Last updated: June 2026
Use this titration calculator to organise acid-base titration calculations involving concentration, volume and mole ratio. It is written for O-Level, IGCSE, A-Level and Singapore H2 Chemistry students who need to calculate the concentration of an unknown acid or base from titration data.
What is a titration?
A titration is an experimental method used to find the concentration of a solution. In a typical acid-base titration, a solution of known concentration is reacted with a measured volume of another solution whose concentration is unknown.
The solution of known concentration is often placed in the burette. The solution of unknown concentration is usually measured using a pipette and placed in a conical flask. An indicator is then added so that the endpoint can be detected by a colour change.
Acid-base titration method
In an acid-base titration, hydrogen ions from the acid react with hydroxide ions from the base to form water.
H+ + OH− → H2O
The important idea is not simply “acid volume equals base volume”. The reacting amounts depend on the mole ratio. For example, hydrochloric acid and sodium hydroxide react in a 1:1 ratio, but sulfuric acid and sodium hydroxide react in a 1:2 ratio.
Neutralisation in titration
Neutralisation is the reaction between an acid and a base to form salt and water. For school chemistry, the core ionic equation is:
H+ + OH− → H2O
In titration calculations, neutralisation is not treated as a separate calculator. Instead, it is built into the mole ratio from the balanced equation.
At the equivalence point:
moles of H+ = moles of OH−
This does not always mean the final pH is exactly 7. Strong acid with strong base gives an equivalence point close to pH 7 at 25°C, but weak acid with strong base gives an equivalence point above pH 7, while strong acid with weak base gives an equivalence point below pH 7.
Titration formula
The basic mole formula used in titration is:
n = cV
where n is the amount of substance in mol, c is concentration in mol dm−3, and V is volume in dm3.
Once the moles of one solution are known, the balanced equation is used to find the moles of the other solution.
Volume conversion in titration
Titration volumes are usually measured in cm3. However, concentration is usually given in mol dm−3. This means volume must be converted into dm3 before using n = cV.
1000 cm3 = 1 dm3
So 25.0 cm3 becomes 0.0250 dm3, while 24.60 cm3 becomes 0.02460 dm3.
Using the mole ratio
The mole ratio comes from the balanced equation. This is the step students often miss.
For hydrochloric acid and sodium hydroxide:
HCl + NaOH → NaCl + H2O
The mole ratio of HCl to NaOH is 1:1.
For sulfuric acid and sodium hydroxide:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
The mole ratio of H2SO4 to NaOH is 1:2. This means one mole of sulfuric acid reacts with two moles of sodium hydroxide.
Using H+ and OH− donated
Another way to think about acid-base titration is to compare moles of H+ and OH−.
For an acid:
moles of H+ = moles of acid × number of H+ donated per acid molecule
For a base:
moles of OH− = moles of base × number of OH− donated per base formula unit
At the equivalence point:
moles of H+ = moles of OH−
Worked examples
Example 1: HCl and NaOH, 1:1 ratio
25.0 cm3 of hydrochloric acid is neutralised by 20.0 cm3 of 0.100 mol dm−3 sodium hydroxide. Calculate the concentration of the hydrochloric acid.
HCl + NaOH → NaCl + H2O
The mole ratio of HCl to NaOH is 1:1.
Convert the sodium hydroxide volume:
20.0 cm3 = 0.0200 dm3
Calculate moles of NaOH:
n = cV = 0.100 × 0.0200 = 0.00200 mol
Since the ratio is 1:1, moles of HCl = 0.00200 mol.
Convert the acid volume:
25.0 cm3 = 0.0250 dm3
Calculate concentration of HCl:
c = n / V = 0.00200 / 0.0250 = 0.0800 mol dm−3
Example 2: H2SO4 and NaOH, 1:2 ratio
25.0 cm3 of sulfuric acid is neutralised by 30.0 cm3 of 0.200 mol dm−3 sodium hydroxide. Calculate the concentration of the sulfuric acid.
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Convert the NaOH volume:
30.0 cm3 = 0.0300 dm3
Calculate moles of NaOH:
n = cV = 0.200 × 0.0300 = 0.00600 mol
From the balanced equation, 2 mol of NaOH reacts with 1 mol of H2SO4.
moles of H2SO4 = 0.00600 / 2 = 0.00300 mol
Convert the acid volume:
25.0 cm3 = 0.0250 dm3
Calculate concentration of H2SO4:
c = 0.00300 / 0.0250 = 0.120 mol dm−3
Example 3: Using H+ and OH−
20.0 cm3 of 0.150 mol dm−3 Ca(OH)2 neutralises 25.0 cm3 of HCl. Calculate the concentration of HCl.
Convert the base volume:
20.0 cm3 = 0.0200 dm3
Moles of Ca(OH)2:
n = cV = 0.150 × 0.0200 = 0.00300 mol
Each Ca(OH)2 provides 2 OH−, so:
moles of OH− = 2 × 0.00300 = 0.00600 mol
At the equivalence point, moles of H+ = 0.00600 mol. Since HCl provides 1 H+ per molecule, moles of HCl = 0.00600 mol.
Convert acid volume:
25.0 cm3 = 0.0250 dm3
concentration of HCl = 0.00600 / 0.0250 = 0.240 mol dm−3
Common exam traps
Frequently Asked Questions
What does a titration calculator calculate?
A titration calculator helps calculate the concentration of an unknown acid or base using concentration, volume and mole ratio information from a titration.
Why is there no separate neutralisation calculator?
Neutralisation is already the chemical idea behind acid-base titration. Since school chemistry usually handles it through balanced equations, mole ratios and n = cV, a separate normality-based neutralisation calculator is not needed here.
What formula is used in titration?
The most common formula is n = cV. After calculating moles of one solution, the balanced equation or H+ to OH− relationship is used to find the moles of the other solution.
Why must I use the balanced equation?
The balanced equation gives the mole ratio between acid and base. Without it, you may wrongly assume that the reacting ratio is always 1:1.
What is a titre?
A titre is the volume delivered from the burette during a titration. In exam questions, the average titre is usually calculated from concordant titres.
Should I use the endpoint or equivalence point?
The endpoint is the observed colour change of the indicator. The equivalence point is the theoretical point where the reacting amounts match exactly. In school calculations, the endpoint is usually treated as close enough to the equivalence point.