Enter the known values and select which quantity to solve for. The calculator supports 1:1 and non-1:1 stoichiometric ratios.
Titration is a volumetric analysis technique used to determine the unknown concentration of a solution by reacting it precisely with a solution of known concentration, called the standard solution. It is one of the most important practical skills assessed at both GCSE and A-Level.
The reaction proceeds until the equivalence point is reached — the exact stoichiometric point where the moles of acid and base have reacted in the correct ratio. In practice, an indicator signals this point via a permanent colour change, which is called the end point.
Titrations are used extensively in the pharmaceutical, food, and water treatment industries to ensure product quality and safety.
| Indicator | Acid Colour | Alkali Colour | pH Range |
|---|---|---|---|
| Phenolphthalein | Colourless | Pink / Magenta | 8.2 – 10.0 |
| Methyl orange | Red | Yellow | 3.1 – 4.4 |
| Litmus | Red | Blue | 5.0 – 8.0 |
| Bromothymol blue | Yellow | Blue | 6.0 – 7.6 |
where n = moles (mol), C = concentration (mol/dm³), V = volume in dm³. Convert cm³ to dm³ by dividing by 1000.
Valid only for 1:1 molar ratio reactions (e.g. HCl + NaOH → NaCl + H₂O).
For H₂SO₄ + 2NaOH: n₁=1, n₂=2. So C(NaOH) = 2 × C(H₂SO₄) × V(H₂SO₄) / V(NaOH)
Given: 25.0 cm³ of 0.100 mol/dm³ NaOH is pipetted into a conical flask. It is titrated with HCl solution. The mean titre is 22.5 cm³.
Step 1: Convert volumes to dm³: V(NaOH) = 25.0 ÷ 1000 = 0.0250 dm³
Step 2: n(NaOH) = C × V = 0.100 × 0.0250 = 0.00250 mol
Step 3: At equivalence, n(HCl) = n(NaOH) = 0.00250 mol (1:1 ratio)
Step 4: V(HCl) = 22.5 ÷ 1000 = 0.0225 dm³
Step 5: C(HCl) = n/V = 0.00250 ÷ 0.0225 = 0.111 mol/dm³
Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Given: 20.0 cm³ of 0.200 mol/dm³ NaOH, titre = 18.4 cm³ H₂SO₄
Step 1: n(NaOH) = 0.200 × 0.0200 = 0.00400 mol
Step 2: Molar ratio H₂SO₄ : NaOH = 1:2, so n(H₂SO₄) = 0.00400 ÷ 2 = 0.00200 mol
Step 3: C(H₂SO₄) = 0.00200 ÷ 0.0184 = 0.109 mol/dm³
A back titration is performed when direct titration is not possible — for example, when the analyte is insoluble or reacts too slowly with the titrant.
Add excess HCl to dissolve the CaCO₃. Back-titrate unused HCl with NaOH. Calculate moles of HCl consumed and hence % CaCO₃.
A primary standard must satisfy all of the following criteria:
| Standard | Formula | Mr | Used to standardise |
|---|---|---|---|
| Anhydrous Na₂CO₃ | Na₂CO₃ | 106 | HCl, H₂SO₄ |
| Potassium hydrogen phthalate | KHC₈H₄O₄ | 204.2 | NaOH |
| Oxalic acid | H₂C₂O₄·2H₂O | 126 | KMnO₄, NaOH |
Every measurement in a titration carries an uncertainty. You must combine these to find the total uncertainty in your calculated concentration.
For a burette with absolute uncertainty ±0.05 cm³: a titre of 22.50 cm³ has a % uncertainty of (0.05/22.50) × 100 = 0.22%.
Note that a titre reading involves two burette readings (start and end), so the combined burette uncertainty is 2 × 0.05 = ±0.10 cm³ when added in quadrature, or simply ±0.10 cm³ if treating as maximum error.
| Equipment | Absolute Uncertainty | Notes |
|---|---|---|
| 50 cm³ Burette | ±0.05 cm³ | Per reading; titre uses 2 readings |
| 25 cm³ Pipette | ±0.06 cm³ | Class B; Class A is ±0.03 cm³ |
| 250 cm³ Volumetric flask | ±0.30 cm³ | Used to prepare standard solution |
| Balance (2 d.p.) | ±0.005 g | Per reading; two readings needed |
The total percentage uncertainty in the final concentration is the sum of all individual percentage uncertainties. To minimise error: use large titre volumes and avoid titres below 10 cm³.
Titration is a quantitative analytical technique used to determine the unknown concentration of a solution by reacting it with a solution of known concentration (the standard solution) until the reaction reaches completion, indicated by a colour change from an appropriate indicator. The volume of standard solution used (the titre) allows the concentration to be calculated.
The key formula is n = CV, where n is moles, C is concentration in mol/dm³, and V is volume in dm³. For 1:1 reactions such as HCl + NaOH, C₁V₁ = C₂V₂ applies directly. For non-1:1 ratios, you must account for the stoichiometric ratio from the balanced equation using n₁/n₂ as a correction factor.
Both phenolphthalein and methyl orange are suitable for strong acid-strong base titrations because the pH at the equivalence point changes steeply (from ~4 to ~10 with just one drop of titrant). Phenolphthalein changes from colourless to pink at pH 8.2. Methyl orange changes from red to yellow at pH 4.0. Both fall within the steep pH change region.
A primary standard is a highly pure, stable substance with a high molar mass that can be dissolved precisely to make an accurately known concentration. Examples include anhydrous sodium carbonate (Na₂CO₃, Mr = 106) and potassium hydrogen phthalate (KHP, Mr = 204.2). They are used to standardise solutions whose exact concentration would otherwise be uncertain, such as NaOH (which absorbs CO₂ from air) and HCl (which is volatile).
A back titration is used when the analyte cannot be titrated directly, e.g. because it is insoluble or reacts too slowly. An excess of a known reagent is added to react completely with the analyte. The unreacted excess is then titrated with a second standard solution. The moles consumed by the analyte = initial moles added − moles remaining (found from titration), and stoichiometry gives the moles of analyte.
Percentage uncertainty = (absolute uncertainty / measured value) × 100%. For a burette reading of 22.50 cm³ with absolute uncertainty ±0.05 cm³ per reading: % uncertainty = (0.10/22.50) × 100 = 0.44% (using ±0.10 for two readings). Add the percentage uncertainties from each piece of equipment to find the overall uncertainty in the final concentration.
The equivalence point is the theoretical point at which the moles of acid and base have reacted in exactly the stoichiometric ratio — no excess of either reagent remains. The end point is the practical point at which the indicator permanently changes colour. A good indicator is chosen so that its end point coincides as closely as possible with the equivalence point, minimising the titration error.