Volumetric analysis is the process of determining the concentration of solution of unknown concentration with the help of solution of known concentration.

2. Content Area: General and Physical Chemistry
1. Volumetric Analysis
1.1 Introduction to gravimetric analysis, volumetric analysis and
equivalent weight
1.2 Relationship between equivalent weight, atomic weight and valency
1.3 Equivalent weight of compounds (acid, base, salt, oxidizing and
reducing agents)
1.4 Concentration of solution and its units in terms of : Percentage, g/L ,
molarity, molality, normality and formality, ppm and ppb
1.5 Primary and secondary standard substances
1.6 Law of equivalence and normality equation
1.7 Titration and its types: Acid-base titration, redox titration ( related
numerical problems) 8hrs

3. WHAT IS ANALYSIS ?
A detailed examination of any thing is ANALYSIS of that thing. In
chemistry, chemical analysis is performed which is very important.
1. Qualitative analysis-it involves the analysis of a substance to
know its constituent elements or radicals or compounds. For
example, qualitative salt analysis in which acid radicals and basic
radicals are determined.
TYPES OF QUANTITATIVE ANALYSIS
2. Quantitative analysis – it involves analysis of substance to know
the amount of its constituent element/s or radicals.
ANALYSIS is of two types, they are:

4. There is a number of methods for quantitative analysis, some important
methods are- GRAVIMETRIC ANALYSIS, VOLUMETRIC ANALYSIS,
INSTRUMENTAL ANALYSIS etc.
GRAVIMETRIC ANALYSIs
It is the method used in analytical chemistry for the quantitative
determination of an analyte(ion or element which is being analysed) on
the basis of mass i.e., quantitative determination is carried out by
measuring the mass of substance.
VOLUMETRIC ANALYSIS
It is the quantitative analysis carried out by measuring the volume of
solutions. From volume analysis, the concentration & other constants
like pH, pOH, pKa, pKb, Ka, Kb of unknown solution can be determined
with the help of std. solution.

5. Here we are mainly concern with the determination of concentration of
a solution whose concentration is not known. Hence volumetric analysis
is the process of determination of the concentration of unknown
solution either in presence or in absence of indicator with the help of
known solution. In this analysis, two different solutions as reactants are
required to react. The volumetric analysis is now also called titrimetric
analysis OR TITRATION.
TYPES OF VOLUMETRIC ANALYSIS
Volumetric analysis may be divided into different categories depending
upon the nature of the reactions involved between two solutions
(reactants).Some examples are:
a. Acid-base titration or neutralisation titration in which reactants are
acids & alkalis, neutralisation reaction occurs between them. This
titration is further classified as Acidimetry (determination of
concentration of acid with the help of base) & Alkalimetry
(determination of concentration of alkali with the help of acid)
b. Redox titration –in this titration, redox reaction is involved between

6. VOLUMETRIC ANALYSIS
•NEUTALISATION TITRATION
•REDOX TITRATION
•PRECIPITION TITRATION
•COPEXOMETRIC TITRATION
•etc.,

7. CONCENTRATION OF SOLUTION
It is defined as the quantity of solute present in the given quantity or
volume of solution or solvent.
DIFFERENT WAYS OF EXPRESSING THE CONCENTRATION OF
SOLUTION:
1. PERCENTAGE
2. GRAM PER LITER(g/l)
3. Normality(N)
4. Molarity (M)
5. Molality (m)
6. Formality
7. Ppm
8. Ppb

8. 1.PERCENTAGE:
It is defined as the no. of parts of a substance present in
100 parts of solution. The following types of percentage are
in common use.
a. Percentage weight by weight(w/w percent):
Wt. of solute in g Wt. of solute in g
% W/W = × 100= × 100
Wt. of solution in g Sp.gr. × vol. of solution
 gL-1 = %w/w × sp. gr. × 10
%w1/w2=100g/sp.gr. (w2)=vol. in ml contains w1g solute.
1000ml contain w1×sp.gr./100 ×1000 g solute
Therefore, gL-1=%w/w ×sp.gr. ×10

9. b) Percentage weight by volume (w/v percent):
wt. of solute in g
%w/v= ×100
vol. of solution in ml
 gL-1= %w/v × 10
c) Percentage vol. by vol.(v/v percent):
vol. of solute in ml
%v/v = × 100
vol. of sol. in ml
Note:- Simply expression of percentage always refers to
percentage weight by volume (i.e. w/v percent).

