Volumetric Analysis

Calculation of equivalent wt

Master Formula
Equivalent wt = Mw/ Charge

Equivalent wt of acid = Mw/ Number of replaceable hydrogen(basicity)

Equivalent wt of Base= Mw/ Number of replace hydroxide(Acidity)
Equivalent wt of salt = Mw/ Total number of cation or anion
Equivalent wt of oxidizing/reducing agent= Mw/ Change in Oxidation number
Equivalent wt of ion= Ionic wt/Charge of ion

Sustance Molecular wt Equivalent wt

Hydrochloric Acid 36.5 36.5/1=36.5
(HCl)
Nitric Acid 63 =63/1=63
(HNO3)
Sulfuric Acid 98 98/2=49
(H2SO4)
Acetic Acid 12+3x1+12+16x2+1=60 60/1=60
( CH3COOH)
Formic acid 1+12+16x2+1=46 46
( HCOOH)
Oxalic acid anhydrous (12+32+1)x2= 90/2=45
[ (COOH)2 90
Oxalic acid hydrated 90+36 = 126 126/2=63
[ (COOH)2 .2H2O]
Phosphoric Acid 3
(H3PO4)
Sodium Hydroxide 1
(NaOH)
Potassium Hydroxide 1
(KOH)
Ammonium hydroxide
(NH4OH)
Calcium hydroxide
[ Ca(OH)2]
Sodium Chloride
(NaCl)
Potassium Chloride
(KCl)
Sodium Carbonate
( Na2CO3)
Sodium Bicarbonate
( NaHCO3)
Silver nitrate
(AgNO3)
Potassium Nitrate
(KNO3)
Sodium Sulfate
(Na2SO4)
Potassium
Permanganate
( KMnO4)
 Potassium Dichromate
( K2Cr2O7)
Mohr’s Salt
FeSO4.(NH4)2SO4.6H2O
Na+/Na 23 23/1
++
Ca 40 40/2
--
SO4
CO3- - 60 60/2
--
NO3
HCO3- -
H2O2

Expression of Concentration/Strength

Primary and secondary standard substance

Primary Standard substance
Standard solution can be prepared by dissolving a known weight of a pure substance directly into
water with a fixed volume is called primary standard substance
Primary Standard solution:-
The solution prepared by dissolving a weighted amount of primary standard substance in to a definite
volume of solvent is called primary standard solution.

Criteria/Requisites/pre-requisites/properties/characteristics/ of primary standard substance

1. They must be easy obtain, process, purify and should be present in pure state.
2. They should not be hygroscopic ( absorb moisture), efflorescent( moisture release), or deliquescent (
absorb moisture and turn into fused or solution state).
3. They should not be reactive for environmental component [ CO 2, SO2, NO2, Water moisture,
O2( should not get oxidized)]

Examples
1.Sodium carbonate anhydrous ( Na2CO3)
2. Oxalic Acid [ (COOH)2.2H2O]
3.Potassium chloride ( Kcl)
4. Mohr’s salt[ FeSO4.(NH4)2SO4.6H2O]
5. Potassium dichromate ( K2Cr2O7)
6. Potassium Iodate ( KIO3), Potassium Bromate ( KBrO3)
7. Sodium chloride ( NaCl)- Analytical
Secondary standard substances
They are defined as substances whose composition may change either in solid phase or in its solution
phase on long standing and cannot be used to prepare primary standard solutions.
Secondary standard solutions:-
The solution whose strength is to be determined after its preparation with the help of primary
standard solution is called secondary standard solution.
Examples:-
1. Copper sulfate ( CuSO4.5H2O)
2. Sodium hydroxide ( NaOH)
3. Acids HCl ( degraded by light), H 2SO4 ( dehydrating agent, capture moisture very fast),
HNO3( unstable normally and produce yellow colored NO2 gas )
4. Potassium Permagnet ( KMnO4)
5. Ferrous sulfate ( FeSO4)

Preparation of Reagent
If reagent is solid
Mw= 105.99g
Ew= 105.99/2= 52.995g
Q. Prepare 0.25N 250mL Na2CO3 solution.
Calculation:
V= 250mL
N=0.25N
Ew= 52.995
Weight required (wt)= V EN/1000 = [250 x 52.995 x 0.25]/1000= 3.31g
Dissolve 3.31g of sodium carbonate in small volume of water and make final volume upto 250mL in
volumetric flask.

