C06 concentration of solutions and volumetric analysis

Chapter 6
LEARNING OUTCOMES
 Define the term standard solution
 Use results from volumetric analysis to calculate
the number of moles reacting, the mole ratio in
which the reactants combine and the concentrate
and mass concentration of reactants
Concentration of Solutions and Volumetric Analysis

Solute and Solvent
 A solution is made up of two parts:
solute + solvent = solution
 The solute is the substance dissolved in a solution.
 The solvent is the substance in which the solute has
dissolved.
 For example, in a beaker of sugar solution, the sugar is
the solute and the water is the solvent.
Chapter 6

Concentrated or dilute?
 Instead of using the words “strong” and “weak” to
describe coffee, we can use the terms
concentrated and dilute.
 A concentrated solution will contain more solute
dissolved in a certain volume of solution.
 A dilute solution will contain less solute dissolved
in the same volume of solution.
Do you like “strong” or “weak” coffee?
Chapter 6

Concentration of solutions
 In order to standardise the volume of the solution,
chemists use 1 dm3
as the unit for measurement.
The concentration of a solution is the mass of
solute dissolved in 1 dm3
of the solution.
1 dm3
= 1000 cm3
Chapter 6

5
Chapter 6
 Concentrations can be expressed in two ways as:
 grams/dm3
or g/dm3
 moles/dm3
or mol/dm3

Concentration of solution in g/dm3
► Suppose a solution of sodium chloride is made by
dissolving 58.5g of the salt in 1 dm3
of the solution.
The concentration of the sodium chloride solution is
equal to: 58.5 g /dm3
Chapter 6

Concentration of solution in mol/dm3
► Since 58.5 g of sodium chloride is equal to 1 mole of the salt,
 The concentration of the solution is also equal to: 1 mol/dm3
(or 1 M).
 The number of moles per dm3
of a solution is also called the molarity of
the solution.
Chapter 6

Formulae
Concentration = Mass of solute in grams
in g/dm3
Volume of solution in dm3
Concentration = No. of moles of solute
in mol/dm3
Volume of solution in dm3
Mass of solute = Volume of solution in dm3
x Concentration in g/dm3
Chapter 6

Worked example 1
A solution of sodium chloride is made by dissolving 11.7 g of sodium chloride in 500 cm3
of
the solution. Find the concentration of the solution in (a) g/dm3
, (b) mol/dm3
.
Solution
Volume of solution = 500 cm3
= 500 = 0.5 dm3
1000
(a) Concentration = Mass in grams
Volume in dm3
= 11.7 g = 23.4 g/dm3
0.5 dm3
(b) No. of moles = 11.7 g = 11.7 = 0.2 mol
Mr of NaCl 58.5
Concentration = No. of moles
Volume in dm3
= 0.2 mol = 0.4 mol/dm3
0.5 dm3
Chapter 6

A solution of magnesium chloride has a concentration of 23.75 g/dm3
.
(a) What is the concentration of the solution in mol/dm3
?
(b) If 200 cm3
of the solution is evaporated to dryness, what
mass of salt can be obtained?
Solution
(a) Number of moles of MgCl2 in 1 dm3
= 23 g/dm3
= 23.75 = 0.25 mol
Mr of MgCl2 95
Concentration = 0.25 mol = 0.25 mol/dm3
1 dm3
(b) Mass of solute = Concentration x Volume of solution
= 23.75 g/dm3
x 200 dm3
1000 = 4.75 g
Chapter 6
Worked example 2

A solution of sulphuric acid has a concentration of 0.25 mol/dm3
(a) What is the concentration of the solution in g/dm3
?
(b) What mass of acid will be contained in 250 cm3
of the solution?
Solution
Mass of H2SO4 = 0.25 mol x Mr = 0.25 x 98 g = 24.5 g
(a) Concentration = Mass in grams
Volume in dm3
= 24.5 g = 24.5 g/dm3
1 dm3
(b) Mass of acid = Concentration x Volume of solution
= 24.5 g/dm3
x 250 dm3
1000
= 6.125 g
Chapter 6
Worked example 3

25 cm3
of a solution of sulphuric acid of concentration 0.400 mol/dm3
is
neutralised with a solution of sodium hydroxide of concentration 0.625 mol/dm3
.
What is the volume of sodium hydroxide solution required?
Equation of reaction: H2SO4 + 2NaOH  Na2SO4 + 2H2O
From the equation, No. of moles of H2SO4 = 1
No. of moles of NaOH 2
Vol. of H2SO4 x Conc. of H2SO4 = 1
Vol. of NaOH x Conc. of NaOH 2
0.025 dm3
x 0.400 mol/dm3
= 1
Vol. of NaOH x 0.625 mol/dm3
2
Vol. of NaOH = 2 x 0.025 x 0.400 = 0.032 dm3
0.625
3
Solution
Chapter 6
Worked example 4

