Acids and Bases
Objectives
 Theories of acids and Bases
 Define acids and bases according to the Brønsted –
Lowry and Lewis theories
 Deduce whether or not a species could act as a
Brønsted – Lowry and / or a Lewis acid or base
 Deduce the formula of the conjugate acid (or base)
of any Brønsted – Lowry base (or acid)
The history of acid-base
theory
 There have been many attempts to define acids
and bases.
 The sour taste and the effect on vegetable
colourings, such as litmus, characterized acids.
 The soapy feel and detergent power
characterized alkalis.
 Acids seem to react with alkalis and also with
some other compounds to give salts.
 The term base replaced the term alkali as
meaning the opposite of an acid.
 A base was defined as a substance which would
react with an acid to form a salt and water.
 An acid was defined by Liebig in 1838 as a
compound that contains hydrogen that can be
displaced by a metal.
 A big advance was the Ostwald-Arrhenius theory
of electrolytic dissociation in 1880.
 They defined acids as substances that produce
hydrogen ions in solution.
 Bases, they said, produce hydroxide ions in
solution and neutralize acid by the reaction:
H+
+
OHH 2O
 Before long this theory ran into difficulties. It
was noted that, while hydrochloric acid conducts
electricity, pure hydrogen chloride does not.
 What is the significance of conducting
electricity?
 Should hydrogen chloride be classified as an acid
or does it only become an acid on contact with
water?
 Also bases such as ammonia neutralize acids by
picking up hydrogen ions, rather than by providing
hydroxide ions.
NH3
+
H+
NH4+
 As work proceeded, doubt was cast on the very
existence of the hydrogen ion, H+, in solution.
 The proton H+, is very small (10-15m diameter)
compared with other cations (around 10-10m diameter).
 The electric field around it is so intense that it
attracts any molecule with unshared electrons, such as
H 2O
 The reaction
H+
+
H 2O
H3O+
was shown to liberate 1300kJ mol-1.
 As the reaction was so exothermic, it gave evidence
that unhydrated protons can not exist in solution.
 The hydrated proton, H3O+, is called the oxonium ion,
and is also referred to as the hydrogen ion.
The Brønsted-Lowry
definition of an acid
 The Brønsted – Lowry definition of acids and
bases are about competition for protons, H+
 This defines an acid as a substance that can
donate a proton (a H+ ion ) and a base as a
substance that can accept a proton.
 Acids and bases can only react in pairs – one acid
and one base.
Example 1
 Ammonia acts as a base in water by accepting a
proton from a molecule of water, using its lone
pair to form the bond.
 In this case the water is acting as an acid
H
N:
H
H
:
H
O
+
H
H
N+
H
+
: O:
H
H
H
Example 2
 Hydrogen chloride is a covalently bonded gas, but
when it dissolves in water, the water acts as a
base and hydrogen chloride donates a proton to a
molecule of water, acting as an acid.
HCl
+
H2O
H3O+ +
 A proton has no electrons at all, so to
Cl-
form a covalent bond with another
species, that species must have a lone
pair of electrons. This is the case with
water:
H
O
H
Conjugate acid-base pairs
HCl
acid
+
NH3
base
Cl-
+
conjugate base
NH4+
conjugate acid
 The Cl- ion left after HCl has donated a proton is
itself able to accept a proton (to go back to HCl)
and is therefore a base.
 It is called the conjugate base of HCl.
 In the same way, the NH4+ ion is an acid as it can
donate a proton to return to being NH3.
 It is the conjugate acid of ammonia.
 So we have two acid-base pairs.
Water as an acid and a
base
 HCl can donate a proton to water, so that water
acts as a base.
 Write the equation to show this.
HCl
+
H 2O
H3O+
+
NH3
OH-
+
Cl-
 Water may also act as an acid. For example
H 2O
+
NH4+
 Here OH- is the conjugate base and H2O the acid
 Water is described as an amphoteric or amphiprotic
solvent
Quick questions
1.
2.
3.
4.
5.
Hydrogen bromide, HBr, is acidic. What is its conjugate
base?
OH- is a base. What is its conjugate acid?
Identify the conjugate acid/base pairs in:
HNO3 + OH- → NO3- + H2O
Magnesium oxide is described as basic whereas sodium
hydroxide is an alkali. Explain the difference?
What species are formed when the following bases accept a
proton?
a)
b)
c)
d)
OHNH3
H2O
Cl-
The Lewis definition of acids
and Bases
 There are reactions which appear to us, on
common-sense grounds, to be acid-base reactions,
and which do not come within the scope of the
Brønsted-Lowry definition.
 Such reactions are:
CaO
NH3
+
+
SO3
BF3
CaSO4
NH3BF3
 To accommodate reactions of this type, G N Lewis
(during 1930 to 1940) proposed a fresh definition
of acids and bases.
 He described acid-base reactions as reactions in
which an unshared electron pair in the base
molecule is accepted by the acid molecule, with
the formation of a covalent bond.
 For example, in this reaction
H
H
H
N:
H
+
B
F
F
H
N
H
B
F
F
 Ammonia is the base, and boron trifluoride is the
acid
 The Lewis definition of a base includes the
Brønsted-Lowry bases because a species with a
lone pair of electrons will accept a proton from a
Brønsted-Lowry acid.
 Lewis acids, such as BF3 and SO3, are not acids in
the Brønsted-Lowry sense, and acids such as HCl,
H2SO4 and CH3CO2H are not acids according to
the Lewis definition
 The Brønsted-Lowry description of acids and
bases lends itself readily to a quantitative
treatment of the strengths of acids and bases.
 No such quantitative treatment is possible for
Lewis acids and bases.
Objectives
 Strong and weak acids and bases
 Distinguish between strong and weak acids and bases
in terms of the extent of dissociation, reaction with
water and electrical conductivity
 State whether a given acid or base is strong or weak
 Distinguish between strong and weak acids and bases,
and determine the relative strengths of acids and
bases using experimental data
Strong acids
 The acidity of a substance depends on the
concentration of H+(aq). This is what pH measures.
 When acids such as hydrochloric and nitric acid
dissolve into water they break up completely into ions.
 This is called complete dissociation.
