Titration: Concentration of an Unknown

Also known as titration calculation · unknown concentration

Ca=n CbVbVaC_a = \frac{n\,C_b V_b}{V_a}

Worked example: 25.00 mL HCl vs 23.45 mL of 0.1000 M NaOH → 0.09380 M — press Try an example to run it live, then adjust anything.

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Titration: Concentration of an Unknown explained

VbCbnCaVa

At the equivalence point the moles of titrant delivered, CbVbC_b V_b, exactly match the moles of analyte present, CaVaC_a V_a, scaled by the balanced equation's mole ratio n. Everything else in a titration — the burette, the indicator, the swirling — exists only to find that point precisely. Titrate a 25.00 mL aliquot of hydrochloric acid with 0.1000 M sodium hydroxide and take 23.45 mL to reach the endpoint: with n = 1, CaC_a = (1 × 0.1000 × 23.45)/25.00 = 0.09380 M, good to four figures from nothing but glassware.

The mole ratio is where marks are lost. For a diprotic acid such as H₂SO₄ titrated with NaOH, one mole of acid consumes two of base, so n = 0.5: a 25.00 mL aliquot needing 30.00 mL of 0.100 M NaOH is 0.5 × 0.100 × 30.00/25.00 = 0.0600 M. Karl Friedrich Mohr systematised the whole technique in his 1855 Lehrbuch der chemisch-analytischen Titrirmethode, introducing the burette clamp and the pinchcock that made reproducible volumetric analysis possible; his methods still underpin water-hardness and chlorine testing today. Note the difference between the endpoint (where the indicator changes) and the equivalence point (where the stoichiometry balances) — the gap between them is the indicator error, which is why the indicator is chosen to change colour on the steep part of the titration curve.

Titration: Concentration of an Unknown formula

Ca=n CbVbVaC_a = \frac{n\,C_b V_b}{V_a}
Where
  • CaC_a= Analyte concentration (M)
  • VaV_a= Analyte volume (aliquot) (L)
  • CbC_b= Titrant concentration (M)
  • VbV_b= Titre volume delivered (L)
  • nn= Mole ratio (analyte per titrant)

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