All subjects · Chemistry · WCH12 (Unit 2)

Titrations and standard solutions (Core Practicals 3 and 4)

Acid-base titrations find an unknown concentration accurately. Core Practical 3 finds the concentration of hydrochloric acid, and Core Practical 4 makes a standard solution from a solid acid and uses it to find the concentration of sodium hydroxide.

Making a standard solution (Core Practical 4)

A standard solution has an accurately known concentration. Weigh the solid acid (a solid such as sulfamic acid is typical) in a weighing bottle, dissolve it in a beaker of distilled water, transfer it with all the washings to a volumetric flask, make up to the mark so the bottom of the meniscus sits on the line, then stopper and invert to mix.

Titration method (Core Practicals 3 and 4)

Pipette 25.0 cm3 of one solution into a conical flask, add a few drops of indicator, and add the other solution from a burette, swirling, dropwise near the end point. Repeat until you have concordant titres (within 0.10 cm3) and average those.

Indicators: methyl orange (red in acid, yellow in alkali, orange at the end point) or phenolphthalein (colourless in acid, pink in alkali).

Calculation

1. Moles of the known solution: n = c x V.

2. Use the equation ratio (for example H2SO4 + 2NaOH needs 1 : 2).

3. Concentration of the unknown = n / V (V in dm3). For g/dm3, multiply by M.

Uncertainty

Each burette reading is ±0.05 cm3, so a titre (two readings) is ±0.10 cm3. A 25.0 cm3 pipette is typically ±0.06 cm3. Percentage uncertainty = uncertainty / value x 100. Larger titres reduce the percentage uncertainty.

Worked example

25.0 cm3 of NaOH needs 22.50 cm3 of 0.100 mol/dm3 HCl. Find the concentration of the NaOH.

  1. n(HCl) = 0.100 x 0.02250 = 2.25 x 10-3 mol
  2. 1 : 1 ratio, so n(NaOH) = 2.25 x 10-3 mol
  3. c = 2.25 x 10-3 / 0.0250

Answer: 0.0900 mol/dm3

Key idea: n = cV; use the equation ratio; average concordant titres only. Standard solution: weigh, dissolve, transfer with washings, make up in a volumetric flask, invert.

Check yourself

  1. 25.0 cm3 of NaOH needs 22.50 cm3 of 0.100 mol/dm3 HCl. Find the concentration of NaOH in g/dm3. (Na = 23.0, O = 16.0, H = 1.0)
    Show answer

    3.60 g/dm3. c = 0.0900 mol/dm3; x 40.0 = 3.60 g/dm3.

  2. 18.75 cm3 of H2SO4 neutralises 25.0 cm3 of 0.150 mol/dm3 NaOH. H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Find the concentration of the H2SO4.
    Show answer

    0.100 mol/dm3. n(NaOH) = 3.75 x 10-3; n(H2SO4) = 1.875 x 10-3; c = 1.875 x 10-3 / 0.01875 = 0.100 mol/dm3.

  3. 2.43 g of sulfamic acid (M = 97.1 g/mol) is made up to 250.0 cm3. Find its concentration.
    Show answer

    0.100 mol/dm3. n = 2.43 / 97.1 = 0.0250 mol; c = 0.0250 / 0.250 = 0.100 mol/dm3.

  4. A titre of 22.50 cm3 has an uncertainty of ±0.10 cm3. Find the percentage uncertainty.
    Show answer

    0.44 %. 0.10 / 22.50 x 100 = 0.44%.

  5. Titres are 23.40, 22.65, 22.55 and 22.60 cm3. Find the mean of the concordant titres.
    Show answer

    22.60 cm3. 23.40 is not within 0.10 cm3 of the others, so it is left out; (22.65 + 22.55 + 22.60) / 3 = 22.60 cm3.

  6. What colour change is seen at the end point when NaOH is added from a burette to HCl with methyl orange?
    • Yellow to red
    • Colourless to pink
    • Red to orange
    • Pink to colourless
    Show answer

    Red to orange. Methyl orange is red in acid; it turns orange at the end point (yellow in excess alkali).

  7. Why are the beaker and rod rinsed into the volumetric flask when making a standard solution?
    • So that all the solute is transferred and the concentration is accurate
    • To cool the solution
    • To dilute the solution to the right pH
    • To remove the indicator
    Show answer

    So that all the solute is transferred and the concentration is accurate. Any solid left behind would make the concentration too low.

Practise with instant feedback

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Last updated 7 November 2026. Spotted a mistake? Tell us.