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VOLUMETRIC ANALYSIS
Grade XII
Volumetric analysis
Volumetric analysis is defined as the process of the
determination of concentration of unknown solution by
finding its volume which will exactly react with a definite
volume of another solution of known concentration
...
of parts by wt
...
008 parts by wt
...
of oxygen or 35
...
of
chlorine
...

of a substance, i
...
one can have more than one
equivalent weights depending on chemical change
...
of different substances are calculated as
follows:
𝐀𝐭𝐨𝐦𝐢𝐜 𝐰𝐞𝐢𝐠𝐡𝐭

a) Equivalent wt
...


Valency

Equivalent
wt
...
5

1

35
...
of radical (ion) =

𝒘𝒕
...
𝒐𝒇 𝒄𝒉𝒂𝒓𝒈𝒆 𝒐𝒏 𝒊𝒕

element

Mass of
radical
...
of
charge on
radical

Equivalent
wt
...
5

1

35
...
of acid =

𝐛𝐚𝐬𝐢𝐜𝐢𝐭𝐲

element

Mol
...


Basicity

Equivalent
wt
...
5

1

36
...
67

CH3COOH

60

1

60

𝐦𝐨𝐥
...
of base =

𝐚𝐜𝐢𝐝𝐢𝐭𝐲

Base

Mol
...


Acidity

Eq
...


NaOH

40

1

40

KOH

56

1

56

Ca(OH)2

74

2

37

Al(OH)3

78

3

26

e) Equivalent wt
...
𝐩𝐞𝐫 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐞
𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐰𝐞𝐢𝐠𝐡𝐭

𝐭𝐨𝐭𝐚𝐥 𝐜𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐨𝐱𝐢𝐝𝐚𝐭𝐢𝐨𝐧 𝐧𝐨
...
wt
...
wt
...
6

K2Cr2O7

294

6

49

(COOH)2

90

2

45

H2S

34

2

17

Equivalent wt
...
= 7-2 =5
Eq
...
of KMnO4 in acidic medium= 158/5 =31
...
= 7-4 =3
Eq
...
of KMnO4 in neutral medium= 158/ 3= 52
...
= 7-6 =1
Eq
...
of KMnO4 in alkaline medium= 158/1 =158

Concentration of solution
The amount of substance (solute) present in a definite
quantity of the solution is called concentration
...

𝑤𝑡
...
/vol
...
solution
...
/wt
...
solution
...
/vol
...
present in 100 ml
...

c) Normality (N): It is defined as the no
...

Normality(N) =

𝑛𝑜
...
𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒

i) Normal solution: The solution containing 1 g
...

Solute in 1 liter solution is called normal solution
...
eqvt
...

iii) Deci-normal solution: The solution containing 0
...
eqvt
...

iv) Centi-normal solution: The solution containing
0
...
eqvt
...

d) Molarity (M): It is defined as the no
...

Molarity (M) =

𝑛𝑜
...
𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒

Molar solution: The solution containing 1 mole solute
in one liter solution is molar solution
...
of moles of
solute dissolved per kg
...

𝑛𝑜
...


f) Formality (F): It is defined as the no
...
formula wt
...
It is used in
ionic compound
...
So, it is not in common practice
nowadays
...
of moles of
solute or solvent and total moles in solution
...
of moles of solute = n1
the no of moles of solvent = n2
total no
...
) solution contains 50 g
...
corresponding solution contains 5 g
...

By definition, g L-1 =50 & %( w/v) = 5
𝑔 𝑝𝑒𝑟 𝑙𝑖𝑡𝑒𝑟

So, %( w/v) =

%( w/w) =

10
𝑔
...
𝑔𝑟𝑎𝑣𝑖𝑡𝑦

Relation between normality and g L-1
no
...
eqvt
...
of solution in litre
wt
...


=

eqvt
...
of solute

=

x

vol
...
of solute in g
...
wt
...
wt
...


g L-1 = Normality x eqvt
...


