FAD1018 W3 (3) — Ionic Equilibria Part 5-6 — Acid-Base Titrations

Week 3 extra lecture covering acid-base titrations in detail. Source file: W3 (3).pdf (59 slides). Dr. Fauzani Md. Salleh, Chemistry Division, Centre for Foundation Studies in Science, 2024/2025.

Objectives

To plot the titration curve between strong acid and strong base
To plot the titration curve between weak acid and strong base
To plot the titration curve between weak base and strong acid

Key Concepts

Acid-Base Titrations — Quantitative analysis via neutralisation
Titration Curves — pH vs volume of titrant plots
Acid-Base Indicators — Visual detection of equivalence point
Equivalence Point — Stoichiometrically equivalent amounts of acid and base
Buffer Solutions — Resisting pH change during titration

Acid-Base Indicators

Definition: Substances used to signal the equivalence point of a titration.

End point vs Equivalence point:

Equivalence point: stage where amounts of acid and base are stoichiometrically equivalent ([H⁺] = [OH⁻])
End point: point at which an indicator changes colour (meant to indicate equivalence point)

Key properties:

Selection depends on the titration curve
Shows different colours in acidic and basic solutions
Litmus paper reading: one decimal point
pH meter: two decimal places

Common Acid-Base Indicators

Indicator	Colour in Acid	Colour in Base	pH range
Thymol blue	Red	Yellow	1.2 – 2.8
Bromophenol blue	Yellow	Purple	3.0 – 4.6
Methyl orange	Orange	Yellow	3.1 – 4.4
Methyl red	Red	Yellow	4.2 – 6.3
Chlorophenol blue	Yellow	Red	4.8 – 6.4
Bromothymol blue	Yellow	Blue	6.0 – 7.6
Cresol red	Yellow	Red	7.2 – 8.8
Phenolphthalein	Colourless	Pink	8.3 – 10.0

O=C1OC(C2=CC=CC=C2)(C2=CC=CC=C2)C2=CC=C(O)C=C12

O=C1OC(C2=CC=CC=C2)(C2=CC=CC=C2)C2=CC=C([O-])C=C12

O=S1(=O)OC(c2ccccc12)(c3cc(C(C)C)c(O)cc3C)c4cc(C(C)C)c(O)cc4C

O=S1(=O)OC(c2ccccc12)(c3cc(Br)c(O)c(Br)c3)c4cc(Br)c(O)c(Br)c4

CN(C)c1ccc(cc1)N=Nc2ccc(cc2)S(=O)(=O)[O-].[Na+]

CN(C)c1ccc(cc1)N=Nc2ccccc2C(=O)O

O=S1(=O)OC(c2ccccc12)(c3cc(Cl)c(O)c(Cl)c3)c4cc(Cl)c(O)c(Cl)c4

O=S1(=O)OC(c2ccccc12)(c3c(C)c(Br)c(O)c(C(C)C)c3)c4c(C)c(Br)c(O)c(C(C)C)c4

O=S1(=O)OC(c2ccccc12)(c3ccc(O)c(C)c3)c4ccc(O)c(C)c4

Types of Acid-Base Titrations

Four theoretical types; lecture covers three (WA-WB crossed out as not covered):

Strong Acid (SA) – Strong Base (SB)
Weak Acid (WA) – Strong Base (SB)
Strong Acid (SA) – Weak Base (WB)
~~Weak Acid (WA) – Weak Base (WB)~~ (not covered)

Four Stages in Titrations

Initial stage — before titrant is added
Before the equivalence point — any point between initial and equivalence; midpoint where mole of titrant is half that of equivalence
At the equivalence point
After the equivalence point

1. Strong Acid – Strong Base Titration

Example system: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

[Na+].[OH-]

Cl

[Na+].[Cl-]

Titration Curve Characteristics

Equivalence point at pH = 7.0 (neutral salt)
Steep pH change around equivalence
Suitable indicators: bromothymol blue, phenolphthalein

