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ACID-BASE REACTIONS
Learning Outcomes
By the end of this section you should be able to:


recall that acid-base reactions involve proton transfer;
identify the proton donor and proton acceptor in an acid-base reaction;
Key Ideas

There are two approaches to acid-base theory. The approach used in AS
Chemistry is known as the Bronsted-Lowry theory and involves transfer of
hydrogen ions (H+) – sometimes referred to as ‘protons’.

The second approach, known as Lewis theory, is introduced in the second year
of the course. It involves transfer of electron-pairs.
Bronsted-Lowry theory

Acids are defined as HYDROGEN ION DONORS (or proton donors)

Bases are defined as HYDROGEN ION ACCEPTORS (or proton acceptors)

In the AS course we limit our study to reactions taking place in aqueous
solution. In aqueous solution, hydrogen ions become attached to water
molecules. This produces the oxonium ion (H3O+)

Water itself plays an important part in acid-base reactions because it is able
to both donate and accept hydrogen ions (not at the same time!!). Substances
such as water are described as AMPHOTERIC.
Course outlines - Minerals to Elements (1)
Page 1 of 8
IMPORTANT EXAMPLES OF PROTON TRANSFER
Strong acids (such as hydrochloric acid and sulphuric acid)

These acids are completely ionised in aqueous solution.

This means that the resulting solution has a relatively high concentration of
oxonium ions

This produces a relatively low pH (usually less than 2)

HCl(g) + H2O(l)

H2SO4(l) + H2O(l) 

H3O+(aq) + Cl-(aq)
H3O+(aq) + HSO4-(aq)
In these reactions water acts as a BASE – a hydrogen ion acceptor.

Complete the following equation for a strong acid:

HNO3 + H2O

Weak acids (such as ethanoic acid and carbonic acid)

These acids are only slightly ionised in aqueous solution.

This means that the resulting solution has a lower concentration of oxonium
ions than a strong acid with the same concentration of acid.

This produces a higher pH (usually greater than 2)

H2CO3(aq) + H2O(l)

H3O+(aq) + HCO3-(aq)

CH3COOH(l) + H2O(l)

H3O+(aq) + CH3COO-(aq)
Complete the following equations for methanoic acid and phenol (weak acids):

HCOOH(l) + H2O(l)


C6H5OH(s) + H2O(l)

Course outlines - Minerals to Elements (1)
Page 2 of 8
Strong bases (such as sodium hydroxide and barium hydroxide)

These bases are completely ionised in aqueous solution, producing strongly
alkaline solutions.

This means that the resulting solution has a relatively high concentration of
hydroxide ions (OH-)

This produces a relatively high pH
(usually greater than 12)

NaOH(s) + aq

Na+(aq) + OH-(aq)

Ba(OH)2(s) + aq

Ba2+(aq) + 2OH-(aq)
Complete the following equations for potassium hydroxide and calcium hydroxide:

KOH(s) + aq


Ca(OH)2(s) + aq

Weak bases (such as ammonia and ethylamine)

These bases are only slightly ionised in aqueous solution.

The base reacts with water, removing a hydrogen ion from the water
molecule. Thus water acts as an acid.

This produces a solution with a relatively low concentration of hydroxide ions.

This produces a lower pH than that of a strong base (usually less than 12)

NH3(g) + H2O(l)

CH3CH2NH2(g) + H2O(l)

NH4+(aq) + OH-(aq)

CH3CH2NH3+(aq) + OH-(aq)
Complete the following equations for hydrazine and methylamine (weak bases):

N2H4(g) + H2O(l)

CH3NH2(g) + H2O(l)


