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Acid-Base Titrations and Indicators

Master titration curves, equivalence points, and choosing appropriate indicators for acid-base titrations.

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Acid-Base Titrations and Indicators

What is a Titration?

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:

  1. Strong acid + strong base
  2. Weak acid + strong base
  3. Strong acid + weak base
  4. Weak acid + weak base (not common)

Each has distinct titration curve!

Strong Acid - Strong Base Titration

Example: HCl + NaOH

Titration curve features:

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)

Common indicators:

IndicatorpK_aColor changepH range
Methyl orange3.7Red → Yellow3.1-4.4
Bromothymol blue7.0Yellow → Blue6.0-7.6
Phenolphthalein9.4Colorless → Pink8.3-10.0

Choosing an Indicator

Match indicator range to equivalence point pH:

Strong acid + strong base:

  • pH at equivalence = 7
  • Any indicator with range 4-10 works
  • Phenolphthalein, bromothymol blue ✓

Weak acid + strong base:

  • pH at equivalence > 7 (8-10 typical)
  • Need indicator that changes in basic range
  • Phenolphthalein (8.3-10.0) ✓
  • Methyl orange ✗ (too acidic)

Strong acid + weak base:

  • pH at equivalence < 7 (4-6 typical)
  • Need indicator that changes in acidic range
  • Methyl orange (3.1-4.4) ✓
  • Phenolphthalein ✗ (too basic)

Polyprotic Acid Titrations

Multiple equivalence points:

Example: H₂SO₃ (diprotic)

First equivalence: H₂SO₃ → HSO₃⁻

  • pH depends on K_a1

Second equivalence: HSO₃⁻ → SO₃²⁻

  • pH depends on K_a2

Curve shows two distinct pH jumps

Titration Curve Sketching

Key points to plot:

  1. Initial pH (before titration)
  2. Buffer region (gradual rise)
  3. Half-equivalence (pH = pK_a for weak acid)
  4. Equivalence point (sharp rise)
  5. After equivalence (slow rise)

Shape depends on acid/base strength!

📚 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

Answer (a): pH = 1.00

(b) 12.5 mL NaOH (half-equivalence)

Moles OH⁻ added: 0.0125 L × 0.100 M = 0.00125 mol

Neutralization: H⁺ + OH⁻ → H₂O

After reaction:

  • Moles H⁺ remaining: 0.00250 - 0.00125 = 0.00125 mol
  • Moles OH⁻: 0 (limiting)

Total volume: 25.0 + 12.5 = 37.5 mL = 0.0375 L

[H⁺]:

[H+]=0.00125 mol0.0375 L=0.0333 M[H^+] = \frac{0.00125 \text{ mol}}{0.0375 \text{ L}} = 0.0333 \text{ M}

pH:

pH=−log⁡(0.0333)=1.48\text{pH} = -\log(0.0333) = 1.48

Answer (b): pH = 1.48

(c) 25.0 mL NaOH (equivalence point)

Moles OH⁻ added: 0.0250 L × 0.100 M = 0.00250 mol

Exactly neutralizes all H⁺!

After reaction:

  • Only Na⁺ and Cl⁻ (neutral ions)
  • Neither hydrolyzes

Strong acid + strong base → neutral salt

pH = 7.00

Answer (c): pH = 7.00

(d) 30.0 mL NaOH (past equivalence)

Moles OH⁻ added: 0.0300 L × 0.100 M = 0.00300 mol

After reaction:

  • Moles H⁺: 0 (all neutralized)
  • Moles OH⁻ excess: 0.00300 - 0.00250 = 0.00050 mol

Total volume: 25.0 + 30.0 = 55.0 mL = 0.0550 L

[OH⁻]:

[OH−]=0.00050 mol0.0550 L=0.00909 M[OH^-] = \frac{0.00050 \text{ mol}}{0.0550 \text{ L}} = 0.00909 \text{ M}

pOH:

pOH=−log⁡(0.00909)=2.04\text{pOH} = -\log(0.00909) = 2.04

pH:

pH=14.00−2.04=11.96\text{pH} = 14.00 - 2.04 = 11.96

Answer (d): pH = 11.96

Summary:

Volume NaOHpHRegion
0.0 mL1.00Before
12.5 mL1.48Before (halfway)
25.0 mL7.00Equivalence
30.0 mL11.96After

2Problem 2medium

❓ Question:

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=x20.150−xK_a = \frac{x^2}{0.150 - x}

Approximation: x² = K_a × 0.150

x=(1.8×10−5)(0.150)=1.64×10−3x = \sqrt{(1.8 \times 10^{-5})(0.150)} = 1.64 \times 10^{-3}

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−]=0.003750.0500=0.0750 M[CH_3COO^-] = \frac{0.00375}{0.0500} = 0.0750 \text{ M}

This is weak base!

Find K_b:

Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}

Base equilibrium: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

Kb=x20.0750K_b = \frac{x^2}{0.0750}

x=(5.6×10−10)(0.0750)=6.48×10−6x = \sqrt{(5.6 \times 10^{-10})(0.0750)} = 6.48 \times 10^{-6}

[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:

PointpHExplanation
Initial2.78Weak acid
Half-equiv4.74pH = pK_a (buffer!)
Equivalence8.81Weak 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+]=0.002500.0500=0.0500 M[NH_4^+] = \frac{0.00250}{0.0500} = 0.0500 \text{ M}

NH₄⁺ is weak acid!

Find K_a:

Ka=KwKb=1.0×10−141.8×10−5=5.6×10−10K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}

Acid equilibrium: NH₄⁺ ⇌ NH₃ + H⁺

Ka=x20.0500−x≈x20.0500K_a = \frac{x^2}{0.0500 - x} \approx \frac{x^2}{0.0500}

x2=(5.6×10−10)(0.0500)x^2 = (5.6 \times 10^{-10})(0.0500)

x2=2.8×10−11x^2 = 2.8 \times 10^{-11}

x=5.29×10−6x = 5.29 \times 10^{-6}

[H⁺] = 5.29 × 10⁻⁶ M

pH:

pH=−log⁡(5.29×10−6)=5.28\text{pH} = -\log(5.29 \times 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

General rule:

  • Strong/strong: any indicator (pH = 7)
  • Weak acid/strong base: basic indicator (phenolphthalein)
  • Strong acid/weak base: acidic indicator (methyl orange)

Summary:

Titration TypepH at equivIndicator
Strong/strong7Any (4-10)
Weak acid/strong base> 7Phenolphthalein
Strong acid/weak base< 7Methyl orange
Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple ChoiceMCQ6090 min50%✅
Free Response (Long)FRQ369 min30%✅
Free Response (Short)FRQ436 min20%✅

📊 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:

22 mol of H2H_2 reacts with 11 mol of O2O_2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O2H_2 + O_2 \to 2H_2O)

2Determine the limiting reagent
3Calculate moles of product
4Convert moles to grams
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Buffer Solutions and Henderson-Hasselbalch
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Acid-Base Theories and pH ScaleWeak Acids, Weak Bases, and K_a/K_bBuffer Solutions and Henderson-Hasselbalch

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What is Acid-Base Titrations and Indicators?▾
Master titration curves, equivalence points, and choosing appropriate indicators for acid-base titrations.
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