Ionic Product of Water (Edexcel International A Level (IAL) Chemistry): Revision Note

Exam code: YCH11

Stewart Hird

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on

Ionic Product of Water, Kw

  • In all aqueous solutions, an equilibrium exists in water where a few water molecules dissociate into protons and hydroxide ions

  • We can derive an equilibrium constant for the reaction:

Deriving Kw, downloadable AS & A Level Chemistry revision notes
  • This is a specific equilibrium constant called the ionic product for water

  • The product of the two ion concentrations is always 1 x 10-14 moldm-6

  • This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:

[H+] & [OH] Table

[H+] and [OH-] table, downloadable IB Chemistry revision notes

The relationship between Kand pKw is given by the following equation:

pKw = -logKw

pH Calculation of a Strong Base

  • Strong bases are completely ionised in solution

BOH (aq) → B+ (aq) + OH- (aq)

  • Therefore, the concentration of hydroxide ions [OH-] is equal to the concentration of base [BOH]

    • Even strong alkalis have small amounts of H+ in solution which is due to the ionisation of water

  • The concentration of OH- in solution can be used to calculate the pH using the ionic product of water

  • Once the [H+] has been determined, the pH of the strong alkali can be founding using pH = -log[H+]

Finding pH of strong bases, downloadable AS & A Level Chemistry revision notes
  • Similarly, the ionic product of water can be used to find the concentration of OH- ions in solution if [H+] is known, simply by dividing Kby the [H+]

Worked Example

pH calculations of a strong alkali

Question 1: 

Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH

Question 2: 

Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50

Answer

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na+ (aq) + OH- (aq) 

Answer 1:

   The pH of the solution is:

   [H+] = Kw  ÷ [OH-]

   [H+] = (1 x 10-14) ÷ 0.15 = 6.66 x 10-14

   pH = -log[H+]

         = -log 6.66 x 10-14  = 13.17

Answer 2

Step 1: Calculate hydrogen concentration by rearranging the equation for pH

   pH = -log[H+]

   [H+]= 10-pH

   [H+]= 10-10.50

   [H+]= 3.16 x 10-11 mol dm-3

Step 2: Rearrange the ionic product of water  to find the concentration of hydroxide ions

   Kw = [H+] [OH-]

    [OH-]= Kw  ÷  [H+

Step 3: Substitute the values into the expression to find the concentration of hydroxide ions

   Since Kw is 1 x 10-14 mol2 dm-6,

    [OH-]= (1 x 10-14)  ÷  (3.16 x 10-11)

   [OH-]= 3.16 x 10-4 mol dm-3

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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.