Volumetric analysis

• In board exam either 7 or 12 marks will come from this chapter.
• 5 marks numerical will come sure in board exam.

• Volumetric analysis deals with the measurement of
volume of solutions involved in the chemical reactions
which ultimately lead to the determination of the
amount of the constituents of the substance.

Different ways of expressing the concentration
of the solutions
• The amount of the substance present in the definite quantity of the
solution is called concentration of the solutions.
There are many ways of expressing the concentration of the
solution. The concentration of the solution can be expressed in any of
the following ways.

1.Gram per litre:
• The amount of the solute in gram present in one litre of the solution
is known as gram per litre of the solution.
gm/l= wt of the solute in gram
volume of the solution in litre
Eg. If 5 gram of NaOH is dissolved in 500 ml of the solution then its
strength is 10 g/l.

Gram equivalent weight
• The equivalent weight expressed in gram is called gram equivalent
weight.
one gram equivalent weight of HCl is 36.5 gram of HCl
Similarly two gram equivalent weight of HCl is 73 gram
Mathematically,
weight of substance in gram
Number of gram equivalent=
equivalent weight of the substance

2) Normality:
The number of gram equivalent of solute present in a litre of the
solution is known as normality of the solution.
Derivation………..in board

• Normal solution(1 N)
• Seminormal solution (N/2)
• Decinormal solution (N/10)
• Centinormal solution (N/100)
• Pentanormal solution (5N)
• Decanormal solution (10N)

Molarity:
The number of the moles of solute present in a litre of the solutions is
known as molarity of the solutions
Simply …..(in white board)

Relation between normality and molarity
(hseb)
Please see in board
Will you?

Percentage(%):
• Percentage strength means number of parts of solute in 100 parts of
solution which is expressed in three different ways:
a. % in weight by weight (w/W): it may be defined as the amount of solute
in gram present in 100 gram of the solution.
Simply,
% by w/W =( wt of solute in gm/wt of the solution in gm) ×100%
b. % in weight by volume (w/V): it may be defined as the amount of solute
in gram present in 100 ml of the solution.
Simlpy,
% by w/V =( wt of solute in gm/volume of the solution in ml) ×100%

c. % in volume by volume (v/V): it may be defined as the volume of
solute in ml present in 100 ml of the solution.
Simply,
% by v/V =( volume of solute in ml/volume of the solution in ml) ×
100 %

Some formula related to percentage
• In board

Molality:
The number of the moles of solute present in one kg of solvent is
known as molality of the solution.
Molality =(no. of moles of solute )/ (wt of solvent in Kg)

Some terms related and terminology related to
volumetric analysis
• Unknown solution: The solution whose strength is not known i.e. the
solution whose strength is to be determined during the titration is
known as unknown solution.
• Standard solution: The solution whose strength is known is known as
standard solution.

Titration:
The process in which the concentration of unknown solution is
determined with the help of standard solution by using the indicator is
called titration.

Titrant:
The solution of known concentration which is usually taken in a burette
is called titrant.
Titrand: The solution of unknown concentration which is being titrated
and usually taken in a conical flask is called titrand.

End point: (hseb)
• The point in a titration at which reaction between two solution is just
completed and at which indicator can show sharp colour change is
called end point.

Equivalence point or theoretical point:
The stage during the titration at which exactly equivalent amount of
titrant is added to titrand is known as equivalence point.
In theoretical concept, when equivalent amount of titrant is
added to the titrand, the indicator should change the colour and
indicate the end point. But in actual practice, the indicator doesn’t
change its colour at equivalence point. This is because the indicator
changes its colour either in acidic or alkaline medium. Therefore in
practice a small difference occurs between equivalence point and end
point. This difference in titration is known as titration error.(hseb)

Primary and secondary standard substance
Primary standard substance: The substance whose standard solution
can be prepared directly by dissolving the known weight of the
substance in fixed volume the solution is known as primary standard
substance. E.g. anhydrous sodium carbonate, oxalic acid, silver nitrate
etc.
The solution of primary standard substance is known as primary
standard solution.

Requirements for a substance to be primary
standard.(HSEB)
• It must be available in pure form and should be non toxic.
• It should not be hygroscopic( ability to absorb moisture) or
deliquescent(turns to solution).
• It should have high molecular weight or equivalent weight so as to
minimise the weighing error during weights.
• It should be readily dissolve in water.
• It should be stable. In other words, the composition of substance
should not change in solid or in solution state for long time.

Secondary standard substance:
The substance whose solution can be standardise or strength can be
determined by the help of primary standard solution is known as
secondary standard substance and the solution is known as secondary
standard solution. Eg. NaOH, HCl, KMnO4, FeSO4 .

Normality factor: (vvi)
It is defined as the ratio of actual weight of the substance taken to the
theoretical weight of the substance to be taken and usually denoted by
“f”.
Simply,
actual weight of the substance taken
Normality factor(f)=
theoretical weight of the substance to be taken

Thus exact strength of the solution can be determined by multiplying
the proposed strength of the solution with the normality factor.
What will be the normality factor if 1.52 gram of Na2CO3 is taken for
the preparation of 250 ml of N/10 solution? Also find the exact
strength of the solution? Answer 1.15
exact strength = 0.115N

Normality equation(principles of volumetric
analysis)
Volumetric analysis is based on the law of chemical equivalent which
states that “ the substance react in the proportion of their equivalent
weghts” i.e. one gram equivalent of acid would neutralize by one gram
equivalent of the base.
1 gram equivalent of HCl = 1 gram equivalent of NaOH.
1000 cc of 1 N solution of HCl= 1000 cc of 1N solution of NaOH
or 50 cc of 1N solution of HCl =50 cc of 1 N solution of NaOH
Thus volume of acid × normality of acid= volume of base × normality of
base

VA × NA = VB × NB
This is normality equation.
Some other important principles of volumetric analysis are as follows
…………… in board

Volumetric analysis

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