Volumetric Analysis: Acid-Base
Chpt. 13
Quantitative Analysis: is analysis which involves
investigating the quantities or amounts of materials
present.
Gravimetric analysis
(Chpt. 11) –
composition of
substances
determined by careful
weighing
Volumetric analysis
(Chpt. 13) –
composition of
substances determined
by reacting together
volumes of solutions
REMEMBER:
A solution is a mixture of a solute and a solvent
A solvent: is a substance that dissolves other
materials
A solute: is the substance that dissolves in the
solvent
Concentration:
A concentrated solution contains a large amount of
solute per litre of solution e.g. strong coffee
A dilute solution contains a small amount of solute per
litre of solution e.g. weak coffee
Concentration
The concentration of a
solution is the amount of
solute that is dissolved in a
given volume of solution
There are several ways of expressing the
concentration of a solution:
1) Percentage of solute – 3 forms
2) Parts per million (ppm)
3) Moles of solute per litre of solution (MOLARITY)
1. Percentage of solute
(This method of expressing concentration is usually
used in many household solutions and in medicine)
There are 3 ways in which the percentage of solute
in a solution is expressed:
a) Percentage weight per weight (w/w)
b) Percentage weight per volume (w/v)
c) Percentage volume per volume (v/v)
a) Percentage weight per weight (w/w):
This is the number of grams of solute per 100g of
solution e.g.
10% w/w NaCl → 10g of sodium chloride per
100g of solution
2% w/w Arnica ointment → 2g of arnica per
100g of ointment
b) Percentage weight per volume (w/v):
This is the number of grams of solute per 100cm3
of solution e.g.
10% w/v NaCl → 10g of sodium chloride per
100cm3 of solution
5% w/v NaCl → 5g of NaCl in 100cm3 of
solution
*Note: usually the units are grams per litre (g/L)
therefore you would have to bring it to a litre*
c) Percentage volume per volume (v/v):
This is the number of cm3 of solute per 100cm3 of
solution e.g.
5% v/v vinegar solution → 5cm3 of ethanoic acid
per 100cm3 of vinegar
13% v/v wine solution → 13cm3 of ethanol
(alcohol) per 100cm3 of
wine
Complete the following table:
Concentration
Unit
Percentage
weight per
volume (w/v)
Percentage
volume per
volume (v/v)
Percentage
weight per
weight (w/w)
Example
3% (w/v) NaCl
solution
3% (v/v) alcohol
solution
3% (w/w) sugar
solution
*Note: DO NOT take down table
Meaning
The following calculations
involve working with
percentages!!!!
Make sure you understand
the definitions!!!!!
Example 1:
A solution contains 20g of potassium hydroxide in 1 litre
of solution. Express the concentration of the solution in
% (w/v).
Solution:
Example 2:
A bottle of vinegar contains 25cm3 ethanoic acid in
500cm3 of solution. Express the concentration of
ethanoic acid in the solution in % (v/v)
Solution:
Example 3:
A solution contains 10g of sodium carbonate in 40g of
solution. Express the concentration of the solution in %
(w/w).
Solution:
Example 4:
The label on a bottle of wine indicates that the
concentration of alcohol in the wine is 9% (v/v). What
volume of alcohol is there in 250cm3 of the wine?
Solution:
Try the following:
1. A sample of sea water has a mass of 1.3kg. On
evaporating the water, 148g of salt was recovered
from it. Express the concentration of the salt as %
w/w.
2. Some illnesses can upset the salt balance in the
body and it may be necessary to administer salt
intravenously. The solution of salt that is injected
is marked 0.85% w/v. What weight of sodium
chloride is needed to make up 250cm3 of this
solution?
2. Parts per million (ppm)
This method of expressing the concentration of a
solution is only used for very dilute solutions i.e. when
dealing with very low concentrations of substances.
