a) Assay of acetyl salicylic acid in aspirin tablets. b) Assay of sodium salicylate tablets c) Determination of potency of penicillin tablets. d) Non- aqueous assay of phenobarbitone tablets. e) Determination of calcium in solid & liquid dosage form by complexometric titration. f) Assay of promethazine hydrochloride. g) Assay of methamphetamine hydrochloride h) Assay of aluminum hydroxide gel. i) Assay of milk of magnesia j) Assay of magnesium and aluminum from antacid preparation. k) Determination of iodine value, saponification value, acid value and R.M. value of oils and fats

1. Laboratory Manual
Course: 310
Subject: Practical - Analytical Pharmacy
Prepared By
Md. Imran Nur Manik
M.Pharm
Department of Pharmacy
University of Rajshahi
Rajshahi-6205, Bangladesh
Available at
Essential Pharma documents
www.pharmacydocs.blogspot.com

2. 1
Serial
No.
Date Name of the Experiment Page No.
01 26.09.11 Determination of potency of penicillin
tablets.
02-05
02 27.09.11 Assay of salicylic acid in aspirin tablets 06-08
03 24.09.11 Determination of potency of paracetamol
Syrup.
09-10
04 01.10.11 Assay of NaHCO3 from supplied sample. 11-12

3. 2
Experiment No. 01 Date: 26.09.11
Name of the experiment: Determination of potency of penicillin tablets.
Principle:
Penicillin after preliminary hydrolysis with NaOH is converted into Penicillonic
acid. This on treatment with concentrated HCl yields D-penicillamine which is
oxidized by excess iodine to the corresponding disulfide. Excess iodine is back
titrated with N/50 sodium thiosulfate solution. The amount of iodine required to
oxidize penicillamine is equivalent to the amount of penicillin present in the sample.
1ml of N/50 I2 = 0.789 mg of Pen-K
Reagents:
1. N/50 K2Cr2O7 solution 4. Concentrated HCl solution
2. N/50 Na2S2O3 solution 5. 1N NaOH
3. N/50 I2 solution
Procedure:
A. Standardization of N/50 Na2S2O3 solution:
1. 10 ml of N/50 K2Cr2O7 solution was taken in a conical flask. Now 2gm of
NaHCO3 and 3gm of KI was added to it and finally 5ml of conc. HCl was added.
2. Then the conical flask was covered by an inverted watch glass and was kept
for 5 minutes in the dark place.
3. The watch glass was washed down with water & the solution was diluted with
50ml water. Then it was titrated against N/50 Na2S2O3 solution.
4. When light yellow color appears, the addition of thiosulfate solution was
stopped and 1ml starch solution was added. Then it was titrated with
thiosulfate solution until the blue color was disappeared and a colorless
solution was obtained.
5. The experiment was repeated for three times.
Calculation:
Preparation of
N
50
Na2S2O3.5H2O Solution:
The molecular weight of Na2S2O3.5H2O = (232 + 323 + 185) = 248 gm
Thus, the gram equivalent with of Na2S2O3.5H2O =
Molecular wieght of Na2S2O3.5H2O
Number of electron transfered
=
248
1
= 248 gm
For 1000ml 1N Na2S2O3.5H2O solution required = 248 gm of Na2S2O3.5H2O
” 1 ml 1N ” ” ” =
248
1000
gm of Na2S2O3.5H2O
” 250 ml N/50 ” ” ” =
248250
100050
gm of Na2S2O3.5H2O
= 1.24 gm
Md.
Imran
Nur
Manik