 For example:
HCl(g)
H+(aq) + Cl-(aq)
HNO3(aq)
H+(aq) + NO3-(aq)
 Acids that completely dissociate into ions in
aqueous solutions are called strong acids.
 The word strong refers only to how much the acid
dissociates and not in any way to how
concentrated it is.
 Can you think of a reason why hydrogen chloride
dissolved into an organic liquid does not have
acidic qualities?
Weak acids
 Weak acids are not fully dissociated when
dissolved in water.
 Ethanoic acid (the acid in vinegar, also called
acetic acid) is a good example.
 In a 1 mol dm-3 solution of ethanoic acid, only
about 4 in every thousand ethanoic acid molecules
are dissociated into ions.
 We say the degree of dissociation is 4/1000; the
rest remain dissolved as covalently bonded
molecules.
 In fact an equilibrium is set up:
CH3CO2H(aq)
H+(aq)
+
CH3CO2-(aq)
Before
dissociation
1000
0
0
at equilibrium
996
4
4
 Acids like this are called weak acids – weak refers
only to the degree of dissociation.
 We can have a very dilute solution of a strong
acid, or a very concentrated solution of a weak
acid.
 Strength and concentration are independent.
Bases
 In aqueous solutions, bases dissociate to produce
the OH- ion (which can react with a H+ ion by
forming a bond through the electrons of one of
its lone pairs).
 Bases can also be classified as weak or strong.
 Strong bases are completely dissociated into ions
in aqueous solutions, while weak bases are only
partially dissociated.
Dissociation of strong and
weak bases
 Sodium hydroxide is a strong base;
NaOH(aq)
Na+(aq) + OH-(aq)
 A solution of ammonia in water (NH3
(aq))
, also
called ammonium hydroxide, NH4OH, is a weak
base:
NH3(aq) + H2O(l)
NH4+(aq) +OH-(aq)
Or
NH4OH(aq)
NH4+(aq) + OH-(aq)
Examples of strong acids and
bases
 Examples of strong acids and bases include:
Strong acids
Strong bases
HCl – hydrochloric acid
NaOH – sodium hydroxide
HNO3 – nitric acid
KOH – potassium hydroxide
H2SO4 – sulphuric acid
Ba(OH)2 – barium hydroxide
 Note: because one mole of HCl produces one mole of H+ ions it is
known as monoprotic (monobasic). Sulphuric acid is known as a
diprotic (dibasic) acid, as one mole of sulphuric acid produces two
moles of hydrogen ions.
Examples of weak acids and
bases
 Weak acids and bases are only slightly dissociated
(ionized) into their ions in aqueous solution:
Weak acids
Weak bases
CH3COOH - Ethanoic acid
NH3 – ammonia
H2CO3 - Carbonic acid
C2H5NH2 - aminoethane
Experiments to distinguish between
strong and weak acids and bases
 It is important for us to be able to distinguish
between strong and weak acids and bases, ways in
which we can do this are:
 pH measurement
 Conductivity measurement
 Using universal indicator
 Observation from vigour of reaction
 pH measurement
 Because a strong acid produces a higher
concentration of hydrogen ions than a weak acid,
with the same concentration, the pH of a strong
acid will be lower than a weak acid.
 Similarly a strong base will have a higher pH than a
weak base, with the same concentration.
 The most accurate way to determine pH is to use a
pH meter
0.1 moldm-3 HCl(aq)
pH = 1.0
0.1 moldm-3 CH3COOH(aq) pH = 2.9
 Conductivity measurement
 Strong acids and strong bases in solution will give
much higher readings on a conductivity meter than
equimolar (equal concentration) solutions of weak
acids or bases, because they contain more ions in
solution
 Using Universal indicator
 Strong acids produce a red colour with universal
indicator weak acids produce orange/yellow colours
 Strong bases produce a purple colour with U.I. weak
bases produce a blue colour
 Observation from reaction
 Reacting an acid with a carbonate or a hydrogen
carbonate will produce carbon dioxide gas .
 The reaction will be more vigorous with a stronger
acid compared to a weaker acid.
HCl + Na2CO3
2NaCl + H2O + CO2
will be more vigorous than
2CH3COOH + Na2CO3
2CH3COONa + H2O + CO2
Acid – base reactions and Le
Chatelier’s principle
CH3COOH
25cm3
25cm3
NaOH →
CH3COONa +
x cm3
H2O
0.10mol dm-3
0.10mol dm-3
HCl
+
+
0.10mol dm-3
NaOH
x cm3
→
NaCl +
H2O
0.10mol dm-3
 The volume of alkali in both cases will be 25 cm3
 Although there are fewer hydrogen ions in the
ethanoic acid solution, once they react with hydroxide
ions then, according to Le Chatelier’s principle, more
of the acid dissociates to replace them until
eventually all of the acid has been neutralized.
Titrations
 The only difference observed in titrations involves
temperature rather than volume.
 Strong acids and bases in aqueous solution always
release the same amount of heat per mole of hydrogen
ions neutralized, because the only reaction taking
place is the neutralisation of hydrogen ions by
hydroxide ions.
 Titrations involving weak acids or bases will release a
different amount of heat, because heat energy is
required to dissociate the molecules to form ions, and
heat energy is released when these ions become
hydrated, in addition to the heat evolved during
neutralisation.
Objectives
 Outline the characteristic properties of acids and bases in
aqueous solution
 With reactive metals
 With neutralization reactions with bases
 With carbonates
 With hydrogen carbonates
 With indicators
Reactions of acids
 All acids react to form salts and the reactions
may be written as balanced or ionic equations.
 Metals above hydrogen in the reactivity series
react with acids to produce hydrogen gas and the
metal salt
Metal
+
acid
salt
+
hydrogen
 Write the balanced symbol equation for the
reaction between magnesium and hydrochloric acid
Mg (s)
+
2HCl
(aq)
MgCl2 (aq)
+
H2 (aq)
Reactions of acids
 All acids react to form salts and the reactions
may be written as balanced or ionic equations.
 Metals above hydrogen in the reactivity series
react with acids to produce hydrogen gas and the
metal salt
Metal
+
acid
salt
+
hydrogen
 Write the balanced symbol equation for the
reaction between magnesium and hydrochloric acid
 Acids react with bases such as metal oxides and
hydroxides to form a salt and water. For example:
Metal oxide
Metal
hydroxide
+
acid
+
acid
salt
water
+
salt
+
water
 Can you write a balanced symbol equation showing
an example of each of these reactions?
