Presentation 2 chemistry , volumetric analysis A to Z

Class -12 ‘Second Lecture’
Subject- Chemistry
Terms and terminologies
1Tej narayan chapagain 8/25/2020

Equivalent weight
E = A/V
The equivalent weight of an element or radical
is equal to its atomic weight or formula
weight divided by the valency it assumes in
compounds. The unit of equivalent weight is
the atomic mass unit ; the amount of a
substance in grams numerically equal to the
equivalent weight is called a gram equivalent.
Tej narayan chapagain 8/25/2020 2

Equivalent weight

Volumetric Analysis
Solution of accurately known strength is called
standard solution.
Substance for preparing primary standard
solution is called primary standard substance.
Substance whose standard solution can’t be
prepared by direct weighing is called……….

Volumetric Analysis
Process of determine strength of unknown
solution (titrate) with the help of standard
solution (titrant) is called titration.
The difference between end point and
equivalence point is called titration error.
The point at which the indicator changes color
is called the endpoint. So the addition of an
indicator to the analyte solution helps us to
visually spot the equivalence point in an acid-
base titration

Eg. Of titration
Q.1. 25ml of Na2CO3 soln requires 20ml of
N(f=1.1) H2SO4 for complete neutralization .
Find the concentration of Na2CO3.
SOLUTION
Vb=25ml Va=20ml
Nb=? Na=1.1/10
Now, VbNb=VaNa or, Nb=VaNa/Vb
or, Nb=0.088N

The equivalence point is the point in
a titration where the amount
of titrant added is enough to
completely neutralize the analyte solution.
The moles of titrant (standard solution)
equal the moles of the solution with
unknown concentration.
 This is also known as the stoichiometric
point because it is where the moles of acid
are equal to the amount needed to
neutralize the equivalent moles of base

Normality factor
Normality factor is defined as the ratio
of observed wt. of solute to the
theoretical wt. of the solute required to
prepare a solution of desired normality.

Titration curve
 Plot of pH of solution against volume of
base added. The middle steep rise indicates
the pH at which acid and base neutralize
each other.

The process of determining strength of acid
solution volumetrically by titrating with
standard alkali in presence of indicator is
called acidimetry.
The process of determining strength of
alkali against standard acid solution is
called alkalimetry.
Chemical substances (weak bases) which
indicates end point of reaction by changing
their colors is called an indicators.

Contd…
Generally, in acid-base titration, organic
complex chemical substances are taken as
indicator.
Indicator have different colour in ionized and
unionized form.
Different indicators have different pH-range
at which there is sharp change in colour.
Litmus paper, methyl orange,
phenolphthalenin etc. are the common
indicators.

Types of indicators
 Internal indicators- Internal indicator is an indicator
which is dissolved in the solution where main reaction is
taking place.Eg-i. acid – base indicator
(phenolphthalein, methyl orange, litmus etc.),
ii. self indicator(when one of the reactant in titration can it
self act as indicator-eg-KMnO4 in titration with oxalic
acid in acidic medium),
iii. absorption indicator etc.
 External indicator- those indicators which are not added
to solution , but used externally are termed as external
indicators. Eg- potassium- ferricyanide in titration of

Potassium dicromate and ferrous
salt.
Characteristics of good indicator (IMP)
The color should change over a short pH
range.
The color should change at the end point of the
reaction.
The suitability of indicators depend on the
nature of acids and bases involved in titration.
The color change should be sharp and stable.

Type of titration

Acid-Base titration
The titration in which the concentration of an
acid is determined by neutralizing it
completely with standard alkali solution in the
presence of indicator or vice versa are called
acid-base titrations. For example, acidimetry
and alkalimetry.
HCl(aq.) + NaOH(aq.) = NaCl (aq.) + H2O

Redox-titration
Those titrations in which the strength of an
oxidizing agent is determined by standard
solution of reducing agent or vice-versa are
called redox-titration.
 This titration involves change in oxidation
number between the reacting substances. For
example, titration of KMnO4 solution in acidic
medium against oxalic acid solution.

In this titration KMnO4 act as self –indicator
due to its pink colour.
 2KMn+7O4 + 3H2SO4 + 5(COOH)2 → K2SO4 + 2Mn+2SO4 + 8H2O + 10CO2↑
pink colourless
Note- HCl cannot be used for acidification
of KMnO4 as KMnO4 oxidises HCl to Cl2
and create problem in volumetric
analysis.

