Physical Pharmacy-I Lab MANIK

Laboratory Manual
Course Code: PHARM 1206
Prepared By
Md. Imran Nur Manik
Lecturer
Department of Pharmacy
Northern University Bangladesh

Manual: Physical Pharmacy-I Lab
Prepared By: Md. Imran Nur Manik Page 0
Lecturer; Department of Pharmacy; Northern University Bangladesh.
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Sl. No. Date Name of the experiment Page No.
01. Preparation of solutions of different concentrations. 01 – 02
02.
Standardization of NaOH solution by Potassium Hydrogen
Phthalate (KHP). 03 – 05
03. Standardization of HCl Solution by NaOH Solution. 06 – 07
04. Determination of pKa value of weak acid. 08 – 10
05. Preparation of constant pH buffer. 11 – 12
Bibliography and Some commonly used Lab Equipment. 12
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Experiment No. 01 Date:
Name of the experiment: Preparation of solutions of different concentrations.
Principle:
The concentration of a solution is defined as: the amount of solute present in a given amount of solution.
The concentration of a solution can be expressed as molarity, molality normality, mole fraction etc.
Molarity (M): Molarity is the number of moles of solute dissolved per liter of solution. It can be mathematically
expressed as,
solutionofLitre
soluteofMoles
)MMolarity( 
Example: Calculate the molar concentration of sodium sulphate (Na2SO4; MW=142 gm).That contains 2.8 gm of
Na2SO4 in 1500mL solution.
SOLUTION:
)i.........(..........
litre)V(in
n
Mor,
solutionofLitre
soluteofMoles
)MMolarity(That,KnowWe 
Here, Molecular mass of Na2SO4 = (232 + 32 + 164) gm=142 gm
Calculation of moles (n) of Na2SO4:
142 gm Na2SO4= 1mol
2.8 gm Na2SO4= 142
8.2
mol= 0.01972 mol
Calculation of volume (V) in litres:
1000 millilitre =1litre
1500 millilitres = 1000
1500
litre =1.5 litre
Calculation of Molarity:
mole/L0.01315
1.5
0.01972
M
V
n
Mget,we)i(equationtheinvaluesthesePutting 
Thus the solution is 0.01315 M
Problem: Calculate the amount required to prepare 250 mL, 0.25M HCl. Purity of acid is 32%.
Normality (N): Normality is the number of gram equivalent (eq.) weights of solute (solu.) per liter of solution.
Equivalent mass (Eq. M.)/Gram equivalent Wight: Equivalent mass of a substance can be calculated
by dividing its molar mass per the number of active units in one molecule of this substance (k) thus:
The active unit in the acid-base reactions is the number of hydrogen ions liberated by a single molecule of an acid
or reacted with a single molecule of a base.
For e.g. 1 mole of NaOH (m.wt. = 40) can combine with 1 mole of hydrogen ion, therefore the equivalent weight
of NaOH is 40÷1=40 gm. Thus the concentration of 1N NaOH is same as 1M NaOH.
On the other hand H2SO4 has two ionisable hydrogen atoms; its equivalent mass will be, g49
2
98
Eq.M. 
Thus, the concentration of 1M H2SO4 is same as 2N H2SO4.
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Relationship between Normality and Molality
Relationship between molarity and normality for the same solute in the same solution is expressed by the
following equation
N=nM
Where, N=normality M=Molarity and n=Valence
Problem: Calculate the amount required to prepare 0.1N H2SO4.Having density of 1.19 g/mL and 98%
purity.
Concentration can also be calculated by dilution method when the concentration and volume of one of the
two reactants is known.
Dilution: It is the addition of solvent to decrease the concentration of solute. The solution volume changes, but
the amount of solute remains constant.
M1V1 = M2V2
Initial values Final values
Problem: How to made 100 mL 0.1M NaCl from 5M NaCl solution?
Calculation:
All calculations are to be done as per the solutions to be made, needed for the experiment.
