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Q24 · Titration3 marks

10.00 mL of vinegar is diluted to 100.0 mL. 20.00 mL aliquots need a mean titre of 16.85 mL of 0.1025 mol L⁻¹ NaOH. Find the concentration of ethanoic acid in the vinegar.

n(NaOH) = 0.1025 × 0.01685 = 1.727 × 10⁻³ mol✓ 1
n = cV, with the titre in litres
c(diluted) = 1.727 × 10⁻³ ÷ 0.02000 = 0.08636 mol L⁻¹✓ 1
1 : 1 ratio, over the aliquot volume
c(vinegar) = 0.08636 mol L⁻¹0 / 1
Marker: The vinegar was diluted ten-fold, so multiply back: 0.864 mol L⁻¹.

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