Solution concentrations and titration calculationsEdexcel A-Level Chemistry: Revision notes

The concentration of a solution is the amount of solute in 1 dm³ of solution.

c (mol dm⁻³) = n (mol) ÷ V (dm³), so n = c × V. Concentration in g dm⁻³ = c (mol dm⁻³) × M (g mol⁻¹).

Volumes are usually given in cm³, so divide by 1000 to get dm³. For example, 25.0 cm³ = 0.0250 dm³.

Worked example: 4.00 g of NaOH in 500 cm³ of solution: n = 4.00 ÷ 40.0 = 0.100 mol; c = 0.100 ÷ 0.500 = 0.200 mol dm⁻³ (or 8.00 g dm⁻³).

In an acid-base titration, a solution of known concentration is added from a burette to an exact volume of the other solution, measured with a pipette, until the end-point is reached, shown by an indicator.

• Rinse the burette with the solution it will hold and the pipette with the solution it will deliver, never with water alone.
• Do a rough titration first, then accurate ones, adding dropwise near the end-point and swirling.
• Read the bottom of the meniscus at eye level.
• Repeat until you have concordant titres (within 0.10 cm³) and take their mean. Never include the rough titre.

An indicator changes colour near the end-point.

• Phenolphthalein: pink in alkali, colourless in acid. With alkali in the flask and acid added, the end-point is pink to colourless; with acid in the flask, colourless to the first permanent pale pink.
• Methyl orange: yellow in alkali, red in acid, orange at the end-point.

For a strong acid with a strong alkali, such as HCl with NaOH, either indicator is suitable. Methyl orange is usually chosen when the alkali is weak (for example ammonia) and phenolphthalein when the acid is weak (for example ethanoic acid), because the colour change must occur at a pH that matches the end-point.

1. Write the balanced equation.
2. Calculate n of the solution of known concentration: n = c × V (V in dm³), using the mean titre.
3. Use the equation ratio to find n of the other reactant.
4. Divide by its volume in dm³ to get the concentration.
5. Multiply by M for g dm⁻³ if asked.

Worked example: 25.0 cm³ of NaOH needs 21.60 cm³ of 0.100 mol dm⁻³ HCl. n(HCl) = 0.100 × 0.02160 = 2.16 × 10⁻³ mol. NaOH + HCl is 1 : 1, so n(NaOH) = 2.16 × 10⁻³ mol and c = 2.16 × 10⁻³ ÷ 0.0250 = 0.0864 mol dm⁻³.

With a diprotic acid such as H₂C₂O₄ or H₂SO₄, the ratio to NaOH is 1 : 2.

Use a solid acid, such as ethanedioic acid dihydrate, which is stable and has a large molar mass, so weighing errors are small.

1. Weigh the acid by difference.
2. Dissolve it in a beaker in a little distilled water.
3. Transfer to a volumetric flask, rinsing the beaker and funnel into the flask.
4. Make up to the mark with a dropper, stopper and invert to mix.
5. Calculate c = (mass ÷ M) ÷ V.

Then titrate 25.0 cm³ portions of the standard solution with the sodium hydroxide, using phenolphthalein, and calculate the sodium hydroxide concentration.

Titrate the hydrochloric acid, in the conical flask, with sodium hydroxide of known concentration from the burette, using phenolphthalein or methyl orange.

The calculation follows the same steps: n(NaOH) = c × mean titre, n(HCl) = n(NaOH), c(HCl) = n ÷ 0.0250. If the acid has been diluted beforehand, multiply by the dilution factor to find the original concentration.

Must know: c = n ÷ V (dm³); g dm⁻³ = mol dm⁻³ × M; use concordant titres only; match the mole ratio to the equation.