Volumetric analysis part 1

Dr. Damodar Koirala | koirala2059@gmail.com
VOLUMETRIC ANALYSIS (PART 1)
-Dr. Damodar Koirala
Amar Singh Model Secondary School 1

CONTENT
 This file
 Introduction
 Review on equivalent weight
 Concentration of solution
 Unit conversions
Dr. Damodar Koirala | koirala2059@gmail.com
 Unit conversions
 Dilution
 Next file
 Primary and secondary standard
 Titration in volumetric analysis
 Indicator in acid base titration
 Numerical based on titration
2

INTRODUCTION
 Chemicals are present everywhere: foods, ink,
drinks, cosmetics, cleaning agent etc
 To detect the presence or absence of some
chemical is qualitative analysis
Dr. Damodar Koirala |
koirala2059@gmail.com
chemical is qualitative analysis
 To detect the amount of some chemical is
quantitative analysis
 Measurement of volume is volumetric analysis
 Measurement of weight is gravimetric analysis
3

INTRODUCTION
Chemical Analysis
Qualitative Quantitative
Gravimetric Titrimetric
Yes / No
1 / 0
On / Off
+ve / -ve
Pass / Fail
Present / Absent
Exact amount
Mass is measured Volume is measured
4

EQUIVALENT WEIGHT
 The number of an element that combines with or
displaces 1.008 parts by weight of hydrogen, or 8
parts by weight of oxygen or 35.5 parts by weight
of chlorine
Molecular weight of acid
Eq wt of acid =
Molecular weight of acid
Basicity of acid
Eq wt of base =
Molecular weight of base
Acidity of base
Eq wt of salt =
Molecular weight of salt
Total charge on cation
5

EQUIVALENT WT OF
OXIDIZING/REDUCING AGENT
 Oxidizing agent: The reagent/reactant that contains an element
whose oxidation number decreases during chemical reaction
 Reducing agent: The reagent/reactant that contains an element
whose oxidation number increases during chemical reaction
whose oxidation number increases during chemical reaction
Equivalent wt of O.A=
Molecular weight
|Change in O.N per mole|
Equivalent wt of R.A=
Molecular weight
|Change in O.N per mole|
6
*Refer to Chemical Arithmetic for practice problems and for more info.

SOLUTION
 Solution is the homogeneous mixture of solute
and solvent
 Solute is the component present in smaller
amount. It gets dissolved in solvent
amount. It gets dissolved in solvent
 Solvent is the component present in larger
amount. It dissolves solute
 The final state of solution is same as the state
of solvent
7

SOLUTION TYPES
Solid-Liquid
Steel
Solid-Liquid Gas-Liquid
Solid-Solid Liquid-Liquid
Steel
8

CONCENTRATION
20L water
30g sugar
10ml water
50mg salt
5g Ca(OH) 2
500ml H2O
9
10 L water
20g sugar
Sweeter ?
0.05 L water
3g salt
Salty?
5g NaOH
800ml H2O
Basic ?

CONCENTRATION
 It is the ratio of the quantity of solute to the
quantity of solution (solvent)
 Quantity can be mass or volume
 Unit of mass: milligram, gram, kilogram, mole,
 Unit of mass: milligram, gram, kilogram, mole,
no. of gram equivalent
 Unit of volume: milliliter, liter, deciliter, cc
Concentration =
Quantity of solute .
Quantity of solution(solvent)
10
 It is the relative term i.e. not absolute

MAINTAINING THE FIXED VOLUME OF SOLUTION
Step 1: Weigh/measure required amount of solute
11
Step 2: Dissolve solute in small volume of solvent
Step 3: Add solvent to get desired volume of solution

UNITS OF CONCENTRATION
Percentage =
Quantity of solute
x 100 %
Quantity of solution
Gram per liter =
Gram of solute
Volume of solution in Liter
12
Gram per liter =
Molarity =
Mole of solute
Normality =
No. of gram equivalent

