36
6. QUANTITATIVE SALT-ANALYSIS
6.1 Volumetric Analysis: Definition and Principle
6.2 Use of chemical Balance
6.3 To prepare the standard solution
6.4 Titration
6.5 Script of titration experiment
6.6 Calculation of titration
37
6.1 Volumetric Analysis : Definition and
Principle
In practical chemistry, based on chemical
analysis the nature and amount of a substance
ion can be determined. This chemical analysis
is of two types’ quantitative analysis and
qualitative analysis. Quantitative analysis
deals with the volumetric and massive analysis
of elements of a substance.
Quantitative analysis is of two types.
1. Quantitative AnalysisIn the quantitative analysis the substance or
radicals, which has to be determined, is
converted into the temporary compound. This
temporary compound is either measure into
its original form or it is burn to convert into
the other compound, which is appropriate for
measurement .Now the weight of the
substance or radicals is determined by the
physical method. The ppt. method is the most
useful method for the quantitative analysis.
Note-Quantitative analysis is not included into
the syllabus of Higher Secondary.
2. Volumetric AnalysisIt is a sub-branch of quantitative analysis. It
is a simple and accurate method of analysis.
This method can determine amount of
substance in terms of volume accurately the,
hence the name volumetric analysis.
In this type of analysis ,a solution of a
substance whose concentration is to be
calculated is taken and a known volume of
another substance whose concentration is
known. The volume of the solution is
measured when the chemical reaction is
completed. The end point of such reaction is
indicated by change in colour or precipitation
etc.Thus in this ,method of calculation, known
volume of two substances are taken and the
concentration of one of them is known then
the concentration of the other is easily
calculated. This method is called Titration.
The description of volumetric analysis of
substance is given into this chapter.
Solution:-The homogenous mixture of any
substance is given into the water or any liquid
in solution.
Concentration or Strength-Amount of a
substance in grams dissolved in a one-litre
solution is called concentration or strength.
Example-If the 2 grams of caustic soda is
dissolved in 1 litre then the concentration of
the solution will be 2gm/litre.
Normality-Amount of a substance in grams equal
to its 1 gm equivalent weight in 1 litre solution is
called Normality of that solution.
Or
Normality can be express by the following
formula, which is a ratio between
concentration of that solution and gram
equivalent weight of that substance. It is
denoted by N. It is a number. It does not have
any limit.
Normality=Concentration /Equivalent weight
Example-(1) Concentration of Caustic Soda
solution is 4 gm/litre and equivalent weight
of Caustic Soda is 40 .Normality of this
solution will be.
=4/40=1/10=-N/10 =0.1N
38
Example-(2) 2N Caustic Soda solution will have
80 gm of it dissolved in 1 litre because:
2xN=2x40 i.e. 80 gm/litre.
Acid : Those substance which ,when dissolved
in water give Hydrogen ion (H+) as the only
positive ion are acids.
Molarity or Gram Molecular Weight :The
amount of substance equal to its gram
molecular weight dissolved in one litre
solution ,is called as Molar Solution. For
example, 20 gm of caustic soda dissolved in
one litre solution will form 0.5 M solution of
Caustic Soda.
Example :
Percentage : Amount of solute dissolved in 100
parts of the solution is called percentage of that
solution.
Example : N/10 HCL is 92% ionized where as
for acidic acid it is 1.3%.Therefore hydrochloric
acid is a strong acid where as acidic acid is a
weak acid.
Example :10 percent solution of NaOH means
10 gram of NaOH dissolved in 100 gram of
NaOH.
Gram Molecular Weight : Amount of a
substance in grams equal to molecular weight
of that substance is called 1gram molecular
weight of that substance.
Example : Molecular weight of Sodium
Carbonate is 106 then its 1 gram molecular
weight is 106 gram and the molecular weight
of 2 gram is 212 gram.
Standard solution : Solution of known
concentration (the solution in which the
quantity of solvent is known in its known
volume) is called a standard solution.
Concentration of standard solution is
expressed in terms of gram per litre or
normality.
Normal Solution : A solution containing 1 gram
equivalent of a solute dissolved in 1 litre of solution
is called a Normal Solution.
Example : In the normal solution of Na2CO3,
53 gram solvent substance is dissolved into
the 1 litre (equivalent weight of Na2CO3 is 53).
HCL=H++ ClH2SO4=2H++SO4—
Strength of Acid : The acid, which gives more
H+ ions its strength, will be more. Two acids
having same normality may have different degree
of ionization and therefore the strengths will also
different.
Relative strength of weak acids :
HCL>HNO3>H2SO4>H3PO4>
CH3COOH
Basisity of acids : Basisity of acids means can
be express its strength to react with bases.
Amount of Hydrogen ions given by one
molecule of an acid is its basisity. It depends
on the number of atoms of the H in acid.
Therefore, number of atom of replaceable
hydrogen of any molecule of acid is basisity
of acid.
HCI,HNO 3,CH 3 COOH are monobasic
acids,H2SO4 is a dibasic acid and H3PO4 is a
tri-basic acid.
Base : Those substances which when dissolved
in water give Hydroxide ion (OH-) as the only
negative ion are called bases.
NaOH=Na+ +OHNH4OH=NH4++OHStrengths of Bases : The base those which give
more OH- ions will have more strength. Two
bases having same Normality may have different
degree of ionization. Therefore, they have
different strength.
Example :
39
For example for same normality of N/10
NaOH IS 84% ionized where as
N/10 NH4OH is only 1.3% ionized.
Relative strength of few bases KOH>NaOH>NH4OH>Ba(OH)2.
Acidity of Bases : Amount of Hydroxide ions
(OH-) present in one molecule of base, is
called the acidity.
Example : NH4OH,NaOH are mono acidic
bases. Ba(OH)2 is a di-acidic base.
Equivalent Weight : Amount of substances
which combines with or displaces 1.008 parts
of hydrogen or 8 part of oxygen or 35.5 parts
of chlorine is called equivalent weight of that
substances.
Gram Equivalent Weight : Equivalent weight
of a substance express in gram is called 1 gm
equivalent weight of the substance.
Example : Equivalent of H2SO4 is 49.So 1 gram
equivalent of H2SO4 will weight 49 gram and 0.1
gm equivalent of H2SO4 acid will weight 4.9
gram.
