Acid-base titrationsEdexcel International A Level Chemistry: Revision notes

Concentration is the amount of solute per unit volume of solution.

• concentration (mol dm⁻³) = amount (mol) ÷ volume (dm³)
• concentration (g dm⁻³) = mass (g) ÷ volume (dm³)
• concentration (g dm⁻³) = concentration (mol dm⁻³) × M (g mol⁻¹)

Convert cm³ to dm³ by dividing by 1000.

Worked example: 1.58 g of H₂C₂O₄·2H₂O (M = 126.1 g mol⁻¹) in 250 cm³: amount = 1.58 ÷ 126.1 = 0.01253 mol, so concentration = 0.01253 ÷ 0.250 = 0.0501 mol dm⁻³, or 1.58 ÷ 0.250 = 6.32 g dm⁻³.

In a titration one solution is added from a burette to a measured volume of the other, in a conical flask, until the end point.

• Methyl orange: red in acid, yellow in alkali, orange at the end point.
• Phenolphthalein: colourless in acid, pink in alkali.

For titrations between a strong acid and a strong base the pH changes sharply through both colour-change ranges, so either is suitable. Phenolphthalein is used for a weak acid such as ethanedioic acid against sodium hydroxide, because the end point is above pH 7.

Calculation steps: balanced equation, amount of the known solution, amount of the unknown using the ratio, then the concentration.

Aim: find the concentration of a hydrochloric acid solution using a standard sodium hydroxide solution.

1. Rinse the burette with the sodium hydroxide and fill it, running some through the tip.
2. Use a pipette to put 25.0 cm³ of the acid into a conical flask and add a few drops of indicator.
3. Do a rough titration first, then add drop by drop near the end point, swirling and using a white tile.
4. Repeat until two titres are concordant (within 0.10 cm³) and calculate the mean.

For 25.0 cm³ of acid needing 21.40 cm³ of 0.100 mol dm⁻³ NaOH: amount = 2.14 × 10⁻³ mol, concentration of acid = 2.14 × 10⁻³ ÷ 0.0250 = 0.0856 mol dm⁻³.

A standard solution has an accurately known concentration. To make one from a solid acid such as ethanedioic acid dihydrate:

1. Weigh the solid accurately in a beaker, by difference.
2. Dissolve it in a little deionised water, then transfer it to a volumetric flask, rinsing the beaker and funnel into the flask.
3. Make up to the mark with deionised water, the last part drop by drop, and invert to mix.

The solution is then used to titrate sodium hydroxide, using H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O, which has a 1 : 2 ratio.

Worked example: 22.60 cm³ of 0.0501 mol dm⁻³ acid neutralises 25.0 cm³ of NaOH: 0.02260 × 0.0501 = 1.132 × 10⁻³ mol, so NaOH = 2.265 × 10⁻³ mol, and concentration = 0.0906 mol dm⁻³ (3.62 g dm⁻³).

Every measurement has an uncertainty. The percentage uncertainty is (absolute uncertainty ÷ measured value) × 100.

• A burette reading has an uncertainty of ±0.05 cm³, so a titre (two readings) has ±0.10 cm³.
• A 25.0 cm³ pipette might have ±0.06 cm³.

Add the percentage uncertainties for the overall percentage uncertainty. For a titre of 24.20 cm³ and a 25.0 cm³ pipette: 0.10 ÷ 24.20 × 100 = 0.41% and 0.06 ÷ 25.0 × 100 = 0.24%, total 0.65%. If c = 0.0968 mol dm⁻³, absolute uncertainty = 0.0006, so c = 0.0968 ± 0.0006 mol dm⁻³.

Minimising uncertainty: use a larger titre, drop-wise additions near the end point, a more precise balance and a larger mass of solid, and repeat to obtain concordant titres.