Common indicators and their pH ranges
Indicator in acidic med. In basic med. pH-range
M.Orange red yellow 3.1-4.4
M.Red red yellow 4.2-6.3
P.Phthalein colorless pink 8.2-10
L. Paper red blue 5.5-8.0
Where, M= methyl
P= phelnol
L= litmus
med = medium pH-range=working pH

Selection of indicator in acid-base titration
If alkali is taken in burette and acid in conical
flask , then the pH of the resulting solution is
gradually increases while adding alkali on
acid.
If pH of the resulting solution is plotted
against the volume of alkali added, then the
plots obtained are called pH-curves or titration
curves.

Contd…
The nature of the pH-curves or sharpness in
the pH curves help to select the suitable
indicator.
The nature of pH curves depends on the acids
and bases taken during titration.
The pH curves are shown in the fig.

i.Titration between S.acid-S.base

When strong acid is titrated against strong
base, then the nature of the pH-curve is shown
above. there is sudden change in pH from 3-
11.
Therefore, the indicators like phenolphthalein
having pH range 8.2-10, methyl orange having
pH range 3.1-4.4 and litmus paper with pH
range 5-8 can be choosen as suitable indicator.

ii. Titration betn s.acid- w.base

When strong acid is titrated against weak base,
then the curve formed is shown above. There
is sudden increase in pH from 3-8.
Therefore , methyl orange having pH range
3.1-4.4 is only suitable indicator.

iii.Titration betn w.acid-s.base

When weak acid is titrated against the strong
base then the curve obtained is shown above.
There is sudden increase in pH from 6-11.
therefore phenolphthalein having pH range
8.2-10 is only suitable indicator.

iv.Titration betn w.acid-w.base

When weak acid is titrated against weak base,
then the curve obtained is shown above.
There is no sharpness in the curve and none of
the indicator can be choosen.
Therefore, the titration between weak acid and
weak base is not carried out.

Q. Why is phenolphthalein suitable indicator
for weak acid Vs strong base titration?
Answer,
– experimentally, the pH curve of weak acid and
strong base is found to range from 6-11. so the
phenolphthalein having pH range 8.2-10 lie in
this interval can detect the end point of
reaction by sharp change in color. that’s why ,
phenolphthalein is suitable indicator for weak
acid and strong base titration.

Q. None indicators are suitable indicators for
weak acid and weak base titration, why?
Answer,
- Experimentally, the pH curve for weak acid
and weak base titration is not found to be
sharp. So, None indicators are suitable
indicators for weak acid and weak base
titration.

Q. Why are both methyl orange and
phenolphthalein suitable indicator for
strong Vs strong base titration?
Answer
- Experimentally, the pH curve of strong acid
and strong base is found to range from 3-11.
so the methyl orange having pH range 3.1-4.4
and phenolphthalein having pH range 8.2-10
lie in this interval can detect the end point of
reaction by sharp change in color . That’s why,
both methyl orange and phenolphthalein are
suitable for strong acid and weak base titration.

Q. Calculate the weight of oxalic acid required to
prepare 250ml of decinormal(N/10) solution.
Solution-
volume of soln to be prepared(V)=250ml
Normality of soln(N) = 1/10
Eq.wt of oxalic acid(E) = 63
Wt. of oxalic acid required(W) =?
since, W=VEN/1000
1.575g

Problem
Q.1. 20ml of sulphuricacid needs 0.106g of
Na2CO3. Calculate the normality of acid.
Soln- 0.1N
Q.2. 0.04 g of pure caustic soda was found to be
required to neutralize 10cc of dilute H2SO4.
calculate the concentration of acid solution in
terms of
a) normality b) g/l c) molarity

Q.4. calculate the volume of conc. Sulphuric
acid required to prepare N/10 solution of it
in 250 ml of water.
Q.5. x g of calcium carbonate required 20ml
of 2N hydrochloric acid. Calculate the value
of x.

Principle of volumetric analysis

Derivation of normality equation
• We know, normality=no. Of gram equivalent÷
litre of solution
• No. Of g equivalent = litre of solution× normality
at the equivalent point of titration
• No. Of g eq. Of acid=No. Of g eq. Of base
• Or, litre of acid solution×normality of acid=litre of
base solution×normality of base
• Or, 1000 ml of acid solution× normality=1000 ml
of base×normality of base
• Or, ml of acid solution×normality of acid=ml of
base solution×normality of base
• Or, V1N1=V2N2 This is normality equation.

