Monitoring reactions: moles, concentration and titrations (Higher)
The mole
One mole = 6.02 × 10²³ particles (Avogadro's constant). The mass of one mole of a substance, in grams, equals its relative formula mass (Mr). Example: 1 mol H₂O = 18 g; 1 mol NaCl = 58.5 g.
n = mass / Mr n = c × V (V in dm³, c in mol/dm³)
Concentration
Concentration is amount of solute per unit volume. Two units appear in OCR papers:
- mass concentration: g/dm³
- molar concentration: mol/dm³
To convert: c (mol/dm³) = c (g/dm³) ÷ Mr.
Remember: 1 dm³ = 1000 cm³ = 1 litre.
Titration (PAG C4)
Titration finds the exact volume of one solution that reacts with a known volume of another, used here for acid–alkali neutralisation.
- Pipette 25.0 cm³ of alkali into a conical flask, add 2–3 drops phenolphthalein (pink in alkali).
- Fill burette with acid, take initial reading to 0.05 cm³.
- Add acid, swirling, until the colour just disappears — this is the end-point.
- Repeat until at least two concordant titres (within 0.10–0.20 cm³).
- Use the mean concordant titre.
✦Worked example
25.0 cm³ of NaOH (0.100 mol/dm³) is exactly neutralised by 22.50 cm³ of HCl. Find c(HCl).
n(NaOH) = 0.100 × (25.0 / 1000) = 0.00250 mol 1:1 ratio in NaOH + HCl → NaCl + H₂O, so n(HCl) = 0.00250 mol c(HCl) = 0.00250 / (22.50 / 1000) = 0.111 mol/dm³ (3 s.f.)
OCR exam tip
Always state significant figures and units. Common loss-of-mark errors: forgetting to convert cm³ → dm³, using the unrounded value once and the rounded value later, or quoting an answer to 4 s.f. when only 3 are justified.
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