1. Equivalent weight of acid : The equivalent
weight of acid is that which has one position
for replaceable hydrogen atom is equivalent
weight of acid. For example, one molecule of
HCL has (gram molecular weight) one
replaceable hydrogen atom. Therefore the
equivalent weight of HCL is 36.5 .Similarly
one molecule of H2SO4 has 2 replaceable
hydrogen atoms (molecular weight 980.
Therefore the equivalent weight of H2SO4 is
49(98/2).
Equivalent weight of acid=Molecular weight
of acid/Basisity of acid
2. Equivalent weight of base : The part of the
base which has one portion for replaceable OH-
radicals or the number of weights which
completely neutralize the one equivalent
weight of any acid. For example in the
following reaction 36.5 gram, HCL (1 gm
equivalent weight) completely neutralized 40
gm NaOH .Therefore the equivalent weight
of NaOH is 40.
NaOH + HCL=NaCl+H2O
40gm
36.5gm
Equivalent weight of base=Molecular weight
of base/Acidity of base
3. Equivalent weight of salt : The equivalent
weight of salt is the sum of the equivalent
weight of the radicals. Generally the
equivalent weight of acid, base or salt can be
determine by the ratio of its molecular weight
to the total positive valancy.
The equivalent weight of Na2+Co3—is 23+60/
2=53 or 106/2=53
Similarly the equivalent weight of HCL=36.5/
1=36.5
4. The equivalent weight of oxidization and
reduction agents : The equivalent weight of any
oxidising agent is that weight which makes
available the 8 weight oxygen for reaction with 1
weight of hydrogen. Similarly the equivalent
weight of the reduction agents is that weight which
makes available one part of hydrogen or reacts
with 8 part of oxygen.
1. KMnO4(Potassium Permanganate)This is react in acidic medium in following
ways2KMnO4+3H2SO4
K2SO4+2MnSO4+3H2O+5O
Therefore, the 2 molecule of the KMnO4
produces 5 atoms of oxygen in acidic medium
i.e.
40
2[39+55=16x4]=5x16=10x1
Or 2[158]=10x8=10x1
Or 316/10=8=1 Or 31.6=8=1
Therefore, the equivalent weight of KMno4
is 31.6 into the acidic medium. In the similar
way.
The equivalent weight of oxidization
agent=Molecular weight of substance x2x
Number of atom of the obtain oxidization
agent.
(Note- Hence 2 is multiplied in formula because
the atomic weight of oxygen is 16 is eight times
more then the equivalent weight of oxygen).
Equivalent weight of KMnO4 in the neutral
medium2KMnO4+H2O=2MnO2+2KOH+3O
2KMnO4=3O=6H
2x158=3x16=6H
316=6x8=6H Or 316/6=52.66=8=1
Therefore the equivalent weight of KMnO4
in the neutral medium is 52.66.
Equivalent weight of KMnO4 in basic
medium2KMnO4+2KOH=2K2MnO4+H2O+O
2KMnO4=O
2x158=16=2x8
Equivalent weight=2x158/2=158.
In this way, the equivalent weight of KMnO4 in
the basic medium is 158.
The example shows that the equivalent weight
of any substance does not remain constant .It
depend upon the reaction and the nature of
the medium.
Ferrous ammonium sulpheta :
FeSO4(NH4)2SO4.6H2O
Or Ferrous Sulphate FeSO4.7H2O
(i) Equivalent weight of iron :
2KMnO4+3H2O=K2SO4+2MnSO4+5O+3H2O
10FeSO4+5H2SO4+5O=5Fe (SO4)3 +5H2O
2KMnSO4+10FeSO4+8H2SO4=K2SO4+
5Fe2(SO4)3+2MnSO4+8H2O
2KMnSO4=5O=10FeSO4.7H2O
Or O=5FeSO4.7H2O
1/2O=FeSO4.7H2O
Therefore the equivalent weight of
FeSO4.7H2O is 278.
Generally Ferrous sulphate is used in place of
Ferrous ammonium sulphate FeSO4. (NH4)2
SO4.6H2O in titration because the ferrous
sulphate oxidized into the ferric sulphate in
the air 1 .Therefore the equilibrium weight of
ferrous ammonium sulphate is 399(the
molecular weight of the ferrous ammonium
sulphate.
The equivalent weight of used compounds in the acid base titration
Name of the compounds
Hydrochloric Acid(HCL)
Nitric Acid(HMO3)
Sulfuric Acid(H2SO4)
Acitic Acid(CH3COOH)
Oxalic Acid(H2C2O42H2O)
Potassium Hydroxide(KOH)
Sodium Hydroxide9NaOH)
Barium Hydroxide(Ba(OH)22H2O)
Amonium Hydroxide(NH4OH)
Sodium Carbonate(Na2CO3)
Table
Molecular
Weight
36.5
63
98
60
126
56
40
315
35
106
Basisity
Acidity
1
1
2
1
1
x
x
x
x
x
x
x
x
x
x
1
1
2
1
2
Equivalent
Weight
36.5
63
49
60
63
56
40
15705
35
53
41
6.2 Use of chemical balance
The important task of volume analysis is to
measure the volume the solution that is taking
part in the given chemical reaction. Therefore
the accuracy in the results of analysis depends
on the pure measurement of the volumes of
the reactants solutions .For measuring these
solutions some special apparatus are used. In
this chapter, an introduction of these apparatus
and method of using them is described.
Apparatus used to prepare the standard
solutions are 1. Chemical Balance
2. Weight Box
3. Weighting tube or bottle
4. Measuring Flask
5. Washing Bottle etc
1. Chemical Balance : The balance is enclosing
in the glass tube. Its body is made of wood or
aluminum alloy. Base is generally made of black
glass or slate. Box can be open from the front by
means of a sliding door. Two more small doors
are present at the right and left side of the box.
They can be open when required.
