Titration calculator
Enter your titration data, get the concentration.
How do you calculate concentration from a titration?
Moles of titrant = molarity × volume used. The balanced equation converts that to moles of the analyte, and dividing by the analyte’s volume gives its molarity.
Example: HCl with NaOH
A 50.00 mL sample of HCl needs 35.23 mL of 0.250 M NaOH. That is 0.03523 L × 0.250 M = 8.81 × 10⁻³ mol NaOH; the acid and base react 1 : 1, so the HCl is 8.81 × 10⁻³ mol ÷ 0.05000 L = 0.176 M (OpenStax Chemistry 2e, Example 4.14).
What if the ratio isn’t 1 : 1?
Enter the coefficients from the balanced equation. Sulfuric acid reacts with 2 NaOH (H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O), so the same volumes with H₂SO₄ give 4.40 × 10⁻³ mol and 0.0881 M.
Do the volumes have to be in liters?
No — enter both volumes in milliliters. The calculator uses the ratio of the two, and moles are shown using liters.
What is the equivalence point?
The point where the titrant added has exactly reacted with the analyte according to the equation. An indicator’s color change (the end point) is chosen to happen very close to it.
How do I find the percent acid in vinegar from a titration?
Titrate a measured sample with a standard base, find the molarity, then multiply by the molar mass of acetic acid, 60.05 g/mol, to get grams per liter. If 5.00 mL of vinegar needs 41.5 mL of 0.100 M NaOH (1 : 1), the acetic acid concentration is 0.830 M, which is 49.8 g/L, or about 5.0 g per 100 mL. That is close to the 5% acidity printed on many vinegar labels, expressed as acetic acid.
How many significant figures should I report?
No more than the least precise measurement allows. In the HCl example the volumes and the molarity have three or four significant figures, so the answer 0.176 M carries three. Carrying extra digits from the calculator display does not add precision.
The calculation, step by step
At the equivalence point, the moles of titrant that were added are chemically equivalent to the moles of analyte in the flask. First find the moles of titrant: molarity × volume in liters. Then convert with the mole ratio from the balanced equation, moles of analyte = moles of titrant × (analyte coefficient ÷ titrant coefficient). Finally divide by the volume of the analyte sample to get its molarity. In the form, the analyte coefficient and the titrant coefficient are the numbers in front of each substance in the balanced equation. The calculation assumes the reaction goes to completion and that the endpoint you observed is the equivalence point.
Worked example: HCl and sulfuric acid with NaOH
A 50.00 mL sample of acid needs 35.23 mL of 0.250 M NaOH. The moles of NaOH are 0.03523 L × 0.250 mol/L = 8.808 mmol. If the acid is HCl (1 : 1), the moles of acid are the same, 8.808 mmol, and the concentration is 8.808 mmol ÷ 50.00 mL = 0.176 M. If it is H₂SO₄, which reacts with 2 NaOH, the moles of acid are half as many, 4.404 mmol, and the concentration is 0.0881 M. Same volumes, half the concentration: the coefficients matter.
Sources of error
Overshooting the endpoint adds extra titrant and makes the result too high, so add the last drops slowly. The indicator should change color near the equivalence point of the particular acid and base: a strong acid with a strong base ends near pH 7, a weak acid with a strong base ends in the basic range. A burette that is not rinsed with the titrant dilutes it. Air bubbles in the tip, reading the meniscus from the wrong angle, or using a titrant whose concentration has changed since it was made all shift the answer. Repeating the titration until volumes agree closely is standard practice.
Common mistakes
Entering the coefficients in the wrong boxes inverts the ratio; the analyte's coefficient goes in one field and the titrant's in the other. Using the volume of the titrant instead of the volume of the sample in the last step is another. Polyprotic acids give more than one equivalence point: one titration to the first point uses the first proton only, and the ratio must match the step you titrated to. Finally, a total volume in the flask (sample plus water added to dilute it) is not the sample volume, because adding water does not change the moles of analyte.
Sources: OpenStax Chemistry 2e, section 4.5 (quantitative chemical analysis: titration, Example 4.14) and section 4.2 (classifying chemical reactions: acid-base reactions); molar mass of acetic acid from standard atomic weights (IUPAC). Related: molarity calculator, pH calculator and dilution calculator.