4. 3
Preparation of K2Cr2O7 solution:
In the same way for 250 ml
N
50
K2Cr2O7 solution 0.245 gm of K2Cr2O7 is required.
B. Standardization of N/50 Na2S2O3 solution against standard K2Cr2O7 N/50
solution:
No. of
observation
Volume
of
K2Cr2O7
Strength
of
K2Cr2O7
Volume of
Na2S2O3
solution (ml)
Difference
(ml)
Mean
(V2
ml)
Strength of
Na2S2O3
S2 =
V1S1
V2Initial Final
1 10 N/50
2 10 N/50
3 0 N/50
So, strength of Na2S2O3, S2 = = F (Let)
C. Assay of tablets:
1. Ten tablets were weighed & average weight of the tablets was determined.
Now 100mg of powdered tablet was accurately weighed & dissolved in water
& diluted the solution into 100 ml.
2. 10ml of solution was transferred to a stoppered bottle & 5ml of 1N NaOH
solution was added and the bottle was allowed to stand for 30 minutes in a
water bath at 30C.
3. The solutions was then acidified with 5ml of 6N HCl acid and 30ml of N/50
iodine was added. The flask was covered with a stopper and allowed to stand
for 15 minutes in a water bath at 30C.
4. The excess of iodine was titrated with standard solution thiosulfate solution
adding 1ml of starch solution near the end point.
5. A blank titration was done by following the above procedure except the
addition of NaOH and HCl acid solution, i.e. by adding N/50 I2 solution (30
ml) and 1 ml of starch solution near the end point in 10ml of blank solution
(distilled water).
The difference between the two titrations represents the amount of iodine reacted
with penicillin.
Calculation:
Preparation of iodine solution:
The amount of iodine is requried for 250 ml
N
50
is 0.634 gm.
Since iodine is slightly soluble in water. So for increasing the solubility of iodine,
excess amount of KI is added which forms a triiodide complex with iodine and this
compound becomes easily soluble in water. I2 + KI  KI3
Preparation of 1N NaOH solution:
In the same way for 100 ml 1N NaOH solution 40 gm of NaOH is required.
Preparation of 6N HCL solution:
For the preparation of 6N HCl solution, 51.025ml concentrated HCl is taken in
100 ml flask and made up to the mark with distilled water.
Md.
Imran
Nur
Manik

5. 4
D. Standardization of blank solution with N/50 Na2S2O3 solution:
No. of
observation
Volume of blank
solution (V2 ml)
Volume of Na2S2O3
solution (ml)
Difference
(ml)
Mean Volume
of Na2S2O3
solution (ml)Initial Final
1 10
2 10
3 10
E. Standardization of sample solution with N/50 Na2S2O3 solution:
No. of
observation
Volume of blank
solution (V2 ml)
Volume of Na2S2O3
solution (ml)
Difference
(ml)
Mean Volume
of Na2S2O3
solution (ml)Initial Final
1 10
2 10
3 10
Difference in Volume of Na2S2O3 = X =
Volume of iodine consumed = X = ml
Weight of each tablet:
Number of tablet Weight of tablet (Mg)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Total weight
Average weight of tablet =
Md.
Imran
Nur
Manik

6. 5
Calculation:
1 ml N/50 Na2S2O3 = 1 ml of N/50 I2
1 ml of N/50  F = 1 ml of N/50  F I2
But, 1ml of N/50 I2 = 0.789 mg of Pen-k
10 ml solution contains = mg of Pen-k
100 ml ” ” = mg
= mg of Pen-k = W1(Let)
Average weight of tablet = y =
As in 100gm sample powdered was dissolved in 100ml solvent.
Thus, 100 mg powder contains = W mg of Pen - k
Y mg ” ” =
W1y
100
=
=
So, % of potency =
=
Result:
The potency of the supplied sample was = %
Comments:
The potency of the supplied sample was which is within BP Limit.
Md.
Imran
Nur
Manik

7. 6
Experiment No. 02 Date: 27.09.11
Name of the experiment: Assay of salicylic acid in aspirin tablet
Principle:
Salicylic acid is a weak organic acid. Since it has acidic property, it is given in the
preparation as Na Salt of salicylic acid or acetyl salicylate. Acetyl salicylate is
converted to Na salt of salicylic acid with excess NaOH solution and the excess
NaOH solution is tirated with HCl acid solution.
The consumed NaOH solution is equivalent to the amount of acetyl slicylate
present. 1ml 0.5 N NaOH solution (0.02 gm NaOH) is equivalent to 0.04504 gm of
acetyl salicylate.
Reaction:
COOH
OCOCH3
COONa
OH
+ 2NaOH + CH3COONa + H2O
Acetyl salicylate Na-salt of salicyli acid Sodium Acetate
NaOH (excess) + HCl NaCl + H2O
Reagents:
1. 96% ethanol (rectified spirit)
2. 0.5N NaOH
3. 0.5N HCl
4. Phenolphthalein
Procedure:
A. Standardization of 0.5N NaOH by 0.5N oxalic acid:
1. 10 ml 0.5 N oxalic acid was taken in a conical flask and 1–2 drops of
phenolphthalein solution was added.
2. It was then titrated by taking 0.5 N NaOH in a burette.
No. of
observation
Volume
of
oxalic
(V1ml)
Strength
of
oxalic
acid (S1)
Volume of
NaOH
solution (ml)
Difference
(ml)
Mean
(ml)
Strength of
NaOH
S2=
V1S1
V2Initial Final
1 10 0.5
2 10 0.5
3 0 0.5
Md.
Imran
Nur
Manik