Balance rod is made of metal. It is place on the
knife-edge of agate stone in the center. The both
end of the rod have knife-edge, each of them is
made of agate stone. The pan of the balance are
suspended from the terminal knife edges by means
of shrimps pointer is attached to the center of the
rod and its another end moves on the scale. When
it reaches on the both side of scale until the equal
number is achieve then the rod is horizontal. At
the bottom of the balance, two horizontal or
parallel screws are attached, so that to keep the
balance in the horizontal position properly. A spirit
level is present in the balance to gauge the plane
of the balance. In some balance to check the
horizontal position a plumbline is suspended with
the base of the rod. A hook is attached by the
rider and a rider is moved with the help of a rider
carrier.
Fine adjustment of screws are present on the two
ends of the beam or rod to mark the rod horizontal
.When the balance is not in use then the pans at
rest on the pan barrier. Base plate at which the
mirror is open having a knob at the front, after
rotating it, the rod and the pans are becomes
independent.
.0002 gram quantity can be measure (at the 4
places of decimal point)
A labeled diagram of chemical balance is
shown below-
Marked Bar
Adjustable Screw
Hollow
Stand
Indicator
Scale
Key
Fig. 6.1 Chemical balance
Weight Box : A weight box containing weights
of various denominations is used. A standard
weight box contain following weights
42
Weight
50 gm.
20 gm.
10 gm.
5 gm.
2 gm.
1 gm.
Weight of Mili Gram
(Fractional Weight)
Number Weight
Number
1
500 gm. 1
2
200 gm. 2
1
100 gm. 1
1
50 gm.
1
2
20 gm.
2
1
10 gm.
1
horizontal rod. If it is not possible then use
the balancing screw for this.
Weight of Gram
Fig. 6.2 Weight Box
Figure:
Rider
Fractional weights are of following shapes
The weight of 500 and 50 mg are have 5 sides,
200 mg. and 20 mg. weight are square or
rectangular (each have 2-2 weight) and
100mg. and 10 mg. weight are triangular
shape.
3. Precaution for the use of chemical
balance: 1.
Clean the balance with soft cloth.
2.
3.
Adjust the balance plane with the help
of adjustment screw. At the position of
adjustment the plumbline is placed in
the line of the points in the stand.
To see that the point goes to the equal
point on the both side of the scale, rotate
the handle or knob and see the
4.
Pick the arrested knob of balance slowly
and put it slowly.
5.
The weight should place right pan and the
substance, which has to be measure keep
in to the left pan.
6.
Do not touch the weight from the handle.
Always use tongs for this.
7.
Always starts the measurement with the big
weight.
8.
When the rod of the balance is upwards
then do not touch the object.
9.
The object should always measure by
the watch glass or the weight it in the
weighting tube.
10.
Weight box should keep only when the
right window is open.
11.
Shut down the window at the time of
weight.
12.
At the time of weighting of these is a
condition when the weight of 10ml is
decreases or increases then use rider and
place it always right side of rod.
13.
Always write down the weight of the
substance is into your practical
notebook.
14.
Do not weight the hot and the cold
things.
4. The method of taking the reading by rider:Use the rider for weighting below 10 ml.
The zero is marked at the middle of the
balance rod of chemical balance and both sides
have 10-10 blocks., which have 1 to 10
number are marked.
43
Weight Box : A weight box containing weights
of various denominations is used. A standard
weight box contain following weights.
To see that the point goes to the equal
point on the both side of the scale, rotate
the handle or knob and see the
horizontal rod. If it is not possible then
use the balancing screw for this.
4
Pick the arrested knob of balance slowly
and put it slowly.
5.
The weight should place right pan and
the substance, which has to be measure
keep in to the left pan.
6.
Do not touch the weight from the
handle. Always use tongs for this.
7.
Always starts the measurement with the
big weight.
8.
When the rod of the balance is upwards
then do not touch the object.
9.
The object should always measure by
the watch glass or the weight it in the
weighting tube.
10.
Weight box should keep only when the
right window is open.
11.
Shut down the window at the time of
weight.
12.
At the time of weighting of these is a
condition when the weight of 10ml is
decreases or increases then use rider and
place it always right side of rod.
13.
Always write down the weight of the
substance is into your practical
notebook.
14.
Do not weight the hot and the cold
things.
Weight of Gram
Weight
50 gm.
20 gm.
10 gm.
5 gm.
2 gm.
1 gm.
Weight of Mili Gram
(Fractional Weight)
Number Weight
Number
1
500 gm. 1
2
200 gm. 2
1
100 gm. 1
1
50 gm.
1
2
20 gm.
2
1
10 gm.
1
3.
Fig 6.2 Weight Box
Fig : Rider
Fractional Weight or mg weight have
following different shapes
The weight of 500 and 50 mg are have 5 sides,
200 mg. and 20 mg. weight are square or
rectangular (each have 2-2 weight) and 100mg.
and 10 mg. weight are triangular shape.
3. Precaution for the use of chemical
balance: 1.
Clean the balance with soft cloth.
2.
Adjust the balance plane with the help
of adjustment screw. At the position of
adjustment the plumbline is placed in
the line of the points in the stand.
4. The method of taking the reading by rider
:
Use the rider for weighting below 10 ml.
44
The zero is marked at the middle of the
balance rod of chemical balance and both sides
have 10-10 blocks., which have 1 to 10
number are marked .Between each two
number there is 5 or 10 blocks are present.
Such type of scale has 0 to 50 or 100 blocks
on one side. i.e. the half portion of the rod of
balance is divided into 50 or 100 small blocks.
The weight of the rider is generally 10 mg
therefore if 10 mg weight from the left pan of
balance and the rider is placed at the right end
of the scale then according to the principle of
balance, after picking up the rod of the balance
the points go to the equal mark on the both
side of the scale i.e.
The value of 50 small blocks = 10mg
The value of one blocks = 10/50 =.2mg
If the balance has 100 small part then the value
of small part is = 10/100 = .0001 gm.
Example : The position of rider is displayed on
the balance rod in the following diagram. The rider
is placed at the third small mark after the fifth
mark point of main scale of the balance rod.
Precaution : And every bigger part is divided
into 5 small parts. Therefore reading will be as
follows.
5 big parts = 5 mg.
3 small parts = 3 × 0.2 mg. = 0.6 mg.
Total weight = 5 mg + 0.6 mg.
= 5.6 mg. = 0.0056 gm
This weight will be added in the weight of the
pan.
