## Acid dissociation constant, Ka and pKa

We can apply the equilibrium expression aqueous solutions of weak acids and weak bases, such as in the reaction below when ethanoic acid dissolves in water :

CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO(aq)

In its simplified form:

CH3COOH(aq) ⇌ H+(aq) + CH3COO(aq)

The equilibrium expression for the above reaction is:

$$K_a\ = \frac{[H^+][CH_3COO^–]}{[CH_3COOH]}$$

Whereby Ka is called the acid dissociation constant. At 298K the value of Kafor the dissociation of ethanoic acid is 1.74 × 10–5mol/dm3.

The units of Ka are deduced the same way as for Kc.

The value of Ka indicates the extent of dissociation of the acid, and therefore:

• a high value of Ka indicates that the position of equilibrium lies to the right, which implies that the acid is almost completely ionised.
• a low value of Ka indicates that the position of equilibrium lies to the left, which implies that the acid is only slightly ionised.

For acids with very low values of Ka (weak acids) we can use pKa values to compare their strengths, whereby

• pKa = –log10Ka

The less positive the value of pKa, the more acidic is the acid.

## Calculating Ka for a weak acid

In order to calculate the value of Ka for a weak acid we need to know:

• the concentration of the acid.
• the pH of the solution.

The general equation of dissociation of a monobasic acid:

HA(aq) → H+(aq) + A(aq)

shows us that for each acid molecule, HA, that ionises produces one
H+ ion and one A ion, which means [H+] = [A].

Hence we can rewrite the equilibrium expression:

$$K_a\ = \frac{[H^+][A^–]}{[HA]}$$

as

$$K_a\ = \frac{[H^+]^2}{[HA]}$$

To calculate Ka we make the following assumptions:

• the concentration of hydrogen ions produced by the ionisation of the water molecules present in the solution is negligible.
• the ionisation of the weak acid is so small that the concentration of undissociated HA molecules present at equilibrium is approximately the same as that of the original acid.

### Example 1 – calculating Ka and pKa of a weak acid

Calculate the value of Ka and pKa for 0.05 mol/dm3 propanoic acid, which has a pH of 3.

### Solution

To calculate Ka we first need to know the value of [H+].

Given the pH of the acid, we calculate [H+] using the formula [H+] = 10-pH.

[H+] = 10-3

= 0.001 mol/dm3

Then:

$$K_a\ = \frac{[H^+]^2}{[HA]}$$

$$K_a\ = \frac{(0.001\ mol/dm^3)^2}{0.05\ mol/dm^3}$$

$$K_a\ = 0.00002\ mol/dm^3$$

$$K_a\ = 2\times 10^{-5}\ mol/dm^3$$

Now to find pKa:

$$pK_a\ = -log_{10}K_a$$

$$pK_a\ = -log_{10}(2\times 10^{-5})$$

$$pK_a\ = 4.7$$

## How to calculate the pH of a weak acid

To calculate the pH value (or [H+]) of a weak acid we need to know:

• the concentration of the acid
• the value of the acid dissociation constant, Ka for the acid.

Then we make the following assumptions:

• the concentration of hydrogen ions produced by the ionisation of the water molecules present in the solution is negligible.
• the ionisation of the weak acid is so small that the concentration of undissociated molecules present at equilibrium is approximately the same as that of the original acid.

### Example 2 – How to calculate the pH of a weak acid given pKa

Calculate the pH of 0.1mol/dm3 ethanoic acid, CH3COOH, given that Ka = 1.74 × 10-5mol/dm3.

### Solution

First we need to determine the equilibrium expression for the dissociation reaction:

CH3COOH(aq) ⇌ H+(aq) + CH3COO(aq)

$$K_a\ = \frac{[H^+]^2}{[HA]}$$

$$K_a\ = \frac{[H^+]^2}{[CH_3COOH]}$$

$$1.74 × 10^{–5}\ = \frac{[H^+]^2}{0.1}$$

$$[H^+]^2 = 1.74 × 10^{–5} × 0.1 = 1.74 × 10^{–6}$$

$$[H^+] = \sqrt{1.74 × 10^{–6}}$$

$$= 1.32 × 10^{–3}\ mol/dm^3$$

Then calculate pH using the formula:

pH = –log10[H+]

= –log10(1.32 × 10–3)

= 2.88