301 Pharmaceutical Analysis

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

INDEX

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

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
5. solution until the blue color was disappeared and a colorless solution was obtained.
6. The experiment was repeated for three times.

Calculation:

Preparation of 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 =

= = 248 gm

For 1000ml 1N Na2S2O3.5H2O solution required = 248 gm of Na2S2O3.5H2O

”       1 ml 1N ”    ”         ”   = gm of Na2S2O3.5H2O

”     250 ml N/50 ”    ”         ”   = ××gm of Na2S2O3.5H2O

  = 1.24 gm

Preparation of K2Cr2O7 solution:

In the same way for 250 ml  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 = ×
			Initial	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 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.

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 =

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         ” ” = × =

=

So, % of potency =

   =

Result:

The potency of the supplied sample was = %

Comments:

The potency of the supplied sample was which is within BP Limit.

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:

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= ×
			Initial	Final
1	10	0.5
2	10	0.5
3	0	0.5

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= ×
			Initial	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

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.

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.

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 = =

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.

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:

××× = 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.

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 =    =    =

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:

×××=

 =

 =

Result :

The percentage of NaHCO3 present in sample was        %

Comments: The potency of the supplied sample was which is within BP Limit.

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