Relation between molarity and g L-1
Molarity =
=

no
...
of solution in litre
wt
...


mol
...
of solute

=

x

1

vol
...
of solute in g
...
of solution in litre

x

1
mol
...
of solute

= gL-1 / mol
...

∴g L-1 = Molarity x mol
...


Relation between Normality &
Molarity
We have, g
...
L-1 Or, Normality x eqvt
...
=
Molarity x mol
...

Or, Normality =

𝑚𝑜𝑙
...


x Molarity

∴N =z x M, where z = valence factor

Summary
1
...
Wt
...
g L-1 = Molarity x mol
...

3
...
𝑝𝑒𝑟 𝑙𝑖𝑡𝑟𝑒

4
...
𝑝𝑒𝑟 𝑙𝑖𝑡𝑟𝑒
10 𝑋𝑠𝑝
...
Normality =
6
...
𝑋 10
𝑒𝑞
...


=

=

𝑒𝑞
...
𝑋 10
𝑚𝑜𝑙
...
N = z M (z =valence factor)
For preparation of solutions
8
...
w =

𝑁𝐸𝑉
1000

where, w= wt
...
; N = normality of solution; E = eqvt
...
of solute; V = vol
...


𝑀𝑀𝑜𝑙
...
of solute in g
...
wt
...
of solution in ml
...
It should be readily available in almost pure form
(99
...

II
...


III
...
It should have fairly high equivalent and molecular
weight to minimize the weighing error
...
2H2O, KMnO4, Na2CO3 etc
...
Standard solution are of two types
a
...
e
...
Solution of anhydrous
Na2CO3, (COOH)2
...

b
...
This type of solution is prepared by the
substances which do not possess the requisites of
primary standard substance
...
g
...


Normality factor:
Taking exact theoretical weight is a tedious job due to
various reasons
...
In such
cases the concentration of solution differs from proposed
concentration by a numerical factor which is called
normality factor
...
It is numerically
equal to the ratio of observed and theoretical weights
...
g
...


𝑤𝑒𝑖𝑔ℎ𝑡 𝑡𝑎𝑘𝑒𝑛
𝑤𝑒𝑖𝑔ℎ𝑡 𝑡𝑜 𝑏𝑒 𝑡𝑎𝑘𝑒𝑛
𝑁
10

Na2CO3 solution, we have to

take exactly;
(E= M/2 = 106/2 = 53)
w=

𝑁𝐸𝑉
1000

=

0
...
325 g
...
330 g
...
330 𝑔
...
325 𝑔
...
004

(f=1
...
standard
decinormal (N/10 or 0
...
wt
...
wt
...
1 𝑥 53 𝑥250
1000

=

= 1
...


Some AR (analytical reagent) grade anhydrous
Sodium carbonate is dried in an electric oven at
260 to 270℃ for an hour and cooled in a desiccator
...
325 g
...
volumetric flask containing little
water
...
The flask is well shaken to make the solution
homogeneous
...
325 g
...


If 1
...
is taken, the concentration of solution =
(f=0
...


𝑁
10

Preparation of 250 ml standard
decinormal solution of oxalic acid
Mol
...
of oxalic acid, (COOH)2
...
Wt
...
2H2O = 126/2 = 63
Now, w =

𝑁𝐸𝑉
1000

=

0
...
575 g
...
575 g
...
volumetric flask containing little
water
...
The flask is well shaken to make the solution
homogeneous
...
575 g
...


If 1
...
is taken, the concentration of solution =
(f=1
...


𝑁
10

Principles of volumetric analysis
The volumetric analysis is based on the law of
equivalents, according to which “all substances react in
the proportion of their equivalent weights i
...
every
substance react in equal no
...
eqvt
...
Acid and alkali of same normality always neutralize
each other in equal volumes
...
g
...
1N HCl solution
= 10 ml
...
1N Na2CO3 solution
2
...
e
...

E
...
1 ml
...
1N NaOH solution
= 10 ml
...


solution
𝑁

100

solution

= 5 ml
...
2N solution

3
...