(a)
Rider
(b)
Fig : 6.3 Use of Rider
Procedure of weighing :
Before taking
weight pan of balance should be cleaned first
and see whether the balance is at equal level. To
bring it in equal level screws then screws on the
edges of the marked rod can be set by rotator if
needed. Rotate the handle and see whether the
indicator goes to and from zero equally or not. If
it does not move equally and screws on the edges
of the marked rod can be set by rotating. The
direction in which indicator covers the more
distance that pan is the lighter one. To make it
right, rotate the screw inside from the heavier pan
or rotate the screw outside from the light is pan.
After checking whether the balance working
properly keep substance in the measuring tube
(clean & dry) in the left pan from the left
window. Keep weight in the right pan from
the right window. First start weighing with
larger weight, then smaller weight are used in
place of larger weight sequentially. After
putting each weight it is cheeked that which
pan heavier.
The direction in which indicated covers more
distance that pan is lighter one. If there is the
need for the weight of less than one gm then
start putting weight of mg. If the weight less
than 10 mg is needed then rider is used. At
first rider is kept on one end point of the right
side of the marked bar and it moves to and
fro from the zero. Observe this situation
carefully this is the exact measurement. Now
take out the weight from the pan & note the
reading.
6. Calculation of weight : Suppose the
following weights are kept on the right pan
for weighing tube & weight of substance.
Gram weight = 5 gm + 2 gm + 1 gm = 8 gm.
Milligram = 500 mg + 200 mg + 200 mg +
20 mg + 10 mg.
= 950 mg = 0.959 gm.
Rider = 3 main parts on the right side + 2
subparts
= 3 × 0.001 gm + 2 × 0.0002 gm.
45
= 0.003 gm + 0.0004 gm
Total weight = 8 gm + 0.95 gm + 0.0034 gm
= 89534 gm
weighing tube + weight of substance = 8.9534gm
Similarly,
the weight of weighing tube = 5.3246 gm.
8.
weight of the substance = (8.9534 – 5.3246) gm
= 3.6288 gm
7.
on the neck. Neck of the flask has the
circular mark. Fill the liquid to the mark, and
liquid will become to the capacity of each
flask is written on it at certain temperature.
Use flask of 250 ml. or 100ml. to prepare a
standard solution in lab. Do not heat the flask
and do not pour any hot liquid in it.
Necessary tube , watch glass and
washing bottle :
These are shown with measuring flask
in the following figure
Measuring flask : Measuring flask is a
flat bottom pot of glass, which has a long
neck and there is a cork of glass. which fixed
(i) Measuring tube
(ii) watch glass
(iii) washing bottle
(iv) measuring flask
Fig. 6.4
6.3 Preparation of Standard Solution
Important standard substance : Some
important substance are obtained in pure
condition. There are substance that react with the
components of atmosphere. These substance are
important standard substances for example :
Oxalic
acid,
Sodium
carbonate.
2.
The substance should not affect with air
and light.
3.
The substance should not give vapour into
the air.
4.
The substance should not be porus
otherwise there will be difference into
the weight of substance .
Characteristic of important standard
Substances :
5.
Substance should not be soluble in
distilled water.
1.
The substance should hundred present
pure.
The following substance do not prepare
standard solution.
46
Acid :
HCl, HNO3 , H 2SO 4 etc.
Base : NaOH because it absorb the air
moisture and CO2 .
Salt : KMnO4 because it decomposed in air &
light. FeSO4 because it oxidized to ferric
sulphate from air.
Secondary Standard Substance : Some
substance are such type which can not be
obtained easily in pure state like KMnO4 ,
H 2SO 4 or NaOH etc, then substance are called
secondary standard substance. The standard
solution of such substance cannot prepare
easily. In such a condition the solution of
substance is prepare whose concentration is
near to the required solution. To find the
acurate. conc. of solution ,it is titrated with
standard solution.
For Example : Before making the standard
solution of NaOH its solution of approximate
solution is prepare and to find its
concentration ,it is titrated with standard
solution of oxalic acid determine by adding
required water in NaOH solution.
solution of derived volume.
Example : To find the required weight of
acid for the preparation of 250 ml. N/10
solution of oxalic acid (Equivalent weight of
acid = 63)
Quantity of substance = Equivalent weight ×
Normality × Volume /1000 gm
= 60 × 1× 250 / 10 × 1000
= 1.575 gm
Method to Prepare Standard Solution.
The method of preparing standard solution ,
is stated by an example.
Objective : To prepare 250ml solution of oxalic
acid of N/10 concentration (Equivalent weight =
63)
Method –
1.
Weight the dry, clean watch glass Now
weight it by adding 1.6 gm oxalic acid.
2.
Cast the substance of watch glass in a
clean beaker according to figure.
To calculate the weight of the substance
required for the preparation of standard
solution :
In lab 250ml standard solution is prepare. For
the 250ml standard flask in used. The required
weight can be calculate with following
formula.
W =
E × N ×V
gm
1000
Fig. 6.5 - Weight the substance in watch glass.
Pour the attached substance of watch glass in
beaker by washing bottle.
When
W = Weight of solote (substance)
E = Equivalent weight of substance
N = Normality
V = Required Volume
From the above formula the required weight
of substance in calculated to prepare standard
Fig. 6.6 - Pour the attached substance in beaker.
47
3.
flask Now remove the funnel from flask.
Now wash the funnel 2 or 3 times. Now
try to drop the substance attached with
the wall of flask neck in the flask with
the help of distilled water. Now pour
more distilled water in flask to fill it 1/3.
Now cover the flask with cork and stir it
properly so that the substance is
completely dissolved. During the
observation it is necessary to observe that
the solution should not comes up to the
neck of flask.
Dissolve the substance of beaker by
stirring and fill the measuring flask by
funnel according to figure
5.
Fig 6.7 Pour the solution in measing flask.
4.
Now stirr the funnel gradually to drop
quantity of substance in the flask. At this
situation neither take out the funnel from
flask nor move it. When the substance
falls in the flask by funnel then wash the
edges of funnel with distilled water,
wash bottle gradually so that the
substance of funnnel reaches into the
Now pour water drop by drop in flask
with the help of a clean and washed
pipette. When the lower level of semi
circular and solution touches the circular
mark then stop pouring water. Remember
that the water should not be more than
the circular surface otherwise the
concentration will be disturbed. It should
be in such a manner that the liquid should
not come out Now press the dot with hand
and stirr the solution so to prepare
homogenous solution. Now standard
solution of oxalic acid has been prepared.
