pH and IONIC
EQUILIBRIA
A guide for A level students
KNOCKHARDY PUBLISHING
2015
SPECIFICATIONS
pH and Ionic Equilibria
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it
may be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are
available from the KNOCKHARDY SCIENCE WEBSITE at...
www.knockhardy.org.uk/sci.htm
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pH and Ionic Equilibria
CONTENTS
• Brønsted-Lowry theory of acids and bases
• Lewis theory of acids and bases
• Strong acids and bases
• Weak acids
• Weak bases
• Hydrogen ion concentration and pH
• Ionic product of water Kw
• Relationship between pH and pOH
• Calculating the pH of strong acids
• Calculating the pH of weak acids
• Calculating the pH of mixtures
• A brief introduction to buffer solutions
• Check list
Acid & Bases
Before you start it would be helpful to…
• know the simple properties of acids, bases and alkalis
ACIDS AND BASES
BRØNSTED-LOWRY THEORY
ACID
proton donor
HCl ——> H+(aq) + Cl¯(aq)
BASE
proton acceptor
NH3(aq) + H+(aq)
——>
NH4+(aq)
ACIDS AND BASES
BRØNSTED-LOWRY THEORY
ACID
proton donor
HCl ——> H+(aq) + Cl¯(aq)
BASE
proton acceptor
NH3(aq) + H+(aq)
Conjugate systems
Acids are related to bases
ACID
Bases are related to acids
BASE
——>
NH4+(aq)
PROTON
+
PROTON
+
CONJUGATE BASE
CONJUGATE ACID
ACIDS AND BASES
BRØNSTED-LOWRY THEORY
ACID
proton donor
HCl ——> H+(aq) + Cl¯(aq)
BASE
proton acceptor
NH3(aq) + H+(aq)
Conjugate systems
Acids are related to bases
ACID
Bases are related to acids
BASE
——>
NH4+(aq)
PROTON
+
PROTON
+
CONJUGATE BASE
CONJUGATE ACID
For an acid to behave as an acid, it must have a base present to accept a proton...
HA
acid
example
+
B
base
CH3COO¯ + H2O
base
acid
BH+
+
A¯
conjugate conjugate
acid
base
CH3COOH
acid
+
OH¯
base
ACID-BASE
PAIR
ACIDS AND BASES
LEWIS THEORY
ACID
lone pair acceptor
BF3
H+
BASE
lone pair donor
NH3
H2O
LONE PAIR DONOR
LONE PAIR DONOR
LONE PAIR
ACCEPTOR
LONE PAIR ACCEPTOR
AlCl3
STRONG ACIDS AND BASES
STRONG
ACIDS
e.g.
completely dissociate (split up) into ions in aqueous solution
HCl ——> H+(aq) + Cl¯(aq)
MONOPROTIC
1 replaceable H
DIPROTIC
2 replaceable H’s
HNO3 ——> H+(aq) + NO3¯(aq)
H2SO4 ——> 2H+(aq) + SO42-(aq)
STRONG ACIDS AND BASES
STRONG
ACIDS
e.g.
completely dissociate (split up) into ions in aqueous solution
HCl ——> H+(aq) + Cl¯(aq)
MONOPROTIC
1 replaceable H
DIPROTIC
2 replaceable H’s
HNO3 ——> H+(aq) + NO3¯(aq)
H2SO4 ——> 2H+(aq) + SO42-(aq)
STRONG
BASES
e.g.
completely dissociate into ions in aqueous solution
NaOH(s)
——> Na+(aq) + OH¯(aq)
WEAK ACIDS
Weak acids
partially dissociate into ions in aqueous solution
e.g. ethanoic acid
When a weak acid dissolves in
water an equilibrium is set up
CH3COOH(aq)
HA(aq) + H2O(l)
CH3COO¯(aq)
HA(aq)
H+(aq)
A¯(aq) + H3O+(aq)
The water stabilises the ions
To make calculations easier
the dissociation can be written...
+
A¯(aq) +
H+(aq)
WEAK ACIDS
Weak acids
partially dissociate into ions in aqueous solution
e.g. ethanoic acid
CH3COOH(aq)
When a weak acid dissolves in
water an equilibrium is set up
HA(aq) + H2O(l)
CH3COO¯(aq)
The weaker the acid
HA(aq)
H+(aq)
A¯(aq) + H3O+(aq)
The water stabilises the ions
To make calculations easier
the dissociation can be written...
+
A¯(aq) +
H+(aq)
the less it dissociates
the more the equilibrium lies to the left.
