Volumetric%20Analysis:%20Acid-Base

1
Volumetric Analysis Acid-Base Chpt. 13
2
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
3
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
4
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
5
Concentration The concentration of a solution is
the amount of solute that is dissolved in a given
volume of solution
6

• There are several ways of expressing the
  concentration of a solution
• Percentage of solute 3 forms
• Parts per million (ppm)
• Moles of solute per litre of solution (MOLARITY)

7

• 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)

8

• 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

9

• 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

10

• 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

11
Complete the following table
Concentration Unit Example Meaning
Percentage weight per volume (w/v) 3 (w/v) NaCl solution
Percentage volume per volume (v/v) 3 (v/v) alcohol solution
Percentage weight per weight (w/w) 3 (w/w) sugar solution
Note DO NOT take down table
12
The following calculations involve working with
percentages!!!! Make sure you understand the
definitions!!!!!
13
Example 1 A solution contains 20g of potassium
hydroxide in 1 litre of solution. Express the
concentration of the solution in
(w/v). Solution
14
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
15
Example 3 A solution contains 10g of sodium
carbonate in 40g of solution. Express the
concentration of the solution in
(w/w). Solution
16
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
17

• Try the following
• 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.
• 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?

18

• 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

19
Example 1 1 gram 1000 milligrams
20
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
21

• 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)

22

• Symbols Used
• - M
• - mol/L or mol L-1

Remember No. of Moles of Mass of
Substance Substance
Molar Mass
23
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
24
Complete the following 1M H2SO4 0.5M
H2SO4 3M H2SO4
25
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
26

• Converting Molarity to Grams per Litre
• 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

Concentration in g/L Molar Mass x Molarity
27
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
28
Example 2 What is the molarity of a solution
that contains 3.68g of NaOH per litre of solution?
29
Example 3 Calculate the concentration in moles
per litre of a solution containing 45 grams of
sulphuric acid per 250cm3 of solution.
30
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
31
Example 2 How many moles of NaOH are present in
25cm3 of 0.55M NaOH
32
Example 3 How many moles of hydrochloric acid
are present in 30cm3 of 0.2M HCl
33
Example 4 What mass of sodium hydroxide is
contained in 25cm3 of a 0.75M solution of sodium
hydroxide?
34
Example 5 What volume of 0.15M sodium hydroxide
solution will contain 5 grams of sodium hydroxide?
35
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.
36
In terms of molecules 2 molecules 1
molecules 1 molecule In terms of Avogadros
number of molecules 2 x 6 x 1023
1 x 6 x 1023 2 x 6 x 1023 molecules
molecules molecules REMEMBER the amount of
a substance which contains the Avogadro number of
particles is called a mole of that substance
2H2 O2 ? 2H2O
2H2 O2 ? 2H2O
37
In terms of Moles of a substance 2 moles 1
mole 2 moles Further examples a) 2Mg
O2 ? 2MgO 2
moles 1 mole 2
moles b) CaCO3 ? CaO CO2
1 mole 1 mole 1
mole c) CH4 2O2 ? CO2
2H2O 1 mole 2 moles
1 mole 2 moles
2H2 O2 ? 2H2O
38
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
39
Example 1 Solution
40
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?
41
Example 2 Solution
42
Concentration of Solutions
1 mole/250cm3
1 mole/L
1 mole/500cm3
1 mole/100cm3
43
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)
44
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
45
When a solution is diluted, more solvent is added
but the quantity of solute is unchanged Moles of
solute Moles of solute before
dilution 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
46
Calculation of the Effect of Dilution on
Concentration MolarityDil x VolumeDil
MolarityConc. x VolumeConc. M1 x
V1 M2 x
V2
47
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
48
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
49
Student QuestionsQuestion 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?
50
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
51
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.
52
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
53
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
54
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
55
Volumetric Analysis - Titrations Definitions Stan
dardise 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
56
Theory regarding apparatus and method involved in
carrying out a titration on handout
57
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
58

• Volumetric Analysis Calculations
• Calculating the unknown concentration of a
  solution from titration data
• Calculating the relative molecular mass and the
  amount of water of crystallisation in a compound
  from titration data.

59
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
VB x MB nA nB VA
volume (cm3) of acid used VB volume (cm3)
of base used MA concentration of acid
MB concentration of base nA no. of moles of
acid in nB no. of moles of base in
balanced eqn for rxn balanced eqn
for rxn
60
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)
61
Example 1 Solution
62
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
63
Example 2 Solution
64
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.
65
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)
66
Example 3 Solution
67
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.
68
Example 4 Solution
69
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
70
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
71
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)
72
Example 1 Solution
73
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)
74
Example 2 Solution
75
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)
76
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
77
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)
78
Example 1 Solution
79

• 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.

80

• 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

81
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
82
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.

83
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 iv) the of water of
crystallisation in the hydrated
sodium carbonate
Na2CO3.xH2O
84
Example 1 Solution
85
Student Questions Book pg. 174 Nos 13.15
3.16