Fig 6.8 To pour the solution in the flask upto last drop.
Fig 6.9 To wash the funnel and fill the measuring flask
Fig 6.10 To fill the mearuring flask with the help of pipette
48
6.4 Titration
6.4.1 Titration
A known volume and concentration of one
substance is reacted with a known volume of
another substance whose concentration is to
be determined this process is called. titration.
Titration is classified into 3 categories depending
upon the nature of the reaction which happen in
the solutions.
(i)
Acid - Base titration or Acidimetry and
Alkalimetry.
(ii)
Oxidation and Reduction titration.
(iii) Precipitation Titration.
(i) Acid - Base Titration :
In this method acid and base are reactants.
They react with each other and neutralize each
other. This titration is called “ Neutralization
Titration “.
Acidimetry : In this method concentration of
an acid in a solution is determined with the help
of an alkali of known concentration.
Alkalimetry : In this method concentration of
an alkali in a solution is determined with the help
of an acid of known concentration.
(ii) Oxidation and Reduction Titration :
In this method, an oxidising agent and a
reducing agent are reacted. In this, one
solution is of oxidation substance and other
is of reducing substance. In this method one
substance is oxidized and another is reduced.
Ex : (i)
(ii)
Titration of KMnO4 with Oxalic
Acid.
Titration
of
KMnO4 with
as
an
FeSO4 KMnO4 acts
oxidising agent in a acidic
medium.
It Oxidizes Oxalic acid & itself gets reduced.
2 KMnO 4 + 3H 2SO4 →
K 2SO4 + 2MnSO 4 + 3H 2O + 50
5 [COOH ] 2 + 50 → 10 CO2 5H 2O
(iii) Precipitation Titration :
In this method of titration concentration of a
substance is determined by its complete
precipitation with the help of another
substance of known concentration.
6.4.2 Indicator
Those chemical substance, which indicate end
point in the titration by changing their colour
are called indicators.
Success of Titration experiment depends
upon the correct measurement of reacting
solutions. For this exact end point is
required. An indicator changes its colour at
the end point. This indicates the physical chemical phase or state of completing the
reaction and do not affect the concentration
of actual solution.
Indicators are of 3 types :
(i)
Internal Indicator
(ii)
External Indicator
(iii)
Self Indicator
(i) Internal Indicator :
Such indicators
are added to the actual solution in the titrating
flask during titration. The other solution is
added from the burette till end point is
reached; are called internal Indicator. In this
type of indicator methyl orange and
phenolphthalein are mostly used.
They are used in acid - Base Titration.
49
(ii) External Indicator :
Indicators that
are not added in the titration solution are called
external indicator. Few drops of such
indicators are kept on a tile and after the
titration, the reacting solution mixture is
treated with external indicator used in the
titration to reach the end point. For example :
Potassium ferrocyanide is an external
indicator used in the titration between
dichromate and Ferrous Ammonium Sulphate.
(iii) Self Indicator : When one of the reacting
solutions during titration changes its colour at end
point then such a solution is called ‘Self
Indicator’. For example: During titration between
Potassium Permangnate and Ferrous Ammonium
Salphate the Potassium Permangnate acts as a
self indicator.
Theory of indicators : Indicators used in acid
base titration are generally coloured organic
compounds. They are weak acid or base and
they have one ‘ionization constant’. In titration
Methyl orange and Phenolphthalein indicators
are used. They change their colour at a specific
pH. They exhibit different colour in ionised
and unionized state.
Methyl Orange : It is orange colour weak base;
it is expressed by the formula MeOH. It ionized
in water as follows.
MeOH
H+
ion to make atom of water and reaction shift
towards right side. More Me + ions are formed
due to which solution becomes pink.
Phenolphtholein : It is a week organic acid it is
denoted by formula HPh. It is ionises as follows.
HpH = H + + Ph −
(Colour less) (Pink)
In base solution the H + ion of phenolphthalein
react with H + ion of base to make water.
Therefore concentration of Ph − ions increases
due to which solution (due to the pink colour
of Ph − ion) turns pink.
In acidic solution the ionization of (HpH) is
affected therefore the Ph − ions do not present
in solution therefore phenolphthalein does not
give any colour.
Choice of indicator : The choice of indicator
depends upon the presence of concentration of
hydrogen in present in solution during the titration.
At least one of the two solutions should be strong
because suitable indicator is not available for
titration between weak acid and weak base.
Suitable indicator can be chosen on the basis of
the following table.
Me + + OH −
Atom (Orange) (Pink) (Colour less)
In basic medium the above equilibrium shifts in
the backward direction (←) and Methyl orange
remains wnionized state. Therefore in Basic
medium the methyl orange do not change any
colour as a result colour of the solution remain
yellow (MeOH) (During the presence of colour
atom of Methyl Orange).
In the acetic medium the OH
− ion acid add with
S.No
Solution
used in Titration
Suitable Indicator
for Titration
1.
Strong Acid and
strong Base
Methyl Orange or
Phenophthelim one
of them
2.
Strong Acid and
Weak Base
Methyl Orange
3.
Weak Acid and
Strong Base
Phenolphthalein
4.
Weak Base and
weak Acid
None of them gives right
result
50
6.4.3 Principle of Titration
Acid base titration is based on the principle
of neutratlization and law of Equivalence
according to the ionc theory. When the base
is added in the conical flask then the
OH − H 2O of base in produced, in the solution.
But practically it is not seen. When the one
drop of acid is add into the neutral solution
the H + ions are produced. So that the colour
of the solution changes by the indicator, this
condition arises when acid in taken in the
conical flask and base in added by burette.
According to the law of Equivalence the
reaction between base and an acid takes place
in the ratio of their equivalent weight.
HCl + NAOH = NaCl + H 2O
Acid
base
Salt
Water
(36.5) (40)
(58.5) (18)
It in clear from the equation that for complete
neutralizes 40 gram of NaOH must react with
36.5 gram of HCl.
i.e one gram of equivalent of NaOH
Completely neutratizes one gram equivalent
of HCl. If Normal Solution of both reagents
are used then one litre of each solution or any
same volume of both solutions will be
required for complete netralisation i.e. end
point
1000 ml = N1 gm. equivalent
V2
ml, =
N1 V1
1000
gm equivalent
Therefore during titration for solution A,
gm equivalent substance is used.