WEAK ACIDS
Weak acids
partially dissociate into ions in aqueous solution
e.g. ethanoic acid
CH3COOH(aq)
When a weak acid dissolves in
water an equilibrium is set up
HA(aq) + H2O(l)
CH3COO¯(aq)
The weaker the acid
HA(aq)
H+(aq)
A¯(aq) + H3O+(aq)
The water stabilises the ions
To make calculations easier
the dissociation can be written...
+
A¯(aq) +
H+(aq)
the less it dissociates
the more the equilibrium lies to the left.
The relative strengths of acids can be expressed as Ka or pKa values
WEAK ACIDS
Weak acids
partially dissociate into ions in aqueous solution
e.g. ethanoic acid
CH3COOH(aq)
When a weak acid dissolves in
water an equilibrium is set up
CH3COO¯(aq)
+
H+(aq)
A¯(aq) + H3O+(aq)
HA(aq) + H2O(l)
The water stabilises the ions
To make calculations easier
the dissociation can be written...
The weaker the acid
HA(aq)
A¯(aq) +
H+(aq)
the less it dissociates
the more the equilibrium lies to the left.
The relative strengths of acids can be expressed as Ka or pKa values
The dissociation constant for the weak acid HA is
Ka =
[H+(aq)] [A¯(aq)]
[HA(aq)]
mol dm-3
WEAK BASES
Partially react with water to give ions in aqueous solution
e.g. ammonia
When a weak base dissolves in water an equilibrium is set up
NH3 (aq)
+ H2O (l)
NH4+ (aq) + OH¯ (aq)
as in the case of acids it is more simply written
NH3 (aq) +
H+ (aq)
NH4+ (aq)
WEAK BASES
Partially react with water to give ions in aqueous solution
e.g. ammonia
When a weak base dissolves in water an equilibrium is set up
NH3 (aq)
+ H2O (l)
NH4+ (aq) + OH¯ (aq)
as in the case of acids it is more simply written
NH3 (aq) +
The weaker the base
H+ (aq)
NH4+ (aq)
the less it dissociates
the more the equilibrium lies to the left
The relative strengths of bases can be expressed as Kb or pKb values.
Hydrogen ion concentration [H+(aq)]
Introduction
hydrogen ion concentration determines the acidity of a solution
hydroxide ion concentration determines the alkalinity
For strong acids and bases the concentration of ions is very much
larger than their weaker counterparts which only partially dissociate.
Hydrogen ion concentration [H+(aq)]
pH
pOH
hydrogen ion concentration can be converted to pH
pH = - log10 [H+(aq)]
to convert pH into hydrogen ion concentration
[H+(aq)] = antilog (-pH)
An equivalent calculation for bases converts
the hydroxide ion concentration to pOH
pOH = - log10 [OH¯(aq)]
in both the above, [ ] represents the concentration in mol dm-3
Ionic product of water - Kw
Despite being covalent, water conducts electricity to a very small extent.
This is due to the slight ionisation ... H2O(l) + H2O(l)
or, more simply
H2O(l)
H3O+(aq) + OH¯(aq)
H+(aq) + OH¯(aq)
Ionic product of water - Kw
Despite being covalent, water conducts electricity to a very small extent.
H3O+(aq) + OH¯(aq)
This is due to the slight ionisation ... H2O(l) + H2O(l)
or, more simply
Applying the equilibrium law
to the second equation gives
[ ] is the equilibrium concentration in mol dm-3
H+(aq) + OH¯(aq)
H2O(l)
Kc
=
[H+(aq)] [OH¯(aq)]
[H2O(l)]
Ionic product of water - Kw
Despite being covalent, water conducts electricity to a very small extent.
H3O+(aq) + OH¯(aq)
This is due to the slight ionisation ... H2O(l) + H2O(l)
or, more simply
Applying the equilibrium law
to the second equation gives
[ ] is the equilibrium concentration in mol dm-3
H+(aq) + OH¯(aq)
H2O(l)
Kc
=
[H+(aq)] [OH¯(aq)]
[H2O(l)]
As the dissociation is small, the water concentration is very large compared with the
dissociated ions and any changes to its value are insignificant; its concentration can
be regarded as constant.
Ionic product of water - Kw
Despite being covalent, water conducts electricity to a very small extent.
