Acid Dissociation Constant (AQA A Level Chemistry): Revision Note

Exam code: 7405

Stewart Hird

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on

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

Equilibria Dissociation of a Weak Acid, downloadable AS & A Level Chemistry revision notes

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

The acid dissociation constant, downloadable AS & A Level Chemistry revision notes
  • 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:

  1. Benzoic acid, C6H5COOH

  2. Carbonic acid, H2CO3

Answer

WE1 Writing Ka Expressions, downloadable AS & A Level Chemistry revision notes

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:

The acid dissociation constant, downloadable AS & A Level Chemistry revision notes
  • 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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Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.