11C5 Acids and Bases Booklet 2025 RT.pdf

Briefly summarize the main points with formulas. must be easy to remember as simple as possible and don't leave out key points and factors

1. Grade 11 Acids and Bases R Toerien Page 1
Grade 11 Chemistry Module C5 Acids and Bases
Introduction
Long ago people observed that some materials have a sour stingy taste, while
other materials were slippery between the fingers. They called these substances
acids and bases. The word ‘acid’ comes from the Latin word for sour. In this
module we are going to investigate the properties of acids and bases and look at
the theories that scientists have formulated to explain acid-base behaviour.
How do we know when something is an acid or a base?
Acids and bases were first recognised by simple properties such as taste.
Acids have a sour taste while bases are bitter, and solutions of bases have a
soapy feel. Some examples of acids are acetic acid in vinegar, citric acid in lemon
juice, and hydrochloric acid which is found in the digestive juices of the stomach.
Bases include aqueous ammonia, which is often found in household cleaners,
magnesium hydroxide which is used to make anti-acid preparations, and sodium
hydroxide which is a component of drain and oven cleaners.
Some acids and bases are corrosive. A corrosive material is a highly reactive substance that causes
damage to living tissue. We are all familiar with the corrosive properties of some laboratory acids, but some
bases are also corrosive. Watch this video to convince you that bases are just as dangerous as acids:
youtu.be/WnPrtYUKke8.
Acids and bases change the colour of some indicators. An indicator is a
chemical compound that changes colour when exposed to certain
conditions and is therefore useful for chemical tests. Litmus, for
example, is an acid-base indicator that becomes red in the presence of
acids and blue in the presence of bases. The table on the right includes
the indicators and colour changes you have learned in Grade 9.
Substances are classified as acids, bases, or neither,
depending on the properties they exhibit. Some acids are
‘strong’ and others are ‘weak’, depending on how
pronounced their properties are. Below are some
laboratory acids and bases that you need to know.
Common laboratory acids and bases
Acids Bases
Perchloric acid HCℓO4(g) Lithium hydroxide LiOH(s)
Sulfuric acid H2SO4(ℓ) Sodium hydroxide NaOH(s)
Hydrochloric acid HCℓ(g) Potassium hydroxide KOH(s)
Hydrobromic acid HBr(ℓ) Barium hydroxide Ba(OH)2(s)
Nitric acid HNO3(ℓ) Calcium hydroxide Ca(OH)2(s)
Magnesium hydroxide Mg(OH)2(s)
Oxalic acid (COOH)2(s)
Sulfurous acid H2SO3(ℓ) Calcium carbonate CaCO3(s)
Phosphoric acid H3PO4(s) Sodium carbonate Na2CO3(s)
Hydrofluoric acid HF(g) Ammonia NH3(g)
Acetic acid CH3COOH(ℓ) Sodium hydrogen carbonate NaHCO3(s)
Carbonic acid H2CO3(g)
Hydrocyanic acid HCN(ℓ)

Name:










Some household
acids and bases
Litmus is blue in a base and red in an acid
Indicator Acid Base
Bromothymol blue Yellow Blue
Phenolphthalein Colourless Pink
Methyl orange Red Yellow
Litmus Red Blue

weak strong

weak strong

2. Grade 11 Acids and Bases R Toerien Page 2
You will notice that all the acids have at least one hydrogen atom in their chemical formula. This is
important to notice as we will be defining acids according to their ability to donate (‘give away’) this
hydrogen atom in the form of a hydrogen ion (H
+
). You should also notice that some acids have one
hydrogen atom, while others have two or three hydrogen atoms in the formulae. Acids that can donate only
one hydrogen ion per molecule are called monoprotic acids, those that can donate two hydrogen ion per
molecule are called diprotic acids and those that can donate three hydrogen ions per molecule are called
triprotic acids. The general name for acids that can donate more than one hydrogen ion per molecule is
polyprotic acids. Note that only the hydrogen ion that can be donated is considered. This is usually written
in the front of the formula, for example, H2SO4. With some acids, the H is part of the -COOH group, for
example, CH3COOH is monoprotic with the H’s which are part of the CH3 group not considered here. This
will be explained later in the module.
Monoprotic acids Diprotic acids Triprotic acids
Hydrochloric acid HCℓ Sulfuric acid H2SO4 Phosphoric acid H3PO4
Nitric acid HNO3 Sulfurous acid H2SO3
Acetic acid CH3COOH Carbonic acid H2CO3
Oxalic acid (COOH)2


Acids and bases form salts
In Grade 9 you learned that acids react with metals and metal compounds to form salts. A salt can be
defined as an ionic substance that is formed from a metal and a non-metal/polyatomic ion. When salts
dissolve in water, positive and negative ions are formed. Examples of salts are copper sulfate (CuSO4),
potassium chloride (KCℓ), or magnesium phosphate (Mg3(PO4)2). Below are four different reaction types
that can be used to form salts:
Metal + acid → salt + hydrogen
Zn + 2HCℓ → ZnCℓ2 + H2
Metal oxide + acid → salt + water
ZnO + 2HCℓ → ZnCℓ2 + H2O
Metal hydroxide + acid → salt + water
Zn(OH)2 + 2HCℓ → ZnCℓ2 + H2O
Metal carbonate + acid → salt + water + carbon dioxide
ZnCO3 + 2HCℓ → ZnCℓ2 + CO2 + H2O

These reactions are very useful when a specific salt needs to be made. For example, if you wanted to
make zinc chloride (ZnCℓ2), you have to use a zinc compound (like Zn(OH)2) and an acid with Cℓ in it,
namely HCℓ. Zn(OH)2 is a base, and HCℓ is an acid, so this would be an acid-base reaction.
Worked example 1
Which acid and base can be used to form BaSO4? Write a balanced equation for this reaction.







Acid proticity
Some acids are able to donate more than one proton.
Monoprotic acids: HCℓ + H2O → H3O
+
+ Cℓ
–
Triprotic acids: H3PO4 + H2O → H2PO4
–
+ H3O
+

Diprotic acids: H2SO4 + H2O → HSO4
–
+ H3O
+
H2PO4
–
+ H2O → HPO4
2–
+ H3O
+

HSO4
–
+ H2O → SO4
2–
+ H3O
+
HPO4
2–
+ H2O → PO4
3–
+ H3O
+

3. Grade 11 Acids and Bases R Toerien Page 3

Exercise 1
1.1 Classify the following acids as monoprotic (M), diprotic (D), and triprotic (T). Use the internet if
needed:
acetic acid, phosphoric acid, perchloric acid, sulfurous acid, arsenic acid, and malonic acid.

1.2 Use the internet to find the colours of the following indicators in acidic and basic solutions:
red cabbage juice, universal indicator, thymol blue, phenol red, and bromocresol green.

1.3 Classify the following substances as acids (A), bases (B), or salts (S). If the substance is an acid,
state whether it is monoprotic (M), diprotic (D), or polyprotic (P).
1.3.1 NH4Cℓ
1.3.2 HBr
1.3.3 MgSO4
1.3.4 CaO
1.3.5 NH4OH
1.3.6 HNO3
1.3.7 ZnSO4
1.3.8 Fe2O3
1.3.9 CH3COOH
1.3.10 H2SO4
1.3.11 KOH
1.3.12 MgO




1.4 Complete the following chemical reactions by writing a balanced chemical equation:
1.4.1 H2SO4 + Mg →
1.4.2 HCℓ + CaCO3 →
1.4.3 H3PO4 + Aℓ(OH)3 →
1.4.4 CH3COOH + NaOH →
1.5 Write four different reactions to form the salt copper sulfate (CuSO4).





1.6 For each of the following salts, suggest which acid and which metal hydroxide were reacted to form
the salt. Include the chemical name and the formula of each substance.

1.6.1 CaBr2
1.6.2 Cu(HCO3)2
1.6.3 (NH4)2SO4
1.6.4 Aℓ(NO3)3
1.6.5 Na3PO4

4. Grade 11 Acids and Bases R Toerien Page 4
Where do acids and bases come from?
When metals and non-metals react with oxygen, the respective oxides are formed. The oxides can dissolve
in water to form acidic or basic (also called alkaline) solutions.
When non-metal oxides dissolve in water, acidic solutions are formed.
SO2(g) + H2O(ℓ) → H2SO3(aq) (Sulfurous acid (H2SO3) is acidic.)
CO2(g) + H2O(ℓ) → H2CO3(aq) (Carbonic acid (H2CO3) is acidic.)

When metal oxides dissolve in water, basic solutions are formed.
MgO(s) + H2O(ℓ) → Mg(OH)2(aq) (Magnesium hydroxide (Mg(OH)2) is basic.)
CaO(s) + H2O(ℓ) → Ca(OH)2(aq) (Calcium hydroxide (Ca(OH)2) is basic.)