If the reagent is liquid

Q. Prepare 0.25N 250mL H2SO4 from stock solution ( concentrated).
Mw= 98.08
Ew = 98.08/2= 49.04
Concentration= 97% w/v => 97g/100mL
wt = 97g V=100mL
Ew=49.04
N = [wt/Ew/V] x 1000
= [ 97/49.04/100]x1000 = 19.77N
Stock solution Dilution solution
C1 = 19.77N C2= 0.25N
V1= ? V2 = 250mL
Dilution formula
C1 x V1 = C2 x V2
or, 19.77 x V1 = 0.25 x 250
or, V1 =3.16mL
Take 3.16mL Conc. Sulfuric acid in pipette and add into volumetric flask already containing excess
water and then dilute upto required volume 250mL and label it.

Normality Equation ( Law of equivalence )

When acid ( acidic salt) and base ( basic salt) completely react/ neutralize/ decompose/ consume/
compell/ digest
No of equivalents of acid = No of equivalents of base.=> Master formula

1. When acid and base both are liquid

No of equivalents of acid = No of equivalents of base.
Na x Va ( L) = Nb x Vb(L)
Na x Va/1000 = Nb x Vb/1000
2. When acid is solid and base is liquid
No of equivalents of acid = No of equivalents of base.
wt/Ew = Nb x Vb/1000
3. When acid is liquid and base is solid
No of equivalents of acid = No of equivalents of base
Na x Va/1000 = wt/Ew

Normality Factor (f)

1. Definition:
It is defined as the ratio of mass taken to the mass to be taken
2. Mathematical express

mass taken
Normality factor (f) =
mass tobe taken
3. Significance:-
Normality factor is used to calculate exact normality of a solution by
Exact normality = Given Normality x Normality factor
=Nxf

4. Example:-
We need to prepare 0.1N 250mL NaOH, then we have to dissolve 1g NaOH in 250mL, but
during weighing, the exact 1g may not be taken .
Suppose, during weighing, we have taken 1.1 g, then
f = 1.1g/1g= 1.1
Actual normality (N) = given normality x f= 0.1N x 1.1 = 0.11N
suppose during weighing, we have taken 0.99g
f= 0.99g/1g= 0.99
Actual normality ( N) = 0.1 x 0.99 = 0.099N
Q.Define molarity of solution. Calculate molality of 1L of 93% H2SO4 ( w/V)( density of the solution is
1.84g/mL)
Ans
H2SO4
Mw= 98
Ew=49
Solute Solution
93% w/V
= 93 g in 100mL volume of solution= 1000mL
= 930g in 1000mL (1L) sp gravity = 1.84g/mL
wt = Vol x density ( sp gravity)
= 1000mL x 1.84g/mL
= 1840g
Then, wt of solvent = wt of solution – wt of solute
= 1840g – 930 g= 910g

No of mole of solute = wt (g)/ Mw = 930g/98g= 9.48mole

no of mole of solute 9.48

molality (m) = x 1000 = x 1000 = 10.42m
wt of solvent (g) 910
Q. 10g of NaOH was added to 200cc of 0.5N ( f= 1.5) H 2SO4, the volume was diluted to 2L. Predict
whether the diluted solution is acidic, basic or neutral? Also calculate the resulting molarity of diluted
solution.
Acid Base
H2SO4 NaOH
Mw= 98 Mw=40
Ew=98/2=49 Ew= 40/1=40
Va= 200cc= 200mL wt = 10g
N= 0.5N Base is “solid”
f= 1.5

Actual normality(Na) = N x f
= 1.5 x 0.5= 0.75N
Acid is “solution”
No of equivalent of acid No of equivalent of base

= Na x Va/1000 = wt/Ew
= 0.75N x 200/1000 = 10g/40g
= 0.15 equivalents = 0.25 equivalents
Since, [ 0.15+0.1]
the no of equivalent of base > No of equivalents of acid
Hence, solution must be basic.