Course outlines - Minerals to Elements (1)
Page 3 of 8
Chemical Ideas
Chapter 8.1
1
Use the Brønsted-Lowry theory to explain what happens when an acid reacts with a
base.
2
Write out each of the following equations and indicate which of the reactants is the
acid and which is the base.
a HNO3 + H2O
H3O+ + NO3f H3O+ + OH2H2O
+
b NH3 + H2O
NH4 + OH
g NH3 + HBr
NH4+ + Brc NH4+ + OHNH3 + H2O
h H2SO4 + HNO3
HSO4- + H2NO3+
d SO42- + H3O+
HSO4- + H2O
i CH3COOH + H2O
CH3COO- + H3O+
e H2O + H
H2 + OH
3
Classify each of the equations which follow as either acid-base or redox.
a NH4+ + CO32NH3 + HCO3c I2 + 2OHI- + IO- + H2O
2+
b H2S + 2OH
S + 2H2O
d Mg + 2H
Mg2+ + H2
4
Write out the following equations and in each case identify the two conjugate acidbase pairs. For example, for hydrochloric acid and the hydroxide ion the answer
would be:
Equation:
HCl + OHCl- + H2O
Conjugate pairs: acid/base
HCl/ClH2O/OH+
+
a NH4 + H2O
NH3 + H3O
c CH3COOH + HClO4
CH3COOH2+ + ClO4+
b H2SO4 + HNO3
HSO4 + H2NO3
5
Write balanced equations for each of the following reactions between acids and
bases. In each case identify the two conjugate acid-base pairs (see problem 4).
a Ethanoic acid and hydroxide ion
b Hydrogencarbonate ion, HCO3-, and hydrochloric acid to produce carbonic acid, H2CO3
c Water reacting with the hydrogensulphate ion, HSO4-.
Course outlines - Minerals to Elements (1)
Page 4 of 8
Mole calculations involving solutions
Making up a standard solution
 The exact mass of solute used to make the solution is converted into moles
using the formula:
moles of solute = mass of solute / Mr
 The molar concentration of solution is then calculated using the formula:
molar concentration of solution = moles of solute / volume of solution (dm3)
Worked Example
Calculate the molar concentration of a solution made up by dissolving 3.000 grams
of anhydrous sodium carbonate (Mr: 106) to make 2 dm3 of solution.

Moles of sodium carbonate = 3.000 / 106
mol
(0.0283 mol)

Molar concentration of solution = 0.0283 / 2 mol dm-3

Molar concentration of solution = 0.0142 mol dm-3
Examples
1.
Calculate the molar concentrations of the following solutions:
(a) Solution X contains 14.85 grams of pure sulphuric acid (H2SO4) in 800 cm3
of solution
(b) Solution Y contains 6.75 grams of potassium hydroxide (KOH) in 250 cm3 of
solution
(c) Solution Z contains 11.64 grams of potassium carbonate (K2CO3) in 750 cm3
of solution
2.
What is the concentration (in mol dm-3) of each of the following solutions:
(a) 10 grams of sodium hydroxide (NaOH) in 1 dm3 of solution
(b) 4 grams of sodium hydroxide in 500 cm3 of solution
(c) 100 grams of calcium bromide (CaBr2) in 100 cm3 of solution
(d) 2 grams of calcium bromide in 25 cm3 of solution
(e) 8 grams of hydrated copper sulphate (CuSO4.5H2O) in 100 cm3 of solution
(f) 40 grams of hydrated copper sulphate in 2 dm3 of solution
[Ar: H:1; C:12; O:16; Na:23; S:32; K:39; Ca:40; Cu:63.5; Br:80)
Course outlines - Minerals to Elements (1)
Page 5 of 8
Processing Titration Data
 In almost all the examples you will meet, the ‘unknown’ is the concentration of
one of the reactants. We will refer to this as c1.
 In order to calculate the value of this concentration you will need to know:
 The volume of this solution (V1)
 The concentration and volume of the other reactant (c2 and V2)
 The balanced equation for the reaction. This gives the mole-ratio in which
the two reactants combine.
 The calculation involves THREE steps:
 Calculate the number of moles of the ‘other reactant’ used in the reaction
(c2 x V2)
 Using the mole-ratio, calculate the number of moles of the ‘unknown’
reactant used in the reaction.
 Hence calculate the ‘unknown’ concentration (c1 = moles / V1)
Worked Example
Calculate the concentration of a sample of hydrochloric acid (c1) given the
concentration of a solution of sodium carbonate (c2) and the volumes of acid (V1)
and carbonate (V2) involved. The balanced equation is:
Na2CO3 (aq) + 2HCl (aq)  2 NaCl (aq) + CO2 (g) + H2O (l)
Moles = c2 x V2
Concentration (c1) = moles of acid / V1
(step 1)
Moles of alkali
Moles of acid (2 x moles of alkali)
(step 2)
Course outlines - Minerals to Elements (1)
Page 6 of 8
Examples
Example
1
2
3
4
5
Calculate
The concentration of
sodium hydroxide
The concentration of
sulphuric acid
The concentration of
calcium hydroxide
The concentration of
nitric acid
The concentration of
phosphoric acid
Information