CONCENTRATION IN PERCENTAGE
Percentage =
Quantity of solute
x 100 %
Quantity of solution
 The ratio of weight of solute to the weight of solution time
100% is percentage (wt/wt)
13
100% is percentage (wt/wt)
 The ratio of weight of solute to the volume of solution time
100% is percentage (wt/vol). Put mass in gram and vol in ml
 The ratio of volume of solute to the volume of solution time
100% is percentage (vol/vol)
 Eg: 10%(w/v) KCl means 10 gram KCl is present in 100 ml
solution

NUMERICAL BASED IN %
i. A solution contains 10 gram of NaCl in 100 gram of
solution, Calculate %(w/w) of the NaCl
ii. Sugar solution has 2 gram sugar in 18 gram
solution. What is the %(w/w)
14
solution. What is the %(w/w)
iii. 3 gram of salt is dissolved in 1L of water. Calculate
the %(w/w).(Density (H2O) = 1 g/ml)
iv. 1000g of NaCl solution has 150 gm of NaCl. What is
the %(w/v). Assume the density of NaCl solution be
1.01 g/ml

CONCENTRATION IN g/L
gram/L =
Gram of solute
 The ratio of gram of solute to the volume of
solution in liter
15
 Or, it is the gram of solute present in 1 liter of
solution
 Unit is gram/liter or gram/L or gram.L-1
 Eg: 2.5 gm/L NaCl means, 2.5 gm NaCl in 1 L
of solution

NUMERICAL BASED IN g/L
 A solution contains 10 gram of NaCl in 100 gram of
solution. Calculate the concentration in g/L if the
density of solution is 1.01 g/ml
 Sugar solution has 2 gram sugar in 180 ml solution.
What is the g/L
16
What is the g/L
 300ml of solution contains 2 mole of HCl. Determine
the concentration in g/L of HCl in the solution
 A NaOH solution is prepared by adding 0.35mole
NaOH to the 100ml of water and then water is added
to make the final solution 800ml. Determine the
concentration in g/L of NaOH in the final solution

CONCENTRATION IN Molarity
Molarity (M) =
Mole of solute
 The ratio of mole of solute to the volume of
solution in liter
17
 Or, it is the number of moles of solute present
in 1 liter of solution
 Unit is mole/L or mole. L-1 or M
 Eg: 4.8 M H2SO4 means, 4.8mole of H2SO4 is
present in 1 L of solution

SIMPLIFYING THE MOLARITY FORMULA
We know,
Also, Mole of solute =
Mass in gram of solute
Gram Molecular Wt
Molarity =
Mole of solute
18
So, Molarity =
Mass in gram of solute x 1 .
Gram Molecular Wt Vol of solution (L)
Mass in gram of solute= Molarity x MWt x Vol(L)
W = M. Mwt. VL
W = M. Mwt. VmL
1000

NUMERICAL BASED IN Molarity
1. What is the molarity of the solution containing 10g
H2SO4 in 300 ml of solution ?
2. 200 ml of solution contains 1.2 mole of NaCl.
Calculated the concentration in molarity ?
19
Calculated the concentration in molarity ?
3. A student wants to prepare 400 ml of 0.4M HCl
solution. What mass of HCl is required ?
4. A student has 3 gm of NaOH. He desires to prepare
2.5M solution. What volume of solution can he
prepare if he completely use available NaOH ?

MOLARITY
 Molar solution : A solution containing 1 mole of solute
dissolved in 1 liter of solution i.e. 1 M solution
 Deci-molar solution: A solution containing 1/10 mole
 Deci-molar solution: A solution containing 1/10 mole
of solute dissolved in 1 liter of solution i.e. 0.1 M or
1/10 M solution
 Semi-molar solution: A solution containing 1/2 mole of
solute dissolved in 1 liter of solution i.e. 0.5 M or 1/2
M solution
20

CONCENTRATION IN Normality
 The ratio of number of gram equivalent of
solute to the volume of solution in liter
Normality (N) =
21
 Or, it is the number of gram equivalent of solute
present in 1 liter of solution
Normality (N) =
 Unit is no. gram eqv/L or no. gram eqv. L-1 or N.
 Eg: 4.3N CaCO3 means, 4.3 gram equivalent of CaCO3 is
present in 1 L of solution

