AQA A Level Chemistry

Revision Notes

5.5.3 The Ionic Product of Water

The Ionic Product of Water

  • 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

pH of Strong Bases

Strong bases

  • 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 Kw by 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+] = K÷ [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]= K÷  [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

Worked Example

What is the pH of a solution of hydroxide ions of concentration 1.0 × 10−3 mol dm−3 ?

Kw = 1 × 10−14 moldm-6

A. 3.00

B. 4.00

C. 10.00

D. 11.00

Answer

The correct option is D.

    • Since Kw = [H+] [OH], rearranging gives [H+]  = Kw ÷ [OH]

The concentration of  [H+] is (1 × 10−14) ÷ (1.0 × 10−3) = 1.0 × 10−11 mol dm−3

[H+]= 10-pH

So the pH = 11.00

Exam Tip

Always give the pH to two decimal places.

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