H3O+(aq) + OH¯(aq)
This is due to the slight ionisation ... H2O(l) + H2O(l)
or, more simply
Applying the equilibrium law
to the second equation gives
H+(aq) + OH¯(aq)
H2O(l)
Kc
[ ] is the equilibrium concentration in mol dm-3
=
[H+(aq)] [OH¯(aq)]
[H2O(l)]
As the dissociation is small, the water concentration is very large compared with the
dissociated ions and any changes to its value are insignificant; its concentration can
be regarded as constant.
This “constant” is combined with
(Kc) to get a new constant (Kw).
Kw = [H+(aq)] [OH¯(aq)] mol2 dm-6
=
1 x 10-14 mol2 dm-6 (at 25°C)
Because the constant is based on an equilibrium, Kw VARIES WITH TEMPERATURE
Ionic product of water - Kw
The value of Kw varies with temperature because it is based on an equilibrium.
Temperature / °C
0
20
25
30
60
Kw / 1 x 10-14 mol2 dm-6
0.11
0.68
1.0
1.47
5.6
H+ / x 10-7 mol dm-3
pH
0.33
7.48
0.82
7.08
1.0
7
1.27
6.92
2.37
6.63
What does this tell you about the equation
H2O(l)
H+(aq) + OH¯(aq) ?
Ionic product of water - Kw
The value of Kw varies with temperature because it is based on an equilibrium.
Temperature / °C
0
20
25
30
60
Kw / 1 x 10-14 mol2 dm-6
0.11
0.68
1.0
1.47
5.6
H+ / x 10-7 mol dm-3
pH
0.33
7.48
0.82
7.08
1.0
7
1.27
6.92
2.37
6.63
What does this tell you about the equation
H2O(l)
H+(aq) + OH¯(aq) ?
• Kw gets larger as the temperature increases
• this means the concentration of H+ and OH¯ ions gets greater
• this means the equilibrium has moved to the right
• if the concentration of H+ increases then the pH decreases
• pH decreases as the temperature increases
Ionic product of water - Kw
The value of Kw varies with temperature because it is based on an equilibrium.
Temperature / °C
0
20
25
30
60
Kw / 1 x 10-14 mol2 dm-6
0.11
0.68
1.0
1.47
5.6
H+ / x 10-7 mol dm-3
pH
0.33
7.48
0.82
7.08
1.0
7
1.27
6.92
2.37
6.63
What does this tell you about the equation
H2O(l)
H+(aq) + OH¯(aq) ?
• Kw gets larger as the temperature increases
• this means the concentration of H+ and OH¯ ions gets greater
• this means the equilibrium has moved to the right
• if the concentration of H+ increases then the pH decreases
• pH decreases as the temperature increases
Because the equation moves to the right as the
temperature goes up, it must be an ENDOTHERMIC process
Relationship between pH and pOH
Because H+ and OH¯ ions are produced
in equal amounts when water dissociates
their concentrations will be the same.
[H+] = [OH¯] = 1 x 10-7 mol dm-3
Kw = [H+] [OH¯] =
1 x 10-14 mol2 dm-6
take logs of both sides
log[H+] + log[OH¯] = -14
multiply by minus
- log[H+] - log[OH¯] = 14
change to pH and pOH
pH + pOH = 14
(at 25°C)
[H+]
100 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
OH¯
10-14 10-13 10-12 10-11 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 10-0
pH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
pOH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
STRONGLY
ACIDIC
WEAKLY
ACIDIC
NEUTRAL
WEAKLY
ALKALINE
STRONGLY
ALKALINE
Relationship between pH and pOH
Because H+ and OH¯ ions are produced
in equal amounts when water dissociates
their concentrations will be the same.
[H+] = [OH¯] = 1 x 10-7 mol dm-3
Kw = [H+] [OH¯] =
N.B.
1 x 10-14 mol2 dm-6
take logs of both sides
log[H+] + log[OH¯] = -14
multiply by minus
- log[H+] - log[OH¯] = 14
change to pH and pOH
pH + pOH = 14
(at 25°C)
As they are based on the position of equilibrium and that varies with
temperature, the above values are only true if the temperature is 25°C (298K)
Neutral solutions may be regarded as those where [H+] = [OH¯].
Therefore a neutral solution is pH 7 only at a temperature of 25°C (298K)
Kw is constant for any aqueous solution at the stated temperature
Calculating pH - strong acids and alkalis
WORKED
EXAMPLE
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentration.
Calculating pH - strong acids and alkalis
WORKED
EXAMPLE
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentration.
