Analyze acid-base titrations and calculate pH at any point in the titration curve
Acid-Base Titration: Calculate pH at various points during titration
Determines equivalence point, initial pH, and pH changes throughout the titration
Titration is a quantitative analytical technique used to determine the concentration of an unknown acid or base solution by reacting it with a solution of known concentration (the titrant). The point at which stoichiometrically equivalent amounts of acid and base have reacted is called the equivalence point.
Example: HCl + NaOH → NaCl + H₂O
Characteristics:
Example: CH₃COOH + NaOH → CH₃COONa + H₂O
Characteristics:
Example: HCl + NH₃ → NH₄Cl
Characteristics:
M₁V₁ = M₂V₂
For monoprotic acids and bases: Moles of acid = Moles of base
Strong Acid-Strong Base:
pH = -log[H⁺] where [H⁺] = (moles acid - moles base) / total volume
Weak Acid-Strong Base (buffer region):
pH = pKa + log([A⁻]/[HA]) (Henderson-Hasselbalch)
Determined by excess titrant (strong base or strong acid)
pH = 14 - pOH where pOH = -log[OH⁻] (excess base)
Indicators are weak acids or bases that change color over a specific pH range. Choose an indicator whose transition range includes the pH at the equivalence point.
| Indicator | pH Range | Color Change | Best For |
|---|---|---|---|
| Methyl orange | 3.1 - 4.4 | Red → Yellow | Strong acid - Weak base |
| Methyl red | 4.4 - 6.2 | Red → Yellow | Strong acid - Weak base |
| Bromothymol blue | 6.0 - 7.6 | Yellow → Blue | Strong acid - Strong base |
| Phenolphthalein | 8.3 - 10.0 | Colorless → Pink | Weak acid - Strong base |
| Thymol blue | 8.0 - 9.6 | Yellow → Blue | Weak acid - Strong base |
Problem: 25.0 mL of 0.100 M acetic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵) is titrated with 0.100 M NaOH. Calculate:
Solution:
1. Initial pH:
[H⁺] = √(Ka × Ca) = √(1.8×10⁻⁵ × 0.100) = 1.34×10⁻³ M
pH = -log(1.34×10⁻³) = 2.87
2. Volume at Equivalence:
M₁V₁ = M₂V₂
Veq = (0.100 × 25.0) / 0.100 = 25.0 mL
3. pH at Equivalence:
Only CH₃COO⁻ present; Kb = Kw/Ka = 10⁻¹⁴/(1.8×10⁻⁵) = 5.56×10⁻¹⁰
[CH₃COO⁻] = 0.0025 mol / 0.050 L = 0.050 M
[OH⁻] = √(Kb × C) = √(5.56×10⁻¹⁰ × 0.050) = 5.27×10⁻⁶ M
pOH = 5.28, pH = 14 - 5.28 = 8.72
4. pH at Half-Equivalence (12.5 mL):
pH = pKa = -log(1.8×10⁻⁵) = 4.74
Answers: Initial pH = 2.87, Veq = 25.0 mL, pH at equivalence = 8.72, pH at half-equivalence = 4.74
Best indicator: Phenolphthalein (transition at pH 8.3-10.0)