## Acid dissociation constant, K_{a} and pK_{a}

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 :

CH_{3}COOH(aq) + H_{2}O(l) ⇌ H_{3}O^{+}(aq) + CH_{3}COO^{–}(aq)

In its simplified form:

CH_{3}COOH(aq) ⇌ H^{+}(aq) + CH_{3}COO^{–}(aq)

The equilibrium expression for the above reaction is:

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

Whereby K_{a} is called the acid dissociation constant. At 298K the value of K_{a}for the dissociation of ethanoic acid is 1.74 × 10^{–5}mol/dm^{3}.

The units of K_{a} are deduced the same way as for K_{c}.

*The value of K _{a} indicates the extent of dissociation of the acid,* and therefore:

- a high value of K
_{a}indicates that the position of equilibrium lies to the right, which implies that the acid is almost completely ionised. - a low value of K
_{a}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 K_{a} (weak acids) we can use pK_{a} values to compare their strengths, whereby

- pK
_{a}= –log_{10}K_{a}

The less positive the value of pK_{a}, the more acidic is the acid.

## Calculating K_{a} for a weak acid

In order to calculate the value of K_{a} 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 K_{a} 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 K_{a} and pK_{a} of a weak acid

Calculate the value of K_{a} and pK_{a} for 0.05 mol/dm^{3} propanoic acid, which has a pH of 3.

### Solution

To calculate K_{a} we first need to know the value of [H^{+}].

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

^{+}] = 10

^{-3}

= 0.001 mol/dm^{3}

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 pK_{a}:

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

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

$$pK_a\ = 4.7$$

## 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, K
_{a}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 pK_{a}

Calculate the pH of 0.1mol/dm^{3} ethanoic acid, CH_{3}COOH, given that K_{a} = 1.74 × 10^{-5}mol/dm^{3}.

### Solution

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

CH_{3}COOH(aq) ⇌ H^{+}(aq) + CH_{3}COO^{–}(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 = –log_{10}[H^{+}]

= –log_{10}(1.32 × 10^{–3})

= 2.88