MgO (s)
+
2HCl
NaOH(aq)
+
HCl
(aq)
(aq)
MgCl2
(aq)
NaCl (aq)
+
H2O
(l)
+
H2O
(l)
 If both the acid and the base are strong,– that is,
they both completely dissociate in water – for
example, the reaction between nitric acid and
sodium hydroxide can be written:
HNO3(aq) + NaOH(aq)
NaNO3(aq) + H2O(aq)
 This can also be written as:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq)
Na+(aq) + NO3-(aq) + H20(l)
 We can remove the spectator ions to leave us with:
H+(aq)
+ OH-(aq)
H20(l)
 The nitrate ions and the sodium ions are ‘spectator
ions’ , and so the only reaction that is actually taking
place is the neutralization of hydrogen ions by
hydroxide ions to form water molecules.
 The enthalpy change for the reaction is -57.3kJ mol-1,
and it will therefore always be the same when a
strong acid is neutralized by a strong base in aqueous
solution.
 A base can be defined as a substance that neutralizes
an acid to form a salt and water, so other acid-base
reactions include the reactions of acids with most
metal oxides.
 All carbonates and hydrogen carbonates will
neutralize acids to produce carbon dioxide and
water. For example:
Metal carbonate
+
acid
salt
+
water
+
carbon dioxide
Metal hydrogen
carbonate
+
acid
salt
+
water
+
carbon dioxide
CaCO3(s) + 2HCl(aq)
2NaHCO3(aq) + H2SO4(aq)
CaCl2(aq) + H2O(l) + CO2(g)
NaSO4(aq) 2H2O(l) + 2CO2(g)
 The carbonate ion (the conjugate base of the
hydrogen carbonate ion, HCO3-) and the hydrogen
carbonate ion (the conjugate base of carbonic acid,
H2CO3) both behave as Brønsted-Lowry bases and
accept protons from the acid
With indicators
 Acid- base indicators can be used to determine whether or
not a solution is acidic.
 Common indicators include:
Indicator
Colour in acidic
solution
Colour in alkaline
solution
Litmus
red
blue
phenolphthalein
colourless
pink
Methyl orange
red
yellow
Characteristic reactions of
bases
 The simplest definition of a base is that it is a
substance that can neutralize an acid.
 A base that is soluble in water is known as an alkali.
 Typical reactions of bases in solution include the
following.
 Neutralization of acids
 Displacement of ammonia from ammonium salts
 With indicators
Displacement of ammonia
from ammonium salts
NH4Cl(s) + NaOH(aq)
NaCl(aq) + NH3(g) + H2O(l)
 Ammonia is very soluble in water, so it may be
formed in solution, or, if small amounts of sodium
hydroxide are added to solid ammonium chloride,
it can be collected as a gas.
 In this reaction the chloride ions and sodium ions
are spectator ions, so the actual reaction taking
place in solution is:
NH4+(aq) + OH-(aq)
NH3(g) + H2O(l)
 The equilibrium lies very much on the right, so a
single arrow has been shown, but technically the
reaction should be written
NH4+(aq) + OH-(aq)
NH3(g) + H2O(l)
 Under the Brønsted-Lowry definition, ammonium
ions (NH4+) and water molecules (H2O) are acting
as acids, and hydroxide ions (OH-) and ammonia
molecules (NH3) are acting as bases.
 As water is a weak acid, its conjugate bases is
strong, so the position of equilibrium lies very
much to the right.
With indicators
 If a base is insoluble in water, for example
copper (II) oxide (CuO), then there will be no
reaction with indicators.
 Indicators actually measure the concentration of
hydrogen ion in aqueous solution, H+(aq), so only
soluble bases (alkalis) that produce hydroxide ions
in water will change the colour of indicators.
 Increasing the concentration of hydroxide ions
will lower the concentration of hydrogen ions
H+(aq) + OH-(aq)
H2O(l)
Objectives
The ionisation of water
 Understand how water ionises and give the ionic
product of water at 298K
AHL
 State the expression for the ionic product of water
(Kw).
 Deduce [H+(aq)] and [OH-(aq)] for water at different
temperatures given Kw values.
The ionisation of water
 Water is slightly ionised:
H2O(l)
H+(aq) + OH-(aq)
 Or this may be written:
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq)
 This equilibrium is established in water and all
aqueous solutions, write the equilibrium expression
for this reaction:
Kc
=
[H+(aq)]eqm [OH-(aq)]eqm
[H2O(l)]eqm
 The equilibrium lies very much over to the left
hand side and so we say the concentration of H2O
stays constant.
 To avoid having two constants (Kc and the
concentration of water) in the same expression, a
new equilibrium constant is defined. This is called
the ionic product for water and is given the
symbol Kw
Kw = [H+(aq)]eqm [OH-(aq)]eqm
 The value of Kw is usually taken to be 1x10-14 mol2
dm-6 at 298K
 Each H2O that dissociates (splits up) gives rise to
one H+ and one OH- so, in pure water at 298K
[H+(aq)]eqm = [OH-(aq)]eqm
So, 1x10-14 = [H+(aq)]2eqm
[H+(aq)] = 1x10-7mol dm-3 = [OH-(aq)]eqm
 This means that only 2 in every 1000,000,000
water molecules is split up into hydrogen and
hydroxide ions!!!
Kw and temperature
 If Kw is 5.13 x 10-13 mol2 dm-6 what is the
concentration of [H+]
Kw = [H+]2
[H+]2 = 5.13 x 10-13
[H+] =
[H+] =
5.13 x 10-13
Objectives
 The pH scale
 Distinguish between aqueous solutions that are
acidic, neutral or alkaline using the pH scale
 Identify which of two or more aqueous solutions is
more acidic or alkaline using pH scales.
 State that each change of one pH unit represents
a 10-fold change in the hydrogen ion concentration
[H+(aq)]
 Deduce changes in [H+(aq)] when the pH of a
solution changes by more than one pH unit
The pH Scale
 The acidity of a solution depends on the concentration
of H+(aq) and is measured on the pH scale.
pH is defined as –log10 [H+(aq)]
 Remember that square brackets, [ ], mean the
concentration in mol dm-3.
