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Acid-Base Titrations

Titration calculations

  • 3 things to know
  • Unit AS 1

Overview

Recall questions on the sequence of steps in a titration calculation, working concentrations for hydrochloric and sulfuric acid against sodium hydroxide, the sodium carbonate reaction, converting between mol dm⁻³ and g dm⁻³, finding a degree of hydration, diluting a vinegar before titration, and the effect of common procedural errors.

What you need to know

Answer each one before you reveal it.

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Process

Describe the general sequence of steps in a titration calculation.

Numbers

25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is neutralised by 22.50 cm³ of hydrochloric acid. Calculate the concentration of the acid.

Key term

Write the equation for the reaction of sulfuric acid with sodium hydroxide and state the mole ratio.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Describe the general sequence of steps in a titration calculation.

    Model answer

    The moles in the known solution are calculated from its concentration and volume, the balanced equation gives the moles of the other reactant, and that value is used with its volume or mass to find the unknown quantity.

  2. 2

    25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is neutralised by 22.50 cm³ of hydrochloric acid. Calculate the concentration of the acid.

    Model answer

    The alkali is 2.50 × 10⁻³ mol and the reaction is one to one, so the acid concentration is 2.50 × 10⁻³ ÷ 0.02250 = 0.111 mol dm⁻³.

  3. 3

    Write the equation for the reaction of sulfuric acid with sodium hydroxide and state the mole ratio.

    Model answer

    The equation is H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l), so one mole of the acid reacts with two moles of the alkali.

  4. 4

    What volume of 0.100 mol dm⁻³ sodium hydroxide is needed to neutralise 25.0 cm³ of 0.0500 mol dm⁻³ sulfuric acid?

    Model answer

    The acid is 1.25 × 10⁻³ mol and needs twice that in alkali, so 2.50 × 10⁻³ ÷ 0.100 = 0.0250 dm³, which is 25.0 cm³.

  5. 5

    Write the equation for the reaction of sodium carbonate with hydrochloric acid.

    Model answer

    The equation is Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g).

  6. 6

    25.0 cm³ of sodium carbonate solution requires 23.40 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the concentration of the carbonate.

    Model answer

    The acid is 2.34 × 10⁻³ mol, so the carbonate is half that at 1.17 × 10⁻³ mol, giving 1.17 × 10⁻³ ÷ 0.0250 = 0.0468 mol dm⁻³.

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