10. 2. gm/lit.(g/L)(also known as strength):
It is defined as the amount of solute in grams
dissolved in one litre of solution.
wt. of solute in g
i.e. g/L =
vol. of solution in litre
wt. of solute in g
= ×1000
vol. of solution in ml

11. 3. Normality(N):
Normality of solution is defined as the no. of
gram equivalent weight of solute present in one
litre of solution.
No. of equivalents of solute
Normality(N) =
Vol. of solution in L
No. of equivalent of solute
= × 1000
Vol. of solution in mL

12. RELATION BETWEEN NORMALITY AND g/L
We have,
No. of equivalents of solute
Normality(N) =
Vol. of solution in L
We know,
Wt. of solute in g
No. of equivalents =
Eq. wt.

13. Therefore,
Wt. of solute in g 1
Normality(N)= ×
Eq. wt. Vol. of solution in L
Or
1 Wt. of solute in g
Normality(N)= ×
Eq. wt. Vol. of solution in L

14. Or
1
Normality(N)= × g/L
Eq. wt.
Hence,
g/L = Normality × Eq. wt.
TYPES OF NORMAL SOLUTION:
 Normal solution(N):
The solution one litre of which contains 1 equivalent of
solute is called normal (1N) solution.

15.  Semi normal solution(N/2):
The solution one litre of which contains 1/2 equivalent of
solute is called semi normal (N/2) solution.
 Decinormal solution(N/10):
The solution one litre of which contains 1/10th equivalent of
solute is called decinormal (N/10) solution.
 Centinormal solution(N/100):
The solution one litre of which contains 1/100 equivalent of
solute is called centinormal (N/100) solution.
 Millinormal solution(N/1000):
The solution one litre of which contains 1/1000 equivalent of
solute is called millinormal (N/1000) solution.
 so on

16. Wt. of solute taken
Normality factor, f=
Wt. of solute to be taken
Or,
Observed wt.
Normality factor, f =
Calculated wt.
Actual normality of solution = proposed normality × f

17. 4. Molarity (M):
Molarity of solution is defined as the no. of
moles of solute present in 1L of solution.
Mathematically,
No. of moles of solute
Molarity(M) =
Volume of solution in L
No. of moles of solute
= × 1000
Volume of solution in mL

18. We know,
Wt. of solute in g
No. of moles of solute =
Molecular wt.
Therefore,
Wt. of solute in g
Molarity(M) = ×
Mol.wt. of the solute
1
Volume of solution in L

19. Or we can write,
1
Molarity(M) = ×
Mol.wt. of the solute
Wt. of solute in g
Volume of solution in L
Or ,
1
Molarity(M) = × g/L
Mol.wt. of the solute

20. Hence,
g/L = Molarity(M) × Molecular Wt.
TYPES OF MOLAR SOLUTION:
 Molar solution(M):
The solution one litre of which contains 1 mole of solute is
called molar(1M) solution.
 Semimolar solution(M/2)
 Decimolar solution(M/10)
Centimolar solution(M/100)
Millimolar solution(M/1000)
 so on
Similarly,

21. 5. Molality(m):
It is defined as the no. of moles of solute
dissolved in one kg of solvent.
Mathematically,
No. of moles of solute
Molality(m) =
Wt. of solvent in kg
No. of moles of solute
= × 1000
Wt. of solvent in g

22. TYPES OF MOLAL SOLUTION
 A molal solution (1m) solution contains one mole of
solute in one kg of solvent.
 A semimolal solution (m/2) solution contains 1/2 mole
of solute in one kg of solvent.
 A decimolal solution (m/10) solution contains 1/10
mole of solute in one kg of solvent.
 A centimolal solution (m/100) solution contains 1/100
mole of solute in one kg of solvent.
 so on

23. 6. FORMALITY(F):
The formality of a solution is defined as the number of
formula mass of any solute dissolved in 1 litre of solution.
Commonly, the term formality is used to express the concentration of
the ionic solids which do not exist as molecules but exist as network of
ions.
Mathematically,
No. of formula mass of solute
Formality(F) =
Vol of solution in L
No. of formula mass of solute
= × 1000
Vol of solution in mL

24. 7. ppm(parts per million):
Ionic compounds exist in the form of the ions rather than in the form of
molecules. So the term formula mass is used instead of molecular mass
and formality for ionic compound in place of molarity. Like molecular
mass is the sum of mass of all atoms in the molecule, formula mass is
the sum of mass of all atoms in the formula. So, mathematically,
molarity and formality will give same values for a given solution.
When the concentration of solution is very much low or the solution is very
dilute, the concentration of such solution is expressed in terms of part ppm.
Concentration of ions like Hg⁺⁺, Pb⁺⁺, etc. in water and traces amount of gas
in atmosphere are also expressed in terms of ppm.
Mass of solute w×106
ppm = ×106 =
Mass of solution wsoln

