Why do we use the 'Moles Bridge' in titration questions?

Because the concentration and volume known for one chemical (the titrant) cannot be directly applied to the unknown chemical. You must calculate the moles of the known chemical first, then use the molar ratio from the balanced chemical equation to 'cross the bridge' and find the moles of the unknown chemical.

O-Level Chemistry: Titration & Moles Calculations

Q: What is the formula linking moles, concentration, and volume?

The formula is n = c × v. 'n' is the number of moles. 'c' is the concentration in mol/dm³. 'v' is the volume. CRITICALLY, volume MUST be in dm³ before you multiply.

Q: How do you convert cm³ to dm³?

Divide the volume in cm³ by 1000. 1 dm³ is equal to 1000 cm³. For example, 25 cm³ becomes 0.025 dm³.

Q: What does concordant results mean in titration?

Concordant results are titre volumes that are within 0.10 cm³ or 0.20 cm³ of each other. In Paper 4, you must only average the concordant results, ignoring any rough trials or anomalies.

A burette delivering acid into an Erlenmeyer flask containing alkali and phenolphthalein indicator.

What is the biggest mistake made in O-Level titration calculations?

Failing to convert cubic centimeters (cm³) into cubic decimeters (dm³). In the formula `Moles = Concentration x Volume`, the volume MUST be in dm³. If the question says 25cm³, you must divide by 1000 to get 0.025dm³ before multiplying. If you don't do this, your mole answer will be 1000 times too large.

1. The Holy Grail Formula (n = cv)
2. The 3-Step Moles Bridge Strategy
3. Worked Exam Question
4. Frequently Asked Questions

Titration math represents the hardest calculation marks in O-Level Chemistry. Whether you are facing Paper 2 theory or Paper 4 practical alternatives, examiners use titration to test your understanding of stoichiometry. This guide from our Ultimate O-Level Chemistry Guide gives you the exact algorithm to solve any titration problem.

1. The Holy Grail Formula (n = cv)

When dealing with solutions (liquids dissolved in water), there is only one formula you need:

n = c × v

n: Number of Moles (mol)
c: Concentration (mol/dm³)
v: Volume. This MUST be in dm³.

💡 Exam tip

The Volume Conversion Trap: Burettes and pipettes measure in cm³. The formula demands dm³. Always immediately divide any cm³ value by 1000. 25 cm³ / 1000 = 0.025 dm³.

2. The 3-Step "Moles Bridge" Strategy

In a titration, you have a KNOWN chemical in the burette, and an UNKNOWN chemical in the flask below. You cannot jump straight from the burette to the flask. You must cross the "Moles Bridge".

Step 1: Focus on the KNOWN chemical. You will be given both its Concentration and Volume. Use n = c × v to find the number of moles.
Step 2: Cross the Bridge using the Balanced Equation. Look at the big numbers in front of the elements. If it takes 2 moles of acid to neutralize 1 mole of alkali, it's a 2:1 ratio. Divide your mole answer from Step 1 by 2. This gives you the number of moles of the UNKNOWN chemical.
Step 3: Focus on the UNKNOWN chemical. You now have its moles (from step 2) and its volume (from the pipette). Rearrange n = c × v to c = n / v to find the missing concentration.

3. Worked Exam Question

Question:

25.0 cm³ of Sodium Hydroxide (NaOH) was neutralised by 20.0 cm³ of Sulfuric Acid (H2SO4) which had a concentration of 0.1 mol/dm³. Calculate the concentration of the Sodium Hydroxide.

2NaOH + H2SO4 → Na2SO4 + 2H2O

Step 1 — Calculate Moles of the KNOWN chemical (H2SO4)

We know everything about the acid.
Volume of acid = 20.0 / 1000 = 0.020 dm³.
Moles of acid (n) = c × v
n = 0.1 × 0.020 = 0.002 moles of H2SO4

Step 2 — Use the balanced equation ratio (The Bridge)

Look at the equation: 2NaOH + 1H2SO4.
The ratio of NaOH to Acid is 2:1. It takes twice as much alkali.
Moles of NaOH = 0.002 × 2 = 0.004 moles of NaOH.

Step 3 — Find the unknown concentration

Volume of NaOH = 25.0 / 1000 = 0.025 dm³.
c = n / v
c = 0.004 / 0.025
Concentration of NaOH = 0.16 mol/dm³

📋 Exam technique note

If the question asks you to convert your final answer from mol/dm³ into g/dm³, you do NOT need volume again. Just multiply your concentration by the Relative Formula Mass (Mr) of the chemical. For NaOH, Mr = 23 + 16 + 1 = 40. So, 0.16 mol/dm³ × 40 = 6.4 g/dm³. That's an easy guaranteed final mark!

Frequently Asked Questions

What is the formula linking moles, concentration, and volume?▼

Moles = Concentration × Volume (n = cv). Remember that volume must be in dm³.

How do you convert cm³ to dm³?▼

Divide the cm³ value by 1000.

Why do we use the 'Moles Bridge'?▼

Because the different chemicals react in specific stoicheometric ratios (e.g. 2:1). You must convert the known chemical's volume to moles, apply the ratio, and then convert back for the unknown chemical.

What does concordant results mean in titration?▼

Titre volumes that are very close to each other (usually within 0.1 or 0.2 cm³). You only average these results in Paper 4, discarding any rough or anomalous trials.

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Titration Calculations: Walking the Moles Bridge