GCSE Chemistry - How to Solve: Concentration (g/dm³ & mol/dm³) and Titration Calculations

How to Solve: Concentration (g/dm³ & mol/dm³) and Titration Calculations

? Introduction

"Mastering concentration and titration calculations unlocks 1 in 10 marks on your GCSE/A-Level Chemistry exam—and real-world jobs like drug dosing, water testing, and food safety. One wrong decimal? You lose 3 marks. Get this right, and you’re already ahead."

? WHAT YOU NEED TO KNOW FIRST

Before starting, you must understand:
1. Moles – The unit for amount of substance (1 mole = 6.02 × 10²³ particles).
2. Molar mass – Mass of 1 mole of a substance (g/mol), found from the periodic table.
3. Volume units – 1 dm³ = 1000 cm³ (like 1 litre = 1000 ml).

If you’re shaky on these, pause and revise them first.

? KEY TERMS & FORMULAS

1. Concentration (g/dm³)

Formula: Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³) - Mass of solute = solid dissolved (grams). - Volume of solution = total liquid volume (dm³). - MEMORISE THIS – Not always given on the exam sheet.

Example: 5 g of salt in 2 dm³ of water → Concentration = 5 ÷ 2 = 2.5 g/dm³.

2. Concentration (mol/dm³)

Formula: Concentration (mol/dm³) = moles of solute (mol) ÷ volume of solution (dm³) - Moles of solute = mass ÷ molar mass. - Volume of solution = total liquid volume (dm³). - MEMORISE THIS – Core formula for titration questions.

Alternative (given on exam sheet): n = c × V - n = moles (mol) - c = concentration (mol/dm³) - V = volume (dm³)

3. Titration Calculations

Key idea: At the endpoint, moles of acid = moles of alkali (if 1:1 ratio). Steps:
1. Write the balanced equation (e.g., HCl + NaOH → NaCl + H₂O).
2. Use n = c × V to find moles of the known solution.
3. Use the mole ratio to find moles of the unknown.
4. Rearrange n = c × V to find the unknown concentration.

MEMORISE: "Moles first, then concentration."

? STEP-BY-STEP METHOD

For g/dm³ Concentration:

1. Write down the mass of solute (g) and volume of solution (dm³).
2. Divide mass by volume: concentration = mass ÷ volume.
3. Check units – volume must be in dm³ (not cm³).

For mol/dm³ Concentration:

1. Find moles of solute: moles = mass ÷ molar mass.
2. Write down the volume of solution (dm³).
3. Divide moles by volume: concentration = moles ÷ volume.
4. Check units – volume must be in dm³.

For Titration Calculations:

1. Write the balanced equation (e.g., HCl + NaOH → NaCl + H₂O).
2. Find moles of the known solution using n = c × V.
3. Volume must be in dm³ (divide cm³ by 1000).
4. Use the mole ratio to find moles of the unknown.
5. Example: If 1 mole HCl reacts with 1 mole NaOH, moles of unknown = moles of known.
6. Rearrange n = c × V to find the unknown concentration.
7. Check units – answer must be in mol/dm³.

✏️ WORKED EXAMPLES

Example 1 – Basic (g/dm³)

Question: 8 g of sodium hydroxide (NaOH) is dissolved in 4 dm³ of water. What is the concentration in g/dm³?

Steps:
1. Mass of solute = 8 g.
2. Volume of solution = 4 dm³.
3. Concentration = mass ÷ volume = 8 ÷ 4 = 2 g/dm³.

What we did and why: - We used the g/dm³ formula because the question asked for grams per dm³. - Volume was already in dm³, so no conversion was needed.

Example 2 – Medium (mol/dm³)

Question: 5.3 g of sodium carbonate (Na₂CO₃) is dissolved in 250 cm³ of water. What is the concentration in mol/dm³? (Molar mass of Na₂CO₃ = 106 g/mol)

Steps:
1. Find moles of Na₂CO₃: - Moles = mass ÷ molar mass = 5.3 ÷ 106 = 0.05 mol.
2. Convert volume to dm³: - 250 cm³ ÷ 1000 = 0.25 dm³.
3. Calculate concentration: - Concentration = moles ÷ volume = 0.05 ÷ 0.25 = 0.2 mol/dm³.

What we did and why: - We converted cm³ to dm³ because the formula requires volume in dm³. - We found moles first because the question asked for mol/dm³.

Example 3 – Exam-Style (Titration)

Question: 25.0 cm³ of sodium hydroxide (NaOH) solution is titrated with 0.100 mol/dm³ hydrochloric acid (HCl). It takes 20.0 cm³ of HCl to reach the endpoint. What is the concentration of the NaOH solution?

Steps:
1. Write the balanced equation: - HCl + NaOH → NaCl + H₂O (1:1 ratio).
2. Find moles of HCl: - Volume of HCl = 20.0 cm³ = 0.020 dm³. - Moles of HCl = c × V = 0.100 × 0.020 = 0.002 mol.
3. Use mole ratio (1:1): - Moles of NaOH = moles of HCl = 0.002 mol.
4. Find concentration of NaOH: - Volume of NaOH = 25.0 cm³ = 0.025 dm³. - Concentration = moles ÷ volume = 0.002 ÷ 0.025 = 0.08 mol/dm³.

What we did and why: - We used the mole ratio to link HCl and NaOH. - We converted cm³ to dm³ before using the formula. - The 1:1 ratio meant moles of NaOH = moles of HCl.

❌ COMMON MISTAKES

? EXAM TRAPS