Apparatus:
1. Electrical balance
2. Pipette and pipette filler
3. Funnel
4. Volumetric flask
5. Beaker
6. Measuring cylinder
7. Spatula
8. Stirrer
9. Permanent marker
Chemicals/Reagents:
Write all the reagents required to prepare the solutions (as per the solutions to be made for the experiment)
Experimental procedures: (Specimen)
Preparation of 1M 50 mL NaOH solution:
1. Weigh out accurately W gm (To be calculated) of NaOH.
2. Transfer it to 50 mL volumetric flask.
3. At first completely dissolve it with small volume of distilled water (DW).
4. Finally adjust the volume up to the 50 mL mark with DW Q.S.
Preparation of 1N 100 mL Na2CO3 solution:
1. Weigh out accurately W gm
(To be calculated) of Na2CO3.
2. Transfer it to 100 mL volumetric flask.
3. At first completely dissolve it with small
volume of distilled water (DW).
4. Finally adjust the volume up to the
mark with DW Q.S.
Fig: Preparation of Solution.
Precautions:
Comments:
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Name of the experiment: Standardization of NaOH solution by Potassium Hydrogen Phthalate (KHP).
Terminologies Used in Titrimetric Analysis
The term 'titrimetric analysis' refers to quantitative chemical analysis. It is carried out by determining the
volume of a solution of accurately known concentration, required to react quantitatively with a measured volume
of a solution of the substance to be determined. The solution of accurately known strength is called the standard
solution.
In titrimetric analysis the reagent of known concentration is called the titrant and the substance being
titrated is termed the titrand or analyte.
The process of adding the standard solution until the reaction is just complete is termed a titration, and
the substance to be determined is titrated. The point at which this occurs is called the equivalence point or the
theoretical (or stoichiometric) end point.
The completion of the titration is detected more usually, by the addition of an auxiliary reagent, known
as an indicator. After the reaction between the substance and the standard solution is practically complete, the
indicator should give a clear visual change (either a colour change or the formation of turbidity) in the liquid
being titrated. The point at which this occurs is called the end point of the titration.
Principle
In titrimetry, chemicals used as reference solutions known as primary standards or secondary standards.
Primary Standards
A primary standard is a compound of sufficient purity from which a standard solution can be prepared by
direct weighing of a quantity of it, followed by dilution to give a defined volume of solution. The solution
produced is called a primary standard solution.
A primary standard should satisfy the following requirements.
1. It must be easy to obtain, purify, dry (preferably at 110-120°C), and to preserve in a pure state.
(This requirement is not usually met by hydrated substances, since it is difficult to remove surface moisture
completely without effecting partial decomposition.)
2. It should be unaltered in air during weighing; that means, it should not be hygroscopic, oxidised by air, or
affected by carbon dioxide.
3. It should maintain an unchanged composition during storage.
4. It should be capable of being tested for impurities by qualitative and other tests of known sensitivity.
(The total amount of impurities should not, in general, exceed 0.01-0.02%)
5. It should be readily soluble under the conditions in which it is employed.
6. The reaction with the standard solution should be stoichiometric and practically instantaneous.
7. The titration error should be negligible, or easy to determine accurately by experiment.
In practice, an ideal primary standard is difficult to obtain, and a compromise between the above ideal
requirements is usually necessary.
The substances commonly employed as primary standards are indicated below:
(a) Acid-base reactions: Sodium carbonate (Na2CO3); sodium tetraborate (Na2B4O7),
potassium hydrogen phthalate (C8H5O4K), constant boiling point hydrochloric acid, potassium hydrogeniodate
KH(IO3)2, benzoic acid (C6H5COOH) etc.
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(b) Complex formation reactions: Silver, Silver nitrate (AgNO3), Sodium chloride (NaCl), various metals
(e.g. Spectroscopically pure Zinc, Magnesium, Copper, and Manganese) and salts, depending upon the reaction
used.
(c) Precipitation reactions: Silver, Silver nitrate (AgNO3), Sodium chloride(NaCl), Potassium chloride (KCl),
and Potassium bromide(KBr) (prepared from potassium bromate).
(d) Oxidation-reduction reactions: Potassium dichromate (K2Cr2O7), Potassium bromate (KBrO3),
Potassium iodate (KIO3), Potassium hydrogeniodate KH(IO3)2, Sodium oxalate (Na2C2O4),
Arsenic(III) oxide (As2O3), and pure Iron.