SIMPLIFYING THE NORMALITY FORMULA
We know,
Also, No. of gram equivalent=
Gram Equivalent Wt
Normality (N) =
22
So, Normality =
Gram Equivalent Wt Vol of solution (L)
Hence, Mass in gram of solute= Normality x Ewt x Vol (L)
W= N.E.VL
W = N. E. VmL
1000

NUMERICAL BASED IN Normality
1. What is the normality of the solution containing 10g
H2SO4 in 300 ml of solution ?
2. 200 ml of solution contains 1.2 mole of NaCl.
Calculated the concentration in normality ?
23
Calculated the concentration in normality ?
3. A student wants to prepare 400 ml of 0.4N Na2CO3
solution. What mass of Na2CO3 is required ?
4. A student has 3 gm of Ca(OH)2. He desires to prepare
2.5N solution. What volume of solution can he
prepare if he completely use available Ca(OH)2 ?

NORMALITY
 Normal solution : A solution containing 1 gram
equivalent of solute dissolved in 1 liter of solution i.e.
1 N solution
Deci-Normal solution: A solution containing 1/10 gram
24
 Deci-Normal solution: A solution containing 1/10 gram
0.1 N or 1/10 N solution
 Semi-Normal solution: A solution containing 1/2 gram
0.5 N or 1/2 N solution

RELATIONSHIP BETWEEN g/L AND Molarity
We know,
Also, Mole of solute =
Gram Molecular Wt
Molarity =
Mole of solute
25
So, Molarity =
Gram Molecular Wt Vol of solution (L)
Hence, Molarity x Gram Molecular Wt=
Vol of solution (L)
M x Mwt = g/L

RELATIONSHIP BETWEEN g/L AND Normality
We know,
Also, No. of gram equivalent=
Gram Equivalent Wt
Normality (N) =
26
So, Normality =
Gram Equivalent Wt Vol of solution (L)
Hence, Normality x Gram Equivalent Wt=
Vol of solution (L)
N x Ewt = g/L

RELATIONSHIP BETWEEN Molarity AND Normality
We know,
Also,
Molarity x Gram Molecular Wt=
Vol of solution (L)
Normality x Gram Equivalent Wt=
Vol of solution (L)
27
So,
Hence, M x Mwt = N x Ewt
Vol of solution (L)
Molarity x Gram Molecular Wt = Normality x Gram Equivalent Wt

RELATIONSHIP BETWEEN Molarity AND Normality
For acid, M x Basicity = N
Basicity =
Mwt
Ewt
M x Mwt = N x Ewt
28
For base, M x Acidity = N
Acidity =
Mwt
Ewt
For salt,
M x (total +ve charge) = N
Total +ve charge =
Mwt
Ewt

RELATIONSHIP BETWEEN UNITS
M x Mwt = g/L
N x Ewt = g/L
29
N x Ewt = g/L
M x Mwt = N x Ewt
M x Mwt = g/L = N x Ewt

CONVERT TO g/L
1) 4.9M H2SO4
2) 1 N Ca(OH)2
3) 1.2 mg of NaOH in 10 mL solution
8 mole L-1 hydrochloric acid
30
4) 8 mole L-1 hydrochloric acid
5) 10 ml of solution contain 2 mole of CaSO4
6) 5 Molar sodium chloride
7) 10 L of 0.3N potassium carbonate
8) 0.05N sodium bicarbonate

CONVERT TO Molarity
1) 0.25N H2SO4
2) 4 g/L of Na2CO3
3) 20L of 1 N Ca(OH)2
31
5) 8 g.L-1 hydrochloric acid
7) 5 Normal sodium chloride
8) 10 L of 0.3N potassium carbonate
9) 0.05N sodium bicarbonate