V1 S1 =V2 S2 OR Va Sa =Vb Sb
Where, V1 = vol
...
of alkali
S2 = normality of alkali
The above equation can also be written as,
N1 V1 = N2 V2 OR Na Va = Nb Vb
Where, N1 = normality of acid
V1 = vol
...
of alkali
Though the principles are explained with the reference
of acid and alkali, they are applicable to all substances
reacted
...
of g
...
Of substance A = No
...
eqvt
...
𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙
...
of g
...
= normality x volume of solution in liter
...
in lit
...
in lit)B
Or, normalityA x vol
...
A = normality x vol
...
B
N1 V1 = N 2 V2
Or
V 1 S1 = V 2 S2
Q
...
Why?
The product of normality and volume in liter (NV) gives the no
...
eq
...
In any
reaction, no
...
eq
...
Thus, N1V1 = N2V2 is always
...
of moles of the substance
...
of moles of the reacting substances may or may not be equal
...
If no
...
of moles of reacting
substances are not equal,
...
eg
...


98 g
...
eq
...
eq
...


1 mol



Na2SO4 + 2H2O
142 g
...


In the above reaction, the no of g
...
of NaOH and H2SO4 are equal, so N1V1 = N2V2 but the no
...
But in the reaction,
NaOH + HCl
40 g
...
5 g
...
eq
...
eq
...


1 mol

The no
...
eq
...
of moles are equal
...


Apparatus used in titration:

Some formulae;
Some important terms
in volumetric analysis
a) Titration: It is the process of
determining the
concentration of unknown
solution with the help of
standard solution
...


For the mixture of solutions,
1
...

2
...

3
...
But, if the value is
negative, the resultant solution is
basic or alkaline
...

d) Alkalimetry: It is the process of determining the
concentration of unknown acid with the help of
standard alkali
...

f) Titrand: The solution of unknown concentration i
...

the solution whose concentration has to be
determined in titration is called titrand
...

h) End point: It is the point in titration at which the
indicator indicates the completion of reaction by its
pronounced color change
...


i) Equivalence point: It is the point at which equivalent
quantity of titrant is added to titrand or vice versa
...

j) Titration error: The difference between end point and
equivalence point is titration error
...

1
...
Redox titration
3
...
Complexometric titration
1
...
In this type of titrations we find
the concentration of acid or base i
...
it is either
acidimetry or alkalimetry
...
g
...
Redox titration:
The titration in which concentration of oxidizing agent
is determined with the help of standard reducing agent
or vice versa is called redox titration
...
g
...

The curve obtained by plotting the graph of pH vs
...
Let us consider we are adding alkali
from burette
...
There is a steep rise in pH of
the solution at the end point
...

The indicator used in acid-base titration are called pH
indicator as they work on the basis of pH
...
The pH range at end point, pH range of

indicators and color of indicators in different medium are
as follows:
Acid

Base

pH jump
at end
point

Indicator

Working
range

Color in
Acid

Color in
Base

strong

strong

3 - 11

MeOH
Hph

3
...
4
8
...
0

Red
colorless

Yellow
pink

strong

weak

3-8

MeOH

3
...
4

red

yellow

weak

strong

6 - 11

Hph

8
...
0

colorless

pink

weak

weak

6-8

No
proper
indicator

………
...
1-4
...
2-10) lies within this pH jump
...

Example; HCl + NaOH  NaCl + H2O
b) Titration of strong acid and weak base:
Titration curve  ABGH
pH jump at the end point  3-8
The working range of only methyl orange (3
...
4) but
not phenolphthalein (8
...

Thus we can use only methyl range as indicator in this
type of titration
...
2-10) but
not methyl orange(3
...
4) lies within this pH jump
...

Example; CH3COOH + NaOHCH3COONa + H2O

d) Titration of weak acid and weak base:
Titration curve EFGH
pH jump at the end point6-8
There is no sharp change in pH of the solution
...
Thus you are suggested not to titrate weak
acid and weak base due to lack of proper indicator in
this level
...
eqvt
...
06 g
...
of 0
...
eqvt
...
eqvt
...

