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Chemical Calculations

Acid-base titration calculations

  • 3 things to know
  • Components 1 and 2

Overview

Titration designed to find, equivalence point and the end point of a titration and standardising a solution.

What you need to know

Answer each one before you reveal it.

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Numbers

25.0 cm³ of sodium hydroxide solution requires 22.5 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the concentration of the alkali.

Key term

Distinguish between the equivalence point and the end point of a titration.

Key term

What is a titre, and what makes a set of titres concordant?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    What is a titration designed to find?

    Model answer

    A titration finds an unknown concentration by measuring exactly what volume of a solution of known concentration reacts with a fixed volume of the unknown.

  2. 2

    Distinguish between the equivalence point and the end point of a titration.

    Model answer

    The equivalence point is where the reactants have combined in exactly the ratio of the balanced equation, and the end point is where the indicator changes colour, an indicator being chosen so that the two coincide.

  3. 3

    What is a titre, and what makes a set of titres concordant?

    Model answer

    A titre is the volume delivered from the burette to reach the end point, and titres are concordant when they agree within 0.10 cm³ of one another.

  4. 4

    Why is the first, rough titration left out of the mean titre?

    Model answer

    It is run quickly to locate the approximate end point and is not accurate enough to be included.

  5. 5

    25.0 cm³ of sodium hydroxide solution requires 22.5 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the concentration of the alkali.

    Model answer

    The acid supplies 0.0225 × 0.100 = 2.25 × 10⁻³ mol, and the ratio is one to one, so the concentration is 2.25 × 10⁻³ ÷ 0.0250 = 0.0900 mol dm⁻³.

  6. 6

    What volume of 0.0500 mol dm⁻³ sulfuric acid neutralises 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide?

    Model answer

    The alkali supplies 2.50 × 10⁻³ mol and two moles of it react with one of the acid, so 1.25 × 10⁻³ mol of acid is needed, a volume of 1.25 × 10⁻³ ÷ 0.0500 = 0.0250 dm³, that is 25.0 cm³.

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