• This is the number of milligrams per litre (mgL-1)
*Note: 1 Litre of water has a mass of 1 million
milligrams*
• So, can say 1mg/L = 1 mg per million mg
= 1 ppm
• Example: the concentration of chlorine in water is
2 ppm this means there are 2 mg of chlorine per
litre of water
Example 1:
1 gram = 1000 milligrams
3) Moles of solute per litre of solution
(MOLARITY)
Remember:
One mole of a substance is the amount of that
substance which contains 6 x 1023 particles (atoms,
ions, molecules) of that substance
Mass of 1 mole of an element = Relative Atomic Mass
in grams
e.g.
1 mole of Na = 23 g
1 mole of Mg = 24g
The most important way of expressing the
concentration of a solution is in terms of moles per
litre of solution (molarity)
Definitions:
• The Molarity of a solution is the number of moles
of solute per litre of solution
• A 1 molar solution is a solution which contains one
mole of solute per litre of solution
also,
- a solution which contains 2 moles of
solute in a litre of solution is said to be 2
molar (2M)
- a solution which contains 0.5 moles of
solute in a litre of solution is said to be 0.5
molar (0.5M)
• Symbols Used:
-M
- mol/L or mol L-1
Remember:
No. of Moles of
Substance
=
Mass of Substance
Molar Mass
Concentration Examples:
-1 mol/L NaOH = 40g NaOH (Mr NaOH = 40) per litre of
solution
- 2mol/L NaOH = 80g NaOH per litre of solution
- 0.5mol/L NaOH = 20g NaOH per litre of solution
- 0.1 M (decimolar) NaOH = 4g NaOH per litre of
solution
Complete the following:
1M H2SO4 =
0.5M H2SO4 =
3M H2SO4 =
Calculations Involving Molarity
Three types:
1. Converting Molarity to Grams per Litre
2. Converting Grams per Litre(Volume) to
Molarity
3. Calculation of number of Moles from Molarity
and Volume
1. Converting Molarity to Grams per Litre
Concentration in g/L = Molar Mass x Molarity
Example 1:
What is the concentration in g/L of a 0.1 M H2SO4
Solution?
Example 2:
How many grams of NaCl per litre are present in a solution
marked 0.25 M NaCl.
Example 3:
Calculate the concentration in grams per litre of
bench dilute sulphuric acid whose concentration Is 1.5mol/L
2. Converting Grams per Litre(Volume) to Molarity
Molarity
=
Grams per Litre
Molar Mass
Example 1:
What is the molarity of a NaOH solution containing 4g
of NaOH per litre?
Solution:
Example 2:
What is the molarity of a solution that contains 3.68g
of NaOH per litre of solution?
Example 3:
Calculate the concentration in moles per litre of a
solution containing 45 grams of sulphuric acid per
250cm3 of solution.
3. Calculation of number of moles from
molarity and volume
No. of moles = Volume(L) x Molarity
Example 1:
How many moles are there in 250cm3 of 0.1 M HCl?
Solution:
Example 2:
How many moles of NaOH are present in 25cm3 of
0.55M NaOH
Example 3:
How many moles of hydrochloric acid are present in
30cm3 of 0.2M HCl
Example 4:
What mass of sodium hydroxide is contained in 25cm3
of a 0.75M solution of sodium hydroxide?
Example 5:
What volume of 0.15M sodium hydroxide solution will
contain 5 grams of sodium hydroxide?
Balanced Chemical Equations
A balanced equation tells you the amounts of
substances that react together and the amounts of
products formed.
Consider the following balanced equation for the
reaction between hydrogen gas and oxygen gas to
form water:
2H2 + O2 → 2H2O
This equation can be interpreted in a number of
ways.