Expression of concentration
Concentration measures quantity of solute
present in given volume of solution.
a) Gram per litre(g/l)
gram per litre of a solution is defined as the
weight of solute in gram present present in
one litre of a solution.
gram /litre= Wt. of solute in gm/volume of
solution in litre

Contd….

Molarity(M)
Molarity of a solution is defined as the number
of gram moles of solute present in one litre of
a solution . It is denoted by M. it decreses with
rise of temperature , as it depends upon
volume of solution.
Molarity =NO. of gram moles of solute /
volume of solution in litre

Contd….

Contd…..

When 180 g sugar present in 1L

H/W

Molar solution
If one litre of soln contains one gram moles of a
solute , then it is said to be molar solution.
Eg. 1M soln of H2SO4 means 98g of it present
in one litre soln.
Deci-molar soln(M/10) -The soln in which one
tenth of gram molecular weight of solute
present in one litre of a solution. Eg.M/10 of
NaoH means 4gm of it present in 1L of soln.

Semi-molar solution(M/2) – the soln in which
half gram moles of solute present in one litre
of soln.
Eg. M/2 NaOH soln means 20gram of it present
in 1L of soln.
Note;
No. of moles = wt. in gm
molecular wt.

when 1/2 gram equivalent of the solute is
dissolved in 1 litre of the solution it is called
as seminormal solution. it is denoted by N/2.
 Deci normal solution : When one-tenth gram
equivalent mass of a substance is present in
one litre of its solution then it is
called decinormal solution .
Note ;
No. of gm eq = wt. in gm
eq.wt

N/100 ( centi normal ) = When 1/100
gm.equivalents of solute are present in one
liter of solution,then solution is centinormal.
N/1000 ( milli normal ) = When 1/10oo
gm.equivalents of solute are present in one
liter of solution,then solution is milli
normal.

Normality Formula
 Normality = Number of gram equivalents × [volume of solution in litres]-1
 Number of gram equivalents = weight of solute × [Equivalent weight of solute]-1
 N = Weight of Solute (gram) × [Equivalent weight × Volume (L) ]-1
 N = Molarity × Molar mass × [Equivalent mass]-1
 N = Molarity × Basicity = Molarity × Acidity.


Calculation of Normality in Titration
 Titration is the process of gradual
addition of a solution of a known
concentration and volume with another
solution of unknown concentration until the
reaction approaches its neutralization. To
find the normality of the acid base titration
 N1 V1 = N2 V2

Where,
N1 = Normality of the Acidic solution
V1 = Volume of the Acidic solution
N2 = Normality of the basic solution
V3 = Volume of the basic solution

Normality Equations
The equation of normality that helps to estimate
the volume of a solution required to prepare a
solution of different normality is given by,
Initial Normality (N1) × Initial Volume (V1) =
Normality of the Final Solution (N2) × Final
Volume (V2)
Suppose four different solutions with the same
solute of normality and volume are mixed;
therefore, the resultant normality is given by;
NR = [NaVa + NbVb + NcVc + NdVd] ×
[Va+Vb+Vc+Vd]-1

If four solutions having different
solute of molarity, volume and
H+ ions (na, nb, nc, nd) are mixed
then the resultant normality is
given by;
NR = [naMaVa + nbMbVb +
ncMcVc + ndMdVd] ×
[Va+Vb+Vc+Vd]-1.

% (w/v)
The number of grams of solute present in
100ml of solution is called % (w/v).
% (w/v) = wt. of solute in gm X 100
vol. of soln in ml

% (w/w)
The number of grams of solute in 100 gm of
solution is called % (w/w)
% (w/w) = wt. of solute in gm x 100
wt. of solution in gm

Molality
Number of moles of solute present in
1000gm of solvent is called molality.
Molality = no. of moles of solute X 1000
wt. of solvent in gm
Note;
No. of moles = wt. in gm
molecular wt.

Formality
No of gm formula wt. of solute present in one
litre solution is called formality.
Formality = no. of gm formula wt
vol. of soln in litre
= no. of gm formula wt X 1000
vol. of soln in ml

Gram per litre and %(w/v)
1. gm/litre = %(w/v) X10
2. Gm/litre = normality X eq.wt
= molarity X molecular wt
= %(w/v) X10 X specific gravity
A.1. A solution of caustic soda contains 5g of
NaOH per litre. Find the normality of
solution.
Ans- 5/40=0.125N

