EN

Titration calculator

This titration calculator finds how concentrated your sample is. Example: 24.6 mL of 0.1 mol/L NaOH neutralise 25 mL of hydrochloric acid, so c = 0.0984 mol/L.

Example: What should be calculated: Concentration of the sample (analyte) · Titrant concentration 0.1 mol/L · Titrant volume used 24.6 mL · Analyte volume 25 mL · Analyte coefficient ν(A) 1 · Titrant coefficient ν(T) 1 → Result: c(analyte) = 0.0984 mol/L. Source: Harris, Quantitative Chemical Analysis, Kap. „Titrations“ (n = c·V, Stoffmengenverhältnis am Äquivalenzpunkt). Updated: .

mol/L
Exactly known concentration of the solution in the burette.
mL
Volume up to the equivalence point or colour change.
mL
The volume placed in the flask.
Factor in front of the analyte in the reaction equation, e.g. 1 for H₂SO₄ + 2 NaOH.
Factor in front of the titrant, e.g. 2 for H₂SO₄ + 2 NaOH.
My ToolboxYour inputs are saved in this browser only.

Result

Result
c(analyte) = 0.0984 mol/L
All representations
QuantityValue
Analyte concentration0.0984 mol/L
Titrant volume24.6 mL
Amount of titrant2.46 mmol
Amount of analyte2.46 mmol
Calculation steps
  • n(T) = c·V = 0.1 mol/L · 24.6 mL = 2.46 mmol; n(A) = n(T) · ν(A)/ν(T) = 2.46 · 1/1 = 2.46 mmol; c(A) = n(A) / V(A) = 2.46 mmol / 25 mL = 0.0984 mol/L.
Valid at the equivalence point for a reaction that goes to completion in a single step. The coefficients come from the balanced equation (mole ratio ν(A) : ν(T)). Average several runs.

How it is calculated

In a titration a solution of unknown concentration (the analyte) is reacted with a standard solution of exactly known concentration (the titrant) up to the equivalence point. The volume used gives the unknown concentration.

Formula

At the equivalence point the amounts of substance are in the ratio of the reaction equation: n(A) / ν(A) = n(T) / ν(T), with n = c · V. Hence c(A) = c(T) · V(T) · ν(A) / (ν(T) · V(A)). To plan the titrant volume instead: V(T) = c(A) · V(A) · ν(T) / (ν(A) · c(T)). With volumes in mL and concentrations in mol/L, the product c·V is directly in millimoles.

Worked example

25 mL of hydrochloric acid is titrated with sodium hydroxide (c = 0.1 mol/L); 24.6 mL flow until the colour change. HCl and NaOH react 1 : 1, so n = 0.1 · 24.6 = 2.46 mmol and c(HCl) = 2.46 / 25 = 0.0984 mol/L. For sulfuric acid (H₂SO₄ + 2 NaOH, ν(A) = 1, ν(T) = 2) the same volume gives only 0.0492 mol/L. A sample of 0.1 mol/L and 25 mL needs exactly 25 mL of titrant at 1 : 1.

Frequently asked questions

What is the equivalence point?

The point at which analyte and titrant have reacted in exactly the ratio of the reaction equation. It is detected by an indicator colour change or a pH measurement.

How do I calculate concentration from the volume used?

With c(A) = c(T) · V(T) · ν(A) / (ν(T) · V(A)). For 24.6 mL of a 0.1 mol/L solution and a 25 mL sample (1 : 1) this gives 0.0984 mol/L.

What do the coefficients ν(A) and ν(T) mean?

The factors from the reaction equation. For H₂SO₄ + 2 NaOH, ν(A) = 1 and ν(T) = 2: twice as much NaOH is used as there is sulfuric acid.

Sources and legal basis

As of: · Editorial policy

Related tools