Worked Example: 50.0 mL of 0.100 M NaOH titrated with 0.100 M HCl

Initial Stage (0 mL HCl added)

$$[NaOH] = [OH^-] = 0.100 \text{ M}$$ $$pOH = -\log(0.100) = 1.00$$ $$pH = 14 - 1.00 = 13.00$$

Before Equivalence Point (25.0 mL HCl added)

$$n_{HCl} = 0.1 \times \frac{25}{1000} = 2.5 \times 10^{-3} \text{ mol}$$ $$n_{NaOH} = 0.1 \times \frac{50}{1000} = 5.0 \times 10^{-3} \text{ mol}$$

NaOH is in excess: $$n_{NaOH \text{ remaining}} = 5.0 \times 10^{-3} - 2.5 \times 10^{-3} = 2.5 \times 10^{-3} \text{ mol}$$ $$V_{total} = 50 + 25 = 75 \text{ mL} = 0.075 \text{ L}$$ $$[NaOH] = [OH^-] = \frac{2.5 \times 10^{-3}}{0.075} = 0.033 \text{ M}$$ $$pOH = -\log(0.033) = 1.48$$ $$pH = 14 - 1.48 = 12.5$$

At Equivalence Point (50.0 mL HCl added)

Using stoichiometry: $$\frac{M_a V_a}{M_b V_b} = \frac{a}{b}$$ $$V_{HCl} = 50 \text{ mL}$$

Neither Na⁺ nor Cl⁻ hydrolyzes. The salt NaCl is neutral. $$pH = 7.0$$

After Equivalence Point (65.0 mL HCl added)

$$n_{HCl} = 0.1 \times \frac{65}{1000} = 6.5 \times 10^{-3} \text{ mol}$$ $$n_{NaOH} = 5.0 \times 10^{-3} \text{ mol}$$

HCl is in excess: $$n_{HCl \text{ remaining}} = 6.5 \times 10^{-3} - 5.0 \times 10^{-3} = 1.5 \times 10^{-3} \text{ mol}$$ $$V_{total} = 50 + 65 = 115 \text{ mL} = 0.115 \text{ L}$$ $$[HCl] = [H^+] = \frac{1.5 \times 10^{-3}}{0.115} = 0.013 \text{ M}$$ $$pH = -\log(0.013) = 1.88$$

2. Weak Acid – Strong Base Titration

Example system: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

CC(=O)O

CC(=O)[O-].[Na+]

Titration Curve Characteristics

Equivalence point at pH > 7 (basic salt — anion hydrolyzes)
Buffer zone before equivalence point
Half-equivalence point: pH = pKa
Suitable indicator: phenolphthalein

Worked Example: 100 mL of 0.1 M CH₃COOH titrated with 0.1 M NaOH (pKa = 4.76)

Initial Stage (0 mL NaOH)

For weak acid: $$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = 10^{-4.76} = 1.74 \times 10^{-5}$$

Assume $x \ll 0.1$: $$1.74 \times 10^{-5} = \frac{x^2}{0.1}$$ $$x = [H^+] = 1.32 \times 10^{-3} \text{ M}$$ $$pH = -\log(1.32 \times 10^{-3}) = 2.88$$

Before Equivalence Point (25.0 mL NaOH added)

$$n_{CH_3COOH} = 0.1 \times 0.100 = 0.01 \text{ mol}$$ $$n_{NaOH} = 0.1 \times 0.025 = 2.5 \times 10^{-3} \text{ mol}$$

Forms a buffer system (CH₃COOH / CH₃COO⁻): $$n_{CH_3COOH \text{ remaining}} = 0.01 - 0.0025 = 7.5 \times 10^{-3} \text{ mol}$$ $$n_{CH_3COO^-} = 2.5 \times 10^{-3} \text{ mol}$$

Using Henderson-Hasselbalch: $$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]} = 4.76 + \log\frac{2.5}{7.5} = 4.76 - 0.477 = 4.28$$

Alternative method (concentration-based ICE table) gives same result.