8. 7
B. Standardization of 0.5N HCl by Standardized NaOH solution:
1. 10ml N NaOH solution was taken in a conical flask & 1/2 drops of
phenolphthalein was added.
2. Then it was titrated by taking 0.5 N HCl in a burette.
No. of
observation
Volume
of
NaOH
(V1 ml)
Strength
of
NaOH
(S1)
Volume of
HCl solution
(ml)
Difference
(ml)
Mean
(V2 ml)
Strength of
HCl
S2=
V1S1
V2Initial Final
1 10
2 10
3 0
C. Assay of tablet:
1. 10 tablets were weighted and average weight was calculated. Then the tablets
were powdered.
2. The powdered tablet equivalent to 1 gm acetyl salicylic acid was taken.
3. This amount of powder dissolve in 10 ml of 96% ethanol and 0.5 N 50 ml
NaOH was added to it. The flask was stoppered and allowed to stand for one
hour.
4. The above solution was titrated with 0.5 N HCl using dilute phenolphthalein
as indicator.
5. The blank titration was carried out according to above process without adding
powdered aspirin.
D. Standardization of the sample to find out the volume of excess NaOH:
No. of
observation
Volume of
sample (ml)
Volume of HCl (ml) Difference
(ml)
Mean (ml)
Initial Final
1 10
2 10
3 10
A. Standardization of blank sample with standard N HCl solution:
No. of
observation
Volume of blank
sample (ml)
Volume of HCl (ml) Difference
(ml)
Mean (ml)
Initial Final
1 10
2 10
3 10
Md.
Imran
Nur
Manik

9. 8
Weight of each tablet:
Number of tablet Weight of tablet (Mg)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Total weight
Average weight of tablet =
Calculation:
The volume of HCl required for the neutralization of NaOH
which is consumed by salicylic acid is = ml
= ml
The volume of NaOH consumed by salicylic acid =
gm powder contains = gm acetyl salicylic acid
1 ” ” ” = gm
” ” ” = gm
= gm
= gm
So, % of potency = %
=
Result:
The potency of the supplied sample was = %
Comments:
The potency of the supplied sample was which is within BP Limit.
Md.
Imran
Nur
Manik

10. 9
Experiment No. 03 Date: 28.09.11
Name of the experiment: Determination of potency of paracetamol syrup.
Principle:
The extent to which radiation is absorbed while passing through a layer of an
absorbing substance is expressed in term of the absorbance A. defined by the
expression; A = log 10 (1/T)
Where, I = Intensity of the radiation passing into the absorbing layer
T = Intensity of the radiation passing out of it
For a solution of an absorbing solute contained in an absorption cell having flat,
parallel optical faces, results are evaluated by the expression:
a(1%, 1cm) = A/cb
Where, a = Specific absorbance
c = concentration of absorbing solute expressed as a percentage w/v
b = Thickness of the absorbing layer in cm
The specific absorbance is therefore the absorbance of a 1 cm layer of a 1% w/v
solution of the absorbing solute, its value of a particular wave length in a give solvent
being a property of the solute.
Apparatus and reagents:
1. Volumetric flask (250 ml, 100 ml and 50 ml) 4. Distilled water
2. Paracetamol syrup 5. Spectrophotometer
3. NaOH
Procedure:
1. A weighted quantity of pedicit elixir (about 2.316 gm of 2ml) was taken in a
250ml volumetric flask.
2. 25 ml of 0.1M sodium hydroxide was added and diluted to 250ml of
volumetric flask.
3. 3ml of that solution was transferred to a 100ml volumetric flask.
4. To that solution, 9.7ml of 0.1M sodium hydroxide was added, with mixing
water added to make final volume 100ml. Filtration done (if necessary)
5. Absorbance of the resulting solution was measured at the maximum at about
257 nm, using a blank in the reference cell.
Preparation of blank solution:
5ml of 0.1M sodium hydroxide was taken in a 50ml volumetric flask, water
added, mixed and volume made up to 50ml.
Calculation of content of Paracetamol is done by taking 715 as the value of ‘a’
(1% 1cm) at 257nm and weight per ml of sample is determined to express the content
of paracetamol as weight per ml.
Md.
Imran
Nur
Manik