Quick check
1. A solution of calcium chloride (CaCl2) contains 37 g of the salt in 250 cm3
of
the solution. Find the concentration of the solution in
(a) g/dm3
, (b) mol/dm3
.
2. 500 cm3
of a solution of sodium nitrate contains 14.7 g of the salt.
(a) Find the concentration of the solution in mol/dm3
.
(b) If 100 cm3
of the solution is evaporated, how much salt can be obtained?
3. A solution of magnesium sulphate has a concentration of 0.25 mol/dm3
.
(a) What is the concentration of the solution in g/dm3
?
(b) What mass of magnesium sulphate is contained in 250 cm3
of the
solution?
4. A solution of nitric acid has an unknown concentration. 25.0 cm3
of the acid
is completely neutralised by 22.5 cm3
of potassium hydroxide solution of
concentration 0.485 mol/dm3
. What is the concentration of the nitric acid?
Solution
Chapter 6

Solution to Quick check
1. (a) Concentration = 37 g = 148 g/ dm3
0.25 dm3
(b) No. of moles = 37 = 0.333 mol
111
Concentration = 0.333 = 1.33 mol/dm3
0.25 dm3
2. (a) No. of moles = 14.7 = 0.173 mol
85
Concentration = 0.173 = 0.346 mol/dm3
0.5
(b) Mass of salt = 0.1 x 0.346 x 85 = 2.94 g
3. (a) Concentration = (0.25 x 120) mol x 1 dm3
= 30 g/dm3
(b) Mass of magnesium sulphate = 0.250 x 30 = 7.5 g
4. Equation: HNO3 + KOH  KNO3 + H2O
No. of moles of nitric acid = 1
No. of moles of KOH 1
25.0 cm3
x Conc. of acid = 1
22.5 cm3
x 0.485 mol/dm3
1
Conc. of nitric acid = 0.437 mol/dm3
Chapter 6
Return

1. http://www.ausetute.com.au/concsols.html
2. http://dl.clackamas.edu/ch105-04/tableof.htm
3. http://en.wikipedia.org/wiki/Concentration
To Learn more about Concentrations of
Solutions, click on the links below!
Chapter 6

16
Introduction
 Volumetric Analysis or VA is a method of finding out
the quantity of substance present in a solid or solution.
 It usually involves titrating a known solution, called a
standard solution, with an unknown solution.
 Based on the equation of reaction, calculations are
then made to find out the concentration of the
unknown solution.
Chapter 6

17
Using a pipette
 A pipette is used to deliver an exact volume, usually
25.0 cm3
of solution into a conical flask.
 The solution in the titrating flask is called the titrate.
 Before using a pipette, it should be washed with tap
water, then rinsed with distilled water and finally with
the liquid it is to be filled.
 For safety reasons, a pipette filler is used to suck up
the solution.
 To use the pipette filler, first fit it to the top of the
pipette, as shown in the diagram.
 Squeeze valve 1 with right index finger and thumb
and squeeze the bulb with the left palm to expel all
the air in the bulb.
 Then place the tip of the pipette below the surface of
the liquid to be sucked up, and squeeze valve 2 to
suck up the liquid.
Chapter 6

18
Using a pipette
 When the liquid rises to a level higher than the
mark, remove the tip of the pipette from the
liquid.
 Gently squeeze valve 3 to release the liquid
slowly until the meniscus of the liquid is
exactly at the mark of the pipette.
 Now place the tip of the pipette into the
titration flask, and squeeze valve 3 to release
all the liquid into the flask.
 When all the liquid in the pipette has run out,
touch the tip of the pipette on the inside of the
flask so that only a drop of liquid is left inside
the tip of the pipette.
Chapter 6

19
Using a burette
 A burette is used to contain and measure the volume
of the liquid, called the titrant used in the titration.
 Before using a burette, it should be washed first with
tap water, then rinsed with distilled water and finally
with the liquid (titrant) it is to be filled.
 The liquid (titrant) in the burette must be released
slowly, a few drops at a time, into the titration flask.
 The readings must be taken accurate to 0.1 cm3
. E.g.
24.0 cm3
, not 24 cm3
.
 Make sure that the clip of the burette is tight and the
liquid is not leaking.
 Also make sure that the burette jet is filled with
liquid, it must not contain any air bubbles.
Chapter 6

20
Using a burette
 The burette should be clamped
to the retort stand in a vertical
position so that the reading will
be accurate.
 When reading the burette, the
eye must be horizontal to the
bottom of meniscus to avoid
parallax error. (See diagram).
Chapter 6

