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Equilibrium and Acid-Base Equilibria

Calculating the pH of a buffer solution

  • 3 things to know
  • Unit A2 1

Overview

Recall questions on the buffer expression and why it uses the ratio of acid to salt, calculating the pH of ethanoic acid and propanoic acid buffers, working from an acid partly neutralised by alkali, why pH equals pKa at half-neutralisation, finding the ratio needed for a target pH, and showing the effect of added alkali.

What you need to know

Answer each one before you reveal it.

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Key term

State the expression used to find the hydrogen ion concentration of an acidic buffer.

Key term

Explain why the pH at half-neutralisation of a weak acid equals its pKa.

Key term

A buffer of pH 4.58 rises only to pH 4.76 when a quantity of alkali is added. Explain how this demonstrates buffering

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the expression used to find the hydrogen ion concentration of an acidic buffer.

    Model answer

    The hydrogen ion concentration equals Ka multiplied by the concentration of the acid divided by the concentration of its salt.

  2. 2

    Explain why the buffer expression uses the ratio of acid to salt rather than their absolute concentrations.

    Model answer

    Both concentrations appear in the same volume of solution, so the volume cancels and only their ratio, or equivalently the ratio of moles, determines the pH.

  3. 3

    Calculate the pH of a buffer containing 0.100 mol dm⁻³ ethanoic acid and 0.100 mol dm⁻³ sodium ethanoate, given Ka = 1.74 × 10⁻⁵ mol dm⁻³.

    Model answer

    The ratio is one, so the hydrogen ion concentration equals Ka at 1.74 × 10⁻⁵ mol dm⁻³ and the pH is 4.76.

  4. 4

    Calculate the pH of a buffer containing 0.200 mol dm⁻³ ethanoic acid and 0.100 mol dm⁻³ sodium ethanoate, given Ka = 1.74 × 10⁻⁵ mol dm⁻³.

    Model answer

    The hydrogen ion concentration is 1.74 × 10⁻⁵ × 2 = 3.48 × 10⁻⁵ mol dm⁻³, giving a pH of 4.46.

  5. 5

    50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid is mixed with 20.0 cm³ of 0.200 mol dm⁻³ sodium hydroxide. Calculate the moles of ethanoic acid remaining.

    Model answer

    There are 0.0100 mol of acid and 0.00400 mol of alkali, so 0.00600 mol of ethanoic acid remains unreacted.

  6. 6

    For that mixture of 0.00600 mol ethanoic acid and 0.00400 mol ethanoate, calculate the pH given Ka = 1.74 × 10⁻⁵ mol dm⁻³.

    Model answer

    The hydrogen ion concentration is 1.74 × 10⁻⁵ × (0.00600 ÷ 0.00400) = 2.61 × 10⁻⁵ mol dm⁻³, giving a pH of 4.58.

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