25. Mass of solute(in g) w×106
Therefore, ppm = ×106 =
vol.of solution(in mL) V
But for very dilute aqueous solution, the amount of solute is very low, so V≈wsoln
and ppm=g/L ×103 =μg/mL(microgram per milliliter)=mg/L

26. Some examples of ppm:
one inch in 16 miles(16miles=1,013,760),
one second in 11.5 days(11.5×24×60×60=993600sec),
one minute in two years(2×365×24×60=1051200min).
• One ppm is equivalent to the absolute fractional amount multiplied by one
million or 1,000,000.
• Concetration in ppm is defined as the mass of solute in one million parts
by mass of solution.
• It also can be expressed as milligrams per liter (mg/L).
• The ppm or mg/L means the same thing.

27. • An even smaller concentration measurement is parts
per billion (ppb).
1. ppb(Parts per billion)
• One ppb is one part in 1 billion.
• One ppb is equivalent to the absolute fractional
amount multiplied by one billion or
1,000,000,000.
• Example : one second in nearly 32 years.
32×365×24×60×60=1,009,152,000sec.

28. EQUIVALENT WEIGHT
Equivalent weight of a substance is the number of parts by
weight of it which combine with or displace directly or
indirectly 1.008 parts by weight of hydrogen or 8 parts by
weight of oxygen or 35.5 parts by weight of chlorine.
EQUIVALENT WEIGHT OF ELEMENT
Every element has a fixed combining weight for all other elements
which is called equivalent weight. It is also known as chemical
equivalent.
The equivalent weight of an element is defined as the number
of parts by weight of it which combines with or displaces
(from a compound) 1.008 parts of hydrogen or 8 parts by

29. weight of oxygen or 35.5 parts by weight of chlorine.
i. Mg + H2 MgH2
Examples:
e.g., 2×1.008g H2 combine with 24g Mg.
Hence, 1.008g H2 combine with 24 ×1.008
2×1.oo8
= 12g
Therefore, equivalent weight of Mg is 12.
24g 2×1.008g
Similarly, in MgCl2 & MgO, equivalent weight of Mg is 12.

30. Mathematically,
equivalent wt. of an element is given by
Wt. of the element in g
= × 1.008
Wt.of hydrogen in g
Wt. of the element in g
= ×8
Wt.of oxygen in g
Wt. of the element in g
= ×35.5
Wt.of chlorine in g

31. RELATION BETWEEN EQUIVALENT WEIGHT, VALENCY
& ATOMIC WEIGHT
Let, atomic weight of an element = A
Equivalent weight = E
Valenvy = V
We know,
valency is given by the no. of hydrogen atoms combined or
displaced by one atom of the element.
Hence,
V atoms of H are combined or displaced by 1 atom of the
element

32. V×1.008 parts by wt. of hydrogen is combined or displaced by
A parts by wt. of element.
Therefore,
A
1.008 ----------------------------------------------- × 1.008
V×1.008
A
= parts
V

33. According definition of equivalent wt.,
Atomic wt.
Equivalent weight =
Valency
This is the relation between equivalent weight(E),
valency(V) and atomic weight(A) of element.
This means, if an element possesses variable valency, it
should have variable equivalent masses.
For example, Cu has two equivalent masses because it
possesses the valencies 1 & 2.

34. For ionic radicals,
Ionic wt.
E =
Valency
24
Equivalent wt of Mg = = 12
2
60
Equivalent wt of CO3= = 30
2

35. 63.5
Equivalent wt of Cu in Cu2O = = 63.5
1
63.5
Equivalent wt of Cu in CuO = = 31.75
2
Similarly, the other elements which exhibit the variable equivalent
weight are Fe, Sn, Pb, Hg, etc.