Secondary Standards
Solutions that are prepared by standardisation against a primary standard are referred to as secondary standards.
It follows that, a secondary standard solution is a solution in which the concentration of dissolved solute has not
been determined from the weight of the compound dissolved, but by the reaction (titration) of a volume of the
solution, against a measured volume of a primary standard solution.
Examples: Sodium tetraborate Na2B4O7.10H2O , Copper sulphate Cu2SO4.5H2O etc.
In the present experiment, NaOH solution is to be used secondary standard chemical and needs to be
standardized by the primary standard chemicals. Here the primary standard chemical is KHP.
This is a 1:1 titration therefore, one mole of base, will titrate one mole of acid.
The end point of the solution would be determined by an indicator, phenolphthalein.
Reaction:
+Na+, K+
Phenolphthalein: Phenolphthalein exists in two tautomeric forms: (i) the benzenoid form which is yellow and
present in basic solution; and (ii) the quinonoid form which is pink and present in acid solution.
Fig. Two forms of Phenolphthalein.
Apparatus:
2. Spatula
3. Volumetric flask
5. Burette
7. Conical flask
8. Beaker
9. Funnel
Chemicals / Reagents:
1. 0.1 N Sodium hydroxide (NaOH) solution
2. 0.1 N Potassium hydrogen phthalate (C8H5O4K)
3. 0.5% Phenolphthalein indicator
4. Distilled Water
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Calculation (Sample): Full calculations required to make the reagents, for the experiment must be written down
Gram Equivalent weight of NaOH = (40÷ 1)= 40 gm
250 mL 0.1N NaOH = {(40250 0.1) ÷1000} gm = 1gm (for 100% pure NaOH)
Gram Equivalent weight of KC8H5O4 = ( 204.22÷ 1) gm = 204.22 gm
50 mL 0.1N KHC8H4O4 = {(204.22500.1) ÷1000} gm = 1.021 gm (for 100% pure KHC8H4O4)
Reagents and their preparations:
1. 0.1 N 50 mL Potassium Hydrogen Phthalate solution: Take 1.021gm Potassium hydrogen phthalate
(KHP) in a 50 mL volumetric flask; now dissolve it completely with small portion of DW. Finally add distilled
water Q.S. to make 50 mL solution.
2. 0.1 N 250 mL NaOH solution: Take 1 gm NaOH in a 250 mL volumetric flask. Completely dissolve
it with small portion of DW (e.g.125 mL). Finally adjusted the volume up to the 250 mL mark by Q.S. of distilled
water.
3. 0.5% 100 mL Phenolphthalein solution: 50 mL Ethanol+ 0.5 gm Phenolphthalein+ 50 mL DW.
Procedures:
1. Standardization of NaOH solution by Potassium Hydrogen Phthalate (KHP) solution:
i. Fill the burette with the prepared 0.1 N NaOH solution.
ii. Take 10 mL standard 0.1N KHP (C8H5O4K) solution in a conical flask.
iii. Add 1-2 drops of Phenolphthalein indicator in the acid solution and titrate it with the NaOH solution,
until the colour of the acid solution changes from colourless to faint pink.
iv. Perform another two titrations and calculate the result.
Table-1: Data for the standardization of NaOH solution:
No. of
observations
Volume of KHP
(V1 mL)
Volume of NaOH solution
(mL)
Difference
(FBR-IBR)
(mL)
Mean
volume
(V2 mL)IBR FBR
1 10
2 10
3 10
2. Calculation of strength of NaOH solution:
We know that, V1S1 = V2S2 Here, Volume of KHP, V1 = 10 mL
S2 =
2
11
V
SV Strength of KHP,S1 = 0.1 N
Volume of NaOH, V2 = mL (Mean)
= Strength of NaOH, S2= ?
= N
Result: The strength of NaOH= N.
Precautions:
Comments:
Md.
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Name of the experiment: Standardization of HCl Solution by NaOH Solution.
Principle
An acid-base titration involves the addition of a titrant solution to an analyte solution. In titrimetry, chemicals used
as reference solutions known as primary standards or secondary standards.
A primary standard is a reagent that is extremely pure, stable, has no water of hydration, and has a high
molecular weight. A secondary standard solution refers to a solution that has its concentration measured by
titration with a primary standard solution.