 This expression is more complicated than simply
stating the concentration of H+(aq), but it does away
with awkward numbers like 10-13, which occur because
the concentration of H+(aq) solutions is so small.
 The minus sign makes almost all pH values positive
(because the logs of numbers less than 1 are negative).
 On the pH scale:
 The smaller the pH, the greater the concentration
of H+ (aq)
 A difference of one pH number means a tenfold
difference in acidity, so that pH 2 is ten times as
acidic as pH 3
 pH measures alkalinity as well as acidity, because
as [H+(aq)] goes up, [OH-(aq)] goes down.
 If a solution contains more H+(aq) than OH-(aq), its
pH will be less than 7 and we call it acidic.
 If a solution contains more OH- (aq) than H+(aq), its
pH will be greater than 7 and we call it alkaline
Finding the pH of strong
acids and bases
 Acids that completely dissociate into ions in
aqueous solutions are called strong acids.
 The word strong refers only to the extent of
dissociation and not in any way to the
concentration.
 So it is perfectly possible to have a very dilute
solution of a strong acid.
 The same argument applies to bases.
 Strong bases are completely dissociated into ions
in aqueous solutions.
pH
[H+]
0
1
1
1x10-1
2
3
1x10-3
4
The pH scale.
 We can work out the concn of hydrogen
ions, [H+], in an aqueous solution if we
know the pH.
 It is the antilog of the pH value
 For example, an acid has pH = 3
5
6
1x10-6
7
1x10-7
8
1x10-8
9
pH = –log10 [H+(aq)]
3 = –log10 [H+(aq)]
– 3 = log10 [H+(aq)]
 Take the antilog of both sides
10
[H+(aq)] =1x10-3 mol dm-3
11
12
13
1x10-13
14
1x10-14
 Complete the [H+] for pH 0,1,6,7,8,13 and
14
[OH-]
pH
[H+]
1x10-14
0
1
1x10-13
1
1x10-1
2
3
1x10-3
4
1x10-8
6
1x10-7
7
1x10-6
1x10-7
1x10-6
8
1x10-8
10 = –log10 [H+(aq)]
 Take the antilog of both sides
[H+(aq)] =1x10-10 mol dm-3
Recall [H+(aq)] [OH-(aq)]= 1x10-14 mol2 dm-6
10
 [OH-(aq)] [1x10-10] = 1x10-14 mol2 dm-6
 [OH-(aq)] = 1x10-14 /1x10-10 =1x10-4moldm-3
12
1
pH = –log10 [H+(aq)]
9
11
1x10-1
the pH of a solution = 10
-10 = log10 [H+(aq)]
5
1x10-4
 With bases we need two steps, suppose
13
1x10-13
14
1x10-14
 Copy and complete [OH+] for
pH = 0, 1, 6,7, 8, 13 and 14
Acid Rain
 Rain is naturally acidic, because it dissolves and reacts
with carbon dioxide as it falls through the air.
 However, carbon dioxide forms only a weak acid in
water, and even if water is completely saturated with
carbon dioxide the pH will only be 5.6.
 Acid rain is therefore defined as precipitation (rain,
snow, hail etc) that has a pH lower than 5.6
 There are two main reasons why acid rain is fomed:
 The release of sulphur dioxide into the atmosphere
from the combustion of fossil fuels
 The release of oxides of nitrogen from jet engines
and internal combustion engines
Sulphur dioxide
 Sulphur dioxide is released into the atmosphere from
the combustion of fossil fuels such as coal, which
contain sulphur, or the smelting of sulphide ores.
 The sulphur dioxide is slowly oxidised in the
atmosphere to form sulphur trioxide, which dissolves
in the rain to form sulphuric acid:
2SO2(g)
SO3(g) +
+
H2O(l)
O2(g) →
→
2SO3(g)
H2SO4(aq)
Oxides of nitrogen
 Oxides of nitrogen are formed in jet engines and in
internal combustion engines where the temperature
reached is high enough to combine with the oxygen in
the air to form nitrogen monoxide:
N2(g) + O2(g) → 2NO (g)
 Oxidation in the atmosphere then occurs to produce
nitrogen dioxide, which then can dissolve and react
with water to form nitric acid and nitrous acid:
2NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq)
 Or be oxidised directly to nitric acid by oxygen in the
presence of water
4NO2(g) + O2(g) + 2H2O(l) → 4HNO3(aq)
Problems with acid rain
 Acid rain is a world wide problem, because it
often precipitates many miles away from its
source as a result of air currents.
 It has a massive impact the planet, for example;
 It causes damage to buildings by reacting with
carbonates in building materials.
 It affects vegetation by leaching out important
minerals from the soil.
 It affects aquatic life by altering the pH of rivers
and lakes.
 It directly affects human health by increasing the
risk of respiratory diseases.
Activity:
 Read through the article on pages 149 – 150 in
Chemistry Companion and answer the questions at
the end.
AHL Objectives
 Calculations involving acids and bases
 Solve problems involving [H+(aq)], [OH-(aq)], pH and
pOH.
 State the equation for the reaction of any weak acid or
base with water, and hence deduce the expressions for
Ka and Kb.
 Solve problems involving solutions of weak acids and
bases using the expressions:
 Ka x Kb = Kw
 pKa + pKb = pKw
 pH + pOH = pKw
 Identify the relative strengths of acids and bases using
values of Ka, Kb, pKa and pKb.
Calculating pH of a strong
monoprotic acid.
 Strong acids are completely dissociated in water
thus the concentration of hydrogen ions for a
solution of a strong monoprotic acid will be the
same as the concentration of the undissociated
acid.
 Example: What is the pH of a 0.10 moldm-3
solution of hydrochloric acid?
pH =
-log10[H+]
=
-log10(0.1)
=
1.0
Calculating the pH of a
strong diprotic acid
 For a strong diprotic acid the hydrogen ion
concentration will be twice the concentration of
the acid:
 Example: what is the pH of 0.10 moldm-3 sulphuric
acid?
1mole of H2SO4 produces 2 moles of H+ thus
pH = -log10[H+]
= -log10(2 x 0.1)
= 0.7
Calculating concentration of a
strong acid from its pH
 Simple reverse the process and convert pH into
hydrogen ion concentration and use the hydrogen
ion concentration to calculate the concentration
of the acid.