Volume of sodium hydroxide = 10.0 cm3
Volume of hydrochloric acid = 13.6 cm3
Concentration of acid = 0.095 mol dm-3
Equation:
NaOH + HCl  NaCl + H2O




Volume of sulphuric acid = 17.40 cm3
Volume of sodium hydroxide = 10.0 cm3
Concentration of alkali = 0.11 mol dm-3
Equation:
2 NaOH + H2SO4  Na2SO4 + 2 H2O




Volume of calcium hydroxide = 10.0 cm3
Volume of hydrochloric acid = 15.9 cm3
Concentration of acid = 0.106 mol dm-3
Equation:
Ca(OH)2 + 2 HCl  CaCl2 + 2 H2O
 Volume of sodium carbonate = 10.0 cm3
 Volume of nitric acid = 21.7 cm3
 Concentration of alkali = 0.045 mol dm-3
 Equation:
Na2CO3 + 2 HNO3  2 NaNO3 + CO2 + H2O




Course outlines - Minerals to Elements (1)
Volume of sodium hydroxide = 10.0 cm3
Volume of phosphoric acid = 8.6 cm3
Concentration of alkali = 0.175 mol dm-3
Equation:
3 NaOH + H3PO4  Na3PO4 + 3 H2O
Page 7 of 8
Examples
1.
The concentration of hydrochloric acid can be found by titrating the solution
with a standardised solution of sodium hydroxide. In such a titration it was
found that 19.00 cm3 of 0.100 mol dm-3 sodium hydroxide were necessary to
react with 25.00 cm3 of dilute hydrochloric acid.
NaOH(aq) + HCl(aq)
(a)
(b)
(c)
2.

How many moles of sodium hydroxide are present in 19.00 cm3 of 0.100
mol dm-3 sodium hydroxide?
How many moles of hydrochloric acid are needed to react with this?
What is the concentration of the acid solution (in mol dm-3) ?
In a titration, 25.00 cm3 of a solution of sodium hydroxide were pipetted into
a conical flask. Sulphuric acid (0.100 mol dm-3) was run from a burette into the
flask. A suitable indicator in the flask changed colour when exactly 22.00 cm3
of the acid had been added.
H2SO4 (aq) + 2NaOH (aq) 
(a)
(b)
(c)
3.
Na2SO4 (aq) + 2 H2O (l)
How many moles of sulphuric acid was added from the burette?
How many moles of sodium hydroxide must have been in the flask?
What is the concentration of the sodium hydroxide in mol dm-3 ?
15.50 cm3 of nitric acid (0.00100 mol dm-3) reacts with 10.00 cm3 of calcium
hydroxide solution. Calculate the concentration of the calcium hydroxide
solution in mol dm-3.
Ca(OH)2 (aq) + 2HNO3 (aq) 
4.
NaCl(aq) + H2O(l)
Ca(NO3)2 (aq) + 2H2O(l)
A solution of sodium hydroxide (0.200 mol dm-3) is used to neutralise a sample
of sulphuric acid (20.00 cm3 of a 0.110 mol dm-3 solution).
H2SO4 (aq) + 2NaOH (aq)
(a)
(b)
(c)

Na2SO4 (aq) + 2H2O (l)
Calculate the number of moles of sulphuric acid that are neutralised.
Calculate the number of moles of sodium hydroxide that must have been
added to neutralise the acid?
What volume of sodium hydroxide solution would be needed?
Course outlines - Minerals to Elements (1)
Page 8 of 8