• Q.1 . 25 ml of soln contain 0.106g of Na2CO3.
calculate normality and molarity.
• Q.2. Commercial sulphuric acid is 98% by wt
and its specific gravity is 1.84. calaulate the
molarity and normality of commercial
sulphuric acid.
• Q.3. A soln of NaOH is found to contain 20g of
NaOH in 250ml. Calculate the concentration of
solution in gm/l and percentage.
• Ans= 1) 0.08N ,0.04M 2) 18.4M,36.8N 3)80,8%

Differences Between Normality and Molarity
Normality
Also known as equivalent concentration.it is
defined as the number of gram equivalent per
litre of solution..It is used in measuring the
gram equivalent in relation to the total volume
of the solution. The units of normality are N or
eq L-1

Molarity
 Known as molar concentration.It is defined
as the number of moles per litre of solution.It
is used in measuring the ratio between the
number of moles in the total volume of the
solution.The unit of molarity is M or Moles L-1

Normality Problems and Examples
Question 1. In the following reaction calculate
and find the normality when it is 1.0 M H3PO4
H3AsO4 + 2NaOH → Na2HAsO4 + 2H2O
Question 2. Calculate the normality of 0.321 g
sodium carbonate when it is mixed in a 250 mL
solution.
 Question 3. What is the normality of the
following?
0.1381 M NaOH
0.0521 M H3PO4

Question 4. What will the concentration of
citric acid be if 25.00 ml of the citric acid
solution is titrated with 28.12 mL of 0.1718
N KOH?
Question 5. Find the normality of the base if
31.87 mL of the base is used in the
standardization of 0.4258 g of KHP (eq. wt =
204.23)?
Question 6. Calculate the normality of acid if
21.18 mL is used to titrate 0.1369 g
Na2CO3?

1
Solution:
If we look at the given reaction we can identify
that only two of the H+ ions of H3AsO4 react
with NaOH to form the product. Therefore, the
two ions are 2 equivalents. In order to find the
normality, we will apply the given formula.
N = Molarity (M) × number of equivalents
N = 1.0 × 2 (replacing the values)
Therefore, normality of the solution = 2.0.

2
Solution:
First, you have to know or write down the
formula for sodium carbonate. Once you do this
you can identify that there are two sodium ions
for each carbonate ion. Now solving the problem
will be easy.
N of 0.321 g sodium carbonate
N = Na2CO3 × (1 mol/105.99 g) × (2 eq/1 mol)
N = 0.1886 eq/0.2500 L
N = 0.0755 N

3/4
 Solution:
 3/a. N = 0.1381 mol/L × (1 eq/1mol) = 0.1381 eq/L = 0.1381
N
 3/b. N = 0.0521 mol/L × (3 eq/1mol) = 0.156 eq/L = 0.156 N
 4/Solution:
 Na × Va = Nb × Vb
 Na × (25.00 mL) = (0.1718N) (28.12 mL)
 Therefore, the concentration of citric acid = 0.1932 N.

5
Solution:
0.4258 g KHP × (1 eq/204.23g) × (1 eq
base/1eq acid):
= 2.085 × 10-3 eq base/0.03187 L = 0.6542 N
Normality of the base is = 0.6542 N.

6
Solution:
0.1369 g Na2CO3 × (1 mol/105.99 g) × (2 eq/1
mol) × (1 eq acid/1 eq base):
= 2.583 × 10-3 eq acid/0.02118 L = 0.1212 N
Normality of the acid = 0.1212 N.
⇒ Try this:

1. What volume of 6M HCl and 2M HCl
should be mixed to get one litre of 3M
HCl?
• 2. How much volume of 10M HCl should be
diluted with water to prepare 2.00L of 5M
HCl.

Solution(q.1)
Suppose the volume of 6M HCl required to obtain 1L
of 3M HCl = XL
Volume of 2M HCl required = (1-x)L
Applying the molarity equation
M1V1 + + M2V2 = = M3V3
6MHCl+ 2MHCl= 3MHCl
6x+2(1-x) = 3x1
6x+2-2x = 3
4x = 1
x = 0.25L
hence, volume of 6M HCl required = 0.25L
Volume of 2M HCl required = 0.75L

2
• Solution
• N1V1 = N2V2
• 10N HCl = 5N HCl
• 10xV1 = 5 x 2.00
• V1 = (5 x 2.00 )/ 10
• = 1.00L
•

Objective
• Question 1: Which of the following types of titration is not
a simple titration?
• a. Acid-base titrations
• b. Back titration
• c. Precipitation titrations
• d. Complexometric titrations
• Question 2: According to the law of equivalents,
• a.N2V1 = N1V2,
• b.N1V2 = N2V1,
• c.N1V1 = N
• d. V = N2V2,

