Doc
Brown's Chemistry
Theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5.6
5.6 Definition of a weak base, theory and examples of K_{b}, pK_{b}, K_{w} weak
base calculations
What is a weak base? What is the Kb of a weak base – base
dissociation constant? How do we calculate the pH of a weak base–alkali
solution? What is the pKb of a base? Why do we need to use Kw in weak
base pH calculations? How do we write equilibrium expressions to show
the dissociation–ionisation of a weak acids? How do we calculate the Ka
of a weak base? All of these terms are defined and explained below with
suitable worked out examples. What is a weak acid? How do we write
equilibrium expressions to show the dissociation–ionisation of a weak
acids? What is a weak acid's pKa? How do we calculate the pH of a
solution of a weak acid? How do we calculate the Ka of a weak acid?
GCSE/IGCSE
reversible reactions–equilibrium notes
*
GCSE/IGCSE notes on acids and bases
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–K_{a}–pK_{a} calculations * 5.5
strong bases–examples–pH calculations
* 5.6 weak bases– examples & pH–K_{b}–pK_{b} calculations *
5.7 A level notes on Acids, Bases, Salts,
uses of
acid–base titrations – upgrade from GCSE!
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principle–rules * Part 2. K_{c} and K_{p} 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, K_{b}, pK_{b}
and K_{w} 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 :

NH_{3(aq)}
+ H_{2}O_{(l)} NH_{4}^{+}_{(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:

K_{b} =

[NH_{4}^{+}_{(aq)}] [OH^{–}_{(aq)}] 
––––––––––––––––––––––––––––– 
[NH_{3(aq)}] 

K_{b} is
the base ionisation/dissociation constant (mol dm^{–3})
for any base i.e.

B: + H_{2}O_{(l)}
BH^{+}_{(aq)} + OH^{–}_{(aq)}

Note [H_{2}O_{(l)}]
is omitted from the K_{b} expression, i.e. incorporated into
K_{b} in a similar manner to that for
weak acid equilibrium
expression above.

pK_{b}
= –log(K_{b}/mol dm^{–3})

The bigger K_{b}
or the smaller the pK_{b} value, the stronger the base. more on comparison with SB

note, sometimes
the pK_{b} isn't quoted, but the pK_{a} for the
conjugate acid is!

i.e. pK_{a}
for BH^{+}_{(aq)} B:_{(aq)} + H^{+}_{(aq)}

In which case it
is useful to know that pK_{a} + pk_{b} = 14
or pK_{b} = 14 – pK_{a}

* The weak base – water
interaction can be expressed in terms of the acidity of the
conjugate acid e.g.

K_{a} =

[NH_{3(aq)}]
[H_{3}O^{+}_{(aq)}] 
––––––––––––––––––––– 
[NH_{4}^{+}_{(aq)}] 

Note that:
K_{a–conj. acid} x K_{b–base} = K_{w}
and pK_{a} + pK_{b} = pK_{w}, 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–NH_{2(aq)}
+ H_{2}O_{(l)} R–NH_{3}^{+}_{(aq)} + OH^{–}_{(aq)}
(R = alkyl = CH_{3}, CH_{3}CH_{2} 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 K_{b} is small.

e.g. ammonia:
NH_{3(aq)}
+ H_{2}O_{(l)} NH_{4}^{+}_{(aq)} + OH^{–}_{(aq)}

a few % ionised
because K_{b} = 1.8 x 10^{–5} mol dm^{–3} ,
pK_{b} = 4.8

Strong bases are
virtually ionised completely to form the hydroxide ion and
corresponding cation and the K_{b} is large.

e.g. sodium
hydroxide: NaOH_{(s)} + aq ==> Na^{+}_{(aq)}
+ OH^{–}_{(aq)}

virtually 100%
ionised because K_{b} is very large, pK_{b} 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,

K_{b} =
[NH_{4}^{+}_{(aq)}] [OH^{–}_{(aq)}]_{/}[NH_{3(aq)}]

As in the case
of weak acids, for simple calculations we assume

[NH_{4}^{+}_{(aq)}]
= [OH^{–}_{(aq)}], ignoring any OH^{–}
from water

[NH_{3(aq)}]_{initial
base} = [NH_{3(aq)}]_{equilibrium} since the
weak base is only a few % ionised.

So we can then
write:

K_{b} =
[OH^{–}_{(aq)}]^{2}_{/}[NH_{3(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.

K_{w} =
[H^{+}_{(aq)}] [OH^{–}_{(aq)}] = 1 x
10^{–14} mol^{2} dm^{–6}, so

[H^{+}_{(aq)}]
= K_{w}_{/}[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
= pK_{w} – 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 K_{b} and pK_{b}.