The secondary standard chemicals are not absolute pure and may contain some impurities that come during their
synthesis. Therefore the secondary standard chemical should standardize by using the primary pure chemicals.
In the present experiment, NaOH solution is to be used secondary standard chemical and needs to be
standardized by the primary standard chemicals. Here the primary standard chemical is Oxalic acid. After being
standardized by the Oxalic acid, NaOH is to be used for the standardization of HCl.
The end point of the solution will be determined by an indicator, phenolphthalein.
Reaction:
1. H2C2O4 (aq) + 2NaOH (aq) ⇌ Na2C2O4 (aq) + 2H2O (l)
2. HCl (aq) + NaOH (aq) ⇌H2O (l) + NaCl (aq)
Apparatus:
2. Spatula
3. Volumetric flask
5. Burette
7. Conical flask
8. Beaker
9. Funnel
2. 0.1 N Oxalic acid (dihydrate) [H2C2O4.2H2O]
3. 0.1 N Hydrochloric acid (HCl) solution
5. Distilled Water
Calculation (Sample): Full calculations required to make the reagents, for the experiment must be written
down.
Gram equivalent weight of H2C2O4 = (126.08 ÷ 2) gm = 63.040 gm
100 mL 0.1N H2C2O4 = {(63.0401000.1)÷ ÷1000 }gm = Y gm (for 100% pure H2C2O4)
Reagents and their preparations:
1. 0.1 N 100 mL Oxalic acid solution: Completely dissolve 0.636 gm of Oxalic acid (99%) in a 100 mL
volumetric flask with a small volume of DW. Finally adjust the volume up to 100 mL mark by adding
distilled water Q.S.
2. 0.1 N 1000 mL NaOH solution: Completely dissolve 4.12 gm of NaOH (97%) in a1000 mL volumetric
flask with a small volume of DW. Finally adjust the volume up to 1000 mL mark by adding distilled water
Q.S.
3. 0.1N 100 mL HCl solution: Take 1.14 mL of HCl (32%) and completely dissolve it in 100 mL
volumetric flask in 100 mL DW.
4. 0.5% 100 mL Phenolphthalein solution: 50 mL Ethanol+ 0.5 g Phenolphthalein+ 50 mL DW.
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Procedures:
1. Standardization of NaOH solution by Oxalic acid Solution:
a) Fill the burette with the prepared 0.1 N NaOH solution.
b) Take 10 mL standard 0.1N Oxalic acid solution in a conical flask.
c) Add 1-2 drops of Phenolphthalein indicator in the acid solution and titrate it with the NaOH solution,
until the colour of acid solution changes from colourless to faint pink.
d) Perform another two titrations and calculate the result.
Table-1: Data for standardization of NaOH solution:
No. of
observations
Volume of Oxalic acid
(V1 mL)
(mL)
Difference
(FBR-IBR)
(mL)
Mean
volume
(V2 mL)IBR FBR
1 10
2 10
3 10
We know that, V1S1 = V2S2 Here, Volume of Oxalic acid, V1 = 10 mL
S2 =
2
11
V
SV Strength of Oxalic Acid, S1= 0.1 N
= Strength of NaOH, S2 = ?
= N
The strength of NaOH= N.
3. Standardization of HCl solution by NaOH solution:
a) Fill the burette with the standardized NaOH solution.
b) Take 10 mL prepared 0.1N HCl solution in a conical flask.
until the colour of acid solution changes from colourless to faint pink.
Table-2: Data for standardization of HCl solution:
No. of
observations
Volume of HCl
(V1 mL)
(mL)
Difference
(FBR-IBR)
(mL)
Mean volume
(V2 mL)
IBR FBR
1 10
2 10
3 10
4. Calculation of strength of HCl solution:
We know that, V1S1 = V2S2 Here, Volume of HCl, V1 = 10 mL
S1 = 1
22
V
SV
Strength of HCl, S1 = ?
= Strength of NaOH, S2 = N (Known Earlier)
= N
Result: The strength of HCl= N.
Precautions:
Comments:
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Name of the experiment: Determination of pKa value of weak acid.