In terms of molecules:
2H2
+
2 molecules
O2
→
2H2O
1 molecules
1 molecule
In terms of Avogadro’s number of molecules:
2H2
2 x 6 x 1023
molecules
+
O2
1 x 6 x 1023
molecules
→
2H2O
2 x 6 x 1023
molecules
*REMEMBER: the amount of a substance which contains
the Avogadro number of particles is called a mole of that
substance
In terms of Moles of a substance:
2H2
2 moles
+
O2
1 mole
→
2H2O
2 moles
Further examples:
a)
2Mg +
2 moles
O2
1 mole
b)
CaCO3
1 mole
c)
CH4 + 2O2
→ CO2 +
1 mole 2 moles
1 mole
→
→
2MgO
2 moles
CaO
+
1 mole
CO2
1 mole
2H2O
2 moles
Reactions between a solution and a solid
In a number of chemical reactions solids react with
solutions. You may be asked to calculate the mass of
metal which reacts with a given volume of acid
Example 1:
What mass of magnesium will react with 50cm3 of
0.5M H2SO4. The balanced equation for the reaction
is:
Mg + H2SO4 → MgSO4 + H2
Example 1 Solution:
Example 2:
Sodium carbonate, Na2 CO3, reacts with dilute
hydrochloric acid according to the equation:
Na2CO3 +
2HCl
→
2NaCl + CO2 + H2O
What volume of hydrochloric acid of concentration
0.75M would be needed to neutralise 7.5g of anhydrous
sodium carbonate?
Example 2 Solution:
Concentration of Solutions
1 mole/L
1 mole/250cm3
1 mole/500cm3
1 mole/100cm3
If in each volumetric flask one mole of solute is
dissolved then as the volume becomes smaller, the
concentration increases.
In the case of a coloured solution , as the
concentration increases, the intensity of the colour
also increases (see diagram pg. 148 book)
Dilution of Solutions
To save space in our prep room we buy solutions in
concentrated form, i.e. 18M HCl (18 mol L-1). We call
these stock solutions
Definition:
The process of adding more solvent to a solution is
called dilution.
A typical dilution involves determining how much water
must be added to an amount of stock solution to achieve
a solution of the desired concentration
When a solution is diluted, more solvent is added but
the quantity of solute is unchanged:
Moles of solute
before dilution
=
Moles of solute
after dilution
Since the volume of the solution increases and the
number of moles present remains the same, then the
concentration of the solution must decrease
*Note: diluting a coloured solution results in a
lightening of the colour of the solution i.e. colour
intensity is proportional to concentration
Calculation of the Effect of Dilution on Concentration
MolarityDil x VolumeDil =
M1
x
V1
=
MolarityConc. x VolumeConc.
M2
x
V2
Example 1:
If 20cm3 of a 3M hydrochloric acid solution is diluted
to a volume of 1 L with water, what is the
concentration of the diluted acid?
Solution:
Example 2:
What volume of a 2M sodium hydroxide solution is
needed to make up 100 cm3 of a 0.1 M sodium
hydroxide solution
Student Questions:
Question 1:
If 12cm3 of a 0.1M sodium hydroxide solution is
diluted to a volume of 500cm3 with water, what is the
concentration of the diluted solution?
Question 2:
What volume of 1M NaOH solution is needed to make
300cm3 of 0.05M solution?
Standard Solutions
Definition:
A standard solution is a solution whose concentration is
accurately known
e.g. a solution containing 10 grams of NaCl per
litre is a standard solution
In the determination of the concentration of an acid a
standard solution of an alkali is used and to determine the
concentration of an alkali a standard acid would be used.
However, before any determinations can be made a
starting accurately standardised solution is required –
from which to find the exact concentration of other
solutions
A standard solution is prepared by weighing out a
sample of solute, transferring it completely to a
volumetric flask, and adding enough solvent
(usually deionised water) to bring the volume up to
the mark on the neck of the flask.
Due to the fact that many substances can not be
obtained in a high degree of purity standard solutions of
common laboratory acids and bases cannot be prepared
directly e.g.:
- cannot weigh out 1 mole of sulphuric acid as it
absorbs moisture from the air
- cannot weigh out 1 mole of nitric acid as it is
volatile
- cannot weigh out 1 mole of iodine as it
sublimes at room temperature
In order to make up standard solutions substances which
can be obtained in a highly pure state and which are
stable in air are required
Primary Standard
Definition:
A primary standard is a substance which can be
obtained in a stable, pure and soluble solid form so that
it can be weighed out and dissolved in water to give a
solution of accurately known concentration
Primary Standard Solution = Pure 100% Soluble Stable
once made up
Examples of Primary Standards:
- Anhydrous sodium carbonate Na2CO3
- Sodium Chloride NaCl
- Potassium Dichromate K2Cr2O7
Mandatory Expt. 13.1:
To prepare a standard solution of sodium
carbonate
Note: You must have a clear understanding of all the
steps you undertake in this experiment and be able
to explain the importance of each step
Volumetric Analysis - Titrations
Definitions:
Standardise: means to find the concentration of a
solution using titration
A Titration: is a laboratory procedure where a
measured volume of one solution is added to a
known volume of another solution until the reaction
is complete.