[H^{+}_{(aq)}]
= 10^{–pH} = 10^{–9.5} = 3.16 x 10^{–10}
mol dm^{–3}

K_{w} =
[H^{+}_{(aq)}] [OH^{–}_{(aq)}] = 1 x
10^{–14} mol^{2} dm^{–6}, so

[OH^{–}_{(aq)}]
= K_{w}_{/}[H^{+}_{(aq)}] = 1 x 10^{–14}
/ 3.16 x 10^{–10 }= 3.16 x 10^{–5} mol dm^{–3}

so, using the
simplified expression

K_{b}
= [OH^{–}_{(aq)}]^{2}_{/}[B_{(aq)}] =
(3.16 x 10^{–5})^{2} / 0.50 = 2.00 x 10^{–9}
mol dm^{–3}

pK_{b}
= –log(2.00 x 10^{–9}) = 8.70

Calculation
example 5.6.4c

The pK_{b}
value for ethylamine is 3.27

(a) Give
the ionisation equation for ethylamine in water and corresponding
equilibrium expression.

(b)
Calculate K_{b}.

(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

K_{w} =
[H^{+}_{(aq)}] [OH^{–}_{(aq)}] = 1 x
10^{–14} mol^{2} 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

5.6.4d is an example of
approaching weak base pH calculations from the point of view of the
K_{a} of the conjugate acid of the weak base.

The pK_{a}
of the conjugate acid of the aromatic weak base phenylamine is 4.62

(a) Give
an ionic equation to show what happens when phenylammonium chloride
is dissolved in water and explain why the solution is acidic.

C_{6}H_{5}NH_{3}^{+}_{(aq)}
+ H_{2}O_{(l)}
C_{6}H_{5}NH_{2(aq)} + H_{3}O^{+}_{(aq)}

In aqueous media the
solution becomes acidic because
hydrogen ion/oxonium ions are formed, so lowering the pH by
proton donation from the conjugate acid to the water molecules,
which in this case act as the base.

(b)
Calculate the value of K_{b} for the original phenylamine
base and use the information to justify the classification of
phenylamine as a very weak base.

pK_{a–conj.acid}
+ pK_{b–orig.base} = pK_{w} = 14

pK_{b}
= 14 – 4.62 = 9.38

A relatively
high pK_{b} value means a very weak base and a stronger
conjugate acid (relatively low pK_{a}), but the 'weakness' of
the base is best appreciated by students if the value of K_{b} is worked
out.

so in terms of
the equilibrium

C_{6}H_{5}NH_{2(aq)}
+ H_{2}O_{(l)}
C_{6}H_{5}NH_{3}^{+}_{(aq)} + OH^{–}_{(aq)}

Note, that if given a pK_{a}
for the conjugate acid of a weak base, its easy to calculate the pK_{b},
then K_{b} and then perform pH and concentration
calculations as exemplified by 5.6.4a–c,

but, equally, you can
readily calculate the pH of a salt solution of the salt of a weak
base and strong acid using a weak acid
calculation (section 5.4) – in the example below, it is
essentially a hydrolysed salt situation, which shows that some
'neutral' salts can be quite acid in aqueous media!

e.g. What is the pH
of a 0.100 mol dm^{–3} solution of phenylammonium chloride?

pk_{a} = 4.62,
so K_{a} = 10^{–4.62} = 2.40 x 10^{–5}
mol dm^{–3}

In general for a weak
acid

K_{a} =

[H^{+}_{(aq)}] [A^{–}_{(aq)}] 
––––––––––––––––– 
[HA_{(aq)}] 

so, making the
assumptions described in section 5.4,

2.40 x 10^{–5} =

[H^{+}_{(aq)}]^{2} 
–––––––––––––––– 
0.100 

[H^{+}_{(aq)}]^{2}
= 2.40 x 10^{–5} x 0.100

[H^{+}_{(aq)}]^{2}
= (2.40 x 10^{–5} x 0.100) = 2.40 x 10^{–6}

[H^{+}_{(aq)}]
= √( 2.40 x 10^{–6}) = 1.55 x 10^{–3}
mol dm^{–3}

pH = –lg(1.55 x
10^{–3}) = 2.81

so, (i) very definitely
an acid solution!, and,

(ii) if you did a
theoretical pH calculation on a 0.1 molar phenylamine solution (like
5.6.4c), you would get a pH value above 7, but not that high!

–
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–K_{a}–pK_{a} calculations * 5.5
strong bases–examples–pH calculations
* 5.6 weak bases– examples & pH–K_{b}–pK_{b} calculations *
5.7 A level notes on Acids, Bases, Salts,
uses of
acid–base titrations – upgrade from GCSE!