36. Equivalent weight of compounds
The concept of eq. wt. is not restricted to elements. It is
defined as the no. of parts by wt. of the compound reacts
completely with 1g equivalent of any other substance.
OR
The equivalent weight of compound is the sum of equivalent
wt. of cations and anions formed after the ionisation of
compounds irrespective of coeficient of ions.
For example, to find eq. wt. of Na2CO3
Na2CO3 2Na+ + CO3

37. 23
Eq. wt. of sodium ion = = 23
1
Eq. wt. of CO3 =60/2 = 30
Therefore,
Eq. of Na2CO3 = 23 + 30 = 53.
The other methods of determining the eq. wt. of some of the
compounds are described below:-

38. 1. Eq. wt. of an acid:- Eq. wt. of acid is defined as the no. of
parts by wt. of an acid that can supply 1.008 parts by wt.
of hydrogen ion.
Mathematically,
Molecular wt. of the acid
Eq. wt. of an acid =
Basicity
Basicity of acid is defined as the no. of replaceable hydrogen atom/s
present in one molecule of the acid. Examples-
Acids Basicity Molecular wt. Eq. wt.
HCl(Monobasic acid) 1 36.5 36.5/1= 36.5

39. Acids Basicity Molecular wt. Eq. wt.
H2SO4(dibasic acid) 2 96 96/2= 49
HNO3(monobasic) 1 63 63/1= 63
(COOH)2.2H2O(dibasic) 2 126 126/2= 63
(Oxalic acid)
CH3COOH(monobasic) 1 60 60/1= 60
(Acetic acid)
H3PO4(tribasic acid) 3 98 98/3= 32.66
H3PO3(H-PO(OH)2)-dibasic 2 82 82/2= 41
(Phosphorous acid)
H4P2O7
(Pyrophosphoric acid) 4 178 178/4= 44.5
Tetrabasic
HPO3(Metaphosphoric acid) 1 80 80/1= 80
Monobasic

40. 2. Eq. wt. of base:- It is defined as the no. of parts by wt.
of a base that can neutralize one gram equivalent of an acid.
Mathematically,
Molecular wt. of the base
Eq. wt. of a base =
Acidity(No. of replaceable OH-)
Acidity of base is defined as the no. of replaceable hydroxyl (or
hydroxide) ions present in one molecule of the base. Examples-
Bases Acidity Molecular wt. Eq. wt.
NaOH(Monoacidic base) 1 40 40/1= 40

41. Bases Acidity Molecular wt. Eq. wt.
NaOH(Monoacidic base) 1 40 40/1= 40
Ca(OH)2(diacidic) 2 74 74/2= 37
KOH(monoacidic) 1 56 56/1= 56
NH4OH (,,) 1 35 35/1= 35
CaO (diacidic) 2 56 56/2= 28
Mg(OH)2 2 58 58/2= 29
Ba(OH)2 2 171 171/2= 85.5
3. Eq. wt. of salt:-
Molecular wt. of the salt
Eq. wt. of salt=
Total no. of positive charge on cation or total no. of negative
charge on anion

42. Examples-
Salts Total positive
or negative charge Molecular wt. Eq. wt.
NaCl Na+ + Cl- 1 58.5 58.5/1= 58.5
Na2CO3 2Na+ + CO3 2 106 106/2= 53
CaCO3 Ca++ + CO3 2 100 100/2= 50
KCl K+ + Cl- 1 74.5 74.5/1= 74.5
Al2(SO4)3 2Al+++ + 3SO4 6 342 342/6= 57
(2×27+3×96=342)
NaHCO3 Na+ + HCO3 1 84 84/1= 84
AgNO3 Ag+ + NO3 1 169.8 169.8/1= 169.8

43. 4. Eq. wt. of oxidising agent(oxidant) and reducing
agent(reductant):-
Mathematically,
Mol. wt.
Eq. wt. of oxidant or reductant =
Change in O.N.(Total no of electron
lost or gained during reaction)
Some examples of equivalents of oxidizing agents are:-
Oxidizing agents Change in O.N. Molecular wt. Eq. wt.
KMnO4(In acidic medium)
2KMn+7O4+3H2SO4+5(COOH)2
K2SO4+2Mn+2SO4+10CO2+8H20 7-2= 5 158 158/5= 31.6
Or, 4KMn+7O4+6H2SO4
2K2SO4+4Mn+2SO4+6H2O+502

44. Oxidizing agents Change in O.N. Molecular wt. Eq. wt.
KMnO4(In alkaline medium)
2KMn+7O4+2KOH
2K2Mn+6O4+H20+ O 7-6=1 158 158/1= 158
KMnO4(In neutral medium)
2KMn+7O4+H2O 2Mn+4O2+2KOH+3 O 7-4=3 158 158/3= 52.6
K2Cr+6
2O7 +4 H2SO4 +3H2S
K2SO4 + Cr+3
2(SO4)3 +7H2O + 3S 12-6=6 294 294/6= 49

45. Reducing agents Change in O.N. Molecular wt. Eq. wt.
2Fe+2SO4 + H2SO4 + [O]
Fe+3
2(SO4)3 + H2O 1 392 392/1= 392
Some examples of eq. wt. of reducing agent(reductant):-
Mohr’s salt[FeSO4(NH4)2SO4.6H2O], oxalic acid, ferrous sulphate, hydrogen sulphide,
etc. are examples of reducing agent.