21
Other tips on safety and accuracy
 When filling or reading the burette, it should be lowered to a
suitable height. Do not attempt to read it by climbing onto a stool.
 Make sure that the tip of the pipette is always kept below the
surface of the liquid when it is being filled, otherwise air bubbles
will get into the pipette.
 After filling a burette, the small funnel should be removed from
the top of the burette, otherwise drops of liquid may run down into
the burette during a titration and affect the reading.
 The titration flask should be placed on a white tile or paper so
that the colour of the indicator can be seen easily.
 Use the wash bottle to wash down the insides of the conical
flask towards the end of the titration.
Chapter 6

22
Use of Indicators
IndicatorIndicator Colour inColour in
acidsacids
Colour at endColour at end
pointpoint
Colour in alkalisColour in alkalis
Methyl orangeMethyl orange redred orangeorange yellowyellow
Screened methyl orangeScreened methyl orange redred greygrey greengreen
LitmusLitmus redred purplepurple blueblue
PhenolphthaleinPhenolphthalein colourlesscolourless pinkpink redred
Chapter 6

23
 In a normal titration, candidates are usually advised to carry out at least one
rough and two accurate titrations.
 You should record your readings in a table like this.
Chapter 6
Titration readings
 In general, you should carry out as many titrations as needed to obtain two or more consistent
volumes.
 If no consistent volumes are obtained, the average value should be calculated.
Titration numberTitration number 11 22 33 44
Final burette readingFinal burette reading
/cm/cm33 25.225.2 24.824.8 33.333.3 24.924.9
Initial burette readingInitial burette reading
/cm/cm33 0.00.0 0.00.0 7.47.4 0.10.1
Volume of NaOH usedVolume of NaOH used
/cm/cm33 25.225.2 24.824.8 25.925.9 24.824.8
Best titration results (√)Best titration results (√)
√√ √√

24
Chapter 6
 Suppose that in an experiment, you are asked to find the concentration of a solution of sulphuric acid by titrating 25.0 cm3
of the acid against a standard solution
of sodium hydroxide of concentration 0.100 mol/dm3
, using phenolphthalein as an indicator.
 First set up the apparatus as shown in the diagram and then carry out the titration, repeating it as many times as necessary to obtain a set of consistent results.
Titration of a known acid with an alkali

25
Mean volume of sodium hydroxide used = 24.8 cm3
Chapter 6
Titration numberTitration number 11 22 33 44
Final burette readingFinal burette reading
/cm/cm33 25.225.2 24.824.8 33.333.3 24.924.9
Initial burette readingInitial burette reading
/cm/cm33 0.00.0 0.00.0 7.47.4 0.10.1
Volume of NaOH usedVolume of NaOH used
/cm/cm33 25.225.2 24.824.8 25.925.9 24.824.8
Best titration results (√)Best titration results (√) √√ √√
Suppose the following readings are obtained:
Results

26
Titration of a known acid with an alkali
 You are then asked to calculate the concentration of the sulphuric acid from
your results.
 The equation for the reaction is:
H2SO4 + 2NaOH  Na2SO4 + 2H2O
 From the equation,
No. of moles of H2SO4 = 1
Vol. of H2SO4 x Conc. of H2SO4 = 1
25.0 x Conc. of H2SO4 = 1
24.8 x 0.100 mol/dm3
2
Therefore, Conc. of H2SO4= 1 x 24.8 x 0.100 mol/dm3
2 x 25.0
= 0.0496 mol/dm3
Chapter 6

27
Va x Ma = x
Vb x Mb y
where Ma, Mb are the concentrations of the acid and base and
Va, Vb are the volumes of the acid and base used in the titration.
Chapter 6
In general if x moles of an acid reacts with y moles of a base, then
 No. of moles of acid = x
No. of moles of base y
 Vol. of acid x Conc. of acid = x
Vol. of base x Conc. of base y
Hence, it can be shown that :
Acid-base titration

28
Aim:
You are provided with a solution containing 5.00 g/dm3
of the acid H3XO4.
You are to find the relative molecular mass of the acid by titrating 25.0 cm3
portions of the acid with the standard (0.100 mol/dm3
) sodium hydroxide
solution, and hence find the relative atomic mass of element X.
The equation for the reaction is:
H3XO4 + 2NaOH  Na2HXO4 + 2H2O
Titration No. 1 2 3 4
Final reading/ cm3
25.4 25.5 25.6 35.8
Initial reading/ cm3
0.0 0.0 0.0 10.0
Volume of NaOH/ cm3
25.4 25.5 25.6 25.8
 Average volume of NaOH used = 25.5 cm3
Results:
Chapter 6
Titration of an unknown acid with an alkali