Similarly Solution B,
N 2 V2
1000
gm. equivalent
substance must be used. (because Normality
N2 and used volume = V2 ml.)
Therefore according to law of equivalent equal
gm. equivalent of both must react completely.
N1 V1 N 2 V2
=
1000
1000
N1 V1
=
N 2 V2
Normality of solution ‘A’ × volume of solution
‘A’ = Normality of solution ‘B’ × Volume of
solution ‘ B’
6.4.4 Apparatus of Titration
1. Burette and its use :
It is long cylindrical glass tube of uniform
diameter
36.5 gm HCl = 40 gm. NaOH
1000 ml N. HCl = 1000 ml . NaOH
V ml . N. HCl = V . ml N. NaOH
Therefore it is clear that according to these
law for complete reaction if required quantity
of the substance is expressed in their gram
equivalents then they must be equal. On this
basis, calculation of titration is calculated as
under.
Suppose solution A of N, normality and V,
ml . Volume for complete reaction requires
solution β of N2 normality and V2 ml .
Volume. Then for solution A
N1 V1
1000
Fig 6.11
(a) Stopper of burette.
(b) pinch cock burette.
51
Divide its length into equal parts and make
many marks Q on it. After every 10 mark, the
no, 0, 1, 2 ........ 50 or 100 are marked from
upto bottom repecivety, which include the
inner volume of burette in cubic cm or ml. In
the same way the volume is read upto 1 no. of
decimal. (0.1 qubic cm). The tube get the jet
shape into its bottom. The stop cork or
pinckcork in fixed on the jet. So that the stop
cork in rotate as required. To drop the solution
of burette drop - by - drop with help of jet.
Method to use burette :
1.
Wash the burette with water. If the
burette in dirty then wash it with uramil
acid and then wash it with water. If the
inner surface of burette in flat then wash
it with NaOH solution, wash it with
water. At last wash it with distilled
water. Then since it with the solution,
which to fill in it. The rinsing process is
shown in figure.
Fig 6.12: Rinse the burette.
2.
Rinse the burette and make it empty, and
fill the used solution upto the zero mark.
The bubble of jet of burette is remove
by pressing pitch rightly upside and then
law it. Now again fill the burette upto
zero after removing air bubbles. Now
burette is ready for titration.
Fig. 6.3 Fill the solution in the burette and measure it .
52
3.
Now take the known volume of the
solution into the beaker in the conical
flask with the help of a pipette & put it
below the burette & note the meniscus
of the surface the burette’s solution.
Now press the pinch cock to pour the burette
solution into the beaker by the left hand finger
& hold the beaker into the right hand shape is
slowly. Now again note down the meniscus
of the surface of burette’s solution when the
neutral point is obtained & determine the
difference between first and the 2nd reading.
This is the volume of the liquid of burette in
titration.
4.
For colourless solution lower meniscus
of half moon the solutions read. At the
time of reading the position eye placed
at the point of the lower surface of half
moon Neither above nor below. Correct
way of reading of burette must be
observed in the following diagram.
5.
It must be checked that the burette is not
leaking. In case of leakage some grease
must be applied at the glass stopper and
replaced the hard pinch cock of the burrete
to another.
6.
Before the experiment if any drop solution
is hanging at the tip of the end point of jet,
then it should be removed by using a filter
paper. But if the drop in hanging after the
addition of solution into the titration flask
then it should be removed by touching it
with the wall of the titration flask.
Pipette and its use
It is a glass tube having cylindrical part is the
middle. It is provided with a jet at its lower end,
It is used for correct measurement of definite
volumes of solution. The volume of pipette is
marked in its middle cylindrical part. It is 10ml,
20ml, 25ml, 50ml. Its upper side tube has a mark.
On filling the solution upto that mark the liquid
becomes equal to the marked volume on the
pipette.
Fig : 6.15 Use of Pipette.
53
pressure on the index finger. Now place
the pipette in the titration flask vertically
and remove the index finger and pour
the complete solution into the titration
flask.
Method of using pipette :
1.
Wash pipette with water.
2.
Rinse the pipette with the given solution.
Fill the liquid into the pipette, put it
horizontal & rotate, & take out the liquid
outside. This whole all procedure is shown
in figure 6.13.
3.
Now put the given solution in pipette
according to fig 6.14, close the upper
end of the pipette by the index figure.
To allow the solution to its fall solwly
upto the marked point in pipette. At the
marked point put the surface of water
in pipette constant by increasing
4.
For removing the left solution in jet of
pipette, touch the end of the jet of pipette
to the surface of the wall of titration
flask, & close the upper end by index
finger. In case some drops are left into
the jet, don’t try to remove them by any
other method. Because the marked
pipette is calibrated taking this liquid
in to account.
Fig : 6.16 (i) To fill the solution in the pipette (ii) To stop the solution with finger (iii) To bring the solution upto the
marked point (iv) To pour the solution in the conical flask (v) To remove the last drop of the pipette.
(Record of Titration-Experiment)
Correct result is obtained after completing
Volumetric analysis with accuracy. It is to be
carefully recorded in the practical note book.
Here the method are given for recording the
practical work related with titration.
1. Single titration : In this titration the unknown
concentration of solution of substance is
determined with the help of standard solution of
another substance. This kind of titration in
performed.
54
between the solution of opposite nature (such
as acid & base), there solutions mutually
naturalise one another during titration. In such
case the titration is performed only once,
therefore it is called single titration.
2. Double Titration : It is also based on
titration between two solutions of opposite
nature. If the solution A and B are of same
type and concentration or normality of one
solution (suppose A) is known then the
concentration of solution B can not be
determined directly after titration. In such a
case a solution having opposite nature is used,
is called intermediate solution & the titration
of both solutions is performed by this.
At first the titration is performed between A,
E (standard solution) and C (intermediate
solution). So that the concentration of solution
C in determined. Now the titration is
performed between C and B and the
concentration of B is determined by
calculation. This kind of titration is called
Double titration.