Ionisation and dissociation
Acids are usually covalent compounds. The atoms are held together by
covalent bonds and molecules are formed. When these molecules dissolve
in water they form ions in a process called ionisation.
Hydrochloric acid: HCℓ(g) + H2O(ℓ) → H3O
+
(aq) + Cℓ
–
(aq)

Sulfurous acid: H2SO3(s) + 2H2O(ℓ) → 2H3O
+
(aq) + SO3
2–
(aq)
As a result, hydronium ions (H3O
+
) are formed. The presence of
hydronium ions in a solution gives the solution acidic properties.
Bases are often metal hydroxides and are ionic compounds. The compounds are three-dimensional lattice
structures made of ions that are held together by ionic bonds (electrostatic forces between positive and
negative ions). When ionic substances dissolve in water the positive and negative ions separate in a
process called dissociation. The water molecules do not take part in the reaction and merely acts as a
solvent. In a chemical equation, it is written above the reaction arrow.
NaOH(s)
H2O
→ Na
+
(aq) + OH
–
(aq)
Mg(OH)2(s)
H2O
→ Mg
2+
(aq) + 2OH
–
(aq)
The presence of hydroxide ions gives the solution basic properties and
allows us to identify the substance as a base. In other words, magnesium
hydroxide is a base because it can form hydroxide ions in solution.
You can watch a video of the dissociation process here: youtu.be/EBfGcTAJF4o
Ammonia is a base that will not dissociate. It is a covalent compound which will ionise according to the
following reaction:
NH3(g) + H2O(ℓ) → NH4
+
(aq) + OH
–
(aq)


Ionisation is the process where molecular substances react with
water to form ions: HBr(g) + H2O(ℓ) → H3O
+
(aq) + Br
–
(aq)
Dissociation is the process where ionic substances ‘break apart’
into ions that were present in the lattice structure:
KOH(s)
H2O
→ K
+
(aq) + OH
–
(aq)

5. Grade 11 Acids and Bases R Toerien Page 5
Bases and alkalis
The term ‘base’ and ‘alkali’ are often used interchangeably, but there is a difference between the two terms.
An alkali is a soluble base, in other words, a base that can dissolve in water. Not all bases are alkalis, but
all alkalis are bases. The broader term ‘base’ or ‘basic’ is used in this module.
Group 1 oxides dissolve in water to form basic solutions, for example, sodium oxide will form sodium
hydroxide when dissolved in water: Na2O(s) + H2O(ℓ) → 2NaOH(aq)
Group 2 oxides and hydroxides are only partially soluble in water, for example, CaO(s)+ H2O(ℓ) → Ca(OH)2(aq)
Some metal oxides, like copper(II) oxide (CuO) or iron(III) oxide (Fe2O3), are insoluble.
Worked example 2
2.1 Sulfuric acid is a diprotic acid. Write balanced equations to show the ionisation of sulfuric acid in
water.



2.2 Phosphoric acid is a white solid at room temperature. Phosphoric acid is dissolved in water. Write a
balanced equation, including state symbols, showing the overall effect of the dissolution of phosphoric
acid.

2.3 Will an aqueous solution of calcium hydroxide be ACIDIC or BASIC? Explain the answer with a
balanced equation.



Exercise 2
2.1 What is the difference between dissociation and ionisation?


2.2 What is the difference between an alkali and a base?


2.3 Write a balanced equation when each of the following acids or bases dissolves/reacts with water:

2.3.1 KOH(s)

2.3.2 HCℓO4(ℓ)

2.3.3 H2CO3(g)

2.4 Will each of the following solutions be ACIDIC of BASIC? Explain the answer by referring to a
balanced chemical equation.

2.4.1 Mg(OH)2(s)

2.4.2 HNO3(ℓ)

6. Grade 11 Acids and Bases R Toerien Page 6
The development of acid-base models
What we know about chemistry today were not all known very long ago. What we know today, scientists
have developed over time, and even today we are expanding our knowledge of chemistry. Just like the
model of the atom developed over time, so did what we know about chemical bonds. The more scientists
knew about atoms, the more they could understand about chemical bonding, which in turn helped them
understand other aspects of chemistry like acids and bases. The more people understood about atoms,
what they are made of, and how they react with other substances, the better they could explain why some
substances have acidic properties, and others have basic properties.
The very first scientists thought all acids and bases contain oxygen, and that the oxides of non-metals were
the acids, and the oxides of metals were bases. Later they found that there are substances that do not
contain oxygen but still had acidic properties. This meant that their initial ideas about acids and bases may
not have been comprehensive enough.
The Arrhenius acid-base model (1887)
In 1887 a Swedish scientist, Svante Arrhenius, defined acids as substances that
produce hydrogen ions (H
+
) in solution and bases as substances that produce
hydroxide ions (OH
–
) in solution. He received the Nobel Prize for Chemistry in 1903 for
his work on ions in solutions.
According to the Arrhenius model, all acid-base reactions must take place in water.
Acids increase the H
+
concentration in solution, and bases increase the OH
–
concentration. When an acid and a base react, a salt and water are formed.
acid + base → salt + water
Scientists also discovered that when you add an acid to a base, their properties cancel
out. The acid no longer has acidic properties, and the base no longer has basic properties. They called this
process neutralisation. According to the Arrhenius theory, neutralisation is defined as a reaction between
hydrogen ions and hydroxide ions to form water:
H
+
(aq) + OH
–
(aq)

⇌ H2O(ℓ)
Hydrogen chloride (HCℓ) is an example of an Arrhenius acid. It is a gas, but when dissolved in water it
forms H
+
ions in solution. Sodium hydroxide (NaOH) can be classified as an Arrhenius base as it forms OH
–

ions when dissolved in water. However, chemists at the time identified some reactions exhibiting similar
properties to acid-base reactions, but since the reactions were not taking place in water, they were not
considered acid-base reactions. For example, when ammonia gas reacts with a hydrochloric acid solution it
is classified as an acid-base reaction.
NH3(g) + HCℓ(aq) ⇌ NH4Cℓ(aq)
However, when ammonia gas and hydrogen chloride gas are mixed, the same reaction takes place and the
same product forms.
NH3(g) + HCℓ(g) ⇌ NH4Cℓ(s)
This reaction was not classified as an acid-base reaction because it is not taking place in water. Although
the Arrhenius acid-base model was a useful theory, there were some shortcomings, and there was a need
to expand the model.


Arrhenius acid is a substance that produces hydrogen ions (H
+
) when dissolved in water.
Arrhenius base is a substance that produces hydroxide ions (OH
–
) when dissolved in water.
Neutralisation according to Arrhenius is a reaction where an acid and a base form a salt and water.

Svante Arrhenius

7. Grade 11 Acids and Bases R Toerien Page 7
The Brønsted-Lowry acid-base model (1923)
Johannes Nicolaus Brønsted and Thomas Martin Lowry formulated the same new
theory for acids and bases in 1923 independent of one another. According to their
theory, acids donate hydrogen ions (H
+
, also referred to as protons), and bases
accept hydrogen ions. A Brønsted-Lowry acid will donate a proton to form a
conjugate base. For example, HCℓ, an acid, will donate an H
+
to form the conjugate
base Cℓ
–
.
Acid-base reactions are therefore not defined in terms of the formation of hydrogen
or hydroxide ions in solution, but rather in terms of the proton transfer to form
conjugate acid-base pairs.
In the reaction: NaOH + HCℓ  NaCℓ + H2O
HCℓ is a Brønsted-Lowry acid. A proton (H
+
) is transferred and the conjugate ion, Cℓ
–
is
formed. HCℓ and Cℓ
–
are a conjugate acid-base pair.
OH
–
is a Brønsted-Lowry base. It accepts a proton (H
+
) and H2O is formed. OH
–
and H2O are a conjugate
acid-base pair. Na
+
is a spectator ion and does not take part in the reaction.
According to the Brønsted-Lowry model, ‘neutralisation’ is seen as the formation of
conjugate acid-base pairs as a result of proton transfer.
Acids and bases react to form, not salts and solvents, but new acids and bases.
Since Brønsted-Lowry acids and bases react independently of the solvent (not just
water), it is a more encompassing theory. According to Brønsted-Lowry, acid-base
reactions are described as follows. HA is the acid, and B is the base:
HA + B ⇌ BH
+
+ A
–

acid base conjugate acid conjugate base


We will discuss conjugate acid-base pairs in the next section.
Note: When we use the term ‘proton’ here, we refer to the hydrogen atom that has lost an electron to
become a hydrogen ion (H
+
). This ion has only one particle – a proton in the centre of the nucleus. It is
therefore a proton, that is why we use the term here.
These two models are not the only ones being used in chemistry. A third model, the Lewis acid-base
model is based on electron transfer was also proposed at the same time at the Brønsted-Lowry model. The
Lewis model is not studied at school level, and you will learn more about this if you study chemistry at
university.
You can watch the following video on acid-base models: youtu.be/DupXDD87oHc


Brønsted acid is a proton (H
+
) donor.
Brønsted base is a proton (H
+
) acceptor.
Neutralisation according to Brønsted-Lowry is the proton transfer from an acid to a base to form the
respective conjugate base and conjugate acid.
Johannes Brønsted
Thomas Lowry

8. Grade 11 Acids and Bases R Toerien Page 8
Conjugate acid-base pairs
According to the Brønsted-Lowry model, an acid is a proton donor and a base is a proton acceptor. Note
that in the Brønsted-Lowry model, acids and bases can be ionic or molecular substances.
Consider the following reaction where ammonia gas is dissolved in water:
NH3(g) + H2O(ℓ) ⇌ OH
–
(aq) + NH4
+
(aq)
When ammonia (NH3) dissolves in water, it acts as a base by accepting a proton (H
+
) from the water to form
the ammonium ion (NH4
+
). NH4
+
can also lose a proton to form NH3. Note that NH3 and NH4
+
differ by one
proton only. Thus, NH3 becomes NH4
+
by accepting a proton while NH4
+
becomes NH3 by donating a proton.
The species NH4
+
and NH3 are a conjugate acid-base pair. Every acid has a conjugate base and every
base has a conjugate acid. In a reaction, we indicate the pairs by labelling them as shown below. a1 and b1
form a pair, and a2 and b2 form another pair. In this reaction water (H2O) donates a proton so acts as a
Brønsted-Lowry acid (a2 in the reaction below). The conjugate base (b2) which is formed when H2O donates
an H
+
ion is the hydroxide ion (OH
–
). H2O and OH
–
therefore also form an acid-conjugate base pair.