At Equivalence Point (100 mL NaOH added)

All CH₃COOH converted to CH₃COONa: $$n_{CH_3COONa} = 0.01 \text{ mol}$$ $$V_{total} = 200 \text{ mL} = 0.2 \text{ L}$$ $$[CH_3COO^-] = \frac{0.01}{0.2} = 0.05 \text{ M}$$

Hydrolysis of acetate: $$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$$

$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}$$

$$5.75 \times 10^{-10} = \frac{x^2}{0.05}$$ $$x = [OH^-] = 5.37 \times 10^{-6} \text{ M}$$ $$pOH = -\log(5.37 \times 10^{-6}) = 5.27$$ $$pH = 14 - 5.27 = 8.73$$

[!note] Lecture result pH = 8.73 at equivalence point. The solution is a basic salt.

After Equivalence Point (125 mL NaOH added)

$$n_{NaOH} = 0.1 \times 0.125 = 0.0125 \text{ mol}$$ $$n_{NaOH \text{ excess}} = 0.0125 - 0.01 = 2.5 \times 10^{-3} \text{ mol}$$ $$V_{total} = 225 \text{ mL} = 0.225 \text{ L}$$

Since OH⁻ is much stronger base than CH₃COO⁻, neglect acetate hydrolysis: $$[NaOH] = [OH^-] = \frac{2.5 \times 10^{-3}}{0.225} = 0.0111 \text{ M}$$ $$pOH = -\log(0.0111) = 1.95$$ $$pH = 14 - 1.95 = 12.05$$

Titration Curve Features

Half-equivalence point: pH = pKa = 4.76 (at 50 mL NaOH)
Buffer zone: region around half-equivalence where pH changes slowly
Equivalence point: pH = 8.73

3. Weak Base – Strong Acid Titration

Example system: NH₃(aq) + HCl(aq) → NH₄Cl(aq)

[Cl-].[NH4+]

Titration Curve Characteristics

Equivalence point at pH < 7 (acidic salt — cation hydrolyzes)
Buffer zone before equivalence point
Suitable indicator: methyl red

Worked Example: 40 mL of 0.1 M NH₃ titrated with 0.1 M HCl (Kb = 1.8 × 10⁻⁵)

Initial Stage (0 mL HCl)

For weak base: $$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5}$$

Assume $x \ll 0.1$: $$1.8 \times 10^{-5} = \frac{x^2}{0.1}$$ $$x = [OH^-] = 1.34 \times 10^{-3} \text{ M}$$ $$pOH = -\log(1.34 \times 10^{-3}) = 2.87$$ $$pH = 14 - 2.87 = 11.13$$

Before Equivalence Point (25.0 mL HCl added)

$$n_{NH_3} = 0.1 \times 0.040 = 4.0 \times 10^{-3} \text{ mol}$$ $$n_{HCl} = 0.1 \times 0.025 = 2.5 \times 10^{-3} \text{ mol}$$

Forms a basic buffer system (NH₃ / NH₄⁺): $$n_{NH_3 \text{ remaining}} = 4.0 \times 10^{-3} - 2.5 \times 10^{-3} = 1.5 \times 10^{-3} \text{ mol}$$ $$n_{NH_4^+} = 2.5 \times 10^{-3} \text{ mol}$$ $$V_{total} = 65 \text{ mL} = 0.065 \text{ L}$$

$$[NH_3] = \frac{1.5 \times 10^{-3}}{0.065} = 0.023 \text{ M}$$ $$[NH_4^+] = \frac{2.5 \times 10^{-3}}{0.065} = 0.038 \text{ M}$$

Using Henderson-Hasselbalch for basic buffers: $$pOH = pK_b + \log\frac{[NH_4^+]}{[NH_3]} = 4.745 + \log\frac{0.038}{0.023} = 4.745 + 0.218 = 4.96$$ $$pH = 14 - 4.96 = 9.04$$

[!note] Lecture result pH = 9.04. This is a basic buffer solution.

