Worked Example — Non-1:1 Mole Ratio
Part of Titrations (HT) — GCSE Chemistry
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 appears regularly enough that it should still be part of a steady revision cycle. 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.
Topic position
Section 6 of 14
Practice
22 questions
Recall
20 flashcards
🧮 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.