NH3(g) + H2O(ℓ) ⇌ NH4
+
(aq) + OH
–
(aq)
base acid conjugate acid conjugate base
b1 a2 a1 b2

Worked example 3
Identify the conjugate acid-base pairs in each of the following reactions:
3.1 HCℓ + OH
–
⇌ H2O + Cℓ
–




3.2 HPO4
2–
+ NH4
+
⇌ NH3 + H2PO4
–





You can watch the following videos on conjugate acid-base pairs: youtu.be/4441EyWBPt8 or
youtu.be/fUhPLe0NAvA


O
H H
ammonia water hydroxide ion ammonium ion
H O
–
+ N H
H
H + N H
H
H
H +
Conjugate acid-base pairs differ by a proton, for example, in the reaction:
HNO3 + H2O ⇌ NO3
–
+ H3O
+

HNO3 and NO3
–
is a conjugate acid-base pair, and H2O and H3O
+
forms another conjugate acid-base
pair.

9. Grade 11 Acids and Bases R Toerien Page 9
Exercise 3
3.1 Classify each of the following statements as A (Arrhenius) or BL (Brønsted-Lowry).
3.1.1 A base is a proton (hydrogen ion) acceptor.
3.1.2 Bases are substances that produce hydroxide ions in solution.
3.1.3 H
+
(aq) + OH
–
(aq)  H2O(ℓ) is a neutralisation reaction.
3.1.4 Ammonium and ammonia are conjugate acid-base pairs.
3.1.5 An acid is a proton (hydrogen ion) donor.
3.1.6 Acids are substances that produce hydrogen ions in solution.
3.1.7 NH3(g) + HCℓ(g) ⇌ NH4Cℓ(s) is an acid-base reaction.
3.1.8 Hydroxide ions are bases because they accept hydrogen ions from acids and form water.
3.1.9 Ammonia gas is a base because it accepts a proton.
3.1.10 Neutralisation is the proton transfer between an acid and its conjugate base.
3.2 Give the conjugate bases for each of the following:
3.2.1 H2O
3.2.2 HCℓO4
3.2.3 NH3
3.2.4 OH
–

3.3 Give the conjugate acids for each of the following:
3.3.1 Br
–

3.3.2 HC2O4
–

3.3.3 NH3
3.3.4 PO4
3–
3.4 Give the conjugate acid/base pairs in the following reactions:

3.4.1 HSO4
–
+ NH3 ⇌ NH4
+
+ SO4
2–

3.4.2 HCO3
–
+ OH
–
⇌ CO3
2–
+ H2O
3.4.3 H2O + CH3COO
–
⇌ CH3COOH + OH
–

3.4.4 CℓO
–
+ H2O ⇌ HCℓO + OH
–

3.4.5 NO3
–
+ H2SO4 ⇌ HNO3 + HSO4
–

3.4.6 O
2–
+ H2O ⇌ OH
–
+ OH
–

3.4.7 NH3 + HBr ⇌ NH4
+
+ Br
–

3.4.8 HPO4
2–
+ H2O ⇌ H2PO4
–
+ OH
–

10. Grade 11 Acids and Bases R Toerien Page 10
The auto-ionisation of water
Protolytic reactions are reactions where protons are transferred. According to Brønsted-Lowry, all acid-
base reactions are protolytic reactions. The reaction below shows two water molecules reacting in a
protolysis reaction. The one water (H2O) molecule (in red) accepts a proton to form the hydronium ion
(H3O
+
). The water molecule acts as a base with the hydronium ion as the conjugate acid. The other water
molecule (in blue) acts as an acid by donating a proton to form the hydroxide ion. A proton is transferred
from water to water! Water is ionising itself and we describe this as the auto-ionisation of water. It is an
example of an auto-protolysis reaction – a proton transfer from water to itself.

H2O (ℓ) + H2O(ℓ) ⇌ H3O
+
(aq) + OH
–
(aq)
base acid conjugate acid conjugate base
b1 a2 a1 b2
A substance that can accept or donate a proton is described as amphiprotic and is called an ampholyte.
An ampholyte is a substance that can act as both an acid and a base. Water is an ampholyte. The
hydrogen carbonate ion (HCO3
–
) is also an ampholyte as shown below:
The hydrogen carbonate ion can act as an acid, donating a proton: HCO3
–
+ H2O ⇌ CO3
2–
+ H3O
+

The hydrogen carbonate ion can act as a base, accepting a proton: HCO3
–
+ H2O ⇌ H2CO3 + OH
–

Other ampholytes include HPO4
2–
, H2PO4
–
or HSO4
–
.
You can watch a video about the auto-ionisation of water (first 6 min of the video): youtu.be/NUyYlRxMtcs

Worked example 4
Show, by means of chemical equations, how H2PO4
–
acts as an ampholyte.










⇌
Protolysis: It is a reaction where protons are transferred.
Ampholyte: It is a substance that can act as either an acid or a base.
Amphiprotic: A substance is amphiprotic when it can act as either an acid or a base, in other words
it can donate or accept a proton.
Auto-ionisation: The auto-ionisation of water is the reaction of water with itself to form H3O
+
ions
and OH
–
ions.

11. Grade 11 Acids and Bases R Toerien Page 11
Strength of acids and bases
Consider the following equation for an acid ‘HA’ dissolving in water. Each chemical species is represented
with a diagram indicated below the equation.
HA + H2O(ℓ) ⇌ H3O
+
(aq) + A
–
(aq)

The A in HA does not stand for a particular element, but for the acid radical part of the molecule. So, for
example, in hydrochloric acid HA would be HCℓ, and A
–
would be Cℓ
–
, whilst in ethanoic acid HA would be
CH3COOH, and A
–
would be CH3COO
–
.
Strong acids and bases
Strong acids ionise completely. The diagram shows an example of a strong
acid HA in water. Since this is a strong acid there are no HA molecules in
solution. All the HA molecules have become H3O
+
and A
–
. An example of a
strong acid is HCℓ, H2SO4, or HNO3.
The ionisation of strong acids can be represented by balanced chemical
equations as follows:
H2SO4(ℓ) + 2H2O(ℓ) ⥂ 2H3O
+
(aq) + SO4
2–
(aq)

HNO3(ℓ) + H2O(ℓ) ⥂ H3O
+
(aq) + NO3
–
(aq)

Note that we use an equilibrium arrow, but the forward arrow is longer than the reverse arrow. This shows
that the forward reaction is favoured and the equilibrium lies towards the products.
Strong bases dissociate completely. Examples of strong bases are Group 1 and 2 hydroxides like NaOH,
KOH, Mg(OH)2, and Ca(OH)2.
The dissociation of strong bases can be represented by balanced chemical equations as follows:
NaOH
H
2O
→ Na
+
(aq) + OH
–
(aq)
Ca(OH)2(s)
H
2O
→ Ca
2+
(aq) + 2OH
–
(aq)
Weak acids and bases
Weak acids do not ionise completely. The diagram on the right shows an
example of a weak acid HA in water. A weak acid does not ionise much in
water and only some acid molecules will be present as H3O
+
and A
–
. Most
of the acid molecules will be present as HA. Can you see that in the
diagram below there are six HA molecules and only two H3O
+
and A
–
ions.
In the previous diagram of a strong acid, there were no HA molecules
present. Acetic acid (ethanoic acid) and oxalic acid are examples of weak
acids. The ionisation of ethanoic acid can be shown as follows:
CH3COOH(ℓ) + H2O(ℓ) ⥄ CH3COO
–
(aq) +

H3O
+
(aq)
Note that we use an equilibrium arrow, but the reverse arrow is longer than the forward arrow. This shows
that the reverse reaction is favoured and the equilibrium lies towards the reactants.
Weak bases do not dissociate completely. Examples of weak bases are Na2CO3, NH3, or NaHCO3. If the
base is a covalent compound then ionisation takes place, for example in the case of ammonia (NH3):
NH3 + H2O ⥄ NH4
+
(aq) + OH
–
(aq)

+ ⇌ +

Strong acids ionise completely

Weak acids partially ionise

12. Grade 11 Acids and Bases R Toerien Page 12
Concentrated and dilute acids and bases
Concentrated acids have a high ratio of solute
to solvent compared to a dilute acid where the
ratio is small.
The diagram on the left shows a concentrated
solution, compared to the diagram on the right
which shows a dilute solution with only a few
acid molecules in solution. A concentrated acid
could have a concentration of 2 moldm
–3
whereas a dilute acid could have a
concentration of 0,002 moldm
–3
. A dilute acid does not mean that it is a weak acid. Strong or weak acids
and bases can be either concentrated or dilute depending on how much solvent is present.
You can watch an animation (youtu.be/NeHl5e7yuv8) and a video (youtu.be/DupXDD87oHc) on strong and
weak acids.
Conductivity and reaction rates
Stronger acids and bases have higher conductivity and faster reaction rates since the concentrations of
ions in solution are higher due to their ability to ionise or dissociate.

Image: opentextbc.ca



Strong acid: A strong acid is an acid that ionises completely in water to form a high concentration of
H3O
+
ions.
Weak acid: A weak acid is an acid that ionise incompletely/partially in water to form a low
concentration of H3O
+
ions.
Strong base: Strong bases dissociate completely in water to form a high concentration of OH
–
ions.
Weak base: Weak bases dissociate/ionise incompletely/partially in water to form a low
concentration of OH
–
ions.
Concentrated acid: A concentrated acid contains a large amount (number of moles) of acid in
proportion to the volume of water.
Concentrated base: A concentrated base contains a large amount (number of moles) of base in
proportion to the volume of water.
Dilute acid: Dilute acids contain a small amount (number of moles) of acid in proportion to the
volume of water.
Dilute base: Dilute bases contain a small amount (number of moles) of base in proportion to the
volume of water.


