3.4.3 Acids and Bases - The ionic product of water, Kw
Students should:
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Water equilibrium
Water is in equilibrium with its dissociated ions (hydrogen and hydroxide).
The equilibrium:
H2O
H+ + OH-
Can be expressed according to the equilibrium law:
Kc =
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[H+][OH-]
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[H2O]
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However, as the concentration of the water effectively remains constant on both sides of the equilibrium then the [H2O] term can be removed to a very close approximation and the equilibrium constant denoted as Kw (sometimes called the ionic product of water).
This gives:
Kw = [H+][OH-]
The constant Kw remains unchanged at constant temperature (as all good constants should!).
At 25ºC the value of Kw = 1 x 10-14 mol2 dm-6
As the concentration of the hydrogen ions equals the concentration of the hydroxide ions then the concentration of hydrogen ions in pure water at 25ºC = the square root of the ionic product of water:
[H+] = 1 x 10-7 mol dm-3
All equilibrium constants are temperature dependent (and this one is no exception):
The dissociation of water molecules into ions is bond breaking and is therefore an endothermic process (energy must be absorbed to break the bonds). Endothermic processes are favoured by an increase in temperature and so as the temperature rises the equilibrium moves further to the right hand side and Kw gets larger.
As Kw gets larger, so do the values of the hydrogen ion concentration and the hydroxide ion concentration.
As pH is a measure of the hydrogen ion concentration (pH = -log[H+]) then as the temperature increases the pH gets lower - i.e. the water becomes more acidic.
Variation of Kw with temperature
The equilibrium
H2O
H+ + OH-
involves the breaking of bonds and is therefore endothermic - energy must be applied to break one of the the H-O-H bonds to give the ions. Consequently, according to Le Chatelier, an increase in temperature favours the forward reaction - i.e. the position of equilibrium shifts towards the right hand side and Kw becomes larger.
However, as the ratio of hydrogen ions to hydroxide ions in pure water must remain 1:1, then if we know the value of Kw, it is a simple matter to calculate the value of either H+ or/and OH- to obtain the concentrations and hence the values of pH and pOH.
Example: Calculate the pH when Kw = 6.5 x 10-14 mol2 dm-6 As... Kw = [H+][OH-] and... [H+] =[OH-] Then... Kw = [H+]2 Therefore... [H+] = √ Kw [H+] = √ 6.5 x 10-14 [H+] =2.55 x 10-7 pH = 6.59 The pOH value is the same as [H+] =[OH-] |
pH of strong bases
Once we have established that the dissociation constant of water is a fixed value at constant temperature, it allows us to calculate the pH of bases by finding the hydroxide ion concentration and then applying the kw equilibrium equation.
Example: Find the pH of a solution of sodium hydroxide of concentration 0.02 mol dm-3 NaOH is a strong base and dissociates 100% in solution NaOH Therefore the concentration of hydroxide ions = the original concentration of NaOH = 0.02 kw = [H+][OH-] = 1 x 10-14 Hence, [H+] = 1 x 10-14/0.02 = 5 x 10-13 pH = -log[H+] = 12.3 |