29
 From the equation,
No. of moles of H3XO4 = 1
Vol. of H3XO4 x Conc. of H3XO4 = 1
25.0 x Conc. of H3XO4 = 1
25.5 x 0.100 mol/dm3
2
Therefore, conc. of H3XO4 = 1 x 25.5 x 0.100
2 x 25.0
= 0.0510 mol/dm3
 Since 1 dm3
of the acid contains 5.00 g of the acid, therefore
0.0510 x Mr of H3XO4 = 5.00 g
Mr of H3XO4= 5.00 = 98.0
0.0510
 Calculate the relative atomic mass of X:
1x3 + X + 16x4 = 98
X = 98 – 67
= 31
Chapter 6
Titration of unknown acid with an alkali

30
25.0 cm3
portions of hydrogen peroxide solution (H2O2) was titrated with
standard (0.020 mol/dm3
) potassium manganate(VII) solution.
Introduction:
Oxidising agents can be titrated with reducing agents. Hydrogen
peroxide is a reducing agent and can be titrated against acidified
potassium manganate(VII), an oxidising agent.
No indicator is required for this titration as potassium manganate(VII)
solution is purple in colour and is decolourised by the hydrogen
peroxide solution when the reaction is complete.
Chapter 6
Titration of hydrogen peroxide with
potassium manganate(VII) solution

31
 Volume of KMnO4 used = 25.2 cm3
A. Calculate the number of moles of KMnO4 used.
No. of moles of KMnO4 = Volume in dm3
x Conc.
= 25.2 dm3
x 0.020 mol/dm3
1000
= 0.000504 mol
Chapter 6
Results:
Titration No. 1 2 3
Final reading/ cm3
25.1 25.2 25.2
Initial reading/ cm3
0.0 0.0 0.0
Volume of KMnO4/ cm3
25.1 25.2 25.2

32
B. If 1 mole of KMnO4 reacts with 2.5 moles of H2O2,
(a) Calculate the number of moles of H2O2 that react with the KMnO4.
(b) Find the concentration of H2O2 solution.
(a) No. of moles of H2O2 = 2.5 x 0.000504 mol = 0.00126 mol
(b) Concentration = 0.00126 mol = 0.0504 mol/dm3
0.025 dm3
C. If 2 moles of H2O2 decompose during the reaction to give 1 mole of oxygen, calculate
the volume of oxygen given off during the titration.
No. of moles of O2 given off = 1 x 0.00126 mol
2
= 0.00063 mol
Therefore, Volume of O2 = 0.00063 x 24000 cm3
= 15.1 cm3
Chapter 6

33
1. After washing the pipette, it should be rinsed with ________.
(A) distilled water (B) the titrate
(C) the titrant (D) tap water
2. After washing the titration flask, it should be rinsed with ________.
3. After washing the burette, it should be rinsed with ________.
4. A titration flask contains 25.0 cm3
of sodium hydroxide and a few drops of
phenolphthalein as indicator. It is titrated against hydrochloric acid contained
in a burette. What colour change would you observe when the end point is
reached?
(A) colourless to light pink (B) light pink to colourless
(C) red to colourless (D) blue to pink
Solution
Chapter 6
Quick Check

34
25.0 cm3
samples of sodium hydroxide solution are titrated against hydrochloric
acid which has a concentration of 0.225 mol/dm3
.
The results obtained are shown in the table below.
Final burette reading/ cm3
24.4 48.9 23.6 48.0
Initial burette reading/ cm3
0.0 24.4 0.0 23.6
Volume of HCl/ cm3
Best titration result (√)
Chapter 6
Solution
Quick Check
(a) Complete the table above.
(b) Calculate the concentration of the sodium hydroxide solution.

35
1. http://www.tele.ed.nom.br/buret.html
2. http://www.chem.ubc.ca/courseware/154/tutorials/exp6A/
To learn more about titration, click on the
links below!
Chapter 6

36
1. After washing the pipette, it should be rinsed with
2. After washing the titration flask, it should be rinsed with
3. After washing the burette, it should be rinsed with
4. A titration flask contains 25.0 cm3
of sodium hydroxide and a few drops of
phenolphthalein as indicator. It is titrated against hydrochloric acid contained in a
burette. What colour change would you observe when the end point is reached?
(A) colourless to light pink (B) light pink to colourless
(C) red to colourless (D) blue to pink
Return
Chapter 6

37
5.
Average volume of HCl used = 24.4 cm3
Equation: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
25.0 x Conc. of NaOH = 1
24.4 x 0.225 mol/dm3
1
Conc. of NaOH = 24.4 x 0.225 mol/dm3
25.0
= 0.220 mol/dm3
Chapter 6
Return
Final burette reading/ cm3
24.4 48.9 23.6 48.0
Initial burette reading/ cm3
0.0 24.4 0.0 23.6
Volume of HCl/ cm3
24.4 24.5 23.6 24.4
Best titration result (√) √ √

C06 concentration of solutions and volumetric analysis

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