moles CH₃COO⁻: 0.001875 mol
Equal moles HA and A⁻!
At half-equivalence: pH = pK_a
pH = 4.74
Answer (b): pH = 4.74
Note: This is a key point - always true for weak acid/strong base!
(c) At equivalence point
Moles NaOH needed: 0.00375 mol (same as acid)
Volume NaOH: 0.00375 mol ÷ 0.150 M = 0.0250 L = 25.0 mL
After reaction:
All CH₃COOH → CH₃COO⁻
Moles CH₃COO⁻: 0.00375 mol
Total volume: 25.0 + 25.0 = 50.0 mL
[CH₃COO⁻]:
[CH3COO−]=0.0500
This is weak base!
Find K_b:
Kb=K
Base equilibrium: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
Kb=0.0750x2
x=(5.6×10−10)(0.0750)
[OH⁻] = 6.48 × 10⁻⁶ M
pOH = -log(6.48 × 10⁻⁶) = 5.19
pH = 14.00 - 5.19 = 8.81
Answer (c): pH = 8.81
Key observation: Equivalence point pH > 7 for weak acid/strong base!
Summary:
Point
pH
Explanation
Initial
2.78
Weak acid
Half-equiv
4.74
pH = pK_a (buffer!)
Equivalence
8.81
Weak base (A⁻)
3Problem 3hard
❓ Question:
For the titration of 25.0 mL of 0.100 M NH₃ (K_b = 1.8 × 10⁻⁵) with 0.100 M HCl: (a) What is the pH at equivalence? (b) Which indicator is appropriate: phenolphthalein (8.3-10.0) or methyl orange (3.1-4.4)?
💡 Show Solution
Given:
25.0 mL of 0.100 M NH₃ (weak base)
K_b = 1.8 × 10⁻⁵
Titrant: 0.100 M HCl (strong acid)
This is strong acid + weak base titration!
(a) pH at equivalence
Initial moles NH₃: 0.0250 L × 0.100 M = 0.00250 mol
At equivalence:
All NH₃ → NH₄⁺ (conjugate acid)
Moles NH₄⁺: 0.00250 mol
Volume HCl added: 25.0 mL (same M and mol)
Total volume: 25.0 + 25.0 = 50.0 mL
[NH₄⁺]:
[NH4+]=0.05000.00250
NH₄⁺ is weak acid!
Find K_a:
Ka=K
Acid equilibrium: NH₄⁺ ⇌ NH₃ + H⁺
Ka=0.0500−xx
x2=(5.6×10−10)(0.0500)
x2=2.8×10−11
x=5.29×10−6
[H⁺] = 5.29 × 10⁻⁶ M
pH:
pH=−log(5.29×10−6)=5.28
Answer (a): pH = 5.28
Note: pH < 7 at equivalence (acidic!)
(b) Choose indicator
Equivalence point pH = 5.28
Phenolphthalein:
Range: 8.3-10.0 (basic range)
Changes color at pH 8-10
Too high! Won't change at pH 5.28 ✗
Methyl orange:
Range: 3.1-4.4 (acidic range)
Changes color at pH 3-4
Close to equivalence pH ✓
Answer (b): Methyl orange is appropriate
Better choice: Bromocresol green (3.8-5.4) if available
Reasoning:
For weak base + strong acid:
Equivalence point is acidic (pH < 7)
Need acidic-range indicator
Methyl orange changes in acidic region
Phenolphthalein wouldn't change until past equivalence
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.