AQA A Level Chemistry复习笔记5.5.3 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:

• 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 mol2 dm-6
• This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:

[H+] & [OH–] Table

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+]

• 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+

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

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