 Example: What is the concentration of some
hydrochloric acid whose pH is 1.60?
antilog the pH to give [H+] = 0.01251 moldm-3
1mole of hydrochloric acid gives 1 mole of
hydrogen ions thus concentration of hydrochloric
acid = 0.01251 moldm-3
Activity:
 Complete problems 4 and 5 from page 187
Calculations in A ’Level chemistry
Calculating pOH of a strong base
 For a strong base such as an aqueous solution of
sodium hydroxide, the hydroxide ion concentration will
be the same as the concentration of the sodium
hydroxide.
 Example; what is the pOH of a 0.10moldm-3 solution
of sodium hydroxide?
 We know;
[H+] x [OH-] = Kw = 1.0 x 10-14 mol2dm-6 at 298K
If [OH-] = 0.10 moldm-3
So pOH = -log10[OH-]
= -log10(0.1)
=1
Using pOH to calculate pH
 We can calculate pH from pOH using the following
relationship:
pOH + pH = pKw
(Remember p just means –log10)
We know that Kw = 1.0 x 10-14 thus pKw = 14
So....
pH + pOH = 14
Example; calculate the pH for a 0.10 moldm-3 solution of
sodium hydroxide
pOH
Thus pH
= -log10(0.1)
=1
= pKw – pOH
= 14 – 1
= 13
Calculations involving weak
acids
 A weak acid is one which does not ionise to a very
large extent in solution. Typical weak acids include
most organic acids and some inorganic ones like
hydrofluoric acid, HF.
 One of the commonest weak acids is ethanoic acid,
CH3COOH. Depending on its concentration, about 1%
of the ethanoic acid molecules are ionised in solution.
An equilibrium is set up:
 Write the equation for this reaction;
CH3COOH + H2O
CH3COO- + H3O+
The dissociation constant
Ka
 The equilibrium constant for this is called the
dissociation constant, and is given the symbol, Ka
 Write the equilibrium expression for this
reaction.
Ka
=
[CH3COO-] [H3O+]
[CH3COOH]
or Ka
=
[CH3COO-] [H+]
[CH3COOH]
 In general the equilibrium expression for all weak
acids is:
Ka
=
[H+] [A-]
[HA]
mol dm-3
Using Ka to calculate the pH of
a weak acid
 We can use the Ka expression to find the pH of a
solution of a weak acid.
 For example; calculate the pH of 0.10 moldm-3 ethanoic
acid (CH3COOH(aq)) given that the acid dissociation
constant, Ka for CH3COOH = 1.8 x 10-5 moldm-3 at 298K
CH3COOH
CH3COO- + H+
0.10
x
0
x
At
At eqm
start
+ 0
the acid is only very slightly dissociated so it is
assumed that the concentration of the acid at
equilibrium is the same as the initial concentration
so Ka
=
[CH3COO-] [H+]
[CH3COOH]
becomes
Ka ≈
x2
0.10
Putting the numbers in...
 If...
Ka ≈
 Then... 1.8 x 10-5 ≈
x2
=
x2
=
1.8 x 10-6
x
=
1.8 x 10-6
=
1.8 x 10-5
0.00134
x
x2
0.10
x2
0.10
0.10
Rearrange the equation
To find x we need to find the square root
x = [H+] thus pH = -log10[H+] = 2.87
Activity:
 Calculations in A’Level Chemistry page 201
complete problem 14 a – c.
pKa
 The values for Ka, for weak acids are very small – for
example, Ka for ethanoic acid is 1.74 x 10-5 moldm-3. To
tidy these numbers up, pKa is often used instead.
pKa = -log10Ka
 The relationship
between pKa and Ka is
exactly the same as that
between pH and [H+].
 If
Ka
= 1.74 x 10-5
pKa = -log10(1.74 x 10-5)
= 4.76
 Going the other way, you
need to “unlog” the pKa
value;
 If
pKa
=4.76
-log10Ka
= 4.76
log10Ka
=-4.76
Ka
= 1.74 x 10-5
Activity:
 Calculations in A’Level Chemistry page 201
complete problems 13, 14 (d-e) and 15
The pH of weak bases
 Ammonia is a typical weak base. It reacts with water
to produce ammonium ions and hydroxide ions.
 Write the equation for this reaction
NH3(aq) + H2O(l)
NH4+
+
OH
(aq)
(aq)
 The ammonia is acting as a base because it is
combining with a hydrogen ion from the water.
 Because most of it remains in solution as unreacted
ammonia molecules, ammonia is described as a weak
base.
 The stronger the base, the further the position of
equilibrium lies to the right.
The dissociation constant
Kb
 The equilbrium constant for this is called the
dissociation constant, and is given the symbol, Kb
 Write the equilibrium expression for this
reaction.
Kb =
[NH4+] [OH-]
[NH3]
Using Kb to calculate the pH of
a weak base
 We can use the Kb expression to find the pH of a
solution of a weak base.
 For example; calculate the pH of 0.10 moldm-3 ammonia
solution (NH3(aq)) given that the base dissociation
constant, Kb for NH3 = 1.8 x 10-5 moldm-3 at 298K
NH3
NH4+ +
OH0.10
0
x
0
x
At
At eqm
start
+
the base is only very slightly dissociated so it is
assumed that the concentration of the base at
equilibrium is the same as the initial concentration
so Kb =
[NH4+][OH-]
[NH3]
becomes
Kb ≈
x2
0.10
Putting the numbers in...
 If...
Kb ≈
 Then... 1.8 x 10-5 ≈
x2
=
x2
=
1.8 x 10-6
x
=
1.8 x 10-6
=
1.8 x 10-5
0.00134
x
x2
0.10
x2
0.10
0.10
Rearrange the equation
To find x we need to find the square root
x = [OH-] thus pOH = -log10[OH-] = 2.87
pKw = pOH + pH therefore pH = 14 – 2.87 = 11.13
pKb
 The values for Kb, for weak bases are very small – for
example, Kb for ammonia is 1.8 x 10-5 moldm-3. To tidy
these numbers up, pKb is often used instead.
pKb = -log10Kb
 The relationship
between pKb and Kb is
exactly the same as that
between pH and [H+].
