This worked example covers Worked Example — Non-1:1 Mole Ratio within Titrations (HT) for GCSE Chemistry. Revise Titrations (HT) in Quantitative Chemistry for GCSE Chemistry with 22 exam-style questions and 20 flashcards. This topic shows up very often in GCSE exams, so students should be able to explain it clearly, not just recognise the term. It is section 6 of 14 in this topic. Treat this as a marking guide for what examiners are looking for, not just a fact list.
🧮 Worked Example — Non-1:1 Mole Ratio
Question: 25.0 cm³ of 0.10 mol/dm³ NaOH solution is neutralised by 12.5 cm³ of sulfuric acid. Calculate the concentration of H₂SO₄.
Equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
n = c × V (V must be in dm³)
n = 0.10 × (25.0 ÷ 1000)
n = 0.10 × 0.0250
n = 0.0025 mol NaOH
NaOH : H₂SO₄ = 2 : 1 (from the balanced equation)
So moles H₂SO₄ = 0.0025 ÷ 2 = 0.00125 mol
This is the critical step! The ratio is 2:1, not 1:1 — always check the balanced equation first.
c = n ÷ V
c = 0.00125 ÷ (12.5 ÷ 1000)
c = 0.00125 ÷ 0.0125
c = 0.10 mol/dm³
Why this is harder than the 1:1 example: With H₂SO₄ and NaOH, you need two moles of NaOH for every one mole of H₂SO₄. If you forget to apply the ratio and use moles NaOH = moles H₂SO₄, you get 0.20 mol/dm³ — double the correct answer. The balanced equation always tells you the ratio.
Practice questions for Titrations (HT)
What is the purpose of a titration?
Explain why the burette should be rinsed with the acid solution before filling it for a titration.