Calculate the pH of 0.02M HCl
H+ + Cl¯
HCl completely dissociates in aqueous solution
HCl
One H+ is produced for each HCl dissociating so
[H+]
= 0.02M
= 2 x 10-2 mol dm-3
pH
= - log [H+]
= 1.7
Calculating pH - strong acids and alkalis
WORKED
EXAMPLE
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentration.
Calculate the pH of 0.02M HCl
H+ + Cl¯
HCl completely dissociates in aqueous solution
HCl
One H+ is produced for each HCl dissociating so
[H+]
= 0.02M
= 2 x 10-2 mol dm-3
pH
= - log [H+]
= 1.7
Calculate the pH of 0.1M NaOH
Na+ + OH¯
NaOH completely dissociates in aqueous solution
NaOH
One OH¯ is produced for each NaOH dissociating
[OH¯] = 0.1M
The ionic product of water (at 25°C)
therefore
Kw
= [H+][OH¯]
= 1 x 10-1 mol dm-3
= 1 x 10-14 mol2 dm-6
[H+] = Kw / [OH¯] = 1 x 10-13 mol dm-3
pH
= - log [H+] = 13
Calculating pH - weak acids
• can’t be calculated by just knowing the concentration
• need to know... the original concentration
the extent of the ionisation
The extent of the ionisation is given by …
Ka The dissociation constant for a weak acid
and
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
Applying the equilibrium law we get
[ ] is the equilibrium concentration in mol dm-3
Kc =
[H3O+(aq)] [A¯(aq)]
[HA(aq)] [H2O(aq)]
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
Applying the equilibrium law we get
[ ] is the equilibrium concentration in mol dm-3
Assumptions
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
For a weak acid there is little dissociation
Kc =
[H3O+(aq)] [A¯(aq)]
[HA(aq)] [H2O(aq)]
[HA(aq)]equil ~ [HA(aq)]undis
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
Applying the equilibrium law we get
[ ] is the equilibrium concentration in mol dm-3
Assumptions
This assumption
becomes less
valid for stronger
weak acids where
there is more
dissociation.
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
For a weak acid there is little dissociation
Kc =
[H3O+(aq)] [A¯(aq)]
[HA(aq)] [H2O(aq)]
[HA(aq)]equil ~ [HA(aq)]undis
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
Applying the equilibrium law we get
Kc =
[ ] is the equilibrium concentration in mol dm-3
Assumptions
For a weak acid there is little dissociation
In dilute solution, the concentration of water
is large compared with the dissociated ions
and any changes to its value are insignificant;
its concentration can be regarded as ‘constant’.
[H3O+(aq)] [A¯(aq)]
[HA(aq)] [H2O(aq)]
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
Applying the equilibrium law we get
Kc =
[ ] is the equilibrium concentration in mol dm-3
Assumptions
[HA(aq)] [H2O(aq)]
For a weak acid there is little dissociation
In dilute solution, the concentration of water
is large compared with the dissociated ions
and any changes to its value are insignificant;
its concentration can be regarded as ‘constant’.
Combine this ‘constant’ with Kc
to get a new constant (Ka)
where Ka = Kc [H2O(l)]
[H3O+(aq)] [A¯(aq)]
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
[HA(aq)]
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
Applying the equilibrium law we get
Kc =
[ ] is the equilibrium concentration in mol dm-3
Assumptions
[HA(aq)] [H2O(aq)]
For a weak acid there is little dissociation
In dilute solution, the concentration of water
is large compared with the dissociated ions
and any changes to its value are insignificant;
its concentration can be regarded as ‘constant’.
Combine this ‘constant’ with Kc
to get a new constant (Ka)
where Ka = Kc [H2O(l)]
[H3O+(aq)] [A¯(aq)]
[HA(aq)]equil ~ [HA(aq)]undis
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
A simpler way to express the dissociation is
[HA(aq)]
HA(aq)
H+(aq) + A¯(aq)
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
Applying the equilibrium law we get
Kc =
[ ] is the equilibrium concentration in mol dm-3
Assumptions
[HA(aq)] [H2O(aq)]
[HA(aq)]equil ~ [HA(aq)]undis
For a weak acid there is little dissociation
In dilute solution, the concentration of water
is large compared with the dissociated ions
and any changes to its value are insignificant;
its concentration can be regarded as ‘constant’.
