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:

The-pH-Scale-equation

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

8.1.8-H-and-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

Answer:

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

 

 

 

 

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