# IB DP Chemistry: SL复习笔记8.2.4 The Ionic Product of Water

### The Ionic Product of Water

#### pH of water

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

H2O(l) ⇌ H+(aq) + OH-(aq)

• The equilibrium constant for this reaction is: Kc x [H2O] = [H+] [OH-]

• Since the concentration the H+ and OH- ions is very small, the concentration of water is considered to be a constant, such that the expression can be rewritten as:

Kw = [H+] [OH-]

Where Kw (ionic product of water)=Kc x [H2O]=10-14 moldm-6 at 298K

• The product of the two ion concentrations is always 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 #### 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 moldm-6

A. 3

B. 4

C. 10

D. 11

The correct option is D.

• Since Kw = [H+] [OH-] , rearranging gives [H+]  = Kw ÷ [OH-]
• The concentration of  [H+] is (1.0 × 10−14) ÷ (1.0 × 10−3) = 1.0 × 10−11 mol dm−3
• So the pH = 11 