Combine this ‘constant’ with Kc
to get a new constant (Ka)
where Ka = Kc [H2O(l)]
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
[HA(aq)]
A simpler way to express the dissociation is
The dissociation constant Ka is then
[H3O+(aq)] [A¯(aq)]
Ka =
HA(aq)
H+(aq) + A¯(aq)
[H+(aq)] [A¯(aq)] mol dm-3
[HA(aq)]
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
H3O+(aq) + A¯(aq)
HA(aq) + H2O(l)
Applying the equilibrium law we get
Kc =
[ ] is the equilibrium concentration in mol dm-3
Assumptions
This assumption
becomes less
valid for stronger
weak acids where
there is more
dissociation.
[HA(aq)] [H2O(aq)]
[HA(aq)]equil ~ [HA(aq)]undis
For a weak acid there is little dissociation
In dilute solution, the concentration of water
is large compared with the dissociated ions
and any changes to its value are insignificant;
its concentration can be regarded as ‘constant’.
Combine this ‘constant’ with Kc
to get a new constant (Ka)
where Ka = Kc [H2O(l)]
[H2O(l)] is ‘constant’
Ka = [H3O+(aq) ] [A¯(aq)] mol dm-3
[HA(aq)]
A simpler way to express the dissociation is
The dissociation constant Ka is then
[H3O+(aq)] [A¯(aq)]
Ka =
HA(aq)
H+(aq) + A¯(aq)
[H+(aq)] [A¯(aq)] mol dm-3
[HA(aq)]
The dissociation constant for a weak acid - Ka
A weak monobasic acid (HA)
dissociates in water thus.
HA(aq) + H2O(l)
The dissociation constant Ka is then
Ka =
H3O+(aq) + A¯(aq)
[H+(aq)] [A¯(aq)] mol dm-3
[HA(aq)]
The weaker the acid
• the less it dissociates
• the fewer ions you get
• the smaller Ka
The stronger the acid
• the more the equilibrium lies to the right
• the larger Ka
pKa
• very weak acids have very small Ka values
• it is easier to compare the strength as pKa values
The conversion is carried out thus...
To convert pKa into Ka
pKa
= - log10 Ka
Ka = antilog (-pKa) or 10-Ka
Calculating pH – weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
Applying the Equilibrium Law
Ka
H+(aq) + A¯(aq)
=
[H+(aq)] [A¯(aq)]
mol dm-3
(1)
(2)
[HA(aq)]
The ions are formed in equal amounts, so
therefore
[H+(aq)]
=
Ka =
[A¯(aq)]
[H+(aq)]2
(3)
[HA(aq)]
Rearranging (3) gives
therefore
[H+(aq)]2
=
[H+(aq)]
=
pH
[HA(aq)] Ka
[HA(aq)] Ka
= [H+(aq)]
mol dm-3
Calculating pH – weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
Applying the Equilibrium Law
Ka
H+(aq) + A¯(aq)
=
[H+(aq)] [A¯(aq)]
mol dm-3
(1)
(2)
[HA(aq)]
The ions are formed in equal amounts, so
therefore
[H+(aq)]
=
Ka =
[A¯(aq)]
[H+(aq)]2
(3)
[HA(aq)]
Rearranging (3) gives
therefore
[H+(aq)]2
=
[H+(aq)]
=
pH
ASSUMPTION
[HA(aq)] Ka
[HA(aq)] Ka
mol dm-3
= [H+(aq)]
HA is a weak acid so it will not have dissociated very much. You can assume that
its equilibrium concentration is approximately that of the original concentration.
WORKED
EXAMPLE
Calculating pH – weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
HX(aq)
H+(aq)
+
X¯(aq)
WORKED
EXAMPLE
Calculating pH – weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
Dissociation constant for a weak acid
H+(aq)
HX(aq)
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
WORKED
EXAMPLE
Calculating pH – weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
Dissociation constant for a weak acid
H+(aq)
HX(aq)
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in
equal amounts and then rearrange equation
[H+(aq)]
=
[HX(aq)] Ka
mol dm-3
WORKED
EXAMPLE
Calculating pH – weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
Dissociation constant for a weak acid
H+(aq)
HX(aq)
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in
equal amounts and the rearrange equation
[H+(aq)]
=
[HX(aq)] Ka
mol dm-3
ASSUMPTION
HA is a weak acid so it will not have dissociated very much. You can assume that its
equilibrium concentration is approximately that of the original concentration
WORKED
EXAMPLE
Calculating pH – weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
H+(aq)
HX(aq)
Dissociation constant for a weak acid
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
[H+(aq)]
Substitute for X¯ as ions are formed in
equal amounts and the rearrange equation
=
[HX(aq)] Ka
mol dm-3
ASSUMPTION
HA is a weak acid so it will not have dissociated very much. You can assume that its
equilibrium concentration is approximately that of the original concentration
[H+(aq)]
ANSWER
=
0.1 x 4 x 10-5 mol dm-3
=
4.00 x 10-6
mol dm-3
=
2.00 x 10-3
mol dm-3
pH = - log [H+(aq)]
=
2.699
CALCULATING THE pH OF MIXTURES
The method used to calculate the pH of a mixture of an acid and an alkali depends on...