Record of titration experiment in practical
note book :
In practical note book on both side of the page
titration experiment in recorded in following
method is popular, acceptable and
systematically.
Left page
1.
2.
3.
4.
Experiment No.1
Object
Formula
Calculation
N/10 standard solution of oxalic acid.
Apparaturs: Burette, pipette, conical flask,
burette stand funnel.
Reagent : Standard solution of Oxalic acid,
Solution of NaOH Phenolphthalein (indicator)
Principle : This titration is performed between
strong base and weak acid they are
Phenolphthalein is. a indicator & NaOH is
taken into the burette. The neutralization
process in completed in following way:
COOH + 2NaOH → COONa + 2H O
as an indicator
Method :
1.
Clean all the apparatus with water. Wash
the burette with given NaOH and fix it
to the stand, fill it with acid. In case, if
there is a air bubbles in the rubber of
pinch cock then remove them and put
the surface of base at zero.
2.
Wash the pipette with standard oxalic
acid, take 25 ml. acid unit and pour it
into the conical flask and. add 1-2 drop
of phenethicillin indicator in this
solution.
3.
Now drop the base from the burette into
the flask slowly and carefully and
determined the end point. Note this
point of the burette.
4.
Repeat this experiment 3-4 times and
find the two same reading of burate for
the above value of oxalic acid.
Right page
1.
2.
3.
4.
5.
6.
7.
8.
Experiment no & dt.
Object
Apparatus & Reagent
Principle or equation
Method (not
necessary to units)
Observation table
Result
Precaution
Experiment No – 1.
Object : To find out the concentration of
hydroxide. solution by titrating it against the
| phenol-ph+nalein
COONa
|
COOH
Observation Table :
Table number – 1
S.No.
Volume
of acid
Reading of
Burate
Starting
1.
2.
3.
25 ml
25 ml
25 ml
0.01 ml
0.01 ml
0.01 ml
20.3 ml
20.3 ml
20.3 ml
Volume
End
of base
20.0ml
55
Formula : N1 V1 = N 2 V2
Standard solution = Solution of unknown
concentration.
Calculation
N1 V1 = N 2 V2
(acid) (base)
N
× 25 = N 2 × 20
10
N 25 N
N2 =
×
=
10 20 8
Normality of given base = 0.125 N
Concentration of base = Normality ×Gram
equivalent weight base
0.125 N × 40 = 5 Gram / liter
Result : The normality of given NaOH is 0.125
N & the concentration of given NAOH is 5gm /
litre.
Precaution :
1.
Wash the instrument with water before
use.
2.
Rinse the burette and pipette with
respective solution.
3.
Solution in the burette should be filled
with the help of a funnel. Remove the
funnel.
4.
Air bubble from the pinch cock must
be removed.
5.
The indicator should be used in very
little quantity (1, or 2 drop)
Experiment No. 2
Object : Make the standard solution of sodium
carbonate (about N/8) and determine the
concentration of Unknown H 2SO4 with the
above solution by using titration method.
Principle : This titration is performed between
conc. acid and weak base. Therefore Methyl
orange indicator is used. The base solution is
taken into the beaker & acid solution is taken
into the burette. Therefore the red colour is
produced at the end point.
Suppose that a standard solution has to prepare
into the 250 ml titration flask therefore weight
of Na2 CO3 is 53. Therefore weight
53 250
×
1.6562
8 1000
gm Na2CO3 for preparing
the solution of N/P normality.
Now take a weighing tube in which a cock is
fixed & note down its weight. Take 1.6562
gm. Na2CO3 and note down its weight. Now
take Na2CO3 into the beaker and wash it with
some water. Fit the cock into the weightube
weight it with the left Na2CO3 and note its
weight.
Pour the solution of beaker is the 250ml of
titration flask with the help of funnel. Wash
the beaker many times with water in the
titration flask. Now fill the water into the
titration flask upto the marked point carefully
and make a homogeneous solution by shaking
the flask.
Observation :
1.
Weight of the weighting tube = 6.888
gm.
2.
Weight tube + weight of
Na2CO3 = 8.5442 gm.
3.
Other weight of the weighing tube =
6.8872 gm.
Weight of Na2CO3 = (8.5442 − 6.8872 ) gm
= 1.8570 gm.
Concentration of
Na2CO3 = (1.6570 × 4gm / littre)
Concentration of
Na2CO3 = (1.6570 × 4gm / littre)
Precaution : Now wash the burette with the
solution of H 2SO4 & fill it with H 2SO4 , and
fixed to the burette stand.
Take 20ml solution of Na2CO3 with the help
of pipette. into the titration beaker or flask
Now add 2-3 drop methyle orange indicator
to perform the titration. The colour of the
solution becomes light pink at the last drop of
solution. Repeat the titration process until the
Volume of H 2SO 4 becomes 2 times.
56
Table No. 1
S.No
Volume
Volume of
Na2CO3
Reading of Burrete
Start
End
of H 2SO4
1.
2.
20 ml
20 ml
1.7 ml
3.2 ml
22.9 ml
24.3 ml
21.2 ml
21.1 ml
3.
20 ml
6.2 ml
24.3 ml
21.1 ml
Calculation : N1 V1 = N 2 V2
Normality=Concentration × equivalent weight
Apparatus : As per practical No. 1
Method :
1.
Rinse the burette with the HCl acid and
then fill the burrete with acid upto 01
ml mark.
2.
Take 25 ml solution of Sodium
Carbonate into a conical flask or beaker
with the help of pippete.
3.
Pour 2-3 drop of Methyl orange into
the solution. And pour the acid solution
from the burette into the conical flask
slowly. Shake the solution.
= N1 = 1.6570 × 4 N = 0.1257
53
V1 = 20 ml
N2 = ?
V2 = 2101 ml
∴
0.125 N × 20 = N 2 × 21.1
∴
N 2 = 0.125 N ×
=
20
20.1
2.50
= 0.123 N
20.1
Note the reading of the burette at which
the pink colour is obtained of the drop.
4.
Repeat this experiment .3 to 4 times, So
as to obtain two consecutive readings.
Concentration=Normality × equivalent weight
=
2.50
× 4.9
20.1
=
1225
201
Observation table
S.No.
