Titration Curve Formula
Graph of pH vs volume of titrant added - shows equivalence point and buffer regions
Key Points on Titration Curve
Equivalence Point
Point where moles of acid = moles of base (stoichiometric completion)
Strong Acid + Strong Base
pH = 7.00 (neutral)
Weak Acid + Strong Base
pH > 7 (conjugate base formed)
Strong Acid + Weak Base
pH < 7 (conjugate acid formed)
Weak Acid + Weak Base
pH depends on Ka and Kb
Half-Equivalence Point (Weak Acid/Base)
pH = pKa
At V = ½Veq: [HA] = [A⁻], maximum buffer capacity
pH Calculation Formulas
Before Equivalence Point (Weak Acid + Strong Base)
pH = pKa + log([A⁻]/[HA])
[A⁻] = moles base added / total volume
[HA] = (initial moles acid - moles base) / total volume
At Equivalence Point (Weak Acid + Strong Base)
pH = 7 + ½(pKa + pKw)
or pH = ½(pKw + pKa + log C)
C = concentration of conjugate base
After Equivalence Point
pH determined by excess base
[OH⁻] = (moles base - moles acid) / total volume
pOH = -log[OH⁻]
pH = 14 - pOH
Curve Characteristics
Strong Acid + Strong Base
- • Low initial pH (<2)
- • Sharp vertical rise at equivalence
- • Equivalence point pH = 7.00
- • Large pH jump (~6 units)
- • No buffer region
Weak Acid + Strong Base
- • Higher initial pH (2-6)
- • Buffer region before equivalence
- • Equivalence point pH > 7
- • Smaller pH jump (~3-4 units)
- • pH = pKa at half-equivalence
Polyprotic Acids
- • Multiple equivalence points
- • One per ionizable H⁺
- • Multiple buffer regions
- • pH = pKa at each half-equivalence
- • Example: H₂SO₄ has 2 steps
Indicators
- • Choose indicator with pKa ≈ pHeq
- • Color change at pH ≈ pKIn
- • Strong-strong: phenolphthalein (pH 8-10)
- • Weak-strong: varies by equivalence pH
- • Endpoint ≈ equivalence point
Worked Examples
Example 1: Find Equivalence Point Volume
Problem: 25.0 mL of 0.100 M HCl titrated with 0.100 M NaOH
Solution:
At equivalence: n(HCl) = n(NaOH)
M₁V₁ = M₂V₂
(0.100)(25.0) = (0.100)(V₂)
V₂ = 25.0 mL
Equivalence at 25.0 mL, pH = 7.00
Example 2: pH at Half-Equivalence
Problem: 50.0 mL of 0.10 M acetic acid (Ka = 1.8×10⁻⁵) titrated with 0.10 M NaOH. Find pH at 25.0 mL added.
Solution:
Equivalence volume = 50.0 mL
At 25.0 mL: half-equivalence point
pH = pKa = -log(1.8×10⁻⁵)
pH = 4.74
This is maximum buffer capacity
Example 3: pH Before Equivalence
Problem: Same titration as Example 2. Find pH after 40.0 mL NaOH added.
Solution:
Initial moles CH₃COOH = 0.10 × 0.050 = 0.0050 mol
Moles NaOH added = 0.10 × 0.040 = 0.0040 mol
Moles CH₃COOH remaining = 0.0050 - 0.0040 = 0.0010 mol
Moles CH₃COO⁻ formed = 0.0040 mol
Total volume = 50.0 + 40.0 = 90.0 mL
pH = pKa + log([A⁻]/[HA])
pH = 4.74 + log(0.0040/0.0010)
pH = 4.74 + log(4.0)
pH = 4.74 + 0.60
pH = 5.34
Example 4: pH After Equivalence
Problem: Same titration. Find pH at 60.0 mL NaOH added.
Solution:
Past equivalence (50.0 mL) by 10.0 mL
Excess NaOH = 0.10 × 0.010 = 0.0010 mol
Total volume = 50.0 + 60.0 = 110.0 mL
[OH⁻] = 0.0010 / 0.110 = 9.1×10⁻³ M
pOH = -log(9.1×10⁻³) = 2.04
pH = 14 - 2.04
pH = 11.96
Common Mistakes
Confusing Endpoint with Equivalence Point
Endpoint = indicator color change. Equivalence = stoichiometric completion. Should be close but not identical.
Assuming Weak Acid Equivalence pH = 7
Weak acid + strong base: pH > 7 at equivalence (conjugate base is basic!)
Forgetting Total Volume
Total volume = initial volume + titrant added. Use this for concentration calculations.
Buffer Region Shortcut
In buffer region, pH changes slowly. At half-equivalence, pH = pKa exactly.