• whether the acids and alkalis are STRONG or WEAK
• which substance is present in excess
STRONG ACID and STRONG BASE - EITHER IN EXCESS
pH of mixtures
Strong acids and strong alkalis (either in excess)
1. Calculate the initial number of moles of H+ and OH¯ ions in the solutions
2. As H+ and OH¯ ions react in a 1:1 ratio; calculate unreacted moles species in excess
3. Calculate the volume of solution by adding the two original volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess the ion in mol dm-3
6. Convert concentration to pH
If the excess is
H+
pH = - log[H+]
If the excess is
OH¯
pOH = - log[OH¯]
then
pH + pOH = 14
or use Kw = [H+] [OH¯] = 1 x 10-14 at 25°C
[H+] = Kw / [OH¯]
pH = - log[H+]
then
therefore
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
25cm3 of
20cm3 of
0.1M NaOH
0.1M HCl
2.5 x 10-3
moles
2.0 x 10-3
moles
moles of OH ¯
= 0.1 x 25/1000
= 2.5 x 10-3
moles of H+
= 20 x 20/1000
= 2.0 x 10-3
WORKED
EXAMPLE
pH of mixtures
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
25cm3 of
20cm3 of
0.1M NaOH
0.1M HCl
2.5 x 10-3
moles
2.0 x 10-3
moles
The reaction taking place is…
or in its ionic form
HCl + NaOH
H+ +
OH¯
NaCl
H2O
+
H2O
(1:1 molar ratio)
WORKED
EXAMPLE
pH of mixtures
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
25cm3 of
20cm3 of
0.1M NaOH
0.1M HCl
2.5 x 10-3
moles
5.0 x 10-4
moles of OH¯
2.0 x 10-3
moles
The reaction taking place is…
HCl + NaOH
H+ +
or in its ionic form
UNREACTED
OH¯
NaCl
+
H2O
H2O
(1:1 molar ratio)
2.0 x 10-3 moles of H+ will react with the same number of moles of OH¯
this leaves
2.5 x 10-3
-
2.0 x 10-3 = 5.0 x 10-4 moles of OH¯ in excess
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
the volume of the solution is 25 + 20 = 45cm3
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
the volume of the solution is 25 + 20 = 45cm3
there are 1000 cm3 in 1 dm3
volume = 45/1000 = 0.045dm3
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
[OH¯]
= 5.0 x 10-4 / 0.045
= 1.11 x 10-2 mol dm-3
WORKED
EXAMPLE
pH of mixtures
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
6. As the excess is OH¯ use
or
[OH¯]
[H+]
pH
= 5.0 x 10-4 / 0.045
= Kw / [OH¯]
= - log[H+]
pOH = - log[OH¯]
Kw
= [H+][OH¯]
then
pH + pOH = 14
so [H+] = Kw / [OH¯]
= 1.11 x 10-2 mol dm-3
= 9.00 x 10-13 mol dm-3
= 12.05
Kw = 1 x 10-14
mol2 dm-6 (at 25°C)
Buffer solutions - Brief introduction
Definition
“Solutions which resist changes in pH when
small quantities of acid or alkali are added.”
Acidic Buffer (pH < 7)
made from
Alkaline Buffer (pH > 7) made from a
Uses
a weak acid + its sodium or potassium salt
ethanoic acid
sodium ethanoate
weak base
ammonia
Standardising pH meters
Buffering biological systems (eg in blood)
Maintaining the pH of shampoos
+
its chloride
ammonium chloride
REVISION CHECK
What should you be able to do?
Recall the definition of acids and bases in the Brønsted-Lowry system
Recall the definition of acids and bases in the Lewis system
Recall and explain the difference between strong and weak acids
Recall and explain the difference between strong and weak bases
Recall the definition of pH
Recall the definition of the ionic product of water
Explain how and why pH varies with temperature
Recall the relationship between pH, [H+], [OH¯], pOH and Kw
CAN YOU DO ALL OF THESE?
YES
NO
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relevant topic(s) again
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pH and IONIC
EQUILIBRIA
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