46. GRAM-EQUIVALENT WEIGHT
The equivalent wt. of any substance expressed in gram is called gram eq.
wt. For example,
Eq. wt. of oxgyen = 8
Therefore, 1 g eq. wt. of oxygen =8g
Similarly, 1 g eq. wt. of hydrogen=1.008g
1 g eq. wt. of hydrochloric acid=36.5g
1 g eq. wt. of oxalic acid=63g
1 g eq. wt. of H2SO4=49g
Wt. of substance in g
No. of gram equivalent =
Eq. wt.

47. PRIMARY AND SECONDARY STANDARD SUBSTANCES
STANDARD SOLUTION:
A solution whose strength is known is called a standard solution.
Standard solution contains calculated or known amount of solute in
definite volume of solution. These are two types:
1. Primary std. solution &
2. Secondary std. solution
The chemical substances used to prepare the
standard solution are also two types:
1. primary standard substances &
2. secondary standard substances.

48. PRIMARY STANDARD SOLUTION
The standard solution prepared by dissolution of exactly
known weight of a solute of sufficient purity in a definite
volume of solution is known as primary standard solution. In
other words, the solution whose exact strength is known is
primary standard solution. And the reference substance used
to prepare primary standard solution is known as primary
standard substance.
The requisites of primary standard substance are as below:
i. It should be easily available in a pure state.
ii. It should be stable i.e., it should not be hygroscopic, efflorescence,
oxidized by air, attacked by CO2 and its composition should remain
same in solid or in solution for long time.

49. iii. It should be possible to dry these substances at 1100 to 1200C.
iv. It should have high eq. wt. in order to minimize error during weight.
v. It should be readily soluble in given solvent to give a stable solution.
Examples of primary standard substances:
Anhydrous Na2CO3, oxalic acid crystals[ (COOH)2.2H2O],
Mohr’s salt[ ferrous ammonium sulphate, FeSO4.
(NH4)2SO4.6H2O], K2Cr2O7, KCl, AgNO3, NaCl, KBr, KBrO3, KIO3,
etc.
SECONDARY STANDARD SOLUTION
The standard solution prepared by dissolution of calculated
wt. of sec. std. substance in definite volume is known as

50. secondary std. solution. But, the concentration of the
secondary standard solution is not exactly known &
concentration of such solution changes with time as well.
 Exact strength of the secondary standard solution is
determined with the help of primary standard solution.
 Since the concentration of secondary standard solution
changes with time, it must be standardised with the help of
pr. std. solution when it is required in accurate concentration.
 Sec. std. substances may not obey the requisites of primary
standard substances.
 Examples of sec. std. substances: HCl, H2SO4, NaOH,
KMnO4, Na2S2O3, CuSO4.5H2O, HNO3, FeSO4 etc.
 The substances used to prepare sec. std. solution are
known as sec. std. substances.

51. WHY CAN NOT WE PREPARE SECONDARY STANDARD
SOLUTION OF ACCURATE CONCENTRATION ?
Required amount of secondary standard substances cannot
be weighed out accurately. Therefore, approximate weight of
these substances are taken to prepare definite volume of
solution.
LAW OF EQUIVALENCE AND NORMALITY EQUATION
According to the law of chemical equivalence, the two substances react
in their equivalent proportion. This means for two substances say A and
B,
No. of equivalents of A = No. of equivalents of B
Let, volume of solution of substance, A = V1
Normality of the solution, A = N 1
Principles of volumetric analysis

52. Similarly,
Let, volume solution of substance, B = V2
Normality of the solution, B = N2
Then, V1 × N1
No. of equivalents of A =
1000
V2 × N2
No. of equivalents of B =
1000
Since, No. of equivalents of solute
normality(N) = × 1000
Vol. of solution in mL(V)

