Acid-Base Titrations and Indicators

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

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:

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:

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:

| Indicator | pK_a | Color change | pH range | |-----------|------|--------------|----------| | Methyl orange | 3.7 | Red → Yellow | 3.1-4.4 | | Bromothymol blue | 7.0 | Yellow → Blue | 6.0-7.6 | | Phenolphthalein | 9.4 | Colorless → Pink | 8.3-10.0 |

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:

Initial pH (before titration)
Buffer region (gradual rise)
Half-equivalence (pH = pK_a for weak acid)
Equivalence point (sharp rise)
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⁺]:

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

pH:

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

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⁻]:

$[OH^-] = \frac{0.00050 \text{ mol}}{0.0550 \text{ L}} = 0.00909 \text{ M}$

pOH:

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

pH:

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

Answer (d): pH = 11.96

Summary:

| Volume NaOH | pH | Region | |-------------|-----|--------| | 0.0 mL | 1.00 | Before | | 12.5 mL | 1.48 | Before (halfway) | | 25.0 mL | 7.00 | Equivalence | | 30.0 mL | 11.96 | After |

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:

$K_a = \frac{x^2}{0.150 - x}$

Approximation: x² = K_a × 0.150

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

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⁻]:

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

This is weak base!

Find K_b:

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

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

$K_b = \frac{x^2}{0.0750}$

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

[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:

| 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⁺]:

$[NH_4^+] = \frac{0.00250}{0.0500} = 0.0500 \text{ M}$

NH₄⁺ is weak acid!

Find K_a:

$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

Acid equilibrium: NH₄⁺ ⇌ NH₃ + H⁺

$K_a = \frac{x^2}{0.0500 - x} \approx \frac{x^2}{0.0500}$

$x^2 = (5.6 \times 10^{-10})(0.0500)$

$x^2 = 2.8 \times 10^{-11}$

$x = 5.29 \times 10^{-6}$

[H⁺] = 5.29 × 10⁻⁶ M

pH:

$\text{pH} = -\log(5.29 \times 10^{-6}) = 5.28$

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 Type | pH at equiv | Indicator | |----------------|-------------|-----------| | Strong/strong | 7 | Any (4-10) | | Weak acid/strong base | > 7 | Phenolphthalein | | Strong acid/weak base | < 7 | Methyl orange |

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