Potentiometric Titration: Finding the Endpoint by the First and Second Derivative
In a potentiometric titration you do not wait for an indicator to change colour. You measure the electrode potential after each addition of titrant and find the equivalence point from the data. This works for coloured or cloudy solutions where an indicator is useless. The skill examiners test is simple: given a table of E (or pH) against volume, find the endpoint by the first-derivative or second-derivative method, then calculate the unknown concentration.
The idea
Near the equivalence point, a tiny addition of titrant causes a very large jump in potential. The curve of E against volume is S-shaped, and the equivalence point is at the steepest part (the inflection point). Reading that point by eye is rough, so we use differences.
Second derivative: Δ(ΔE/ΔV)/ΔV, plotted at the midpoint of each pair of first-derivative values
Endpoint: the volume where ΔE/ΔV is a maximum, or where Δ²E/ΔV² crosses zero
- E is the measured potential, usually in mV (for a pH electrode the same method works with pH instead of E).
- V is the volume of titrant added, in mL.
- Δ means “difference between neighbouring readings”.
Worked examples
Example 1: first and second derivative. A 20.00 mL sample of Fe2+ is titrated with 0.1000 M Ce4+. The (practice) readings are:
| V (mL) | E (mV) | Midpoint V (mL) | ΔE/ΔV (mV mL−1) |
|---|---|---|---|
| 24.00 | 200 | — | — |
| 24.50 | 215 | 24.25 | 30 |
| 24.80 | 236 | 24.65 | 70 |
| 24.90 | 262 | 24.85 | 260 |
| 25.00 | 330 | 24.95 | 680 |
| 25.10 | 402 | 25.05 | 720 |
| 25.20 | 420 | 25.15 | 180 |
| 25.50 | 438 | 25.35 | 60 |
| 26.00 | 452 | 25.75 | 28 |
Example row: between 24.90 and 25.00 mL, ΔE/ΔV = (330 − 262) / (25.00 − 24.90) = 68 / 0.10 = 680 mV mL−1.
First-derivative method: the largest value is 720 at a midpoint of 25.05 mL, so the endpoint is near 25.05 mL. Because the step sizes are not equal, this is only a rough guide.
Second-derivative method: take the change in ΔE/ΔV between neighbouring midpoints.
- Between 24.95 and 25.05: (720 − 680) / (25.05 − 24.95) = 40 / 0.10 = +400, placed at 25.00 mL.
- Between 25.05 and 25.15: (180 − 720) / (25.15 − 25.05) = −540 / 0.10 = −5400, placed at 25.10 mL.
The second derivative changes sign between 25.00 and 25.10 mL. Interpolate linearly to the zero: fraction = 400 / (400 + 5400) = 0.069, so Veq = 25.00 + 0.069 × 0.10 = 25.007 ≈ 25.01 mL.
Example 2: concentration from the endpoint. Ce4+ reacts with Fe2+ in a 1 : 1 ratio.
Moles of Ce4+ = 0.1000 mol L−1 × 25.01 mL = 2.501 mmol, so Fe2+ = 2.501 mmol.
[Fe2+] = 2.501 mmol / 20.00 mL = 0.1251 M (0.12505 before rounding).
Example 3: a 1 : 5 ratio. A 25.00 mL solution of Fe2+ needs 18.40 mL of 0.0200 M KMnO4 (acid medium) at the equivalence point found from the derivative curve.
MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Moles of MnO4− = 0.0200 × 18.40 = 0.368 mmol
Moles of Fe2+ = 5 × 0.368 = 1.840 mmol
[Fe2+] = 1.840 / 25.00 = 0.0736 M
Check by n-factor: KMnO4 normality = 5 × 0.0200 = 0.100 N; Fe2+ normality = 0.100 × 18.40 / 25.00 = 0.0736 N, and for Fe2+ (n = 1) this is 0.0736 M. ✓
Practical tips for good data
- Add titrant in large steps at first, then in equal small steps (for example 0.10 mL) near the jump. Equal steps make the derivative calculation cleaner.
- Wait for a steady reading after each addition, and stir well.
- Record at least a few points on each side of the jump.
Common mistakes
- Reporting the volume at the end of a step, not at its midpoint, when plotting ΔE/ΔV.
- Reading the endpoint from the largest E value instead of the largest ΔE/ΔV.
- Using the second derivative without noting that the zero crossing lies between two listed points, so interpolation is needed.
- Forgetting the stoichiometric ratio (1 : 5 for MnO4− and Fe2+).
- Dividing by the titrant volume instead of the sample volume when finding the unknown concentration.
Quick revision: why derivative methods work
Think of the titration curve as a hill with a very steep side. Before the equivalence point the potential changes slowly because each addition only changes the ratio of oxidised to reduced forms of the analyte by a small amount. Right at the equivalence point that ratio changes enormously with one drop of titrant, so the potential jumps. After the equivalence point it flattens again because the excess titrant now controls the potential.
The first derivative ΔE/ΔV measures the steepness, so it peaks at the jump. The second derivative measures how the steepness itself changes: positive while the curve is getting steeper, zero at the steepest point, and negative afterwards. That is why a sign change in the second derivative marks the endpoint, and why interpolating to the zero is more precise than choosing the largest number in a column.
In your answer, show a small table of volume, ΔE/ΔV and Δ²E/ΔV², state the endpoint volume with units, and then complete the stoichiometry. Marks are usually given for each of these steps.
Where this is used in exams
| Topic | How it appears |
|---|---|
| Electroanalytical methods | Finding the endpoint from E–V or pH–V data. |
| Redox titration calculations | Stoichiometry applied after the endpoint is found. |
| Data-handling questions | First- and second-derivative tables. |
Check the official syllabus of your exam for the level of detail expected on instrumental methods.
Use the Scientific Calculator for the differences, ratios and interpolation in the endpoint calculation.
Open the Chemistry Calculator Suite →Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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