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

Exam code: 9CHO

Last updated

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

pKa

  • The range of values of Ka is very large and for weak acids, the values themselves are very small numbers

Table of Kvalues

Table of Ka values, downloadable AS & A Level Chemistry revision notes
  • For this reason it is easier to work with another term called pKa 

  • The pKa  is the negative log of the Ka value, so the concept is analogous to converting [H+] into pH values

pK= -logKa

  • Looking at the pKa  values for the same acids:

Table of pKvalues

Table of pKa values, downloadable AS & A Level Chemistry revision notes
  • The range of pKa values for most weak acids lies between 3 and 7

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

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na+ (aq) + OH- (aq) 

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

Answer 1:

   The pH of the solution is:

   [H+] = fraction numerator K subscript 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 space mol space dm to the power of negative 3 end exponent end fraction = 6.66 x 10-14

   pH = -log[H+]

         = -log 6.66 x 10-14  = 13.17

Worked Example

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na+ (aq) + OH- (aq)  

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

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-] = fraction numerator K subscript w over denominator open square brackets H to the power of plus close square brackets end fraction

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-]= fraction numerator 1 space cross times 10 to the power of negative 14 end exponent over denominator 3.16 cross times 10 to the power of negative 11 end exponent end fraction

 [OH-]= 3.16 x 10-4 mol dm-3

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