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Chemical Calculations

Topic 5 of 6

Concentration and titration

Volume tells you moles only when concentration is known.

O-Level 6092 (2026) / SEC G3 K324 (2027)

Concentration and titration

Volume tells you moles only when concentration is known.

Molar concentration c = n/V, with V in dm3. Hence n = cV. Mass concentration in g dm-3 = mass/volume; dividing this by molar mass converts to mol dm-3. Always convert cm3 to dm3 by dividing by 1000 before using n = cV.

Worked example

A titration with a 1:2 ratio

25.0 cm3 of sulfuric acid requires 20.0 cm3 of 0.150 mol dm-3 NaOH. Find the acid concentration.

  1. H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l).
  2. n(NaOH) = 0.150 x 0.0200 = 0.00300 mol.
  3. n(acid) = 0.00300/2 = 0.00150 mol.
  4. c(acid) = 0.00150/0.0250 = 0.0600 mol dm-3.
Answer

0.0600 mol dm-3. A direct cV = cV shortcut would miss the factor of two.

In a real titration, calculate titre from final minus initial burette reading. Repeat until sufficiently close titres are obtained and average concordant results, excluding the rough run. For example, 19.85 and 19.95 cm3 average to 19.90 cm3. Use the supplied equation for an unfamiliar reaction; the calculation path stays the same.

Check your understandingA solution contains 8.0 g NaOH per dm3. Find its molar concentration; M = 40 g mol-1.Think it through, then reveal the answer
8.0/40 = 0.20 mol dm-3. The units show that grams cancel when mass concentration is divided by molar mass.