Principle
A buffer solution is one which resists (or buffers) a change in its pH. If acid is added then, within reason, the pH
does not fall; if base is added, the pH does not rise. Buffers are usually composed of a mixture of weak acids or
weak bases and their salts and function best at a pH equal to the pKa of the acid or base involved in the buffer.
The equation that predicts the behaviour of buffers is known as the Henderson–Hasselbalch equation
(named after chemists Lawrence Joseph Henderson and Karl Albert Hasselbalch). It is derived as follows, by
considering a weak acid that ionises in solution.
[Acid]
[Salt]
logpHpKaor,
[Acid]
[Salt]
logpKapH


Where, pKa=–log Ka; and Ka=Decomposition constant of acid.
An example of a buffer is a mixture of acetic acid and sodium acetate, which ionises and acts as follows:
CH3COOH⇌CH3COO– +H+
CH3COO–Na+⇌CH3COO– +Na+
Apparatus:
2. Spatula
3. Volumetric flask
5. Burette
6. Pipette and pipette filler Fig. Mechanism of Buffer action of an acid buffer.
7. Conical flask
8. Beaker
9. Funnel
2. 1% Acetic acid (CH3COOH) solution
3. 0.1 N Oxalic acid (dihydrate)[ H2C2O4.2H2O]
5. Distilled Water
Calculation: Full calculations required to make the reagents, for the experiment must be written down.
Preparation of Reagents:
A. 1% 100 mL Acetic Acid Solution: Take 1 mL glacial acetic acid (100%) in a 100 mL volumetric flask, and
then adjust the volume up to 100 mL mark by adding distilled water.
B.0.1 N 1000 mL NaOH Solution: Completely dissolve 4.12 gm NaOH (97%) pellets in a 1000 mL volumetric
flask at first with a small portion of DW, and finally adjust the volume up to the mark by adding DW Q.S.
C.0.1 N 100 mL Oxalic Acid Solution: Completely dissolve 0.636 gm oxalic acid (99%) in a 100 mL volumetric
flask at first with a small portion of DW, and finally make the volume up to mark by adding distilled water Q.S.
D.0.5% Phenolphthalein: 50 mL Ethanol + 0.5 gm phenolphthalein + 50 mL distilled water.
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Procedures:
1. Standardization of NaOH solution by Oxalic acid Solution:
a) Fill the burette with the prepared 0.1 N NaOH solution.
b) Take 10 mL standard 0.1N Oxalic acid solution in a conical flask.
Table-1: Data for standardization of NaOH solution:
No. of
observations
Volume of Oxalic acid
(V1 mL)
(mL)
Difference
(FBR-IBR)
(mL)
Mean
volume
(V2 mL)IBR FBR
1 10
2 10
3 10
We know that, V1S1 = V2S2 Here, Volume of Oxalic acid,V1 = 10 mL
S2 =
2
11
V
SV Strength of Oxalic acid, S1 = 0.1 N
= Strength of NaOH, S2= ?
= N
The strength of NaOH = N.
3. Determination of the strength of CH3COOH solution by NaOH solution:
a) Fill the burette with the standardized NaOH solution.
b) Take 10 mL prepared Acetic acid (CH3COOH) solution in a conical flask.
Table-2: Data for standardization of CH3COOH solution:
No. of
observations
Volume of CH3COOH
(V1 mL)
(mL)
Difference
(FBR-IBR)
(mL)
Mean
volume
(V2 mL)IBR FBR
1 10
2 10
3 10
4. Calculation of strength of Acetic acid solution:
We know that, V1S1 = V2S2 Here, Volume of CH3COOH, V1 = 10 mL
S1 =
1
22
V
SV Strength of CH3COOH, S1 = ?
= Strength of NaOH, S2 = N (Known Earlier)
= N
The strength of CH3COOH = N.
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5. Determination of pH of Buffer solution:
1) Take 40 mL 1% Acetic acid solution and 20 mL standardized NaOH solution in a 100 mL beaker and
mix them very well.
2) Then measure the pH of the buffer solution using a pH meter.
