Some acids ionise in solution to a greater extent than others, making it useful to have information on the position of equilibrium for a given acid and to be able to compare the degree of ionisation – acid strength. The position of equilibrium can be described by the equilibrium constant, K_a.

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[Explaining the acid dissociation constant. Animation with voice over explaining is the clearest way of doing this]

What assumption has been made in the expressions above?

The concentration of undissociated molecules [HA] is assumed to be equal to the concentration of the acidic solution, due to very few molecules dissociating in a weak acid.

Calculate the concentration of H⁺ ions in a solution of a weak monoprotic acid with a concentration of 0.01 mol dm^–3. K_a of the acid is 1.8 x 10^–4.

[H⁺_(aq)]= √K_a x c = √ 1.8 x 10^–4 x 0.01 = 1.34 x 10^–3 mol dm^–3

A solution of ethanoic acid has a hydrogen ion concentration of 1.30 x 10^–3 mol dm^–3. Calculate the concentration of ethanoic acid using a value of 1.7 x 10^-5 for the K_a of ethanoic acid.

c = [H⁺_(aq)]² / K_a = (1.3 x 10^–3)² / 1.7 x 10^–5  = 1.7 x 10^–6 / 1.7 x 10^–5 = 0.1 mol dm^–3

Another useful expression is pK_a = – log₁₀ K_a. This expression is analagous to the familiar pH and [H+] expression. Most school or exam data books give a list of pK_a values alongside the K_a values. These numbers can be more manageable than their coressponding K_a values.

Consider how an expression can also be derived that usefully links pH and pK_a?

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Using the expression pH = ½ pKa – ½ log₁₀ c, calculate the pH of a solution of benzoic acid that has a concentration of 0.20 mol dm^–3. (benzoic acid pKa = 4.20)

pH = ½ pK_a – ½ log₁₀ c = ½(4.2) – ½(–0.70) = 2.45

The pH of a solution of propanoic acid is measured as pH 4.30. Calculate the concentration of the acid (propanoic acid pKa = 4.87)

4.23 = ½(4.87) – ½(log10c) therefore 2pH = pKa – log10c;

– log10 c = 2pH – pKa; 8.60–4.87 = 3.73

c = 1.86 x 10^-4 mol dm^–3