1
...
in litre
=
=
= 0
...
wt
...
1 x
given wt
...
of moles =
mol
...

= 0
...
06
=
= 0
...

No
...
in litre
= 0
...
01

How many kg of wet NaOH containing 12% water are
required to make 120 L of a 0
...

Solution,
We have, wt
...
25 x 40 x 120 x 1000
1000

= 1200 g = 1
...
of wet NaOH be y gm
...
2 kg
...
363 kg
...
363 Kg
...
of

𝑵
𝟐𝟎

(f=2
...
12)

HNO3 are mixed together, calculate the normality of
mixture
...
06)

N2 =

𝑁
10

(f=1
...
05 x 2
...
1 x 1
...
06 + 3
...
42/50 =0
...
1084N

Find the strength of mixture prepared by mixing 50 ml
...


Solution,

𝑴
𝟐𝟎

H2SO4 and 1 g
...


We have
...
eqvt
...
eqvt
...
eqvt
...
eqvt of NaOH
Or, (N V)m = (N V)HCl + (N V)H2SO4 – (
Or, N x

150
1000

=

1
10

x

50
1000

+

1
10

x

given wt
...
wt
...
15N = 0
...
01 – 0
...
01/0
...
066 (- sign indicates soln is
alkaline
...
066
...
9 atm
...
of a solution of H2SO4
...

Solution
Here, mass of ammonia is calculated from ideal gas
equation
...
9 x 2 = n x 0
...
07236
Mass of ammonia = n x mol
...

= 0
...
23 g
...
eq
...
eq
...


Or, (
Or,

eq
...

1
...
23x1000
17x134

= 0
...
54/2 = 0
...
of a mixture of Mg and Al were dissolved in 250 cc
1N H2SO4 solution
...
5 cc of 0
...
Calculate % composition of Al and Mg
...

Let, the mass of Al be y g
...

Now, the normality of residual acid =

0
...
5
10

=

0
...
eq
...
eq
...

eq
...
of Al
eq
...

2−y
12

+

wt of Mg
eq
...


+ 0
...
25 – 0
...
2014 x 36

+ (N V) residual acid
250

1000

Or, y = 7
...
2504 g
...
2504
2

x 100

= 62
...
52) % = 37
...
1 M H2SO4 solution is titrated against 0
...
Calculate the concentration of acid & salt
formed when the acid is one fourth neutralized
...

Here,
Acid taken = 85 cc of 0
...
1 M = 21
...
1 M
4

Acid left unreacted = 85 – 21
...
75 cc of 0
...
of KOH required for reacting acid =

21
...
2
0
...
5 cc
(Normality of H2SO4 = 2 x Molarity)
Total vol
...
5 = 127
...
75 x 0
...
5

=0
...

1 mol
...
1 x 21
...
1 x 21
...
of mol = Molarity x vol
...
of mol
...
in l
no
...
in cc
0
...
25
127
...
017 M

4 gram of divalent metal was dissolved in 100 ml of 2M
(f=1
...
The excess of
acid required 30 CC of 1N NaOH for complete
neutralization
...

Solution,
Here, for H2SO4,
Normality (N) = M x basicity = 2 x 2 = 4N
Exact normality = N x f = 4 x 1
...
04N
Now, g
...
of H2SO4 = g
...
of metal + g
...
of NaOH
Or, (N V)H2SO4 = (
Or, 4
...


𝑒𝑞
...
69
Finally, atomic wt
...
69 x 2
=21
...
8 gram of divalent metal was dissolve in 100 ml of
1
...
Then 50 ml of the diluted solution required 54
...
22 N NaOH for neutralization
...

Solution,
Here,
g
...
of initial HCl = g
...
of metal + g
...
of diluted acid
Or, (N V)initial HCl = (

𝑤𝑡
...


Or, 1
...
8
𝐸

)metal + (N V)dil HCl

+Nx

200
1000

For strength of dil HCl,
Dil
...
22N
V1 = 50 mL
V2 = 54
...
24N
Now,

0
...
8
𝐸

+ 0
...
= E x V
= 10 x 2
=20 amu

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