CONVERT TO Normality
1) 2.5M H3PO4
2) 4 g/L of Na2CO3
3) 20L of 1 M Ca(OH)2
32
5) 8 g.L-1 hydrochloric acid
7) 5 Moral sodium chloride
8) 10 L of 0.3M potassium carbonate
9) 0.05M sodium bicarbonate

WHICH SOLUTION HAS HIGHER CONCENTRATION?
a) 0.3M NaCl or 5g/L NaCl
b) 80 gm/litre NaOH solution or 3 M NaOH solution
c) 2M Na2SO4 or 3.5N Na2SO4
d) 4g/L KOH or 0.1 M KOH
33
d) 4g/L KOH or 0.1 M KOH
e) 5.3 gm/Litre Na2CO3 or N/10 Na2CO3
f) 5N H2SO4 or 2.5M H2SO4
g) 10 M Na2CO3 or 18 N Na2CO3
h) 1 M NaOH or 35 g/L NaOH
i) 5 M HCl or 5 N HCl
j) 0.3 M H2SO4 or 0.3 M H2SO4

NORMALITY FACTOR (f)
 Measurement of exact required weight is very difficult
 In such case, solution of certain concentration but not
exact concentration is prepared
34
 A factor is introduced in order to show the nearness of
the weight between actual and theoretical
 The factor is called normality factor, denoted by ‘f’
 It is the numerical value and is close to 1

NORMALITY FACTOR (f)
 It is defined as the ratio of measured mass to the
theoretical mass of solute required to prepare the
given volume of solution of desired normality.
Normality factor (f) =
Measured mass of substance
35
Normality factor (f) =
Theoretical mass
 Eg: f=1.05, means the measured mass is 1.05 times the
theoretical mass
 The actual normality is also ‘f’ times the theoretical normality
 0.1 N(f= 0.92) means, the desired normality is 0.1N but the
actual normality is 0.92 * 0.1 = 0.092 N

NORMALITY FACTOR (f) NUMERICAL
 Student wants to prepare 250ml of deci-normal
Na2CO3. She measured 1.400g of Na2CO3. Calculate
the normality factor
 What mass of NaCl is required to prepare 200 ml of
N/10 (f=1.1)
36
What mass of NaCl is required to prepare 200 ml of
N/10 (f=1.1)
 Which solution has higher concentration: 3.6 g/L HCl
or 0.1N(f=0.95) HCl
 Determine the normality factor when you prepared
0.11N solution but desired to prepare deci-normal
solution.

OTHER CONCENTRATION UNITS
 Molarity is the number of moles of solute dissolved in
1 kg of solvent
Molality (m) =
No. of moles of solute
Mass of solvent in kg
37
Mole fraction of solute =
n1
n1 + n2
Mole fraction of solvent =
n2
n1 + n2
 Mole fraction is the ratio of two or more component
present in the solution in terms of moles

DILUTION OF SOLUTION
Dilution
8 N H SO
By adding solvent
38
Dilution refers to the process of adding additional
solvent to a solution to decrease its concentration.
8 N H2SO4
18 N H2SO4
By adding solvent

DILUTION OF SOLUTION
 Diluting solutions is a necessary process in the laboratory, as
stock solutions are often purchased and stored in very
concentrated forms
 This process keeps the amount of solute constant, but
increases the total amount of solution, thereby decreasing its
final concentration
39
 Dilution equation
Sf.Vf = Si.Vi
Where,
 Sf is final strength
 Vf is final volume of solution
 Si is initial strength
 Vi is initial volume of solution
 Also: Solvent added = Vf - Vi
 Note: the units of same quantities must be same

DILUTION NUMERICAL
 300ml of 2N NaCl solution is diluted to 500ml. What is the
concentration of final solution?
 What volume of added should be added to 100ml of 1M
HCl solution to make it exactly semi-molar ?
40
 What volume of water is required to make 100mL of N/10
(f=1.25) HCl solution exactly deci-normal?
 What volume of normal NaOH solution is required to
prepare 500ml of 0.4N NaOH solution?
 x cc of 5 N HCl was diluted to one liter of normal solution.
Calculated the value of x.

Volumetric analysis part 1

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Volumetric analysis part 1