11. 10
Calculation:
We know the relationship: A (1%, 1cm) = abc
Where, A = Absorbance
a = Extinction co-efficient
b = Length of layer of thickness
c = Concentration of percentage, weight in
volume in the experiment condition
Here,
A =
a =
b =
c = ?
Thus, A = abc
C =
A
ab
=
Therefore, 250 ml diulte sample contains = ml of paracetamol
 1 ” ” ” ” = ” ”
 ” ” ” ” = ” ”
Again, ml of paracetamol contains = gm of paracetamol
 1 ” ” ” ” = ” ”
 ” ” ” ” = ” ”
= ” ”
=
Thus the content of Paracetamol = ml =
But, the claimed content is =
Therefore, percentage of claim =
=
Result:
The supplied paracetamo syrup has got % potency.
Comments: The potency of the supplied sample was which is within BP Limit.
Md.
Imran
Nur
Manik

12. 11
Experiment No. 04 Date: 01.10.11
Name of the experiment: Assay of NaHCO3 from supplied sample.
Principle:
Acidimetry is the measurement of the quantity of base in a given sample by
titration with a suitable acid.
Direct titration is conducted by introducing a standard acid solution gradually
from a burette into a solution of the base being assayed until chemically equivalent
amount of each have reacted as shown by some change in some properties of the
mixture, such as end point, which must be close to the stoichiometric point, is made
evident by a change in the color of some indicator or by potentiometric means.
Accurately weigh sodium bicarbonate mix with 25 ml of water and titrate with 1N
sulfuric acid. Methyl orange is the most suitable indicator and NaHCO3 is act as a
base as follows:
2NaHCO3 + H2SO4  Na2SO4 + 2H2O + 2CO2
From the equation it is obvious that 2 moles of NaHCO3 is equivalent to 1 mole of
H2SO4. Therefore 1 mole of NaHCO3 is equivalent to 1
/2 mol of H2SO4 and the
equivalent weight is equal to the gram molecular weight, 84.01 gm. One milliliter of 1
N sulfuric acid or 1 mEq is equivalent to 0.08401 mg of 1 mEq of NaHCO3.
Calculation of percentage purity of NaHCO3:
mla  N  0.084  100
Sample weight
= Percent of NaHCO3
Procedure:
Weight accurately about 3 gm of sodium bicarbonate, mix with 25 ml of water,
add 2 drops of methyl orange and titrate with 1 N sulfuric acid. Each milliliter of 1 N
sulfuric acid is equivalent to 81.01 mg of NaHCO3.
Reagents and Preparation:
1. 3gm sodium bicarbonate
2. 1 N sulfuric acid
3. 1 N Na2CO3
Preparation of sulfuric acid:
ml of sulfuric acid was taken into 100ml distilled water in a 250 ml volumetric
flask. finally volume was adjusted up to mark by adding distilled water.
Preparation of 1 N sodium carbonate:
gm of sodium carbonate was taken into 100ml distilled water in a 250 ml
volumetric flask. Finally volume was adjusted up to mark by adding distilled water.
Md.
Imran
Nur
Manik

13. 12
Standardization of 1 N H2SO4 by Na2CO3:
1. 10ml of Na2CO3 was taken in a conical flask. 1–2 drops of methyl red
indicator was added to it.
2. The titration was performed using 1 N H2SO4 solution which was previously
kept on burette. When the yellow colour disappeared, addition of H2SO4 was
stopped and red colour appeared.
3. The experiment was repeated for three times.
Data for standardization of H2SO4 solution:
No. of
observation
Volume of
Na2CO3 (V1 ml)
Volume of H2SO4
(V2 ml)
Difference
(ml)
Mean (ml)
Initial Final
1 10
2 10
3 10
Calculation of strength of H2SO4 solution:
We know that, V1S1 = V2S2
S2 =
V1S1
V2
=
=
Here, V1 = 10 ml
S1 = 1N
V2 = ml
S2 = ?
Data of assay of NaHCO3:
No. of
observation
Volume of
NaHCO3 sample
(ml)
Volume of H2SO4
(ml)
Difference
(ml)
Mean (ml)
Initial Final
1 25
2 25
3 25
Calculation of percentage of NaHCO3:
mla  N  0.084  100
Sample weigth
=
=
=
Result :
The percentage of NaHCO3 present in sample was %
Comments: The potency of the supplied sample was which is within BP Limit.
Md.
Imran
Nur
Manik