Concentrated weak acid Dilute weak acid
strong acid
faster reaction
weak acid
slower reaction

13. Grade 11 Acids and Bases R Toerien Page 13
Exercise 4
4.1 Four diagrams are given below. Each diagram is meant to represent one of the descriptions in the table.
Match the description to the diagram by writing the corresponding number in the column on the right.

Description Corresponding diagram
A concentrated solution of a strong acid
A dilute solution of a strong acid
A concentrated solution of a weak acid
A dilute solution of a weak acid

4.2 Write balanced chemical equations for each of the following acids ionizing in water and answer the
questions that follow: HNO3 ; H2SO4 ; H3PO4





4.2.1 Identify the ampholytes in the above reactions.
4.2.2 For each ampholyte above, write down its conjugate acid.
4.2.3 For each ampholyte above, write down its conjugate base.
4.2.4 Which substances can only behave as acids?
4.2.5 Which substances can only behave as bases?

4.3 For each of the following reactions, identify the conjugate acid-base pairs.
4.3.1 HCO3
–
(aq) + HF(aq) ⇌ H2CO3(aq) + F
–
(aq)


4.3.2 HCO3
–
(aq) + OH
–
(aq) ⇌ CO3
2–
(aq) + H2O(ℓ)


Diagram 1 Diagram 2 Diagram 3 Diagram 4

14. Grade 11 Acids and Bases R Toerien Page 14
Hydrolysis of salts
Neutral salts
Ionic salts are prepared by reacting an acid with a base, for
example, NaOH + HCℓ → NaCℓ + H2O. When the salt is dissolved
in water, it will dissociate into its ions as shown in the following
equation:
NaCℓ(s)
H
2O
→ Na
+
(aq) + Cℓ
–
(aq)
Note that the H2O is written above the arrow to indicate that the
salt is dissolved in the solvent and that the solvent does not take
part in the chemical reaction. This reaction can be visualised in the diagram. The sodium ions and chloride
ions are surrounded by water molecules (denoted by (aq) in the equation), but they do not react with the
water molecules.
The resulting solution is neutral and at a pH of 7. However, not all salts that are dissolved in water form
neutral solutions. Only the salts of strong acids and strong bases form neutral solutions. NaCℓ is the salt of
a strong base and a strong acid. NaOH is a strong base and HCℓ is a strong acid.
Acidic salts
Ammonium chloride (NH4Cℓ) is a salt formed from a strong acid (HCℓ) and a weak base (NH3). The
dissociation of ammonium chloride is shown by the following equation:
Dissociation of ammonium chloride: NH4Cℓ(s)
H
2O
→ NH4
+
(aq) + Cℓ
–
(aq)
Ammonium ions (NH4
+
(aq)) will react with water molecules in the solution in a process called hydrolysis.
Hydrolysis is the reaction with water. It is an acid-base reaction where a proton is transferred from NH4
+
to
H2O. NH4
+
is acting as an acid by donating a proton and forming the conjugate base NH3, while H2O is
acting as a base by accepting a proton and forming the conjugate acid H3O
+
. The equation is:
Hydrolysis of ammonium ions: NH4
+
(aq) + H2O(ℓ) → H3O
+
(aq) + NH3(aq)
This reaction forms hydronium (H3O
+
) ions in solution, causing the pH to decrease below 7, and thus
resulting in an acidic solution. The presence of H3O
+
ions gives the solution its acidic properties.
Basic salts
Sodium carbonate (Na2CO3) is a salt formed from a weak acid (H2CO3) and a strong base (NaOH). The
dissociation of sodium carbonate is shown by the following equation:
Dissociation of sodium carbonate: Na2CO3(s)
H
2O
→ 2Na
+
(aq) + CO3
2–
(aq)
Carbonate ions (CO3
2–
(aq)) will undergo hydrolysis. Protons will be transferred from H2O to CO3
2–
. H2O is
acting as an acid by donating a proton and forming the conjugate base OH
–
, while CO3
2–
is acting as a base
by accepting a proton and forming the conjugate acid HCO3
–
. The equation is:
Hydrolysis of carbonate ions: CO3
2–
(aq) + H2O(ℓ) → HCO3
–
(aq) + OH
–
(aq)
This reaction forms hydroxide (OH
–
) ions in solution, causing the pH to increase above 7 and thus resulting
in a basic solution. The presence of OH
–
ions gives the solution its basic properties. You can watch the
following video on the hydrolysis of salts: youtu.be/-vIwTn7LZiM.

Strong acids + strong bases form neutral salts
Strong acids + weak bases form acidic salts
Weak acids + strong bases form basic salts

15. Grade 11 Acids and Bases R Toerien Page 15
Worked example 5
5.1 Ammonium sulfate ((NH4)2SO4) is dissolved in water. Is the solution ACIDIC, BASIC or NEUTRAL?
Explain the answer by referring to balanced chemical equations.





5.2 Potassium nitrate is dissolved in water. Is the solution ACIDIC, BASIC or NEUTRAL? Explain the
answer by referring to balanced chemical equations.







Exercise 5
5.1 A few crystals of ammonium nitrate are added to distilled water in a test tube and a solution is formed.
5.1.1 Is the solution ACIDIC, BASIC, or NEUTRAL?
5.1.2 Write down two ionic equations that will explain the answer to QUESTION 5.1.1.

5.2 Each of the following salts is dissolved in water. Predict whether the resulting solution will be ACIDIC,
BASIC, or NEUTRAL.
5.2.1 sodium sulfate (Na2SO4)
5.2.2 potassium chloride (KCℓ)
5.2.3 calcium acetate ((CH3COO)2Ca)
5.2.4 ammonium bromide (NH4Br)

Hydrolysis of salts
Hydrolysis is the reaction of a substance with water. Salts undergo hydrolysis to form acidic, basic or
neutral solutions.
Acid Base Aqueous solution of the salt
Strong Strong Neutral
Strong Weak Acidic
Weak Strong Basic

Example: Ammonium chloride forms an acidic solution.
• NH4Cℓ is the salt of a weak base (NH3) and a strong acid (HCℓ)
• Dissociation of ammonium chloride: NH4Cℓ(s)
H
2O
→ NH4
+
(aq) + Cℓ
–
(aq)
• Hydrolysis of ammonium ions: NH4
+
(aq) + H2O(ℓ) → H3O
+
(aq) + NH3(aq)
• The H3O
+
ions give the solution acidic properties

16. Grade 11 Acids and Bases R Toerien Page 16
Standard solutions
A standard solution is a solution of known concentration. Substances that make good standard solutions
have the following properties.
- They have a high degree of purity.
- They are preferably solid substances as it is easier to measure a solid.
- They are not hygroscopic (absorb water) or efflorescent (give off water).
Good examples of substances for standard solutions are oxalic acid [(COOH)22H2O] or sodium carbonate
(Na2CO3).
A standard solution is prepared by weighing the solid accurately, transferring the solid to a volumetric flask,
and then adding solvent to the mark. This ensures that a known mass and a known volume are obtained,
and thus the concentration can be accurately calculated.


You can watch this video of making a standard solution: youtu.be/iPYyRNjXkgY
Worked example 6
6.1 Calculate the concentration of a sodium carbonate solution where exactly 2,65 g solid was dissolved
in 750 cm
3
distilled water.



6.2 Determine the mass of oxalic acid, (COOH)22H2O needed to prepare 250 cm
3
of a 0,1 moldm
–3

standard solution.





Standard solution is a solution of which the concentration is
known. Concentration is calculated using:
c =
m
MV


Image: saylordotorg.github.io

17. Grade 11 Acids and Bases R Toerien Page 17
Dilutions
Most solutions that are used in laboratories, are bought at high concentration (called stock solutions) and
needs to be diluted to make them suitable for use in the laboratory. The desired number of moles is
removed by pipetting out a fixed volume and transferring it to a second flask, which is then filled to a fixed
volume with a solvent, like distilled water.


For example, if the stock solution has a concentration of 5 moldm
–3
, and 50 cm
3
are removed, the number
of moles removed is n = c V = 5 x
50
1000
= 0,25 mol. If this is now transferred to the new flask which has a
volume of 250 cm
3
, the new concentration is c =
n
V
=
0,25
0,250
= 1 moldm
–3
. You will
notice that the moles taken out (nbefore) is the same as the moles in the new flask
(nafter). We use this principle to solve dilution problems as shown in the worked
examples below. The equation we use is shown below. Situation ‘1’ is used for
before the dilution and situation ‘2’ for after the dilution:
You can watch the following video on diluting solutions: youtu.be/7AqfAqh5oSY
Worked example 6 (cont.)
6.3 A 2 moldm
–3
solution of H2SO4 needs to be diluted to 0,01 moldm
–3
for a titration. What volume of
acid should be added to a 500 cm
3
volumetric flask to make this solution?







6.4 What volume of water should be added to 400 cm
3
of a 0,12 moldm
–3
HCℓ solution to dilute it to
0,1 moldm
–3
?







Image: saylordotorg.github.io
nbefore = nafter
c1V1 = c2V2

18. Grade 11 Acids and Bases R Toerien Page 18

Exercise 6
6.1 Calculate the concentration of an acetic acid (CH3COOH) solution where 4,80 g solid was dissolved
in 500 cm
3
distilled water.



6.2 Determine the mass of phosphoric acid (H3PO4) needed to prepare 750 cm
3
of a 0,3 moldm
–3

standard solution.




6.3 What is the concentration of a 200 cm
3
solution of sodium hydroxide (NaOH) which is prepared from
50 cm
3
of a 2,5 moldm
–3
stock solution?




6.4 What volume of 5 moldm
–3
stock solution is needed to prepare 250 cm
3
of a 0,1 moldm
–3
copper
sulfate (CuSO4) solution?