Objective
• Question 3: Which of the following conditions is nor required for the back
titration to work ?
• a. Compounds ‘A’, ‘B’ and ‘C’ should be such that ‘A’ and ‘B’ react with
each other.
• b. ‘A’ and pure ‘C’ also react with each other but the impurity present in ‘C’
does not react with ‘A’.
• c. Product of ‘A’ and ‘C’ should not react with ‘B’.
• d. Product of ‘A’ and ‘B’ should be ‘C’.
• Question 4: The titration in which an oxidizing agent is made to react with
excess of solid KI and the oxidizing agent oxidizes I– to I2 is known as
• N2(g) + 3H2(g) → 2NH3(g) ?
• a. Iodometry
• b. Iodimetry
• c. Double titration
• d. Back titration

Answer
Q.1 b
Q.2 a
Q.3 d
Q.4 a

Problems
Q.1. A vinegar soln contains 8%by wt. of
acetic acid and has the specific gravity 1.02.
find the normality of soln.
Q.2. The molecular wt of a dibasic acid is 126.
calculate the normality and molarity of the
soln that is prepared by dissolving 0.315g of
the pure acid in water and making the
volume up to 400mL.
Ans- (1.36), 0.0125N,0.00625M

Problem
• Q.3. what volume of 20% sodium carbonate
soln is required to prepare 1L of decimolar
soln?
• Q.4. commercial sulphuric acid is 96.4% by wt
and its sp. Gravity is 1.84. find the normality
and molarity of the soln. what volume of this
acid is required to make 500mL of centinormal
soln.
• Ans- 53.05mL, 36.2N.18.1M, 1.38mL

Problems
• Q.5. what volume of decinormal of sulphuric
acid reacts completely with 40ml of 0.08M
NaOH soln?
• Q.6. 18cc N/2 HCl,20cc of 2N H2SO4 and
12cc of N/10 HNO3 are mixed. Find the
normality of the mixture. Hint -
• (VmixNmix =V1N1+V2N2+V3N3) Ans-1.004N

Q.7. 0.53g of pure sodium carbonate is mixed
with 100 cc of N/10 H2SO4. is the resulting
solution acidic, basic or neutral? Hint-
i. (W/E)acid > (VN/1000)base then res. Acidic
Q.8. 20ml of 0.8NHCl is mixwdwith 4.5ml of
3.5NNaOH solution , and the whole is diluted
to 100ml with water. Calculate the normality
of the resulting solution. (ans-0.0025)
Q.9. 50cc of NaOH soln is mixed with 10cc of
0.8 N HCl soln. it further requires 40cc of 0.5
N H2SO4 for neutralization. Find the normality
of NaOH solution. (0.56N)

Q.10. in an experiment , 0.18 g of a metal was
treated with 100ml of N/2 HCL soln. the
resulting mixture was diluted to 500ml . 25ml
of the diluted acid was completely neutralised
by 17.5 ml of N/10 alkali. Find the eq. wt of
the metal. Ans- 12
Q.11. 0.2016g of metallic oxide is dissolved in
100mL of N/10 H2SO4 and 30mL 0.12N
NaOH is used up to neutralize the remaining
H2SO4. calculate the equivalent weight of
metal oxide. Ans- 31.5

Q.12. 4mL of bench sulphuric acid is diluted to
100mL. If 10mL of the diluted acid is
neutralized by 8.8 mL of standard alkali
N/10(f=0.96). Express the concentration of
bench sulphuric acid in molarity. Ans-1.05M
Q.13. 2.4g of a piece of marble was dissolved in
50ml of normal solution of HCl. The residual
acid was neutralized by 12mL of 1N (f=0.85)
NaOH. Find the percentage of CaCO3 in the
marble. Ans- 81.25%

Home assignment
2g of a mixture of CaCO3 and MgCO3 was
dissolved in 200mL of 0.25N HCl solution.
After the reaction was complete, the resulting
solution was diluted to 250mL, and 10mL of
this solution required 12mL of N/30 NaOH
solution for neutralization. Find the percentage
of CaCO3 in the mixture.
Ans- 28.55%

2. Find the volume of 4N HClwill be required to
liberate 50g soda lime which contains 85%
NaOH and 15% CaO. Ans- 332.5mL

References
1. Google.com
2. Pom lal kharel, Numerical Chemistry
3. A.D mishra, Pioneer chemistry
4. Suk Dev Acharya, Foundation of chemistry
5. Moti kaji sthapit , elementary chemical
calculation
6. Chemistry book, surya publication

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