Titration Calculations — Endpoint, Equivalence Point and the Working
Titration is the one experiment that appears in every chemistry syllabus from Class 11 to postgraduate analysis, and the calculation is worth full marks if you follow a fixed order. Most lost marks come from two places: mixing up the endpoint with the equivalence point, and forgetting the mole ratio when the reaction is not one-to-one. This guide fixes both, with four worked examples.
Endpoint is not the equivalence point
These two terms are used loosely in conversation and precisely in exams.
- The equivalence point is a chemical fact: the moment when exactly enough titrant has been added to react completely with the analyte, in the ratio the balanced equation demands. You cannot see it.
- The endpoint is what you observe: the drop at which the indicator changes colour. You can see it.
The difference between the two is the titration error. A good indicator is one whose colour change happens inside the steep part of the pH curve, so that the endpoint falls within a fraction of a drop of the equivalence point. Choosing the indicator is therefore a real decision, not a ritual.
The formula
For a reaction in which a moles of A react with b moles of B:
M is molarity in mol/L, V is volume, and a and b are the coefficients from the balanced equation. Volumes may be left in millilitres provided both sides use millilitres, because the units cancel — but the moment you need moles as an actual number, you must convert to litres.
When a = b = 1 the formula collapses to the familiar MAVA = MBVB. In the normality system the equivalents are already built into the unit, so NAVA = NBVB always holds — that is the attraction of normality, and its danger, because it hides the chemistry. Pick one system and stay in it for the whole question.
Worked example 1 — a 1 : 1 acid–base titration
Question. 25.00 mL of hydrochloric acid requires 22.40 mL of 0.100 M sodium hydroxide to reach the endpoint. Find the molarity of the acid.
HCl + NaOH → NaCl + H₂O, so a = b = 1.
M(HCl) = (0.100 × 22.40) ÷ 25.00 = 2.24 ÷ 25.00 = 0.0896 M
Sense check: less base than acid was needed, so the acid must be more dilute than the base — and 0.0896 < 0.100 ✓
Worked example 2 — when the ratio is not 1 : 1
Question. 20.00 mL of sulphuric acid requires 25.60 mL of 0.200 M NaOH. Find the molarity of the acid.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so one mole of acid needs two of base.
Step 1 — moles of NaOH: 0.200 × 0.02560 L = 5.120 × 10⁻³ mol
Step 2 — moles of H₂SO₄: 5.120 × 10⁻³ ÷ 2 = 2.560 × 10⁻³ mol
Step 3 — molarity: 2.560 × 10⁻³ ÷ 0.02000 L = 0.1280 M
Skipping step 2 gives 0.256 M — exactly double, and a classic full-marks-to-zero error.
Worked example 3 — a redox titration
Redox titrations use larger ratios, which is why the balanced equation matters even more. Potassium permanganate in acid medium is its own indicator: the first permanent pink tinge is the endpoint.
Question. What volume of 0.0200 M KMnO₄ is needed to titrate 25.00 mL of 0.0500 M oxalic acid in acid medium?
2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O, so the ratio is 2 permanganate to 5 oxalate.
Step 1 — moles of oxalate: 0.0500 × 0.02500 = 1.250 × 10⁻³ mol
Step 2 — moles of MnO₄⁻: (2 ÷ 5) × 1.250 × 10⁻³ = 5.00 × 10⁻⁴ mol
Step 3 — volume: 5.00 × 10⁻⁴ ÷ 0.0200 = 0.02500 L = 25.00 mL
This titration is done warm (about 60 °C) because the reaction is slow at room temperature, and dilute sulphuric acid is used — not hydrochloric, because chloride would itself be oxidised by permanganate and consume extra titrant.
Worked example 4 — percentage purity
The most common "application" version of the question: titrate a weighed impure solid and work back to how much of it was the real thing.
Question. 0.500 g of impure sodium carbonate is dissolved and titrated against 0.250 M HCl using methyl orange; 35.85 mL is required. Find the percentage purity. (M(Na₂CO₃) = 105.988 g/mol.)
With methyl orange both protons react: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
Step 1 — moles of HCl: 0.250 × 0.03585 = 8.9625 × 10⁻³ mol
Step 2 — moles of Na₂CO₃: 8.9625 × 10⁻³ ÷ 2 = 4.48125 × 10⁻³ mol
Step 3 — mass of Na₂CO₃: 4.48125 × 10⁻³ × 105.988 = 0.4750 g
Step 4 — purity: (0.4750 ÷ 0.500) × 100 = 95.0%
Sense check: the answer must lie between 0 and 100%. Anything above 100% means an arithmetic error, a wrong mole ratio, or a wet sample.
Choosing the indicator
| Titration | pH at equivalence | Suitable indicator |
|---|---|---|
| Strong acid vs strong base | 7 | Methyl orange or phenolphthalein — the jump is very steep |
| Weak acid vs strong base | Above 7 | Phenolphthalein (changes in the basic range) |
| Strong acid vs weak base | Below 7 | Methyl orange (changes in the acidic range) |
| Weak acid vs weak base | Near 7, no steep jump | No indicator is reliable — use a pH meter or a conductometric method |
| Permanganate redox in acid | — | Self-indicating: first permanent pink |
Getting the reading right
The calculation can only be as good as the burette reading. Take a rough titration first, then repeat until you have concordant readings — usually two or three agreeing within 0.05 mL — and average only those. Read the bottom of the meniscus with your eye level with it to avoid parallax, and remember the burette scale increases downwards, so the titre is the final reading minus the initial reading.
Common mistakes that cost marks
- Using millilitres where moles are needed. Moles = M × V with V in litres. 25.00 mL is 0.02500 L.
- Ignoring the mole ratio. Always write the balanced equation before touching the calculator, even for a titration you have done ten times.
- Treating endpoint and equivalence point as the same thing in a theory answer. They are related but not identical, and the difference is the titration error.
- Mixing molarity and normality mid-question. N = M × n-factor. Convert once, at the start, and stay in one system.
- Averaging the rough titration along with the accurate ones. The rough run is a guide, not data.
- Using HCl to acidify a permanganate titration. Chloride is oxidised and the titre comes out too high.
- Rounding the titre. Burette readings are recorded to 0.05 mL; keep that precision through the working and round only the final answer.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE / ICSE Class 11–12 practicals | Acid–base and permanganate titrations, purity of a salt |
| JEE / NEET | Stoichiometry numericals, back titration, double indicator problems |
| IIT-JAM / CUET-PG | Volumetric analysis, equivalent weight from titration data |
| GATE / CSIR-NET | Analytical chemistry: titration curves, indicator selection, errors |
Check the one-to-one step instantly. A 1 : 1 titration is arithmetically the same M₁V₁ = M₂V₂ relationship as a dilution, so the Dilution calculator will solve example 1 and every strong-acid-against-strong-base titre for you. For ratios other than 1 : 1, use it after you have divided by the mole ratio yourself.
Open the M₁V₁ = M₂V₂ Dilution Calculator →Practical work and volumetric numericals both carry marks in Class 11–12. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — abcchemistry.in. For one-to-one help at home in Delhi, Noida or Gurgaon, see delhihometutor.com.