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Brown's Chemistry Theoretical
Chemistry - Equilibria - Chemical Equilibrium 5.6
5.6 Definition, theory and examples Kb, pKb, Kw weak
base calculations
Revision notes for GCE Advanced Subsidiary Level AS
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KS4
Science GCSE/IGCSE notes acids and bases *
KS4 Science GCSE/IGCSE notes acid-base
theory
Equilibria Part
5 sub-index:
5.1 Lewis and Bronsted-Lowry acid-base theories * 5.2
self-ionisation of water and pH scale * 5.3
strong acids-examples-calculations *
5.4 weak acids-examples & pH-Ka-pKa calculations * 5.5
strong bases-examples-pH calculations
* 5.6 weak bases- examples & pH-Kb-pKb calculations *
5.7 A level notes on Acids, Bases, Salts,
uses of
acid-base titrations - upgrade from GCSE!
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Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principle-rules * Part 2. Kc and Kp equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * 4.
Partition,
solubility product and ion-exchange * Part 5. pH, weak-strong acid-base theory and calculations * Part 6. Salt
hydrolysis,
Acid-base titrations-indicators, pH curves and buffers * Part 7.
Redox equilibria, half-cell electrode potentials,
electrolysis and electrochemical series
*
Part 8. Phase equilibria-vapour
pressure, boiling point and intermolecular forces
5.6
Definition, examples and pH, Kb, pKb
and Kw calculations of weak
bases
-
5.6.1 Definition
and examples of WEAK BASES
-
A weak base
is only weakly or partially ionised in water e.g.
-
A good example
is ammonia
solution, which is only about 2% ionised :
-
NH3(aq)
+ H2O(l)  NH4+(aq) + OH-(aq)
-
Ammonia is the
base and the ammonium ion is its conjugate acid.
-
Water is the
acid and the hydroxide ion is its conjugate base.
-
This
equilibrium is sometimes referred to as a base hydrolysis.
-
The low % of
ionisation gives a less alkaline solution of lower pH than for strong soluble bases (alkalis),
but pH is still > 7.
-
Again, the
concentration of water is considered constant in a similar manner to that for
weak acid equilibrium,
and to solve simple problems, the base ionisation equilibrium
expression is written as:
-
|
Kb =
|
[NH4+(aq)] [OH-(aq)] |
|
----------------------------- |
|
[NH3(aq)] |
-
Kb is
the base ionisation/dissociation constant (mol dm-3)
for any base i.e.
-
B: + H2O(l)
BH+(aq) + OH-(aq)
-
Note [H2O(l)]
is omitted from the Kb expression, i.e. incorporated into
Kb in a similar manner to that for
weak acid equilibrium
expression above.
-
pKb
= -log(Kb/mol dm-3)
-
The bigger Kb
or the smaller the pKb value, the stronger the base. more on comparison with SB
-
note, sometimes
the pKb isn't quoted, but the pKa for the
conjugate acid is!
-
i.e. pKa
for BH+(aq) B:(aq) + H+(aq)
-
In which case it
is useful to know that pKa + pkb = 14
or pKb = 14 - pKa
-
* The weak base - water
interaction can be expressed in terms of the acidity of the
conjugate acid e.g.
-
|
Ka =
|
[NH3(aq)]
[H3O+(aq)] |
|
------------------------------ |
|
[NH4+(aq)] |
-
Note that:
Ka-conj. acid x Kb-base = Kw
and pKa + pKb = pKw, check
it out for yourself.
-
* Apart from the equation, this section is NOT
needed by UK AS-A2 syllabus as far as I can tell?
-
5.6.2 Other
examples of weak bases
-
Aliphatic
amines
-
e.g.
methylamine, ethylamine etc. which are quite soluble in water but
only ionise by a few % like ammonia.
-
R-NH2(aq)
+ H2O(l) R-NH3+(aq) + OH-(aq)
(R = alkyl = CH3, CH3CH2 etc.)
-
5.6.3 Comparison of weak and
strong bases
-
Weak bases are
only partially ionised to give the hydroxide ion and corresponding
cation and the Kb is small.
-
e.g. ammonia:
NH3(aq)
+ H2O(l) NH4+(aq) + OH-(aq)
-
a few % ionised
because Kb = 1.8 x 10-5 mol dm-3 ,
pKb = 4.8
-
Strong bases are
virtually ionised completely to form the hydroxide ion and
corresponding cation and the Kb is large.