53. Hence,
•We can write, V1 N1 = V2 N2
•This equation is called the Normality equation. This equation can be
applied in all reactions & also for the dilution of a solution.
• But the corresponding molarity equation, V1M1 = V2M2 in terms of
molarity cannot be used in all reactions. Why? Because this equation
doesnot follow the law of chemical equivalence for all reactions.
•However, this molarity equation can be used for the dilution for any
solution. Why? Because during dilution, no reaction takes place, so the
law of chemical equivalence is not necessary. As a result this equation
can be used for dilution of any solution.
Therefore, V1 × N1 V2 × N2
=
1000 1000

54. From the law chemical equivalence we can also write the following
relation
Acid and alkali solutions of same normality neutralize each other in
equal volumes,
eg., 10mL of 1NHCl = 10mL of 1NNaOH
= 10mL of 1NNa2CO3
 Equivalent volumes are inversely proportional to their normality.
For example, 1mL of 1NHCl = 10mL of N/10 HCl = 100mL of N/100HCl
= 0.5mL of N/0.5HCl = 0.5mL of 2NHCl
= 0.2mL of N/0.2HCl = 0.2mL of 5NHCL
= 2mL of N/2HCL = 2mL of o.5NHCl
 No. of equivalents of A = No. of equivalents of B
Mass of A, m1 Mass of B, m2
Or, =
Equivalent wt. of A,E1 Equivalent wt of B,E2

55. Hence,
m1 m2
=
E1 E2
 m1 V2 × S2
=
E1 1000
Or,
V1 × S1 m2
=
1000 E2

56. 1.7 TITRATION AND ITS TYPES: ACID-BASE TITRATION,
REDOX TITRATION ( RELATED NUMERICAL PROBLEMS)
TITRATION :
It is a technique for determining the concentration of
unknown solution with the help of primary standard solution
in presence or absence of indicator by measuring volume of
two reacting solutions for their completion of reaction.
 Titrant[ Titrate]: During the titration, the solution which is taken in
burette is known as titrant. Generally, it is the solution of known
concentration.
 Titrand[ or Analyte]: During the titration, the solution which is taken
in titration flask is known as titrand. Generally, it is the solution of
known volume.

58. TYPES OF TITRATION
1. Acidimetry and alkalimetry(Acid – base titration):
Examples, HCl + NaOH NaCl + H2O
H2SO4 + Na2CO3 Na2SO4 + H2O + CO2
Volumetric analysis (TITRATION)may be divided into different categories
depending upon the nature of the reactions involved between two
solutions (reactants).Some examples are:
Strong acid Strong base
Strong acid Weak base
CH3COOH + NaOH CH3COONa + H2O
Weak acid Strong base
M.O. or
Phenolphthalein
Methyl orange
Phenolphthalein

59. 2. REDOX TITRATION:
Titration in which redox reaction takes place is known as redox titration.
Examples:
2KMnO4 + 3H2SO + 5(COOH)2 K2SO4 + 2MnSO4 + 10CO2 + 8H20
Oxidizing agent Reducing agent
K2Cr2O7 + 7H2SO4 + 6FeSO4 K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O
Oxidizing agent Reducing agent

60. EQUIVALENCE POINT(theoretical end point):
The stage during titration at which the two solutions are used
up in their stoichiometric proportions is called equivalence
point.
END POINT:
The equivalence point determined by using an indicator is
called end point. Ideally, end point & equivalence point
should be same. But, the color of indicator generally changes
only after the equivalence point & hence end point differs
slightly from equivalence point.

61. TITRATION ERROR:
In actual practice, a small difference usually occurs between
equivalence point and end point, this is called titration error.
INDICATOR:
To find the end point of the titration, a small quantity of a
third substance is usually used. This third substance is
known as INDICATOR. Indicators are chemicals (usually
organic) used in titration which by change of colour
indicate the end point.

62. NUNERICALS
Q. A sample of NaOH weighing 0.48g dissolved in water and
the solution is made to 100mL. Calculate the normality and
molarity of the solution.
Solution,
Wt. of sodium hydroxide=0.48g
Volume of solution=100mL
Molecular wt. of NaOH= 40
We have,
Mass of solute in g × 1000
Molarity=
Molecular wt. × vol. of solution in mL

63. Therefore, molarity=0.12M
Normality=Molarity × Acidity
= 0.12 × 1 =0.12N
Q. A sample of commercial sulphuric acid solution contains
95% by mass of sulphuric acid. If the specific gravity of the
solution is 1.85, calculate the molarity of the solution.
Solution,
Wehave, g/L = % by mass × specific gravity × 10
= 95 × 1.85 × 10
= 1757.5
g/L = Molarity × Molecular weight
1757.5 = Molarity × 98
Molarity = 17.93M Answer