The no equivalent of base left unreacted= 0.25eq-0.15eq= 0.1eq

Final Volume (V)= 2L=2000mL
So
Normality of resulting solution (N) = [(no of equivalent of base left over/ Vol of solution(mL)] x 1000
= (0.1/2000mL)x1000= 0.05N

Normality ( N) = Molarity ( M) x Valency/acidity/basicity/charge

0.05N = M x 1
M= 0.05M
The molarity of resulting solution will be 0.05M.

Q. Define normality factor. 0.18g of divalent metal was completely dissolved in 250cc of acid solution
containing 4.9g H2SO4 per litre. 50Cc of residual acid solution required 20cc of N/10 alkali for
complete neutralization. Calculate the atomic wt of metal.
Ans
First, we consider reaction between metal and sulfuric acid
Acid Base ( metal)
H2SO4 wt= 0.18g
Mw= 98g Mw= Mw
Ew=98/2=49g Ew= Mw/2 ( divalent)
4.9g/L= 4.9g/1000mL base is solid
N= (wt/Ew/V)x1000
= (4.9g/49g/1000)x1000
N= 0.1N
acid is solution
Let suppose Base react with “V” volume acid
Consider reaction between acid and base
No of equivalent of acid= no of equivalent of base
Na x Va/1000 = wt/Ew
0.1N x V/1000 = 0.18/Mw/2
V= ( 0.18 x 2 x 1000/0.1Mw)= 3600/Mw

Volume left(X) = total volume – V = (250mL – 3600/Mw)mL

of 0.1N H2SO4

the amount of acid in Left volume= Na x Volume

= 0.1 x (250-3600/Mw)/1000
= (250-3600/Mw)/10000

This left solution should have normality of acid taken for second reaction i.e. 0.04N.
N = No of equivalent/Vol x 1000
0.04 = [[(250-3600/Mw)/1000] / (250-3600/Mw) ] x 1000

Considering reaction between Left acid and base

Acid Base
Na=? Nb=(N/10)=0.1N
Va=50mL Vb=20mL

Consider complete neutralization

Na x Va = Nb x Vb
Na x 50 = 0.1 x 20
Na = ( 0.1x20)/50= 0.04N

No of equivalent of acid that has been taken from first solution for reaction
= Na x Va= 0.04N x 50/1000= 0.002 equivalent
Q. 0.315g of a dibasic acid requires 50mL of decinormal NaOH solution for complete neutralization.
Find the molecular mass of the acid.
Ans
Acid Base
Mw=Mw NaOH, Mw=40,
Ew=40
Ew= Mw/2 ( dibasic) Nb= 0.1N
wt = 0.315g Vb = 50mL
acid is solid base is solution
If they react completely during neutralization , then
no of equivalent of acid = no of equivalent of base
wt/Ew = Nb x Vb/1000
0.315/Mw/2 = 0.1 x 50/1000
0.63/Mw= 0.005
Mw= 0.63/0.005 = 126

Q. Calculate the volume of 1M NaOH require to neutralize 200cc of 2M HCl. What mass of NaCl are
produced from the neutralization reaction.
Acid Base
HCl NaOH
Mw=36.5 Mw=40
Ew=36.5 Ew=40
Na=Ma=2M=2N ( monobasic) Nb= Mb= 1M=1N
Va= 200mL Vb= 400mL
No eq= Na x Va/1000 No of eq of base = Nb x Vb/1000
= 2 x 200/1000 = 1 x 400/1000=0.4eqs
=0.4 eq

If they react completely,

Normality equation
Na x Va = Nb x Vb
2 x 200= 1 x Vb
Vb= 400mL
NaOH + HCl======> NaCl + H2O
40g 36.5g 58.5g 18g

Consider base ( NaOH) for calculation.

wt of NaOH for reaction = VEN/1000
V= 400mL, E=40, N=1N
wt = 400 x 40 x 1x /1000 = 16g NaOH
Then
40g NaOH will produce 58.5 g NaCl.
16g NaOH “ “ = (58.5/40)x16= 23.4g NaCl will be produced

Consider Acid ( HCl)

wt = VEN/1000 = 200 x 36.5 x 2/1000= 14.6g HCl

36.5g HCl will produce 58.5g NaCl.