At Equivalence Point (40 mL HCl added)

All NH₃ converted to NH₄Cl: $$n_{NH_4Cl} = 4.0 \times 10^{-3} \text{ mol}$$ $$V_{total} = 80 \text{ mL} = 0.08 \text{ L}$$ $$[NH_4^+] = \frac{4.0 \times 10^{-3}}{0.08} = 0.05 \text{ M}$$

Hydrolysis of ammonium: $$NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$$

$$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.555 \times 10^{-10}$$

$$5.555 \times 10^{-10} = \frac{x^2}{0.05}$$ $$x = [H_3O^+] = 5.27 \times 10^{-6} \text{ M}$$

Check assumption validity: $$%\alpha = \frac{5.27 \times 10^{-6}}{0.05} \times 100% = 0.011% < 10% \quad \text{(valid)}$$

$$pH = -\log(5.27 \times 10^{-6}) = 5.278$$

[!note] Lecture result pH = 5.278 at equivalence point. The solution is an acidic salt.

After Equivalence Point (80 mL HCl added)

$$n_{HCl} = 0.1 \times 0.080 = 8.0 \times 10^{-3} \text{ mol}$$ $$n_{HCl \text{ excess}} = 8.0 \times 10^{-3} - 4.0 \times 10^{-3} = 4.0 \times 10^{-3} \text{ mol}$$ $$V_{total} = 120 \text{ mL} = 0.12 \text{ L}$$

Since H⁺ is much stronger acid than NH₄⁺, neglect ammonium hydrolysis: $$[HCl] = [H^+] = \frac{4.0 \times 10^{-3}}{0.12} = 0.033 \text{ M}$$ $$pH = -\log(0.033) = 1.48$$

Titration Curve Features

Buffer region: around pKa of NH₄⁺ = 9.25 (pKb of NH₃ = 4.745)
Equivalence point: pH = 5.27
Indicator: methyl red (pH range 4.2–6.3) is suitable; phenolphthalein would change colour too early

Summary Table: Titration Types

Titration Type	Equivalence Point pH	Salt Type	Suitable Indicator
Strong Acid – Strong Base	7.0	Neutral	Bromothymol blue, Phenolphthalein
Weak Acid – Strong Base	> 7 (basic)	Basic salt	Phenolphthalein
Weak Base – Strong Acid	< 7 (acidic)	Acidic salt	Methyl red

Related Course Page

FAD1018 - Basic Chemistry II

FAD1018 W3 (3) — Ionic Equilibria Part 5-6 — Acid-Base Titrations

Objectives

Key Concepts

Acid-Base Indicators

Common Acid-Base Indicators

Types of Acid-Base Titrations

Four Stages in Titrations

1. Strong Acid – Strong Base Titration

Titration Curve Characteristics

Worked Example: 50.0 mL of 0.100 M NaOH titrated with 0.100 M HCl

Initial Stage (0 mL HCl added)

Before Equivalence Point (25.0 mL HCl added)

At Equivalence Point (50.0 mL HCl added)

After Equivalence Point (65.0 mL HCl added)

2. Weak Acid – Strong Base Titration

Titration Curve Characteristics

Worked Example: 100 mL of 0.1 M CH₃COOH titrated with 0.1 M NaOH (pKa = 4.76)

Initial Stage (0 mL NaOH)

Before Equivalence Point (25.0 mL NaOH added)

At Equivalence Point (100 mL NaOH added)

After Equivalence Point (125 mL NaOH added)

Titration Curve Features

3. Weak Base – Strong Acid Titration

Titration Curve Characteristics

Worked Example: 40 mL of 0.1 M NH₃ titrated with 0.1 M HCl (Kb = 1.8 × 10⁻⁵)

Initial Stage (0 mL HCl)

Before Equivalence Point (25.0 mL HCl added)

At Equivalence Point (40 mL HCl added)

After Equivalence Point (80 mL HCl added)

Titration Curve Features

Summary Table: Titration Types

Related Topics

Related Course Page