The pH of the solution=
6. Determination of pKa of the solution:
Now calculate pKa value is by the following equation,
volumeTotal
COOHCHofStrengthCOOHCHRemainingofVolume
VolumeTotal
NaOHofStrengthNaOHofVolume
logpHpKa
getw)equation(1thefromThus
VolumeTotal
COOHCHofStrengthCOOHCHRemainingofVolume
[Acid]Acid,ofionconcentratexperimentpresenttheIn
VolumeTotal
COOHCHofStrengthCOOHCHofVolume
[Acid]Acid,ofionconcentrattheAnd
VolumeTotal
NaOHofStrengthNaOHofVolume
[Salt]Salt,ofionconcentrattheHere,
)1..(..............................................................................................................
[Acid]
[Salt]
logpHpKa
33
33
33










e
Result: The pKa value of the solution =
Precautions:
Comments: Md.
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Name of the experiment: Preparation of constant pH buffer.
Principle:
A solution which resists changes in pH when small quantities of an acid or an alkali are added to it, is called
buffer solution. Most of the buffer solutions usually consist of a mixture of weak acid and one of its salts or a
weak base and one of its salts. The resistance to a change in pH is known as buffer action. A buffer solution
contains large and equal concentrations of an acid and its conjugate base. The pH of this solution is
approximately equal to the pKa of the acid. Addition of small amounts of acid or base results in the mopping up
or the release of protons by the conjugate base or the acid as necessary, which keeps the solution pH constant.
TYPES OF BUFFERS
Buffers are classified into following types
1. Simple Buffers:
Simple buffers are categorized into three different ways.
A) Salts of weak acid and a weak base
Example: Ammonium acetate-CH3COONH4, Ammonium cyanide-NH4CN.
B) Proteins and amino acids.
C) A mixture of an acid salt and a normal salt formed from polybasic acid
Example: Na2HPO4 and Na3PO4
2. Mixed Buffers
Acidic Buffer: A solution of weak acid and its salt with strong base.
Example: CH3COOH +CH3COONa, H2CO3 +Na2CO3
Basic Buffer: A solution of weak base and its salt with strong acid.
Example: NH4OH+NH4Cl, NH4OH+NH4NO3
3. Natural Buffers
A solution is said to be naturally buffered if it contains buffering compounds
as it exists in nature. Blood is an example of a naturally buffered solution.
Fig.Citric acid monohydrate
Fig. pH meter
Apparatus:
2. Spatula
3. Volumetric flask
4. Conical flask
5. Beaker
7. pH meter
8. Stirrer
Chemicals/Reagents:
1. 0.1 M Na-Acetate (CH3COONa) solution
2. Citric acid (monohydrate)
3. Distilled Water
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Calculation: Full calculations required to make the reagents, for the experiment must be written down.
Preparation of Reagents:
0.1 M 100 mL CH3COONa Solution: Completely dissolve 1.374 gm CH3COONa (99%) in a 100 mL volumetric
flask with a small amount of DW. Later on adjust the volume up to the mark by adding distilled water Q.S.
Procedures:
1. Take 100 mL 0.1 M CH3COONa solution in a 250 mL beaker.
2. Carefully submerge the detector of pH meter into the CH3COONa solution.
3. Take the reading of pH meter.
4. Now add crystalline citric acid (C6H8O7.H2O) to the CH3COONa solution with continuous stirring by
the stirrer.
5. Carefully submerge the detector of pH meter into the solution.
6. Take the reading of pH meter simultaneously while adding citric acid.
7. Continue the step 4 until the pH is adjusted to 6 or above.
Table-1: Data for pH adjustment:
No. of observation pH of CH3COONa Solution pH of final solution
1
Result: The pH value of the solution=
Precautions:
Comments:
Bibliography:
1. G H Jeffery, J Bassett, J Mendham, and
R C Denney: Vogel's textbook of
quantitative chemical analysis, 5th edn,
( 1989). Longman Group UK Limited,
Longman Scientific & Technical.
2. Arun Bahl, B.S Bahl, and G.D Tuli:
Essentials of Physical Chemistry, Multi
colour edn. (2011). New Delhi, S.Chand
and Company Limited.
3. Donald Cairns: Essentials of
Pharmaceutical Chemistry, 3rd edn.
(2008). Pharmaceutical Press, UK.
Md.
Imran
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