-
e.g. sodium
hydroxide: NaOH(s) + aq ==> Na+(aq)
+ OH-(aq)
-
virtually 100%
ionised because Kb is very large, pKb very
negative.
-
5.6.4 Weak base
calculations - calculating the pH of a weak base
-
Calculation example 5.6.4a
-
Calculate the
expected hydroxide and hydrogen ion concentrations and the pH of a
0.40 mol dm-3 solution of ammonia,
-
Kb =
[NH4+(aq)] [OH-(aq)]/[NH3(aq)]
-
As in the case
of weak acids, for simple calculations we assume
-
[NH4+(aq)]
= [OH-(aq)], ignoring any OH-
from water
-
[NH3(aq)]initial
base = [NH3(aq)]equilibrium since the
weak base is only a few % ionised.
-
So we can then
write:
-
Kb =
[OH-(aq)]2/[NH3(aq)] =
1.78 x 10-5 = [OH-(aq)]2
/ 0.40
-
[OH-(aq)]
= √(0.40 x 1.78 x 10-5) = 2.67 x 10-3 mol
dm-3
-
In base
calculations you need to use the ionic product of water expression
to calculate the H+ ion concentration.
-
Kw =
[H+(aq)] [OH-(aq)] = 1 x
10-14 mol2 dm-6, so
-
[H+(aq)]
= Kw/[OH-(aq)] = 1 x 10-14/2.67
x 10-3 = 3.74 x 10-12 mol dm-3
-
pH =
-log(3.74 x 10-12) = 11.4
-
Note: pOH
= pKw - pH = 14 - 11.4 = 2.6
-
Calculation example 5.6.4b
-
A 0.50 mol dm-3
aqueous solution of a very weak base B, has a pH of 9.5.
-
Calculate the
hydrogen and hydroxide ion concentrations in the solution and the
value of the base dissociation constant Kb and pKb.
-
[H+(aq)]
= 10-pH = 10-9.5 = 3.16 x 10-10
mol dm-3
-
Kw =
[H+(aq)] [OH-(aq)] = 1 x
10-14 mol2 dm-6, so
-
[OH-(aq)]
= Kw/[H+(aq)] = 1 x 10-14
/ 3.16 x 10-10 = 3.16 x 10-5 mol dm-3
-
so, using the
simplified expression
-
Kb
= [OH-(aq)]2/[B(aq)] =
(3.16 x 10-5)2 / 0.50 = 2.00 x 10-9
mol dm-3
-
pKb
= -log(2.00 x 10-9) = 8.70
-
Calculation
example 5.6.4c
-
The pKb
value for ethylamine is 3.27
-
(a) Give
the ionisation equation for ethylamine in water and corresponding
equilibrium expression.
-
(b)
Calculate Kb.
-
(c)
Calculate the pH of a 0.25 mol dm-3 aqueous solution of
ethylamine.
-
substituting in
the Kb expression:
-
|
5.37
x 10-4 =
|
[OH-(aq)]2 |
|
--------------- |
|
0.25 |
-
therefore: [OH-(aq)]
= √(5.37 x 10-4 x 0.25) =
0.0116
-
Kw =
[H+(aq)] [OH-(aq)] = 1 x
10-14 mol2 dm-6, so rearranging
-
[H+(aq)]
= 1 x 10-14/0.0116 = 8.62 x 10-13 mol dm-3
-
pH =
-lg(8.62 x 10-13) = 12.1
-
Calculation
example 5.6.4d
-
The pKa
of the conjugate acid of phenylamine is 4.62 (comparable to ethanoic
acid).
-
(a) Give
an ionic equation to show what happens when phenylammonium chloride
is dissolved in water and explain why the solution is acidic.
-
C6H5NH3+(aq)
+ H2O(l)
C6H5NH2(aq) + H3O+(aq)
-
Acidic aqueous
hydrogen/oxonium ions are formed, so lowering the pH.
-
(b)
Calculate the value of Kb for the original phenylamine
base and use the information to justify the classification of
phenylamine as a very weak base.
-
pKa-conj.acid
+ pKb-orig.base = pKw = 14
-
pKb
= 14 - 4.62 = 9.38
-
A relatively
high pKb value means a very weak base and a stronger
conjugate acid (lower pKa) and perhaps the 'weakness' of
the base is best appreciated if the value of Kb is worked
out.
-
Kb =
10-9.38 = 4.17 x 10-10 mol dm-3
-
so in terms of
the equilibrium
-
C6H5NH2(aq)
+ H2O(l)
C6H5NH2(aq) + OH-(aq)
-
there isn't very
much on the RHS!
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