14.6g HCl “ “ = (58.5/36.5)x14.6=23.4g NaCl will be produced.
Q. Commercial sample of sulfuric acid has specific gravity 1.8. 10mL of this acid is diluted upto 1L
with water. 10ML of diluted acid require 30mL of decinormal NaOH for complete neutralization.
Calculate the percentage purity of H2SO4 in commercial sample.
Ans
Considering neutralization between acid and bases
Acid Base
H2SO4, Mw=98, Ew=49 NaOH, Mw=40, Ew=40
Na= ? Nb= 0.1N
Va= 10mL Vb= 30mL
Consider complete neutralization,
Na x Va = Nb x Vb
or, Na x 10mL/1000 = 0.1 x 30mL/1000
or, Na= (0.1x30)/10= 0.3N

The acid after dilution should have normality of 0.3N in 1L.

Then, wt in 1000ml diluted sulfuric acid solution,
wt = VEN/1000= (1000 x 49 x 0.3) /1000= 14.7g in 1000mL

1000mL contains 14.7g Acid

According to question
specific gravity = 1.8g/mL
10 mL contains =1.8g/mL x 10= 18g/10mL==> diluted=> 18g/1000mL
Experimental wt= 14.7g
Theoretical wt = 18g
% purity = (Experimental wt/Theoretical wt)x1000= (14.7/18)x100%
= 81.67% pure H2SO4

Q.
Before dilution
After dilution
H2SO4
Purity = 97% purity =
15%
Density = 1.84g/cc ( with impurity) density = 1.1g/cc
Volume ( V1)= ?? Volume(V2) =
100mL
Actual concentration ( C1) Actual concentration ( C2)
= 1.84x 97% g / cc = 1.1 x 15% g /mL
= 1.78g/mL = 0.165g/mL

then, according to law of dilution,

C1 x V1 = C2 x V2
or, V1= (C2 x V2/C1 )= 0.165 x 100/ 1.78 = 9.26 mL
volume of water that has to be added= (100mL – 9.26mL) = 90.73mL
Hence, 9.26mL should be taken from 97% sulfuric acid ( sp gravity=1.84) and 90.73mL of water should
be added to prepare 15% ( sp gravity = 1.1) 100mL solution.
Q. 12g of commercial zinc is made to react with excess dilute H2SO4. The total volume of Hydrogen gas
liberated was found to be 4.2L, at 570mmHg and 279K. Determine percentage purity of zinc. ( At
mass of zinc= 65)
Ans

Zn + H2SO4====> ZnSO4 + H2
65g 98g 161g 2g

given,
Volume ( V) = 4.2L
Pressure ( P) = 570mmHg= (570/760)atm = 0.75atm
Temperature ( T) = 279K R= 0.0821L.atm.K-1mole-1
Using Ideal gas equation to calculate the amount of hydrogen liberated,
PV = nRT
n = PV/RT= [ 0.75 x 4.2/(279 x 0.0821)] = 0.1375 mole.
So, wt of hydrogen = no of mole x Mw = 0.1375 x 2 = 0.275g of H2
From stoichiometric equation ,
2g H2 can be produced by 65g Zinc.
0.275g can be “ = (65/2)x0.275 = 8.93g Zinc is required.

Theoretical value = 12g

Experimental value = 8.93g
% Purity = [Experimental Value / theoretical value ] x 100
= (8.93/12)x100 = 74.47%
Hence, the percentage purity of zinc was found to be 74.47% .

Q. 1g of NaOH is added to 2L of xM H 2SO4 solution , so that the pH of the resulting solution is 7

( there is complete neutralization), find the value of x.

Ans
Base
Acid
NaOH, Mw=40, Ew=40 H2SO4, Mw=98, Ew=49
wt= 1g Molarity = xM
base is solid Normality ( NA)= Molarity x acidity or basicity
= “ x” x 2 = 2x
Volume ( VA) = 2L
acid is solution
If the pH of the solution is 7, means to say the neutralization is complete, and
No. of equivalents of base = No of equivalents of acid
( wt/Ew) = NA x VA( L)
or, (1/40) = 2x . 2L
or x = 0.00625M sulfuric acid is there.