 If
Kb
= 1.8 x 10-5
pKb = -log10(1.8 x 10-5)
= 4.74
 Going the other way, you
need to “unlog” the pKb
value;
 If
pKb
=4.74
-log10Kb
= 4.74
log10Kb
=- 4.74
Kb
= 1.8 x 10-5
Ka values for weak bases
 The conjugate acid of ammonia is the ammonium
ion NH4+(aq).
 It is a weak acid as it dissociates to form
ammonia and hydrogen ions.
 Write the equation for this reaction.
NH4+(aq)
NH3(aq) + H+(aq)
 Write the equilibrium expression for this
reaction:
Ka
=
[NH3] [H+]
[NH4+]
The relationship between
Ka and Kb
 So if...
Kb =
[NH4+][OH-]
and Ka
[NH3]
=
[NH3] [H+]
[NH4+]
 We can cancel out the ammonium ions and
ammonia to give us.....
Kb x Ka
=
[OH-]
x [H+]
Where have we seen this
before?
 So......
Kb x Ka = Kw take it further pKb + pKa = pKw
Using this expression to find
the pH of a base
 Calculate the pH of a 1.0 x 10-3 moldm-3 aqueous
solution of ethylamine, C2H5NH2, given that pKa
value for ethylamine is 10.73 at 298K
 Write the equation for the dissociation of
ethylamine in water
C2H5NH2(aq) + H2O(l)
C2H5NH3+(aq) + OH-(aq)
 Write the Kb equilibrium expression for this:
Kb =
[C2H5NH3+(aq)] [OH-(aq)]
[C2H5NH2(aq)]
Putting the numbers in....
 Rearrange
So ....
the equation
 And remember...
[OH-]2 =pKa
5.37
+ pKb
x 10=-4pKw
x 1.0 x 10-3
[C2H5NH3+(aq)] [OH-(aq)]
Kb =
 Thus = 5.37 x 10-7
[C2H5NH2(aq)]
 Therefore
the square root
pKb = pKw – taking
pKa
2
[OH
]
(aq)
-4
= 14
– 10.73
[OH ] = 7.32
x 10
5.37 x 10-4 =
pKb
= 3.27
[C2H5NH2(aq)]
We
can
use the expression..
pOH + pH = pKw

Kb therefore
Therefore
5.37 x 10-4 ≈
-] – 3.27
= antilog
pOH = -log10[OH
-4
=3.14 = 5.37 x 10
So pH = pKw – pOH = 14 – 3.14 = 10.86
[OH-(aq)]2
1.0 x 10-3
Relative strengths of acids and
bases using Ka, Kb, pKa and pKb
 Remember that Ka and Kb are just equilibrium
constants....
 The greater the equilibrium constant the greater
the extent of dissociation, i.e. the equilibrium lies
to the right (products).
 So the greater the value of Ka the greater the
dissociation of the acid therefore = a stronger
acid.
 pKa and pKb are logarithms of Ka and Kb thus the
greater the value of Ka and Kb the lower the value
of pKa and pKb.
Acid dissociation constants
of some weak acids
Compound
Equilibrium expression
Hydrofluoric
acid, HF
HF + H2O
Ethanoic acid,
CH3COOH
CH3CO2H + H2O
Benzoic acid,
C6H5COOH
C6H5CO2H + H2O
Phenol,
C6H5OH
C6H5OH + H2O
Ka /
moldm-3
pKa
H3O+ + F-
6.3 x 10-4
3.2
H3O+ + CH3COO-
1.6 x 10-5
4.8
H3O+ + C6H5COO-
6.3 x 10-5
4.2
H3O+ + C6H5O-
1.3 x 10-10
9.9
Base dissociation constants
of some weak bases
Compound
Ammonia, NH3
Equilibrium expression
NH3 + H2O
Ka /
moldm-3
NH4+ + OH-
pKa
1.6 x 10-5
4.8
Methylamine,
CH3NH2
CH3NH2 + H2O
CH3NH3+ + OH-
4.0 x 10-4
3.4
Phenylamine,
C6H5NH2
C6H5NH2 + H2O
C6H5NH3+ + OH-
4.0 x 10-10
9.4
Phenylmethyl
amine,
C6H5CH2NH2
C6H5CH2NH2 + H2O
2.5 x 10-5
4.6
C6H5CH2NH3++ H3O+
Objectives
 Salt Hydrolysis
 Deduce whether salts form acidic, alkaline or
neutral aqueous solutions.
Salt hydrolysis
 Salt hydrolysis is just the interaction of a salt in
water. You would assume that all salts are neutral
but this is not true, only some salts are neutral.
 Salts are ionic, and are therefore already
completely dissociated.
 When they dissolve in water they are strong
electrolytes. For example:
H2O(l)
Na+ ClNa+(aq) + Cl-(aq)
 Salts like sodium chloride, which are derived from
a strong acid (hydrochloric acid) and a strong
base (sodium hydroxide), form neutral solutions
when they dissolve in water.
Alkaline salts
 Salts that are derived from a weak acid and a strong
base produce alkaline solutions when they are
dissolved in water.
 An example is sodium ethanoate, which can be formed
from the reaction between ethanoic acid and sodium
hydroxide.
 When the salt dissolves in water the ethanoate ions
combine with hydrogen ions from the water to form
mainly undissociated ethanoic acid.
 This leaves an excess of hydroxide ions in the
solution, which do not combine with the sodium ions,
because sodium hydroxide is a strong base.
The hydrolysis of sodium
ethanoate
CH3COONa(s)
H2O(l)
Na+(aq)
+
CH3COO-(aq)
+
H2O(l)
OH-(aq) +
H+(aq)
CH3COOH(aq)
The hydrolysis of ammonium
chloride
 Similarly salts such as ammonium chloride which are
derived from a strong acid and a weak base, will be acidic
in solution.
NH4Cl(s)
H2O(l)
H2O(l)
NH4+(aq) +
Cl-(aq)
+
OH-(aq) +
H+(aq)
NH3(aq) +
H2O(l)
 The acidity of salts also depends on:
 The size of the cation
 The charge of the cation
 Aluminium chloride reacts vigorously with water to give
a strongly acidic solution:
AlCl3(s) + 3H2O(l) →Al(OH)3(s) + 3HCl(aq)
 The tripositive (+3) charge is spread over a very small
ion, which gives the aluminium ion, Al3+, a very high
charge density.
