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 # Phenolphthalein (acid form, colourless)
O=C1OC(C2=CC=CC=C2)(C2=CC=CC=C2)C2=CC=C([O-])C=C12 # Phenolphthalein (base form, pink)
O=S1(=O)OC(c2ccccc12)(c3cc(C(C)C)c(O)cc3C)c4cc(C(C)C)c(O)cc4C # Thymol blue
O=S1(=O)OC(c2ccccc12)(c3cc(Br)c(O)c(Br)c3)c4cc(Br)c(O)c(Br)c4 # Bromophenol blue
CN(C)c1ccc(cc1)N=Nc2ccc(cc2)S(=O)(=O)[O-].[Na+] # Methyl orange
CN(C)c1ccc(cc1)N=Nc2ccccc2C(=O)O # Methyl red
O=S1(=O)OC(c2ccccc12)(c3cc(Cl)c(O)c(Cl)c3)c4cc(Cl)c(O)c(Cl)c4 # Chlorophenol blue
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 # Bromothymol blue
O=S1(=O)OC(c2ccccc12)(c3ccc(O)c(C)c3)c4ccc(O)c(C)c4 # Cresol red
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-] # NaOH (sodium hydroxide)
Cl # HCl (hydrochloric acid)
[Na+].[Cl-] # NaCl (sodium chloride)
O # H2O (water)
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 # CH3COOH (acetic acid)
CC(=O)[O-].[Na+] # CH3COONa (sodium acetate)
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)
N # NH3 (ammonia)
[Cl-].[NH4+] # NH4Cl (ammonium chloride)
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 Topics
- Ionic Equilibria — Comprehensive concept page
- Buffer Solutions — Henderson-Hasselbalch derivation and applications
- FAD1018 W3 — Buffer Solutions — Part 4 of Ionic Equilibria
- FAD1018 W2-W3 — Ionic Equilibria — Parts 1-6 overview
Related Course Page
- FAD1018 - Basic Chemistry II