The Ionic Product of Water, Kw (Oxford AQA International A Level Chemistry)

Revision Note

Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Stewart Hird

Ionic Product of Water Calculations

  • In all aqueous solutions, some water molecules dissociate into protons and hydroxide ions

  • An equilibrium is established:

H2O (l) ⇌ H+ (aq) + OH- (aq)

  • Kw is the ionic product of water

    • It is the equilibrium constant for the dissociation of water at 298 K

    • Its value is 1.00 x 10-14 mol2 dm-6

  • For the ionisation of water, the equilibrium expression to find Kw is:

H2O (l) ⇌ H+ (aq) + OH- (aq)

Kwfraction numerator stretchy left square bracket H to the power of plus stretchy right square bracket bold space stretchy left square bracket OH to the power of minus stretchy right square bracket over denominator stretchy left square bracket H subscript 2 O stretchy right square bracket end fraction

  • As the extent of ionisation is very low, only small amounts of H+ and  OH- ions are formed

  • The concentration of H2O can therefore be regarded as constant and removed from the Kw expression

  • The equilibrium expression therefore becomes:

Kw = [H+] [OH-]

  • As the [H+] = [OH-] in pure water, the equilibrium expression can be further simplified to:

Kw = [H+]2

Worked Example

Calculate the concentration of H+ in pure water, using the ionic product of water.

Kw = 1 × 10−14 moldm-6

Answer:

  • Step 1: Write down the equation for the partial dissociation of water:

    • In pure water, the following equilibrium exists:

H2O (l) ⇌ H+ (aq) + OH- (aq)

  • Step 2: Write down the equilibrium expression to find Kw:

    • Kwfraction numerator stretchy left square bracket H to the power of plus stretchy right square bracket bold space stretchy left square bracket OH to the power of minus stretchy right square bracket over denominator stretchy left square bracket H subscript 2 O stretchy right square bracket end fraction

  • Step 3: Simplify the expression:

    • Since the concentration of H2O is constant, this expression can be simplified to:

    • Kw = [H+] [OH-]

  • Step 4: Further simplify the expression:

    • The ratio of H+ to OH- is 1:1

    • The concentration of H+ and OH- is, therefore, the same and the equilibrium expression can be further simplified to:

    • Kw = [H+]2

  • Step 5: Rearrange the equation to find [H+]:

    • [H+] = square root of bold K subscript bold w end root

  • Step 6: Substitute the values into the expression to find Kw:

    • [H+] = square root of 1.00 cross times 10 to the power of negative 14 end exponent end root

    • [H+] = 1.00 x 10-7 mol dm-3

The effect of temperature on Kw

  • The dissociation of water to form hydrogen and hydroxide ions is an endothermic process so absorbs heat energy

H2O (l) ⇌ H+ (aq) + OH- (aq)

  • If temperature is increased, the forward reaction will be favoured to counteract the change and lower the temperature

  • Th equilibrium will shift to the right and more H+ and OH- ions will be formed causing the value of Kw to increase

  • So, as temperature increases the value of Kw increase

  • If the value of Kw increases, the pH decreases as shown below:

    T (°C)

    Kw (mol2 dm-6)

    pH

    0

    0.114 x 10-14

    7.47

    10

    0.293 x 10-14

    7.27

    20

    0.681 x 10-14

    7.08

    25

    1.008 x 10-14

    7.00

Calculating the pH 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 found using pH = -log[H+]

stretchy left square bracket H to the power of plus stretchy right square bracket bold equals fraction numerator bold K subscript bold w over denominator stretchy left square bracket OH to the power of minus stretchy right square bracket end fraction

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bold pH bold space bold equals bold space bold minus bold log bold left square bracket bold H to the power of bold plus bold right square bracket

  • 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 Kw by the [H+]

Worked Example

Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH at 25 °C.

Kw at 25 °C = 1 × 10−14 moldm-6

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na+ (aq) + OH- (aq) 

Answer:

The pH of the solution is:

  • [H+] = fraction numerator K w over denominator open square brackets O H to the power of minus close square brackets end fraction

  • [H+] = fraction numerator 1 space cross times 10 to the power of negative 14 end exponent over denominator 0.15 end fraction = 6.66 x 10-14

  • pH = -log[H+]

  • pH= -log 6.66 x 10-14  = 13.17

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Alexandra Brennan

Author: Alexandra Brennan

Expertise: Chemistry

Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom at SME.

Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Lead

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.