6.5 A solution of sodium carbonate (solution A) is prepared by dissolving 21,2 g of Na2CO3 in a 100 mℓ
volumetric flask.
6.5.1 Calculate the concentration of Solution A.




6.5.2 What volume of Solution A will be required to prepare 500 mℓ of a dilute sodium carbonate
solution with a concentration of 0,25 moldm
–3
?





6.6 A 10 mℓ sample of a 1,07 moldm
–3
solution of potassium hydrogen phthalate (KHP, molar mass =
204,22 gmol
–1
) is diluted to 250 mℓ.
6.6.1 What is the concentration of the final solution?





6.6.2 How many grams of KHP are in the 10 mℓ sample?

19. Grade 11 Acids and Bases R Toerien Page 19
Performing an acid-base titration
In this experiment, you will determine the concentration of acetic acid in vinegar using a standardised
sodium hydroxide solution.
Aim: Determining the concentration of acetic acid in household vinegar.
Apparatus: You will need the following:
Preparation of a standard solution Apparatus for the titration

Chemicals needed:
• Sodium hydroxide
• Distilled water
• Household vinegar
• Phenolphthalein indicator
Method: Part A Making a standard solution
1. Calculate the mass of sodium hydroxide needed to prepare 250 cm
3
of a 0,10 moldm
–3
solution.
2. Accurately weigh off the required mass in a small beaker. Write down the mass:
3. Transfer the solid to a 250 cm
3
volumetric flask.
4. Add distilled water to dissolve the solid.
5. Fill the volumetric flask to the mark.
6. Calculate the exact concentration from the mass used. The solution can now be used in Part B.
Method: Part B Titration
1. Pour 10 - 15 cm
3
of the NaOH solution into the burette, rinse it thoroughly and then pour out the
NaOH into the waste beaker.
2. Fill the burette to above the zero mark, place a beaker under the tap and open the tap fully for a
moment to fill the tip of the burette with solution.
3. Clamp the burette in the stand and zero it by slowly running the solution out until the bottom edge of
the meniscus is on the zero mark. (Make sure that your eyes are level with the zero mark.)
4. Pipette exactly 25,00 cm
3
of the vinegar solution into a clean Erlenmeyer flask.
5. Add 2 - 3 drops of phenolphthalein.
6. Run the NaOH solution at a brisk rate into the Erlenmeyer, manipulating the tap with the left hand
whilst swirling the solution with a wrist movement of the right hand until a colour change seems
about to take place.
7. Now titrate dropwise - but still fairly fast - up to a definite colour change to pink, lasting for at least
15 seconds.

20. Grade 11 Acids and Bases R Toerien Page 20
8. Take the reading and write it down in the table. This serves as a trial run to give you an idea of the
endpoint of the titration.
9. Fill and zero the burette again.
10. Rinse the Erlenmeyer thoroughly and pipette another 25,00 cm
3
of vinegar into it and add 2 - 3
drops of indicator.
11. Run in NaOH at a brisk rate to within 0,5 cm
3
of the endpoint as obtained in the "trial run" and then
carefully, drop by drop, to the first 15 seconds pink colour.
12. Take and record the reading (Vb).
13. Do another titration and compare the two readings. If they do not agree to within 0,1 cm
3
a further
titration must be done.
14. Record your readings on the table.
15. Calculate the concentration of acetic acid.
Results:
1. Draw the following table in your workbook:
Volume of acetic acid (Va) in cm
3
Volume of sodium hydroxide (Vb) in cm
3
25,00 Trial:



Average: Average:

2. Write down the concentration of NaOH from Part A:
3. Do the necessary calculations to determine the concentration of acetic acid in the vinegar sample.
Conclusion:
Write a conclusion for the experiment.

You can watch the following video to show you how a titration is done. The video also discusses how to do
this experiment with high accuracy. youtu.be/YqfvRBJ-iPg
Exercise 7
For this exercise, you need to open the following website: http://www.rsc.org/learn-
chemistry/resources/screen-experiment/titration/experiment/2/5
You will be doing a titration using a simulation.
You DO NOT have to register or log in.
Go to the ‘Quickstart’ option.
Choose ‘Titration level 1’
Follow the prompts and answer all the
questions.
You do not need to write any comments to the
teacher, only make a note of your final score.
Write your final score here:
Only titration level 1 is required for this exercise, but you can continue with the other titrations if you like.

21. Grade 11 Acids and Bases R Toerien Page 21
Choosing an indicator for a titration
An acid-base indicator is a chemical compound that changes colour depending on the pH of the solution.
We use indicators to help us visualise when the pH is at a certain value. For example, phenolphthalein is a
chemical substance that changes from colourless to pink at a pH value between 8,2 and 10,0. If we add
phenolphthalein to an acid-base reaction that changes from acidic to basic in this pH range, then we will be
able to ‘see’ when the change over from acidic to basic takes place. The endpoint of a titration is the point
where the indicator changes colour.
The strength of the acid and base, and thus the salt formed during a titration, will determine the pH range
where neutralisation takes place. Neutralisation is when the equivalence point is reached and not when
the pH is 7. The pH at the equivalence point depends on the hydrolysis of the salt that is formed. There are
four possible scenarios for titrations, depending on the strength of the acid and base used:
1. The reaction between a strong base and weak acid will have its equivalence point in the basic pH
range (pH > 7).
2. The reaction between a strong base and strong acid will have its equivalence point around pH = 7.
3. The reaction between a weak base and strong acid will have its equivalence point in the acidic pH
range (pH < 7).
4. For the reaction between a weak base and weak acid, it is difficult to predict where the equivalence
point will be, as it depends on the relative strengths of the acid and base used.
The table below summarises the colours of a few indicators in acidic and basic solutions and the pH range
of the indicator.
Indicator Colour in
acid
Colour at
equivalence
point
Colour in
base
pH range at
equivalence
point
Appropriate
for which
titration

Litmus Red Blue Not used

Methyl orange Red Orange Yellow 3,1 – 4,4
Strong acid
and weak
base

Bromothymol
blue
Yellow Green Blue 6,0 – 7,6
Strong acid
and strong
base

Phenolphthalein Colourless Pale Pink Pink 8,2 – 10,0
Strong base
and weak
acid

Universal
indicator
Red-
orange
Green
Blue-
purple
Not used



0 7 14
pH

0 7 14
pH

0 7 14
pH

0 7 14
pH

0 7 14
pH

Equivalence point: The equivalence point of a titration is the point at which the acid and the base
has completely reacted with each other.
Endpoint: The endpoint of a titration is the point where the indicator changes colour.
pH scale: The pH scale is a scale of numbers from 0 to 14 used to express the acidity or
alkalinity of a solution.

22. Grade 11 Acids and Bases R Toerien Page 22
Solving Titration problems
The following equation can be used to solve titration problems:


Worked example 8
8.1 In a titration experiment, 25 cm
3
of a 0,1 moldm
–3
sodium hydroxide (NaOH) solution was placed in
an Erlenmeyer flask. The endpoint was reached when 42,6 cm
3
of sulfuric acid (H2SO4) was added to
the sodium hydroxide solution.
8.1.1 Suggest an appropriate indicator for this reaction. Provide a reason for the choice of the indicator.




8.1.2 Calculate the concentration of the sulfuric acid solution.












8.2 A student was titrating 25 mℓ of a NaOH solution with an HCℓ solution of concentration 0,28 moldm
–3
.


The student ran out of the HCℓ solution after having added 32,5 mℓ, so he borrowed an HCℓ solution
that was labelled as 0,32 moldm
–3
. An additional 11,5 mℓ of the second solution was needed to
complete the titration. What was the concentration of the NaOH solution?
HCℓ + NaOH → NaCℓ + H2O
32,5 cm
3
; 0,28 moldm
–3
? moldm
–3
11,5 cm
3
; 0,32 moldm
–3
25 cm
3











caVa
c
bV
b
=
na
n
b


na: moles of acid
nb: moles of base
ca: concentration of acid
cb: concentration of base
Va: volume of acid
Vb: volume of base

23. Grade 11 Acids and Bases R Toerien Page 23
Exercise 8
8.1 Concentrated HCℓ is usually sold with a concentration of 10 moldm
–3
. What volume of this acid would
you need to make up 500 mℓ of a 0,1 moldm
–3
solution?






8.2.1 Calculate the volume of 0,15 moldm
–3
NaOH(aq) neutralised by 22,5 cm
3
of 0,07 moldm
–3
HCℓ(aq).
The balanced equations is as follows: NaOH + HCℓ → NaCℓ + H2O





8.2.2 Calculate the volume of 0,12 moldm
–3
KOH(aq) neutralised by 27 cm
3
of 0,25 moldm
–3
H2SO4(aq).
The balanced equation is as follows: 2KOH + H2SO4 → K2SO4 + 2H2O





8.2.3 Calculate the volume of 0,1 moldm
–3
Na2CO3(aq) neutralised by 10 cm
3
of 0,2 moldm
–3
HNO3(aq).
The balanced equation is as follows: Na2CO3 + 2HNO3 → 2NaNO3 + CO2 + H2O





8.2.4 Calculate the concentration of 25 cm
3
LiOH(aq) neutralised by 20 cm
3
of 0,23 moldm
–3
(COOH)2(aq).
The balanced equation is as follows: 2LiOH + (COOH)2 → (COOLi)2 + 2H2O

24. Grade 11 Acids and Bases R Toerien Page 24
8.3 25 cm
3
of a dilute HCℓ solution is neutralised by 18,8 cm
3
of a solution of NaOH with a concentration
of 0,12 moldm
–3
. The chemical equation for the reaction is NaOH + HCℓ → NaCℓ + H2O. Calculate:
8.3.1 the concentration of the HCℓ.
8.3.2 the concentration of the solution obtained if 50 cm
3
of the NaOH solution is diluted to 250 cm
3
.




