Volume of Reading
the base of Start
Burette
End
1.
2.
3.
25 ml
25 ml
25 ml
26.7 ml
26.4 ml
26.4 ml
= 6.09 litre.
Experiment no. 3 :
Objective:To find out the concentration of HCl
solution with the help of standard solution of N/
10. Sodium Carbonate,
Principle : When HCl is added into the sodium
carbonate then the reaction is completed into two
steps :
(i)
Na2CO3 + HCl → NaHCO3 + NaCl
(ii)
Na2CO3 + HCl → NaCl + H 2O + CO2
In the first step the Carbonate is converted in
to the bicarbonate. In the 2nd step the bi carbonate reacts with acid and gives carbondie-oxide.
In this titration methyle orange is used as a
indicator and acid solution is be taken in to
the burette.
0.0 ml
0.0 ml
0.0 ml
Used
Volume
of acid
26.4 ml
Calculation : Sodium Carbonate Hydrochloric acid
N1 V1 = N 2 V2
N
× 23 = N2 × 26.4
10
N2 =
N
25
25
×
=
N
10 264 264
Normality of HCl solution = 0.094 N
Concentration=Normality × equivalent weight
= 0.094 × 36.5
= 3.421 gm/Litre.
Result : Concentration of HCl = 3.421 gm/
litre
Precaution : As per the experiment no . 1
56
57
6.6 Calculations of Single Titration
Example 1: 1.2 gram of a solution is dissolved
in 200 ml solution then calculate the strength of a
solution
Example 3: Calculate amount of caustic soda
in gram in a deci normal solution of it prepared
in one litre
Solution 1: When 200 ml of solution contains
1.2 grams solute.
Solution. 3: Normality of Deci normal solution
∴ 1000
=
ml of solution will contain
N
10
Strength = equivalent weight X Normality
= 1000 ×
1.2
= 6 gm
200
= 40 N = 4 gram /litre
10
Strength of solution 6 gm / litre
∴2
Example 2: 53 gm of sodium carbonate is
dissolved in 100 ml solution find strength and
normality of the solution . Equivalent weight
of sodium carbonate is 53
litre solution will contain
4 × 2 = 8 gram
Result -2 litre solution of caustic soda contains
8 gram solvent
Solution 2:
∴ 100 ml of solution contains .53 grams
solution
Example 4 : Find the amount of KOH dissolved
in 500 ml of 0.085 N KOH solution
∴
Solution 4 : Strength = Normality × Equivalent
weight
100 ml solution will contain
=
= 0.085 × 56 = 4.76 gram/litre
0.53 × 1000
= 5.3 gm
100
1000 ml of solution contain 4.76 gram solution
Strength = 5.3 gram / litre
Normality =
Strength
Equivalent weight
=
5.3
53
=
1
N
=
10 10
Result(1) Strength of solution of sodium
carbonate is 5.3 gram/litre
(2)
Normality of solution of sodium
carbonate = N/10
∴
500 ml solution contain will contain
=
4.76 × 500
= 2.38 gm
1000
Result : 500 ml solution contain 2.38 gm. KOH.
Example 5 : To find the quantity substance in
gm. to prepare 500 ml N/10 Sodium Carbonate.
Solution 5 : Strength = Normality × equivalent
weight
58
N2 =
N
25
2.5
×
=
10 23.8 23.8
or
=
2.5
238 N
= N1 V1 = N 2 V2
= base
= 0.105 N
Result : Normality of HCl in 0.105 N
Example 7 : 2.65 gram Na2CO3 is dissolved
in 250 ml water to prepare standard solution.
It neutrilize 25 ml of this solution with
H 2SO 4 , 27.3 ml H 2SO4 was required. Find
the normality and strength of H 2SO4 .
Solution : ∵ 500 ml solution contain 2.65 gm.
Solvent
∴ 1000ml solution will contain = 5.3 gm.
Strength
=
N
× 25 = N 2 × 27.3
10
N2 =
or
acid
25
N
×
10 27.3
= 0.091 N
Strength = Normality × Equivalent weight
= 0.091 × 49
= 4.459 gram / litre
= 4.46 gram / litre
Normality = Equivalent weight
Result : (1) Normality of H 2SO4 = 0.091 N
=
(2) Strength of H 2SO4 = 4.46 gram/ litre
5.3
1
N
=
=
53 10 10
Exercise Question
Q.1
What do you understand by titration?
Q.2
Define Normality, Molarity and Strength?
Q.3
Explain the function of indicators in Acid Base titeration?
Q.4
What is the difference between dinormal & decimolar solution of Sulphuric acid?
Q.5
Which indicator is suitable for titration between oxalic acid and caustic soda
solution?
Q.6
Explain giving reasons, Acidimetry & Alkalinimetry with example?
Q.7
What is the normality of Na2CO3 solution prepared by dissolving 10.6 gram in 500 ml
flask?
Q.8
How much NaOH is required to prepare 0.01N solution in 500 ml solution?
Q.9
How much Na2CO3 is required to prepare 1 N solution in 500 ml.
Q.10
How much water must be added in 6 NHNO3 acid solution so that it becomes 2 N
concentration?
59
Q.11 How much water must be added in 1 litre 0.125 N NaOH solution so as to make on
0.100 N solution ?
Q.12 Find the amount of KOH dissolved in it was completely neutralized by 25 ml of N/10 HCl
Solution ?
58
Q.13 0.5 N HCl Solution completely neutralized by 22.5 ml Caustic Soda Solution, find the amount
of Caustic Soda dissolved in the Solution ?
Q.14 To neutralized 35 ml of H 2SO4 of unknown strength 28.6 ml. N/10 NaOH was used. Find
the normality of H 2SO4 .
Q.15 20 ml of N/20 was completely neutralized by N/10. Calculate the used quantity of acid ?
Q.16 31.00ml of 1 base was neutralized by 25 ml of N/10 HCl solution. If strength of base is 4.5
gm/litre then find the equivalent weight of base.
Answer
(7) 0.4 N
(8) 2 gm
(11) 250 ml
(12) 0.14 N
(15) 100 ml
(16) 55.8
(9) 26.5 gm
(13) 0.45 gm
(10) 1 : 2
(14) 0.0817 N