 This makes it an excellent lewis acid, and it will attract
a non-bonding pair of electrons on six water molecules
to form the hexahydrated ion.
 Can you draw this moelcule?
The hydrolysis of hexahydrated
aluminium (III) ion
3+
OH2
H2O
OH2 - H+
[Al(H2O)5OH]2+
Al
H2O
OH2
OH2
- H+
[Al(H2O)4OH]+
- H+
[Al(H2O)3OH3]
 The non-bonding pair of one of the six water molecules
surrounding the ion is strongly attracted to the ion and loses a
hydrogen ion in the process.
 This process will continue until three hydrogen ions have been
released and aluminium (III) hydroxide is formed.
 The equilibrium can be moved further to the right by adding
hydroxide ions, or back to the left by adding hydrogen ions,
which explains the amphoteric nature of aluminium hydroxide.
Other highly charged ions
 Similar reactions occur with other small, highly
charged ions such as the iron (III) ion, Fe3+.
 Even magnesium chloride, MgCl2, is slightly acidic
in aqueous solution for the same reason.
ion
charge
Ionic radius/nm
Aqueous
solution
Na+
+1
0.098
neutral
Mg2+
+2
0.065
acidic
Al3+
+3
0.045
acidic
Objectives
 Buffer solutions
 Describe the composition of a buffer solution and
explain its action.
 Solve problems involving the composition and pH of
a specified buffer system.
Buffers
 A buffer solution is defined as an aqueous solution
that resists a change in pH when a small amount
of acid, base or water is added to it.
 There are two types of buffer;
 Acidic buffers with a pH less than 7
 Alkaline buffers with a pH more than 7
 Buffers are formed from an acid, a base and the
salt made from the acid – base reaction.
Acidic buffers
 An acidic buffer can be made in the laboratory by mixing
a weak acid together with the salt of that acid and a
strong base (in other words by combining a weak acid
with a salt containing its conjugate base) e.g. ethanoic
acid and sodium ethanoate.
 A practical method of doing this is to add excess
ethanoic acid to sodium hydroxide solution. Once all the
sodium hydroxide has been neutralised to form the salt,
the remaining solution contains the salt and the excess
acid.
NaOH(aq) + CH3COOH(aq) →CH3COONa(aq) + H2O(l) + CH3COOH(aq)
limiting reactant
salt
excess weak acid
How an acidic buffer works
 The explanation of buffer action lies in the fact that
the weak acid is only slightly dissociated in solution,
but the salt is fully dissociated into its ions, so the
concentration of ethanoate ions is high.
CH3COONa(aq)
CH3COOH(aq)
Na+(aq)
+
CH3COO-(aq)
CH3COO-(aq)
+
H+(aq)
How an acidic buffer works
 If an acid is added, the extra hydrogen ions from the
added acid combine with ethanoate ions to form
undissociated ethanoic acid, and so the hydrogen ions
remain unaltered
+
CH3COONa(aq)
Na+(aq)
CH3COOH(aq)
CH3COO-(aq)
CH3COO-(aq)
+
H+(aq)
 If an alkali is added, the hydroxide ions from the alkali
are removed by their reaction with the undissociated
acid to form water, so again the hydrogen ion
concentration again stays constant.
CH3COOH(aq)
+
OH-(aq)
CH3COO-(aq) +
H2O(aq)
How an alkaline buffer works
 An alkaline buffer is made by combining a weak base
with the salt of that base with a strong acid.
 An example is ammonia with ammonium chloride:
NH4Cl(aq)
NH3(aq) +
H2O(aq)
NH4+(aq)
NH4+(aq)
+
Cl-(aq)
+
OH-(aq)
 If hydrogen ions are added, they will combine with
hydroxide ions to form water, and more of the ammonia
will dissociate to replace them.
 If more hydroxide ions are added, they will combine
with ammonium ions to form undissociated ammonia.
 In both cases the hydroxide ion concentration and the
hydrogen ion concentration will remain constant.
Uses of buffer solutions
 Buffer solutions have many applications in the
laboratory: for example, they are used to
standardize a pH meter.
 They are also extremely important in biological
systems, because enzymes function efficiently
only within a very narrow pH region.
 Outside this region their structure may rapidly
become distorted or broken down (denatured).
Blood
 Blood, is responsible for carrying oxygen around the
body, contains several different buffering systems.
 One of the components of the system is that the oxygen
adds on reversibly to the haemoglobin (Hb) in the blood:
HHb(aq) + O2(g)
H+(aq) + HbO2-(aq)
 If the pH increases ([H+(aq)] falls), the equilibrium will
move to the right, and the oxygen will tend to be bound
to the haemoglobin more tightly.
 If the pH decreases ([H+(aq)] increases), the oxygen will
tend to be displaced from the haemoglobin and released.
 Both of these processes are potentially life-threatening.
Buffer calculations
 Calculations involving buffer solutions are very similar to




the calculations done previously for weak acids and bases.
However, there is one very important difference.
When a weak monoprotic acid on its own in aqueous solution
dissociates, the concentration of the hydrogen ions will be
the same as the concentration of the acid anions.
In a buffer solution the concentration of the acid anions
will be much greater than the hydrogen concentration ,
because the salt that produces the acid anions is
completely dissociated.
In fact, the assumption is made that the concentration of
the acid anions is due completely to the concentration of
the salt, because the anions from the dissociation of the
weak acid make a negligible contribution.
Acid buffer calculation
 Example: calculate the pH of a buffer solution
containing 3.00g of sodium ethanoate in 100cm3 of
0.050 moldm-3 ethanoic acid at 298K (given that Ka
for ethanoic acid = 1.8 x 10-5 moldm-3 at 298K.