Q. 7.35 g of a dibasic acid was dissolved in water and diluted to 250mL . 25 mL of this solution was
neutralized by 15mL of 1N NaOH solution. What is the equivalent wt and molecular wt of that acid.
Ans.
Acid Base
Mw=Mw NaOH,
Mw=40, Ew=40
Ew= Mw/2( dibasic ) VB= 15mL, NB= 1N
wt = 7.35g
Volume ( V) = 250mL
Normality ( NA) = ( wt/Ew/V) x 1000
= ( 7.35/ Mw/2/250) x1000
= (58.8/Mw)N
Consider reaction between 25mL ( 58.8/Mw)N acid with 15mL 1N NaOH
NA= 58.8/Mw VB=15mL NB= 1N
VA=25mL
If the reaction is complete neutralization, then
NA VA= NB V B
or, 58.8/Mw x 25 = 1 x 15
or, Mw = 98
Again, Ew=Mw/2=98/2=49

Titration:-
1. Definition:-
The method of quantitative chemical analysis which involves the determination of volume of
standard( Burette) solution required to react with measured volume of solution of unknown
strength( flask) is called titration.
Titrant:-
The solution of known strength ( standard solution ) is called TITRANT(BURETTE).

Titrand:-
The solution whose strength in unknown is called TITRAND ( FLASK)

INDICATOR:-
A. Definition:-
To indicate the completion of the reaction between titrant and titrand
need an auxiliary( extra) chemical substance is added which is called indicator.

B. Function:-
a. To indicate whether a substance is acidic or basic.
b. The completion of titration is detected by some
physical change produce by one of the solution itself
or by indicator.

C. Mechanism ( working principle) of indicator

Most indicators show their color changes either in acidic medium or in alkaline.
When they get complementary medium, they ionizes and produces color.
Example:-
Phenolpthalein ( Acid)====> Base==> Phenolpthalein ionizes==> produces pink color in base

Methyl Orange ( Base) ====> Acid==> Methyl orange ionizes ==> produces red color in acid

End point:-
The point at which indicator show the completion of reaction between titrant and titrand by sharp
visual change or the appearance of turbidity is called end point of the titration.

Equivalence point/ theoretical end point/ Stoichiometric end point

During titration, a point will appear at which equivalent quantity of titrand is neutralized by titrant
is called equivalent point.
In the ideal titration, end point coincides with equivalence point but in practice the value of end point
is slightly different than the value of stoichiometric end point.
Titration error:
The difference between end point and equivalence point of titration is called titration error.
Q. How to minimize titration error?
Titration error can be minimized by following ways
1. Keen observation and operation.
2.Indicator should be chosen in such a manner that it would avoid or minimize the titration error.

Types of Titration
1. Acid-Base titration ( Neutralization titration)
2. Redox-Titration ( Oxidation-Reduction titration)
3. Complexometric Titration ( Complex Formation titration)
4. Precipitation titration
5. Potentiometric titration.

1. Acid-Base titration ( Neutralization titration)

A. Definition: The titration which involve the reaction of free acid
and free base or salt ( weak acids and bases) is
called acid-base titration.
B. Example:
HCl + NaOH======> NaCl + H2O
HCl + Na2CO3======> NaCl + CO2+ H2O
C. Indicators:-
End point is detected by the use of acid-base
indicator or pH indicator.

2. Redox-Titration ( Oxidation-Reduction titration)

A. Definition: The titration involving oxidizing agents and reducing agent is called redox titration
B. Example
a. Oxidizing agent=> KMnO4,K2Cr2O7, KIO3,KBrO3, H2SO4,I2 etc
b. Reducing agent=> Mohr’s salt [ FeSO4. (NH4)2SO4.6H2O],
Sodium thiosulfate ( Na2S2O3),
 Oxalic acid [ (COOH)2]
[Permanganometric titration]
KMnO4 + (COOH)2+ H2SO4 ==> K2SO4+ MnSO4+ CO2+ H2O
C. Indicator:
No external or internal indicator is required because disappearance of pink color indicate the
completion of reaction, so KMnO4 acts as self indicator.

3. Complexometric Titration ( Complex Formation titration)

Definition: The titration which involve the formation of complex compound is called complexometric
titration.