8.4 The concentration of a solution of sodium carbonate is exactly 0,05 moldm
–3
. 20 cm
3
of this solution
neutralises exactly 24,5 cm
3
of a solution of sulfuric acid. Calculate the concentration of the sulfuric
acid. The unbalanced equation for the reaction is shown below:
Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2

25. Grade 11 Acids and Bases R Toerien Page 25
8.5 2,5 grams of sodium hydroxide are dissolved in water. The volume of the solution is 250 cm
3
. Of this
solution, exactly 25 cm
3
are used to react completely with 15 cm
3
of sulfuric acid. Calculate the
concentration of the sulfuric acid. The unbalanced equation for the reaction is shown below:
NaOH + H2SO4 → Na2SO4 + H2O
















8.6 During a titration of a 0,1 moldm
–3
sodium hydroxide solution with oxalic acid, learners find that
25,1 cm
3
of the NaOH(aq) neutralises exactly 14,2 cm
3
of the (COOH)2(aq). The balanced equation for
the reaction is as follows:
2NaOH(aq) + (COOH)2(aq) → (COO)2Na2(aq) + 2H2O(ℓ)

8.6.1 Calculate the concentration of the oxalic acid solution.









The following indicators are available for the titration:
Indicator pH range
A 3,1 – 4,4
B 6,0 – 7,6
C 8,3 – 10,0

8.6.2 Which ONE of the indicators above is most suitable for this titration? Give a reason for the
answer.

26. Grade 11 Acids and Bases R Toerien Page 26
Solving titration excess problems
Worked example 9.1
9.1 A learner needs to determine the percentage of magnesium oxide in a health tablet. The tablet is
dissolved in 0,05 dm
3
of 0,8 moldm
–3
hydrochloric acid.
All of the magnesium oxide in the tablet reacts with the acid as shown below:
MgO(aq) + 2HCℓ(aq) → MgCℓ2(aq) + H2O(ℓ)
Not all the hydrochloric acid reacts. The learners titrate the excess hydrochloric acid with a solution of
sodium hydroxide. It takes 0,02 dm
3
of 0,5 moldm
–3
sodium hydroxide to neutralise the excess
hydrochloric acid, as shown in the equation below:
NaOH(aq) + HCℓ(aq) → NaCℓ(aq) + H2O(ℓ)
9.1.1 Calculate the number of moles of acid present in 0,05 dm
3
of 0,8 moldm
–3
hydrochloric acid solution.



9.1.2 The original mass of the tablet is 0,96 g. Calculate the percentage of magnesium oxide in the tablet.

27. Grade 11 Acids and Bases R Toerien Page 27
Exercise 9a
9.1 Washing soda contains sodium carbonate (Na2CO3). A sample of 5,13 g of washing soda crystals
was dissolved in water to prepare a 250 cm
3
solution. 25 cm
3
of this solution is neutralised by 36 cm
3

of a 0,05 moldm
–3
sulfuric acid solution. The balanced equation for the reaction is:

Na2CO3(aq) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(ℓ)

9.1.1 Calculate the number of moles of sulfuric acid that reacted with the sodium carbonate. (3)
9.1.2 Calculate the mass of sodium carbonate in the sample of washing soda. (4)
9.1.3 What percentage of the washing soda sample is sodium carbonate? (3)

28. Grade 11 Acids and Bases R Toerien Page 28
9.2 Calcium carbonate forms a large percentage of seashells. In order to determine the percentage purity
of the calcium carbonate in seashells, learners react 5 g of powdered seashells with 500 cm
3
of
0,25 moldm
–3
hydrochloric acid. The reaction that takes place is represented by the following
balanced equation:
CaCO3(s) + 2HCℓ(aq) → CaCℓ2(aq) + H2O(ℓ) + CO2(g)
The excess HCℓ is then titrated with 140 cm
3
of a 0,2 moldm
–3
solution of sodium hydroxide.
9.2.1 Calculate the mass of CaCO3 present in the sample of seashells. (8)
9.2.2 Calculate the percentage purity of the calcium carbonate in the seashells. (3)

29. Grade 11 Acids and Bases R Toerien Page 29
Worked example 9.2
9.2 A packet of washing soda crystals has been left open to the atmosphere for some time. The crystals
have formed a white powder which may be represented by the formula Na2CO3xH2O. The value of x
may be found by dissolving a known quantity of the powder in water and titrating the solution with an
acid of known concentration.
In one experiment a learner weighed out 1,59 g of the powder and dissolved it in water. The volume
of the solution was made up to exactly 250 cm
3
. 25 cm
3
of this solution was titrated with hydrochloric
acid of concentration 0,10 moldm
–3
. 25,6 cm
3
of the acid was required for a complete reaction. The
equation for this reaction is Na2CO3 + 2HCℓ → 2NaCℓ + H2O + CO2.
9.2.1 Calculate the mass of sodium carbonate in the 250 cm
3
of solution. (8)
9.2.2 How many moles of water were present in the original powder? (4)
9.2.3 Calculate the value of x. (3)

30. Grade 11 Acids and Bases R Toerien Page 30
Exercise 9b
9.3 0,212 g of an unknown carbonate having a formula X2CO3 reacted completely with 25 cm
3
nitric acid
of concentration 0,2 moldm
–3
. The nitric acid was in excess. The balanced equation is shown below.
X2CO3 + 2HNO3 → 2XNO3 + H2CO3
After the reaction ceased, the excess nitric acid was neutralised with a sodium hydroxide solution of
concentration of 0,1 moldm
–3
. If 10 cm
3
of the sodium hydroxide solution were needed to exactly
neutralise the excess acid.
9.3.1 Calculate the number of moles of nitric acid that was added to the unknown carbonate. (3)
9.3.2 Calculate the number of moles of nitric acid which reacted with the unknown carbonate. (5)
9.3.3 Calculate the molar mass of the element X. (3)
9.3.4 Identify element X. (3)

31. Grade 11 Acids and Bases R Toerien Page 31
9.4 60 cm
3
of distilled water is added to 120 cm
3
of a 0,3 moldm
–3
sodium hydroxide (NaOH) solution.
9.4.1 Calculate the concentration of the DILUTED NaOH solution. (3)
0,48 g of IMPURE ammonium bromide is then heated with 25 cm
3
of the DILUTED NaOH according
to the following balanced equation. The NaOH is in excess.
NH4Br(s) + NaOH(s) → NH3(g) + H2O(ℓ) + NaBr(aq)
The excess NaOH was neutralised by exactly 10 cm
3
of a 0,05 moldm
–3
solution of oxalic acid
according to the following balanced equation.
(COOH)2(aq) + 2NaOH(aq) → (COONa)2(aq) + 2H2O(ℓ)
9.4.2 Calculate the percentage purity of the ammonium bromide. (10)

32. Grade 11 Acids and Bases R Toerien Page 32
The pH concept
As you have seen in the previous section, the presence of hydronium ions (H3O
+
) tells you whether the
solution has acidic or basic properties. However, the concentration of H3O
+
ions can be very small. It is
more convenient to quote the concentration in terms of pH which is defined as:

Quoting the hydronium ion concentration in terms of pH avoids the use of exponents. Below is an example
of how the concentration is converted to pH.

Worked example 10
10.1 The hydronium ion concentration of swimming pool water is 1 x 10
–6
moldm
–3
. Determine the pH of
the water and state whether the swimming pool water is acidic or basic.





10.2 A sulfuric acid solution has a concentration of 0,001 moldm
–3
. Calculate the pH of the solution.








There is a relationship between pH and the hydronium ion concentration. The diagram below shows this
relationship and the pH values for a variety of solutions.



pH = – log [H3O
+
]

33. Grade 11 Acids and Bases R Toerien Page 33
The relationship between Kw and pH
You have learned in a previous section that water undergoes autoprotolysis as shown below:
H2O + H2O ⇌ H3O
+
+ OH
–

In all aqueous solutions the product of the concentrations of H3O
+
and OH
–
has a fixed value of 1 x 10
–14
.
This value is called the equilibrium constant for water (Kw). You will learn more about equilibrium in
Grade 12. The relationship between Kw and the hydronium and hydroxide concentrations are shown below:

This is a very useful equation as you can now calculate the concentration of H3O
+
if you have OH
–
or vice
versa.
Worked example 10 (cont.)
10.3 The pH of a sodium hydroxide (NaOH) solution is 11.
10.3.1 Calculate the hydroxide ion concentration of the solution.









10.3.2 What mass of sodium hydroxide was added to 500 cm
3
to make this solution?








10.4 Calculate the pH of pure distilled water.










Kw = [H3O
+
] [OH
–
]
1 x 10
–14
= [H3O
+
] [OH
–
]



Note:
To calculate [H3O
+
] you press
Shift log (-) 11 =
You will notice that ‘Shift log’
will give you 10
x
.

34. Grade 11 Acids and Bases R Toerien Page 34
Worked example 10 (cont.)
10.5 Calculate the pH of each of the following solutions:
10.5.1 A 0,05 moldm
–3
solution of sodium hydroxide (NaOH).
NaOH(s)
H
2O
→ Na
+
(aq) + OH
–
(aq)








10.5.2 A 0,00125 moldm
–3
solution of barium hydroxide (Ba(OH)2).
Ba(OH)2(s)
H
2O
→ Ba
2+
(aq) + 2OH
–
(aq)








Exercise 10
10.1 Calculate the pH of pure distilled water.



10.2 Calculate the pH of a solution of which the hydronium ion concentration is 0,01 moldm
–3
.



10.3 Calculate the pH of a solution of which the concentration of the hydroxide ion is 0,01 moldm
–3
.

35. Grade 11 Acids and Bases R Toerien Page 35
10.4 0,73 g of hydrogen chloride is dissolved in water to yield a solution of volume 1000 cm
3
. Calculate
the pH of the solution.










