Acid Dissociation Constant (AQA A Level Chemistry)
Revision Note
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 expressionsWrite the expression for the following acids:
Benzoic acid, C6H5COOH
Carbonic acid, H2CO3
Answer
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:
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]
Taking the square roots of each side
[H+] = √(Ka x [HA])
Then take the negative logs
pH = -log[H+] = -log√(Ka x [HA])
Worked Example
pH calculations of weak acidsCalculate 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
Answer
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
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