Equivalence Point (End Point): the stage when the
two solutions just react completely with each other
Theory regarding apparatus and method involved in
carrying out a titration on handout
Mandatory Expt. 13.2:
To use a standard solution of sodium
carbonate to standardise a given
hydrochloric acid solution
Note 1: You must have a clear understanding of all the
steps you undertake in this experiment and be able to
explain the importance of each step
Note 2: You must be able to carry out calculations on
your results – see notes to follow
*Need to calculate concentration of HCl in mol\L & g/L
Volumetric Analysis Calculations
1. Calculating the unknown concentration of a
solution from titration data
2. Calculating the relative molecular mass and the
amount of water of crystallisation in a compound
from titration data.
1. Calculating the unknown concentration of a
solution from titration data
In straightforward titration calculations, where only the
unknown concentration is required, the following formula
can be used:
VA x MA
nA
= VB x MB
nB
VA = volume (cm3) of acid used
MA = concentration of acid
nA = no. of moles of acid in
balanced eqn for rxn
VB = volume (cm3) of base used
MB = concentration of base
nB = no. of moles of base in
balanced eqn for rxn
Example 1:
In a titration, 25cm3 of a 0.05M sodium carbonate
solution required 22cm3 of a hydrochloric acid
solution for complete neutralisation. Calculate the
concentration of the hydrochloric acid solution. The
equation for the reaction is:
2HCl(aq) + Na2CO3(aq)
2NaCl(aq) + H2O(l) + CO2(g)
Example 1 Solution:
Example 2:
Hydrochloric acid and sodium hydroxide react according
to the equation:
HCl + NaOH
NaCl + H2O
25cm3 of a sodium hydroxide solution was titrated
against a 0.2M HCl solution. The average titration figure
was 23.5cm3. Calculate the concentration of the sodium
hydroxide solution in:
a) mol/L
b) g/L
Example 2 Solution:
Using results from your experiment calculate the
concentration of the given hydrochloric acid solution
in mol/L and g/L
*Note:
The first titration you performed was a rough titration
which gave you an idea of where the end point is and so
this result should be neglected.
The remaining two titration results should agree within
0.1cm3 of each other. The average of these results
should be used in your calculation of the concentration
of HCl.
Example 3:
1.45g of sodium carbonate was dissolved in water and
the solution was made up to 250cm3 in a volumetric
flask. 25cm3 of this solution were titrated against a
solution of hydrochloric acid using methyl orange as
indicator. One rough and two accurate titrations were
performed. The titration results and the equation for
the reaction are given below. What is the concentration
of the HCl solution in
a) mol/L
b) g/L
Titration
1
2
3
Volume Acid (cm3)
19.8
19.5
19.6
2HCl(aq) + Na2CO3(aq)
2NaCl(aq) + H2O(l) + CO2(g)
Example 3 Solution:
Example 4:
10.0g of impure sodium hydroxide were weighed out,
dissolved in water and solution made up to 250cm3
in a volumetric flask. 25cm3 of this solution, on being
titrated with 1.1M HCl, required 21.8cm3 of the acid
for neutralisation. Calculate the % purity of the
original sodium hydroxide.
Example 4 Solution:
Definition Secondary Standard:
Make up a solution and then standardise this
solution using a primary standard. This secondary
standard can then be used to standardise other
solutions e.g. HCl standardised and then used to
standardise NaOH
Mandatory Expt. 13.3 (Ordinary Level):
A Hydrochloric Acid/Sodium Hydroxide titration
and the use of this titration in making the salt
sodium chloride
Note 1: You must have a clear understanding of all the
steps you undertake in this experiment and be able to
explain the importance of each step
Note 2: You must be able to carry out calculations on
your results
*Calculate concentration of NaOH in mol/L and g/L
More Difficult Problems
In a calculation involving a standard solution where
more than just the unknown concentration is required,
or where a solid is one of the reactants an alternative
method from first principles should be used.