10.5 4 g of sodium hydroxide is dissolved in water to form a solution of volume 1000 cm
3
. Calculate the
pH of the solution.











10.6 The pH of a solution is 5. Calculate the concentration of the hydronium and hydroxide ions.

36. Grade 11 Acids and Bases R Toerien Page 36
pH and titrations
Worked example 11
11.1 500 cm
3
of a hydrochloric acid solution has a pH of 3. How much water must be added to increase
the pH to 4?
HCℓ + H2O → H3O
+
+ Cℓ
–
500 cm
3
pH = 3







11.2 30 cm
3
of a 0,4 moldm
–3
solution of NaOH are added to 20 cm
3
of a 0,1 moldm
–3
H2SO4 solution at
25°C.
11.2.1 Write down the balanced equation for the reaction that takes place. (3)
11.2.2 Write down the name of a suitable indicator that can be used in this reaction. (1)
11.2.3 Calculate the concentration of the hydroxide ions in the mixture after the two solutions were
mixed. (5)
11.2.4 Calculate the pH of the mixture after the two solutions were added to each other. (5)

37. Grade 11 Acids and Bases R Toerien Page 37
Exercise 11
11.1 6,2 g potassium hydroxide (KOH) is dissolved in 250 cm
3
of water. Calculate the pH of this solution. (7)


















11.2 A factory accidentally spills sulfuric acid into a nearby river. The fish species in the river CANNOT
survive in water with a pH LOWER THAN 5,8. Analysis of water samples from the river shows that
the hydronium ion concentration is 5,6 x 10
–6
moldm
–3
. Show with the aid of the calculation that the
fish will not survive in the river. (3)








11.3 Concentrated hydrochloric acid (HCℓ) with a concentration of 10 moldm
–3
is diluted to form 500 cm
3

of a 0,25 moldm
–3
solution.
11.3.1 Is the dilute hydrochloric acid a STRONG or a WEAK acid? (1)
11.3.2 Calculate the volume, in cm
3
, of the concentrated hydrochloric acid that must be used to
make the dilute solution. (4)
11.3.3 Calculate the pH of the 0,25 moldm
–3
solution of hydrochloric acid. (3)

38. Grade 11 Acids and Bases R Toerien Page 38
11.4 Magnesium hydroxide (Mg(OH)2) is often used as medicine to relieve an upset stomach. The pH of
the HCℓ in one person’s stomach was found to be 1,2.

11.4.1 Calculate the concentration of the hydrochloric acid in the person’s stomach. Round your
answer to three decimal places. (3)
11.4.2 Will the pH in the stomach INCREASE, DECREASE, or STAY THE SAME after taking a dose
of Mg(OH)2? (2)
A person takes a dose of Mg(OH)2.
11.4.3 Write down the balanced equation for the reaction that takes place in the stomach. (3)
11.4.4 What mass of magnesium hydroxide Mg(OH)2 will be needed to completely neutralise 500cm
3

of this HCℓ. (6)

39. Grade 11 Acids and Bases R Toerien Page 39
11.5 A 200 cm
3
solution of hydrochloric acid has a pH of 1.
11.5.1 Calculate the volume of water that must be added to this solution to change the pH to 2. (5)
Sodium carbonate crystals (Na2CO310H2O) is used to neutralise the original hydrochloric acid (with
a pH of 1). The balanced equation for the reaction is shown below:
Na2CO310H2O + 2HCℓ → 2NaCℓ(aq) + 11H2O + CO2
11.5.2 Calculate the mass of sodium carbonate crystals that will be required to neutralise 200 cm
3
of
the hydrochloric acid solution. (6)
11.5.3 Choose from the following table the most suitable indicator for the reaction. (1)
Indicator pH range in which the colour changes
Methyl red 4,8 – 6,0
Neutral red 6,8 – 8,0
Chlorophenol red 7,0 – 8,8
11.5.4 Give a reason for the answer to QUESTION 11.5.3. (2)

40. Grade 11 Acids and Bases R Toerien Page 40
11.6.1 What volume of a 0,978 moldm
–3
solution of NaOH must be added to 25 cm
3
of 0,583 moldm
–3

HCℓ to completely neutralize the acid? (4)
11.6.2 If the endpoint of the titration was accidentally overshot and 20 cm
3
of NaOH was added, calculate
the pH of the resulting solution. (8)

41. Grade 11 Acids and Bases R Toerien Page 41
pH excess calculations
Worked example 12

12.1 When 0,05 moles of HBr(ℓ) are added to 0,005 moles of KOH(aq) in a beaker, the total volume of the
solution is 75 cm
3
. HBr dissociates completely when dissolved in water. The chemical equation for
the reaction is shown below. Calculate the pH of the final solution. (7)
HBr + KOH → KBr + H2O

42. Grade 11 Acids and Bases R Toerien Page 42
Worked example 12 (cont.)

12.2 In a titration, 50 cm
3
of H2SO4 of concentration 0,1 moldm
–3
, is titrated against a NaOH solution of
concentration 0,1 moldm
–3
.
H2SO4 + NaOH → Na2SO4 + H2O
12.2.1 Calculate the volume of NaOH that is needed to neutralise exactly half the acid solution. (4)
12.2.2 Hence, determine the pH of the solution in the flask when only half the acid is neutralised.
(7)

43. Grade 11 Acids and Bases R Toerien Page 43
Worked example 12 (cont.)

12.3 A volume of 0,60 dm
3
of a 0,20 moldm
–3
NaOH solution is added to a beaker containing 0,40 dm
3
of
a H2SO4 solution. It was found that the pH of the final solution is 12,5 at a temperature of 25°C.

12.3.1 Calculate the hydroxide ion (OH
–
) concentration on completion of the reaction. (5)
12.3.2 Write down the balanced equation for the reaction that takes place. (3)
12.3.3 Calculate the concentration of the initial H2SO4 solution. (7)

44. Grade 11 Acids and Bases R Toerien Page 44
Exercise 12
12.1 When 0,08 moles of HCℓ are added to 0,1 moles of Ca(OH)2 in a beaker, the total volume of the
solution is 240 cm
3
. Assume that the Ca(OH)2 dissociates completely when dissolved in water. The
chemical equation for the reaction is shown below. Calculate the pH of the final solution. (10)
2HCℓ + Ca(OH)2 → CaCℓ2(aq) + 2H2O












12.2 A learner dissolves 15 g of copper(II) oxide powder by adding an excess of dilute hydrochloric acid to
it. 850 cm
3
of a 0,6 moldm
–3
HCℓ solution is used.
12.1 Write down a balanced equation for the reaction that takes place between the copper(II) oxide
and hydrochloric acid. (3)
12.2 Calculate the pH of the solution after all the copper(II) oxide has dissolved. (11)

45. Grade 11 Acids and Bases R Toerien Page 45
12.3 Hydrochloric acid is a highly corrosive strong acid with many industrial uses. When 0,02 dm
3
of
sodium hydroxide is added to 0,15 dm
3
of diluted hydrochloric acid of concentration 0,03 moldm
–3
,
the pH of the mixture changes to 4.
12.3.1 Give a reason why hydrochloric acid is classified as a strong acid. (2)
12.3.2 Write down a balanced chemical equation to show the ionisation of HCℓ in water. (3)
12.3.3 Will the final mixture be ACIDIC or BASIC? Give a reason for the answer by referring to the
pH of the mixture. (2)
12.3.4 Calculate the final concentration of hydronium ions in the mixture. (3)
12.3.5 Calculate the original concentration of the sodium hydroxide solution. (7)















12.4 A sulfuric acid solution is prepared by dissolving 7,35 g of H2SO4(ℓ) in 500 cm
3
of water.
12.4.1 Calculate the number of moles of H2SO4 present in this solution. (2)
Sodium hydroxide (NaOH) pellets are added to the 500 cm
3
H2SO4

solution. The balanced equation
for the reaction is:
H2SO4(aq) + 2NaOH(s) → Na2SO4(aq) + 2H2O(ℓ)

After completion of the reaction, the pH of the solution was found to be 1,3. Assume complete
ionisation of H2SO4.
12.4.2 Calculate the mass of NaOH added to the H2SO4 solution. Assume that the volume of the
solution does not change. (9)

46. Grade 11 Acids and Bases R Toerien Page 46
Grade 11 Chemistry Module C5 Worksheet Acids and Bases
QUESTION 1
1.1 HPO4
2–
can act as an ampholyte. In which one of the following reactions does HPO4
2–
act as
a Brønsted-Lowry acid?
A HPO4
2–
+ H
+
⇌ H2PO4
–

B HPO4
2–
+ HPO4
2–
⇌ 2HPO4
2–

C HPO4
2–
+ H2O ⇌ H2PO4
–
+ OH
–

D HPO4
2–
+ H2O ⇌ PO4
3–
+ H3O
+
(2)


1.2 Consider the incomplete reaction equation below:
CH3CO2H(aq) + C5H5N(aq) ⇌ _________ + C5H5NH
+
(aq)
The conjugate base of CH3CO2H is …
A CH3CO2
–

B CH3CO2
+

C C5H6N
D C5H5NH
+
(2)

1.3 When sodium hypochlorite (NaOCℓ) dissolves in water, the pH of the solution is 9,82. Which
one of the following equations offers the best explanation for this pH value?
A 2H2O ⇌ H3O
+
+ OH
–

B NaOCℓ ⇌ Na
+
+ OCℓ
–

C OCℓ
–
+ H2O ⇌ HOCℓ + OH
–

D NaOCℓ + H2O ⇌ NaCℓ + O2 + H2 (2)

