Acid Dissociation Constant

pH of a solution

You might have come across the term pH while studying Chemistry. pH is a measure of the acidic strength of a solution. For the sake of understanding, it can be considered as a scaled quantity of concentration of H⁺ ions in a solution.

$pH=-{\log }_{10}\left(\left[H^{+}\right]\right)$ $=-{\log }_{10}\left[{\mathrm{H}}^{+}\right]$

The pH of a solution can range from 0 to 14. Solutions with pH less than 7 are acidic, while those with a pH above 7 are basic. The characteristic sour taste of lemons is due to the presence of citric acid, giving it a pH of ≈ 2. The black coffee you have in the morning has a pH of ≈ 5. Saliva has a pH of ≈ 6.6. On the basic side, the baking soda you have in the kitchen has a pH of 9.5. Liquid drain cleaners have a pH of ≈ 14 (very strong, which is why we use them with care; they can hurt skin).

Acid dissociation constant of HCl

A common example of a strong acid is hydrochloric acid (HCl). For HCl, the ionisation reaction would be written like this:

$\mathrm{HCl}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{Cl}}^{-}$

The equilibrium constant for the above reaction would be calculated as:

$K_a=\frac{\left[H_3{O}^{+}\right]\left[C{l}^{-}\right]}{\left[HCl\right]\left[H_{2}O\right]}$

The H⁺ ions are transferred to H₂O and form hydronium ions (H₃O⁺). This is because free H⁺ ions do not exist in an aqueous solution. For simplicity, we can replace [H₃O⁺] with [H⁺]. Also, [H₂O] is equal to 1 as the concentration of water is 1. Hence, the equation for K_a becomes:

$K_a=\frac{\left[H^{+}\right]\left[C{l}^{-}\right]}{\left[HCl\right]}$

HCl is a strong acid. This means that it dissociates completely into H⁺ and Cl^- ions in solution. For example, if 1 mol of HCl is added to 1 L of water, it will give out 1 mol of H⁺ ions and 1 mol of Cl^- ions.

This can be represented with the help of an ICE table (Initial, Change, Equilibrium), where we can write the concentrations of the reactants and the products of the reaction at the initial stage (Initial), the change in their concentration from initial stage till equilibrium (Change), and their final concentrations at equilibrium (Equilibrium).

In the ICE table for the ionisation reaction of HCl, we have written the concentrations of the reactants and the products at the initial stage, and the final stage. We can now substitute these values in the formula for K_a for HCl.

$K_a=\frac{1·1}{(verysmallnumber)}$

There is a very small number in the denominator, as there is always a very small amount of undissociated acid at equilibrium. Therefore, the value of K_a for strong acids is very large. K_a for HCl is very high, at 1.3 x 10⁶.

In this subsection, we learnt that strong acids are so-called because they dissociate completely into H⁺ ions and the corresponding conjugate bases.

But what about acids that cannot dissociate completely in a solution? Those acids are called weak acids.

Determination of the dissociation constant of a weak acid

Let us consider an acid HB. The ionisation reaction equation will be written as:

$HB\rightleftharpoons H^{\mathit{+}}+B^{\mathit{-}}$

Consider HB as a weak acid. Weak acids do not dissociate completely, i.e. if 1 mol of HB is added to water, at equilibrium the solution will have less than 1 mol of H⁺ and B^-, and will also contain a certain concentration of undissociated HB at equilibrium (unlike the case of strong acid, which only contained H⁺ and the conjugate base at equilibrium). Due to incomplete dissociation/ionisation of the weak acid, the solution is not as acidic as it would be if an equal quantity of a strong acid was added to the solution.

Let us take the example of acetic acid (CH₃COOH), a weak acid. Consider adding 1 mol of acetic acid to 1 L of water. The concentration of acetic acid would be 1 mol/L or 1M. The equation of the reaction can be written as:

$C{H}_{3}COOH+H_{2}O\mathit{}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+C{H}_{3}CO{O}^{\mathit{-}}$

At equilibrium, the K_a for this equation can be written as:

$K_a=\frac{\left[H^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

Just as we drew an ICE table for the ionisation of HCl, we can draw the ICE table for the ionisation of acetic acid. This will help us better understand the concentrations of the reactants and the products at different stages of the reaction.

Weak acids do not dissociate completely, which is why the change in concentration of CH₃COOH is -x. We use the variable x because we do not know how much the acid will dissociate, and hence how many H⁺ ions or CH₃COO^- ions there will be in the solution.

You know that acetic acid is a weak acid. You have also understood that weak acids do not ionise completely in a solution. Therefore, in the equation of K_a for acetic acid, the concentration of the H⁺ ions and the conjugate base is less than 1M, while the concentration of the reactants is a significant value, as we know there is still undissociated acid at equilibrium (unlike the case of strong acids). So, for weak acids, the numerator while calculating K_a is small, whilst the denominator is big. This implies that the value of K_a for weak acids is small. K_a for acetic acid is 1.8 × 10⁻⁵ .

Examples of other weak acids are:

• Methanoic acid - HCOOH (K_a = 1.78 × 10⁻⁴)
• Benzoic acid - C₆H₅COOH (K_a = 6.3 × 10⁻⁵)
• Hypochlorous acid - HClO (K_a = 2.9 × 10⁻⁸)
• Hydrocyanic acid - HCN (K_a = 6.2 × 10⁻¹⁰)