Example 1:
In a titration, 25cm3 of a 0.12M NaOH solution
required 24cm3 of a H2SO4 solution for complete
neutralisation. Calculate:
i) the number of moles of NaOH consumed
ii) the number of moles of H2SO4 consumed
iii) the concentration of the H2SO4
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H20(l)
Example 1 Solution:
Example 2:
What mass of magnesium will react with 20cm3 of a
0.09M hydrochloric acid solution? The equation for the
reaction is:
2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)
Example 2 Solution:
Applications of Acid-Base Titrations
Mand. Expt.13.4: To determine the percentage of
ethanoic acid in vinegar
Mand. Expt. 13.5: To determine the percentage of
water of crystallisation in hydrated sodium carbonate
(washing soda)
Mandatory Expt. 13.4:
To determine the percentage of ethanoic acid (acetic
acid) in vinegar
Note 1: You must have a clear understanding of all the
steps you undertake in this experiment and be able to
explain the importance of each step
Note 3: You must understand the need for and the use
of the dilution factor
Note 2: You must be able to carry out calculations on
your results
*Calculate concentration of ethanoic acid in the
original vinegar in mol/L, g/L and % w/v
Example 1:
A sample of vinegar was diluted from 25cm3 to 250cm3
with water. In a titration, 25cm3 of a 0.1M NaOH
solution required 30cm3 of the diluted vinegar for
complete neutralisation. Calculate the concentration of
ethanoic acid (CH3 COOH) in the vinegar in:
i) mol L-1
ii) g/L
iii) %w/v
The equation for the reaction is
CH3 COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
Example 1 Solution:
2. Calculating the amount of water of
crystallisation in a compound and the relative
molecular mass of the compound from the
titration data
Definition:
• Water of crystallisation: is water that is chemically
bound in the compound, which gives rise to the
crystalline form.
• Crystals that contain water of crystallisation are said
to be hydrated.
Mandatory Expt. 13.5:
To determine the percentage of water of crystallisation
in hydrated sodium carbonate (washing soda)
• Hydrated sodium carbonate has the formula
Na2CO3.xH2O ( where x = no. of molecules of water of
crystallisation present)
• The purpose of this experiment is to determine a value
for x
Mandatory Expt. 13.5:
To determine the percentage of water of crystallisation in
hydrated sodium carbonate (washing soda)
Note 1: You must have a clear understanding of all the
steps you undertake in this experiment and be able to
explain the importance of each step
Note 2: You must be able to carry out calculations on
your results
*Calculate the molar mass of hydrated sodium
carbonate, the percentage of water of crystallisation
and the value of X in the formula Na2CO3.xH2O
Mandatory Expt. 13.5:
NOTE:
• Washing soda is composed of large translucent
crystals of Na2CO3.10H2O (form of hydrated sodium
carbonate)
• It is important to realise that hydrated sodium
carbonate gradually looses water of crystallisation
over a period of time.
• Thus, if washing soda is used a lower value than the
expected value of 10 for x in Na2CO3.xH2O will be
obtained.
Example 1:
Crystals of hydrated sodium carbonate (Na2CO3.xH2O)
of mass 3.15g were dissolved in water and made up to
250cm3 in a volumetric flask. 25cm3 of this solution
required 15cm3 of a 0.15M HCl solution for complete
neutralisation. The equation for the reaction is:
2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g)
Find the:
i) concentration of the sodium carbonate solution
ii) the molar mass of the sodium carbonate
solution
iii) the value of x ion the formula Na2CO3.xH2O
iv) the % of water of crystallisation in the
hydrated sodium carbonate
Example 1 Solution:
Student Questions:
Book: pg. 174 No’s 13.15 & 3.16