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Topic 12
pH of bases
You can calculate the pH of a strong base if you take into account the ionic product of water, Kw.
Ionic product of water, Kw
Water reacts with itself in an acid–base reaction:
H2O(l) + H2O(l) ⇋ H3O+(aq ) + OH−(aq )
This can be simplified to:
H2O(l) ⇋ H+(aq ) + OH−(aq )
You can write an expression for Kc but for water
this is called the ionic product of water:
[H2O(l)] is a constant
Kw = [H+(aq )][OH−(aq )]
and is not included in
For pure water at 298 K:
the expression.
• Kw = 1.00 × 10−14 mol2 dm−6
In the same way that pKa = −log10Ka:
pKw = −log10Kw
For pure water at 298 K,
• pKw = 14.0
(a) Explain why sodium hydroxide, NaOH, is a
strong base.
(1 mark)
It is fully dissociated in aqueous solution:
NaOH(aq ) → Na+(aq ) + OH−(aq )
(b) Calculate the hydrogen ion concentration in
0.250 mol dm−3 sodium hydroxide at 298 K.
(Kw = 1.00 × 10−14 mol2 dm−6)
(1 mark)
Kw = [H+(aq )][OH−(aq )]
Kw
so [H+(aq )] = _________
​
​  −     
[OH (aq )]
(1.00 × 10−14)
  ​
  
  
[H+(aq )] = ​ ______________
0.250
= 4.00 × 10−14 mol dm−3
(c) Calculate the pH of this solution at 298 K.
Express your answer to 2 decimal places.
(1 mark)
Neutral pH
In pure, neutral water,
[H+(aq )] = [OH−(aq )].
This means that:
• Kw = [H+(aq )___
]2
+
• [H (aq )] = √​  Kw  
​
You can calculate_____________
the pH of water at 298 K:
pH = –log10(​√​  (1.00
  
× 10−14) ​= 7.00
The dissociation of water is endothermic.
This means that as the temperature increases:
Kw increases
pKw decreases
the pH of pure water decreases.
Neutral pH is only 7.00 at 298 K.
in solution.
Strong bases are fully dissociated
dibasic base).
They include KOH and Ca(OH)2 (a
ociated in
Weak bases are only partially diss
e.
bas
k
wea
a
solution. Ammonia is
The temperature is quoted because the value
for Kw varies with temperature.
When you calculate [OH−(aq )]:
•• [OH–(aq)] = [monobasic strong base]
•• [OH–(aq)] = 2 × [dibasic strong base]
You may have to calculate the pH of a strong
base from its concentration. The steps are:
1. Calculate [OH−(aq )] from [base] (see above).
−
2.Calculate [H+(aq )] from Kw and [OH ]
+
3.Calculate pH using [H (aq )].
pH = −log10(4.00 × 10−14) = −(−13.40)
= 13.40
Enthalpy changes of
neutralisation
Strong acids are fully dissociated in solution
and have the greatest magnitude of DH Uneut.
Weak acids are partially dissociated in solution:
• Energy is needed to dissociate them.
• The magnitudes of their DH Uneut are lower.
82
Calculate the pH of these strong bases at 298 K.
Express your answers to one decimal place.
(a) 0.500 mol dm−3 sodium hydroxide.
(2 marks)
(b) 0.500 mol dm−3 calcium hydroxide, Ca(OH)2.
(2 marks)