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A titration is a quantitative technique used to determine the concentration (or purity) of one solution by reacting it with a solution of accurately known concentration (a standard solution, or titrant). For acid-base titrations we monitor the point at which the acid and base have exactly reacted - the end-point, which should match the theoretical equivalence point.
A standard solution is made from a primary standard - a pure, stable, non-hygroscopic solid of known formula. Examples include anhydrous sodium carbonate, Na2CO3, and potassium hydrogen phthalate (KHP).
| Apparatus | Purpose | Uncertainty (typical) |
|---|---|---|
| 25.00 cm3 volumetric pipette | Deliver fixed volume of one reagent | +/- 0.06 cm3 |
| 250.0 cm3 volumetric flask | Make the standard solution | +/- 0.30 cm3 |
| 50.00 cm3 burette | Deliver variable volume of titrant | +/- 0.05 cm3 per reading (so +/- 0.10 cm3 per titre) |
| Analytical balance | Weigh solid | +/- 0.001 g |
| Conical flask | Hold the analyte (colour change observed against white tile) | - |
n = c x V (V in dm3) or n = (c x V)/1000 (V in cm3)
Use the balanced equation to find moles of the other reactant.
For unknown concentration: c = n/V For percentage purity: %purity = (theoretical mass / actual mass) x 100
Q: 25.0 cm3 of a sodium hydroxide solution of unknown concentration is titrated against 0.100 mol dm-3 sulfuric acid. The mean titre is 22.50 cm3. Calculate the concentration of the NaOH solution.
Equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O
Step 1: Moles of H2SO4 = (0.100 x 22.50) / 1000 = 2.25 x 10^-3 mol Step 2: Moles of NaOH = 2 x 2.25 x 10^-3 = 4.50 x 10^-3 mol Step 3: [NaOH] = (4.50 x 10^-3) / (25.0/1000) = 0.180 mol dm-3
Q: 1.25 g of impure sodium carbonate is dissolved in water and made up to 250.0 cm3 in a volumetric flask. 25.0 cm3 of this solution is titrated against 0.100 mol dm-3 HCl, requiring a mean titre of 22.80 cm3. Calculate the percentage purity of the sample.
Equation: Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
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