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Quiz yourself on Titrations (HT)Get your flashcardsBuild my revision planTrack my progress🧮 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.
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