1.4 Which one of the following 0,1 moldm
–3
solutions has the lowest pH?
A HCℓ
B H2SO4
C NaOH
D CH3COOH (2)

1.5 Consider a solution of ammonium chloride (NH4Cℓ) in water.
Which one of the following statements is FALSE?
A [ H3O
+
] < [ OH
–
]
B [ H3O
+
][ OH
–
] = 10
–14

C [ H3O
+
] > 10
–7
moldm
–3

D The pH of the solution is less than 7. (2)
[10]

47. Grade 11 Acids and Bases R Toerien Page 47
QUESTION 2
0,1 moldm
–3
aqueous solutions of the following compounds are prepared.
HNO3 H2SO4 KCℓ NaHCO3 H3PO4 LiOH HCOOH
Choose from the list a:
2.1 Strong base (1)
2.2 Weak polyprotic acid (1)
2.3 Strong monoprotic acid (1)
2.4 Weak monoprotic acid (1)
2.5 Solution with a pH of approximately 8,5 (1)
[5]







QUESTION 3
Sodium sulfite (Na2SO3) is dissolved in water.
3.1 Write down a balanced equation to show the dissociation of sodium sulfite in water. (2)
3.2 Will the solution of sodium sulfite in water be ACIDIC or BASIC? (1)
3.3 Explain the answer to QUESTION 3.2 with the aid of a balanced ionic equation. (1)
3.4 Write down the NAMES of the conjugate acid-base pairs present in the reaction in QUESTION 3.3. (4)
3.5 Give the NAME of the spectator ion when sodium sulfite dissolves in water and undergoes hydrolysis.
(1)
[11]

48. Grade 11 Acids and Bases R Toerien Page 48
QUESTION 4
4.1 Define an acid in terms of the Brønsted-Lowry theory. (2)
4.2 Carbonated water is an aqueous solution of carbonic acid, H2CO3.
H2CO3(aq) ionises in two steps when it dissolves in water.
4.2.1 Write down a balanced equation for the first step in the ionisation of carbonic acid. (3)
4.2.2 Write down the FORMULA of the conjugate base of H2CO3(aq) in the equation in
QUESTION 4.2.1. (1)
4.2.3 The pH of a carbonic acid solution at 25°C is 3,4. Calculate the hydroxide ion concentration in
the solution. (5)
4.3 HX is a monoprotic acid.
4.3.1 State the meaning of monoprotic. (1)
4.3.2 A sample of HX is titrated with a standard sodium hydroxide solution using a suitable indicator.
At the endpoint, it is found that 25 cm
3
of acid HX is neutralised by 27,5 cm
3
of the sodium
hydroxide solution of concentration 0,1 moldm
–3
. Calculate the concentration of acid HX. The
chemical equation for the reaction is HX + NaOH → NaX + H2O (5)

4.3.3 The concentration of H3O
+
ions in the sample of acid HX is 2,4 x 10
-4
moldm
–3
. Is acid HX a
WEAK or a STRONG acid. Explain the answer by referring to the answer in QUESTION 4.3.2.
(3)
[20]

49. Grade 11 Acids and Bases R Toerien Page 49
QUESTION 5
5.1 In a reaction, 25 cm
3
of a barium hydroxide (Ba(OH)2) solution are added to 40 cm
3
of a
0,15 moldm
–3
sulfuric acid (H2SO4) solution. The chemical equation for the reaction is shown below:
Ba(OH)2 + H2SO4

→ BaSO4 + 2H2O
5.1.1 The barium hydroxide has a pH of 13,6 at the start of the reaction. Calculate the concentration
of the barium hydroxide solution used in this reaction. (5)
5.1.2 Calculate the number of moles of barium hydroxide used in the reaction. (3)
5.1.3 Calculate the pH of the final solution. (7)

5.2 The salt barium sulfate (BaSO4) is dissolved in water.
5.2.1 Define the term hydrolysis. (2)
5.2.2 Will an aqueous solution of barium sulfate be ACIDIC, NEUTRAL, or BASIC? (1)
5.2.3 Give a reason for the answer in QUESTION 5.2.2. (2)
[20]

50. Grade 11 Acids and Bases R Toerien Page 50
QUESTION 6
A sample of seawater is treated with 500 cm
3
of a 2,5 moldm
–3
sodium hydroxide solution to remove the
magnesium ions. The reaction that takes place is represented by the following balanced equation:
Mg(NO3)2(aq) + 2NaOH(aq) → Mg(OH)2(s) + 2NaNO3(aq)
After removal of the precipitate, the excess NaOH solution is titrated with exactly 95 cm
3
of a 0,2 moldm
–3

sulfuric acid solution when the equivalence point is reached. The balanced equation for the reaction is:
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ)

6.1 Calculate the number of moles sodium hydroxide added to the seawater. (3)
6.2 Calculate the original mass of magnesium nitrate in the seawater. (6)
[9]

















TOTAL: 75



nbefore = nafter
c1V1 = c2V2


pH = – log [H3O
+
]




Kw = [H3O
+
] [OH
–
]
1 x 10
–14
= [H3O
+
] [OH
–
]




c
aV
a
c
bV
b
=
n
a
n
b



n =
V
V
M
c =
n
V
c =
m
MV
n =
N
N
A





n =
m
M

51. Grade 11 Acids and Bases R Toerien Page 51
Grade 11 Chemistry Module C5 CAPS Acids and Bases
• Describe the general macroscopic properties of acids and bases
• Write the overall equation for simple acid-metal hydroxide, acid-metal oxide, and acid-metal
carbonate reactions and relate these to what happens at the macroscopic and microscopic level
• Use acid-base reactions to produce and isolate salts e.g. Na2SO4; CuSO4 and CaCO3
• Use the acid-base theories of Arrhenius and Bronsted-Lowry to define acids and bases
• Define an acid as an H
+
donor and a base as an H
+
acceptor in reactions
• Identify conjugate acid-base pairs
• Define an ampholyte
• List common acids (including hydrochloric acid, nitric acid, sulfuric acid, and acetic acid) and
common bases (including sodium carbonate, sodium hydrogen carbonate, and sodium hydroxide)
by name and formula
• Write the reaction equations of aqueous solutions of acids and bases
• Give conjugate acid-base pairs for given compounds.
• Determine the approximate pH of salts in salt hydrolysis
• Give the neutralisation reactions of common laboratory acids and bases.
• How do indicators work? What is the range of methyl orange, bromothymol blue, and
phenolphthalein indicators?
• Do simple acid-base titrations
• Do calculations based on acid-base titration reactions
• Explain the auto-ionisation of water
• Distinguish between strong and weak acids. Name some common strong and weak acids and
bases
• Distinguish between concentrated and dilute acids
• Define the concept of Kw
• Explain the pH scale
• Calculate pH values of strong acids and strong bases
• Compare strong and weak acids in terms of pH, conductivity and reaction rate

Grade 11 Chemistry Module C5 Resources Acids and Bases
Below is a list of additional videos that you can watch:
How to make your own pH indicator: youtu.be/I0pVkB30LUc and youtu.be/lEdfYik5iX8.
How to make indicator paper strips: youtu.be/MgCAB6rKzho.
Acids and bases introduction: youtu.be/RP7WOs5P2QI
Acid-base theories: youtu.be/hQLWYmAFo3E
Conjugate acid-base pairs: youtu.be/sUrvEjm3kUg
Neutralisation and salt formation: youtu.be/WKE87E-x-bk
Indicators: youtu.be/aypvZKmGns4
Titration simulation: http://pages.uoregon.edu/tgreenbo/acid_base.html
Overview and acid-base calculations: youtu.be/u3t74x34DPc and youtu.be/SzHh8hjOImM

52. Grade 11 Acids and Bases R Toerien Page 52
Grade 11 Chemistry Module C5 Definitions Acids and Bases
Arrhenius acid: An acid is a substance that produces hydrogen ions (H
+
) when dissolved in water.
Arrhenius base: A base is a substance that produces hydroxide ions (OH
–
) when dissolved in water.
Neutralisation: The reaction where an acid and a base form a salt and water.
(Arrhenius)
Brønsted acid: An acid is a proton (H
+
ion) donor.
Brønsted base: A base is a proton (H
+
ion) acceptor.
Neutralisation: Neutralisation is the proton transfer from an acid to a base to form the
(Brønsted) respective conjugate base and conjugate acid.
Strong acid: A strong acid is an acid that ionises completely in water to form a high concentration
of H3O
+
ions.
Weak acid: A weak acid is an acid that ionise incompletely/partially in water to form a low
concentration of H3O
+
ions.
Strong base: Strong bases dissociate completely in water to form a high concentration of OH
–
ions.
Weak base: Weak bases dissociate/ionise incompletely/partially in water to form a low
concentration of OH
–
ions.
Concentrated acid: A concentrated acid contains a large amount (number of moles) of acid in proportion
to the volume of water.
Concentrated base: A concentrated base contains a large amount (number of moles) of base in proportion
to the volume of water.
Dilute acid: Dilute acids contain a small amount (number of moles) of acid in proportion to the
volume of water.
Dilute base: Dilute bases contain a small amount (number of moles) of base in proportion to the
volume of water.
Ampholyte: It is a substance that can act as either an acid or a base.
Amphiprotic: A substance is amphiprotic when it can act as either an acid or a base, in other
words, it can donate or accept a proton.
Hydrolysis: Hydrolysis is the reaction of a salt with water.
Equivalence point: The equivalence point of a titration is the point at which the acid and the base has
completely reacted with each other.
Endpoint: The endpoint of a titration is the point where the indicator changes colour.
pH scale: The pH scale is a scale of numbers from 0 to 14 used to express the acidity or
alkalinity of a solution.
Auto-ionisation: The auto-ionisation of water is the reaction of water with itself to form H3O
+
ions and
OH
–
ions.