 First we need to write the equilibrium expression
for this buffer:
We can assume this
This is what we are trying
to find out
Ka
=
number is the same as the
concentration of the salt
[H+] [CH3COO-]
[CH3COOH]
We can rearrange this
equation to find [H+]
We can assume this
number is the same as the
concentration of the acid
[H+]
x 10-5
= 1.8Ka
 Ka = 1.8 x
10-5
0.05
[CH3moldm
COOH]-3
=
[CH
0.366
3COO ]
2.46 x 10-6
pH
moldm-3
 [CH3COOH] = 0.05 moldm-3
= - log10 [H+]
= -log10(2.46 x 10-6)
= 5.61
 We need to find [CH3COO-]
 We are given a mass so we can find moles
 We are given the volume it is dissolved in, so we can
find the concentration
Moles
=
=
mass
RMM
RMM
3.00g
82
=
=
12 + 3 + 12 + 32 + 23 = 82
0.0366 mol in
100cm3
Therefore 0.366 mol in
1000cm3
Alkaline buffer calculation
 Example: what mass of ammonium chloride must be
dissolved in 1.00dm-3 of 0.100 moldm-3 ammonia
solution to produce a buffer solution with a pH
equal to 9.00, pKb for ammonia = 4.75?.
 First we need to write the equilibrium expression
for this buffer:
we can find this using the
This is what we are trying
to find out
we can convert
pKb to Kb
Kb
=
expression pOH = pKw - pH
[NH4+] [OH-]
[NH3]
We can rearrange this
equation to find [H+]
We can assume this number
is the same as the
concentration of the alkali
[NH4+]
=
1.78
Kbx 10-5
-3
0.1 [NH
moldm
3]
=
-] -5
[OH
1 x 10
0.176 moldm-3
 Kb = antilog 4.75 = 1.78x10-5 moldm-3
 [NH3] = 0.10 moldm-3
 We need to find [OH-]
 We have been given pH so we can find pOH
 We can convert pOH to OH
pOH = pKw – pH pOH = 14 – 9 = 5
so OH = antilog 5 = 1 x 10-5
Assuming all ammonium ions originate from ammonium chloride
RMM of ammonium chloride
=
14 + 4 + 35.5 = 53.5
Therefore 0.176 mol of NH4Cl will have a mass of 0.176 x 53.5 = 9.42g
Objectives
 Indicators
 Describe qualitatively the action of an acid – base
indicator.
 State and explain how the pH range of an acid –
base indicator relates to its pKa value.
 Identify an appropriate indicator for a titration,
given the equivalence point of the titration and the
pH range of the indicator.
Acid – base indicators
 An indicator is a solution of a weak acid in which the
conjugate base has a different colour from that of the
undissociated acid.
 If we represent an aqueous solution of an indicator by
HIn(aq), then the equation for the dissociation in water is
H+(aq)
HIn(aq)
Colour A
(colour in acid solution)
+
In-(aq)
Colour B
(colour in alkali solution)
what would the equilibrium expression be?
Kin =
[H+(aq)] [In-(aq)]
[HIn(aq)]
Indicator end-points
 Assuming the colour changes when the concentration of
the acid is approximately equal to the concentration of
the conjugate base
([HIn(aq)] = [In-(aq)])
then the end point of the indicator will be when
[H+(aq)] ≈ Ka: that is, when pH ≈ pKa.
 It can be seen that the various indicators have
different Ka values, and so change colour within
different pH ranges.
 Two common indicators are methyl orange and
phenolphthalein, look up their pKa values and compare
with their pH ranges.
Objectives
 Acid – Base titrations
 Sketch the general shapes of graphs of pH against
volume for titrations involving strong and weak
acids and bases, and explain their important
features.
Acid – base titrations
 Acids and bases neutralise each other when mixed
together in aqueous solution.
 An acid – base titration is a technique used to measure
the volumes of acid and base required for neutralisation.
 If the concentration of one solution is known, then the
concentration of the other may be calculated.
 There are four possible combinations of strong and weak
acids and bases that may be considered for a titration:
 Strong acid – Strong base
 Strong acid – Weak base
 Weak acid – Strong base
 Weak acid – Weak base
Strong acid – strong base
A titration curve
 A titration curve shows the pH



14
pH

changes that occur when an
12
aqueous acid reacts with an aqueous
10
base.
8
It is a graph of pH against volume.
6
4
The titration of a strong acid with
2
a strong base causes a large pH
0
change on neutralisation, as shown
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
Volume of NaOH added cm3
in the graph.
The end-point of a titration
The equivalence point of this
titration occurs when the amounts occurs when the indicator
of acid and base are exactly equal changes colour.
We can use suitable indicators to The indicator must be chosen
observe the equivalence point of a with care to ensure that the
end-point coincides with the
titration.
equivalence point.
Choosing an indicator
 The titration curve for a strong acid-stong base
neutralisation shows a large change in pH around
the equivalence point.
 Any indicator with a range between pH 3 and pH11
may be used.
 The titration curves for other combinations of
strong/weak acid/base pairs offer a more limited
change in pH.
 As a result, it is more difficult, if not impossible
to choose a suitable indicator.
Common indicators
Titration of a weak acid
with a strong base
Titration Curve
 In this case, you can see
12
10
8
pH
that the large change in
pH around the
equivalence point occurs
approximately between
pH7 and pH 11.
 What would be a
suitable indicator for
this titration?
14
6
4
2
0
0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30
volume of NaOH added cm3
Titration of a strong acid
with a weak base
 Sketch a graph of the titration curve you would
expect in this acid – base reaction.
 What indicator could you use and what colour
change would you see?
Titration of a weak acid
and a weak base
 There is hardly any change in pH around the
equivalence point in weak acid – weak base
titration.
 No indicator will detect the end-point accurate to
one drop.
 However, this is not a problem, as it is never
necessary to perform this type of titration
because, if the unknown is weak, the standard
solution can always be chosen to be strong.
Summary of titrations
 The end-point of a titration occurs when the indicator





changes colour.
The equivalence point occurs when there are equal
amounts of H3O+ and OH- ions in the titration.
Indicators with ranges between pH 11 and pH 3 are
suitable for strong acid – strong base titrations.
The indicator range for a strong acid – weak base
titration must fall between pH 3 and pH 7. Methyl
orange is a useful choice.
The indicator range for a weak acid – strong base
titration must fall between pH 7 and pH 11.
Phenolphthalein is a useful choice.
Indicators cannot be used to find the equivalence point
of a weak acid – weak base titration.