# 5.6.1 Acid Dissociation Constant

### Acid Dissociation Constant

#### Weak acids

• A weak acid is an acid that partially (or incompletely) dissociates in aqueous solutions
• Eg. most organic acids (ethanoic acid), HCN (hydrocyanic acid), H2S (hydrogen sulfide) and H2CO3 (carbonic acid)
• The position of the equilibrium is more over to the left and an equilibrium is established

The diagram shows the partial dissociation of a weak acid in aqueous solution

• As this is an equilibrium we can write an equilibrium constant expression for the reaction

• This constant is called the acid dissociation constant, Ka, and has the units mol dm-3
• Values of Ka are very small, for example for ethanoic acid Ka = 1.74 x 10-5 mol dm-3
• When writing the equilibrium expression for weak acids, the following assumptions are made:
• The concentration of hydrogen ions due to the ionisation of water is negligible
• The value of Ka indicates the extent of dissociation
• The higher the value of Ka the more dissociated the acid and the stronger it is
• The lower the value of Ka the weaker the acid

#### Worked Example

Writing Ka expressions

Write the expression for the following acids:

1. Benzoic acid, C6H5COOH
2. Carbonic acid, H2CO3

### pH of Weak Acids

#### Weak acids

• The pH of weak acids can be calculated when the following is known:
• The concentration of the acid
• The Ka value of the acid
• From the Ka expression we can see that there are three variables:

(Ka expression)

• However, the equilibrium concentration of [H+] and [A] will be the same since one molecule of HA dissociates into one of each ion
• This means you can simplify and re-arrange the expression to

Ka x [HA] = [H+]2

[H+]2 Ka x [HA]

[H+] = √(Ka x [HA])

• Then take the negative logs

pH = -log[H+] = -log√(Ka x [HA])

#### Worked Example

pH calculations of weak acids

Calculate the pH of 0.100 mol dm-3 ethanoic acid at 298 k with a Ka value of 1.74 × 10-5 mol dm-3

Ethanoic acid is a weak acid which ionises as follows:

CH3COOH (aq) ⇌ H+ (aq) + CH3COO (aq)

Step 1: Write down the equilibrium expression to find Ka

Step 2: Simplify the expression

The ratio of H+ to CH3COO ions is 1:1

The concentration of H+ and CH3COO ions are therefore the same

The expression can be simplified to:

Step 3: Rearrange the expression to find [H+]

Step 4: Substitute the values into the expression to find [H+]

= 1.32 x 10-3 mol dm-3

Step 5: Find the pH

pH = -log[H+]

= -log(1.32 x 10-3)

= 2.88

Close