Example:-
This method is exclusively applied for calcium and Magnesium by ( Titrand) reacting with EDTA
( Ethylene Diamine tetra acetic acid, (Titrant)
Ca++ + disodium EDTA==> Calcium EDTA complex + 2Na+
Indicator:-
Eriochrome Black-T
=========X=================X==========X============
Water+ EBT=> Blue
Ca++ + EBT==> Pink
Mg++ + EBT===> Red
============X===============X======================

Application:-
This type of titration can be applied for estimation of hardness of water, quantification of calcium in
medicines, or in biological fluid like blood , plasma, serum etc

Precipitation Titration
Definition :-
It is the type of titration which involve the formation of precipitate by the combination of ions of
titrant and titrand.
Example :-

NaCl + AgNO3===========> AgCl ( ↓ )+ NaNO3

White precipitate
This type of titration is governed by solubility product principle.

Indicator
In above titration , Potassium Chromate ( K2CrO4) is used as indicator.
Ag+ + CrO4- - ======> Ag2CrO4 ( Red)
Selection of Indicator on Acid – Base titration:-
Phenolpthalein(acid) => 8-10
Methyl orange(Base) => 3.2-4.4

Acid Base PH Indicator

(equivalent point)
Strong Strong 3-11 Phenolpthalein
HCl, H2SO4, HNO3 NaOH, KOH Methyl orange
Strong Weak 3- 8 Methyl orange
NH4OH,
Ca(OH)2
Weak Strong 6-11 Phenolpthalein
H2CO3, HCOOH
CH3COOH
Weak Weak 6-8 No indicator

Titration curve
Definition :-
A curve obtained by plotting a graph of pH of solution versus volume of alkali added is called titration
curve

Nature of Titration Curve

AB==>Smooth increase in pH value of resulting solution, due to addition of base and reaction with
acid.
BC==> Near the end point or equivalence point, there is sudden rise in pH value represented by BC.
CD===> This segment represent that the alkali will be adding in excess, and smooth increase.
The sudden rise or steep rise in pH value during titration indicates the neutralization in acid- base
reaction.
Q. How many mL of 0.1M HCl are required to react completely with 1g mixture of Na2CO3 and
NaHCO3 containing equimolar amount of the two.
Ans
1g mixture is containing Na2CO3 and NaHCO3, considering them separately for preparation of
solution as well as for the reaction with 0.1M HCl.
Total wt of mixture = 1g
Let, “ V” volume is used to prepare equimolar solution of these two solids.
Na2CO3 NaHCO3
Mw=106, Ew= 53 Mw=84, Ew=84/1=84g
Volume (V) Volume ( V)
Amount = “x”g Amount = (1 -x)
Molarity ( M) = Molarity (M)=
=[(wt/Mw)/Volume] x 1000 = [(wt/Mw)/V]x1000
M = [(x/106)/V] x 1000 = [(1-x)/84/V]x1000
If they produce equimolar solution in equal volume, then their molarity should be equal i.e.,
[(x/106)/V] x 1000 = [(1-x)/84/V]x1000
or, x/106 = 1-x/84
or, 84x= 106-106x
or, 84x+106x = 106
or, 190x = 106
or, x = 0.557g ( wt of Na2CO3)
Then wt of NaHCO3= 1-0.557g= 0.442g

Consider titration of both solid separately with 0.1M HCl.

Reaction between Na2CO3 and HCl
Acid Base
HCl Na2CO3 , Mw=106, Ew = 53
NA = 0.1M=0.1N wt = 0.557g
(VA )1 = ?
acid is solution base is solid
Using normality equation
no of equivalent of acid = no of equivalent of base
or, NA x (VA)1/1000 = wt / Ew
or, 0.1 x (VA)1/1000 = 0.557/53
or, (VA)1 = 105.09mL

Considering Reaction between NaHCO3 and HCl

Acid Base
HCl NaHCO3
NA= 0.1N wt = 0.442g
(VA)2 = ? Ew = 84
consider complete neutralization
no of equivalents of acid = no of equivalents of base
NA x (VA)2 /1000= wt/Ew
or, 0.1 x (VA)2 /1000 = 0.442/ 84
or, (VA)2 = 52.61mL
So, total volume of HCl = (VA)1 + (VA)2 = 105.09+52.61= 157.9mL