IBChemistryDPHLRevision NotesWhat are the Mechanisms of Chemical Change?Proton Transfer ReactionsThe Ion Product of Water

The Ion Product of Water (DP IB Chemistry): Revision Note

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

DP Chemistry: HLDP Chemistry: SL

The ion product of water

pH of water

  • An equilibrium exists in water, where a few water molecules dissociate into proton and hydroxide ions

H2O (l) ⇌ H+ (aq) + OH– (aq)

  • The equilibrium constant for this reaction is:

K subscript straight c space equals space fraction numerator left square bracket straight H to the power of plus right square bracket left square bracket OH to the power of minus right square bracket over denominator left square bracket straight H subscript 2 straight O right square bracket end fraction

Kc x [H2O] = [H+][OH–]

  • Since the concentration of the H+ and OH- ions is very small, the concentration of water is considered to be a constant

  • This means that the expression can be rewritten as:

Kw = [H+] [OH-]

  • Where Kw (ion product of water) =  Kc x [H2O] =  1.00 10-14 at 298K

  • The product of the two ion concentrations is always 1.00 x 10–14 

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

[H+] & [OH–] Table

[H+]

[OH–]

Type of solution

0.1

1 x 10–13

acidic

1 x 10–3

1 x 10–11

acidic

1 x 10–5

1 x 10–9

acidic

1 x 10–7

1 x 10–7

neutral

1 x 10–9

1 x 10–5

alkaline

1 x 10–11

1 x 10–3

alkaline

1 x 10–13

0.1

alkaline

Worked Example

What is the pH of a solution of potassium hydroxide, KOH (aq) of concentration 1.0 × 10−3 mol dm−3 ?

 Kw = 1.0 × 10−14 at 298 K

   A. 3

   B. 4

   C. 10

   D. 11

Answer:

The correct option is D

  • Rearrange Kw = [H+] [OH-]

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

  • Calculate the concentration of [H+]

[H+]  = fraction numerator open parentheses 1.0 space cross times 10 to the power of negative 14 end exponent close parentheses over denominator open parentheses 1.0 space cross times 10 to the power of negative 3 end exponent close parentheses end fraction= 1.0 × 10−11 mol dm−3

  • So the pH = 11

How does temperature affect the ion product of water, Kw?

  • The ionisation of water is an endothermic process

2H2O (l) ⇌ H3O+  (aq) + OH- (aq) 

  • In accordance with Le Châtelier's principle, an increase in temperature will result in the forward reaction being favoured

    • This causes an increase in the concentration of the hydrogen and hydroxide ions

    • This leads to the magnitude of Kw increasing

    • Therefore, the pH will decrease

  • Increasing the temperature decreases the pH of water (becomes more acidic)

  • Decreasing the temperature increases the pH of water (becomes more basic)

Graph to show how Kw changes with temperature

Graph showing \(K_w\) (10\(^-^{14}\)) varying exponentially with temperature (°C), ranging from 10 to 100. \(K_w\) increases from 0 to beyond 40.
As temperature increases, Kw increases so pH decreases

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourselfFlashcards
    Previous:The pH ScaleNext:Strong & Weak Acids
    Author
    Reviewer
    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry Content Creator

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

    Richard Boole

    Reviewer: Richard Boole

    Expertise: Chemistry Content Creator

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.