How to do titration calculations

A-Level Chemistry Titration calculations

The route from titre to concentration, a worked acid-base example, and why concordant results matter before you calculate anything.

What the examiner is testing

Two separate things: whether you select the right titres, and whether you can carry a mole ratio through a calculation with the units intact. Both are marked.

The formula:

$$ n = c \times \frac{V}{1000} $$

with \( n \) in moles, \( c \) in mol/dm³ and \( V \) in cm³.

Before any calculation: concordant results

Use only titres within 0.10 cm³ of one another, and never the rough titre. Averaging in a rough run is one of the most common mark losses on the whole topic.

Run Titre (cm³) Use it?
Rough 25.40 No — always discarded
1 24.75 Yes
2 24.80 Yes
3 25.15 No — outside 0.10 of the others

Mean titre: \( (24.75 + 24.80)/2 = 24.78 \text{ cm}^3 \).

The method

  1. Average the concordant titres only.
  2. Write and balance the equation.
  3. Moles of the substance you know everything about.
  4. Cross to the other substance using the balancing numbers.
  5. Divide by its volume in dm³ to get concentration.

Worked example

25.0 cm³ of sodium hydroxide is titrated against 0.100 mol/dm³ hydrochloric acid. The mean concordant titre is 24.78 cm³. Find the concentration of the sodium hydroxide.

Step 2 — equation.

$$ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} $$

The ratio is \( 1 : 1 \).

Step 3 — moles of acid (the substance with both values known):

$$ n(\text{HCl}) = 0.100 \times \frac{24.78}{1000} = 2.478 \times 10^{-3} \text{ mol} $$

Step 4 — cross the equation. With a 1:1 ratio:

$$ n(\text{NaOH}) = 2.478 \times 10^{-3} \text{ mol} $$

Step 5 — concentration of the alkali.

$$ c = \frac{n}{V/1000} = \frac{2.478 \times 10^{-3}}{0.0250} = 0.0991 \text{ mol/dm}^3 $$

Answer: 0.0991 mol/dm³ (3 s.f.).

Sanity check: the titre was slightly less than the 25.0 cm³ pipetted, so the alkali must be slightly less concentrated than the acid. 0.0991 against 0.100 fits.

If the acid were sulfuric

$$ \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} $$

Only step 4 changes: \( n(\text{NaOH}) = 2 \times n(\text{H}_2\text{SO}_4) \). Everything else is identical.

The three mistakes that lose marks

1. Including the rough titre in the mean. Discard it, always.

2. Forgetting the 1000. Working in cm³ throughout gives an answer a thousand times out. Write the units beside every number.

3. Assuming a 1:1 ratio. Check the balanced equation. Diprotic acids and Group 2 hydroxides are where this is deliberately tested.

30-second recap

Concordant titres only. Balance the equation. Moles of the fully-known substance, cross using the balancing numbers, divide by the other volume in dm³. Then ask whether the size of your answer makes sense against the titre.

Common questions

Only the concordant ones — those within 0.10 cm3 of each other. The rough titre is never included, and averaging it in is a guaranteed lost mark.

Because concentration is in mol/dm3 and burette readings are in cm3. There are 1000 cm3 in 1 dm3, so a volume in cm3 must be divided by 1000 before use.

The mole ratio. Sulfuric acid reacts with sodium hydroxide in a 1:2 ratio, so the same titre corresponds to half the acid concentration you would get assuming 1:1.

More revision guides

Written by StudyAI to cover a topic students ask about often. It uses its own worked example — no exam board's questions are reproduced here.