pH of weak acids from Ka and pKa

Overview

Recall questions on the Ka expression and the approximations used, calculating the pH of a weak acid from its Ka and concentration, working back from a measured pH to Ka, the relationship between pH and pKa at half neutralisation, and when the approximations break down.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   State the two assumptions made when the pH of a weak acid is calculated from Ka.

   Model answer

   That [H⁺] equals [A⁻] because the water's own ionisation is negligible, and that the equilibrium concentration of the acid equals the concentration made up because so little of it dissociates.

2. 2

   State the expression used to find the hydrogen ion concentration of a weak acid.

   Model answer

   [H⁺] = √(Ka × [HA]).

3. 3

   Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, for which Ka is 1.75 × 10⁻⁵ mol dm⁻³.

   Model answer

   [H⁺] = √(1.75 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³, so the pH is 2.88.

4. 4

   Calculate the pH of 0.0500 mol dm⁻³ methanoic acid, for which Ka is 1.60 × 10⁻⁴ mol dm⁻³.

   Model answer

   [H⁺] = √(8.00 × 10⁻⁶) = 2.83 × 10⁻³ mol dm⁻³, so the pH is 2.55.

5. 5

   Calculate the pH of 0.200 mol dm⁻³ propanoic acid, for which Ka is 1.30 × 10⁻⁵ mol dm⁻³.

   Model answer

   [H⁺] = √(2.60 × 10⁻⁶) = 1.61 × 10⁻³ mol dm⁻³, so the pH is 2.79.

6. 6

   Calculate Ka for a 0.0800 mol dm⁻³ solution of a weak acid whose pH is 3.60.

   Model answer

   [H⁺] = 10⁻³·⁶⁰ = 2.51 × 10⁻⁴ mol dm⁻³, so Ka = (2.51 × 10⁻⁴)² ÷ 0.0800 = 7.9 × 10⁻⁷ mol dm⁻³.

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