Assignment on Determination of active content of marketed Tetracycline Capsule by Spectrophotometric Method

 Materials

All the chemicals and reagents used in this study were of analytical grade. Their experimental mixtures were prepared in standard volumetric flasks about 30 minutes prior to recording the data. Suitable storage conditions were maintained to store the working chemicals and reagents.

Apparatuses used in the present study

• Volumetric flasks (50ml, 100 ml & 1000 ml)
• Conical flasks (250 ml)
• Beaker (250 ml, 1000ml)
• Burette
• Pipette (1ml, 2ml, &5ml)
• Pipette filler
• Mortar & Pestle
• Stand & Clamp
• Funnel
• Filter Paper
• 

• Fig 3.1.1.: Apparatus used in the study

• • Digital Weight Balance (M-310, Denver Instrument Company, USA)
  • Electronic Shaker/ Lab Rotator (Analog-DSR-2100A/RPM-240, Taiwan)

• L-C Oven (Lab Line)

Fig 3.1.2.3: Electric Dryer (Model-L-C Oven)

• Ø Disintegration Time Test Apparatus (Model-PTZS)

• Ø UV Spectrophotometer ( Model- DR/4000 U)

3.1.3 Reagents used for the study

a)     Hydrochloric acid (HCl)

b)    Sodium Hydroxide (NaOH)

c)     Oxalic Acid

d)    Sodium Carbonate (Na₂CO₃)

e)     Distilled water

f)      Phenolphthalein indicator

g)     Methyl orange indicator

 3.1.4 Active Ingredients used in the study

a)     Pure Tetracycline HCl powder

b)    Tetracycline HCl in solid dosage form

3.1.5 Samples used in the study

• Several brands of Marketed Tetracycline HCl capsule

3.2 Preparation of stock solutions.

3.2.1 Preparation of standard 5 N oxalic acid solutionsAbout 31.5 g pure oxalic acid was weighed out on an electronic balance and transferred into a 100 ml volumetric flask. Then it was dissolved with sufficient water and the final volume was made up to the mark of the volumetric flask with water and stored as a stock solution.

Calculation

Eq. wt. of oxalic acid (C₂H₂O₄.2H₂O) = 126 / 2 = 63 g

1000 ml 1 N solution contains 63 g sodium hydroxide

100    ,,  5 N    ,,            ,,    ( 63 × 5 × 100 ) / 1000  ,,   ,,

= 31.5 g       ,,          ,,

So, 31.5 g of oxalic acid was required to produce 100 ml of 5 N solution.

3.2.2 Preparation of 5 N sodium hydroxide solutions

About 200 g of pure sodium hydroxide was weighed out carefully on an electronic balance and transferred into a 1000 ml volumetric flask. Sufficient amount of distilled water was added to dissolve and after that the volume was made up to the mark with water.

Calculation

Eq. wt. of sodium hydroxide = 40

1000 ml 1 M solution contains 40 g sodium hydroxide

,,        5M    ,,            ,,      40 × 5  ,,          ,,

= 200 g       ,,          ,,

So, 200 g of sodium hydroxide was required to produce 1000 ml of 5 N solution.

3.2.3 Standardization of 5N sodium hydroxide with oxalic acid

The burette was filled with oxalic acid. 10 ml of sodium hydroxide solution was taken in the conical flask and it was diluted with 10 ml of distilled water, 2 drops of phenolphthalein solution was added as an indicator and became pink in color. Then it was titrated with oxalic acid till the color changed to colorless.

Table 3.2.3.1: Standardization data chart for 5N sodium hydroxide

Strength of NaOH in normality

V₁ = volume of oxalic acid = 10ml

S₁ – strength of oxalic acid = 5 N

V₂ = volume of NaOH = 10.53ml

S₂ = strength of NaOH =??

We know,

V₁ × S₁   = V2× S₂

Or, S₂ = (V₁×S₁) / V₂ = (10× 5 N) /10.53 = 4.75 N

3.2.4 Preparation of standard 0.01 N sodium carbonate solutions

About 53 mg pure sodium carbonate was weighed out on an electronic balance and transferred into a 100 ml volumetric flask. Then it was dissolved with sufficient water and the final volume was made up to the mark with water.

Calculation

Eq. wt. of sodium carbonate (Na₂CO₃) = 53 g

1000 ml 1 N solution contains 53 g sodium hydroxide

100    ,,  0.01 N    ,,            ,,    ( 53 × 0.01 × 100 ) / 1000  ,,   ,,

= 0.053 g       ,,          ,,

So, 0.053 g of sodium carbonate was required to produce 100 ml of 0.01 N solutions.

3.2.5 Preparation of 0.01 N hydrochloric acid solutions

About 0.83 ml of pure hydrochloric acid was weighed out carefully on an electronic balance and transferred into a 1000 ml volumetric flask containing 100 ml of distilled water and mixed. Then the volume was made up to the mark with water.

Calculation

Molecular wt. of hydrochloric acid = 36.50 g

Eq. wt. of hydrochloric acid = 36.50 g

To prepare 0.01 N HCl solution 0.365 g HCl needed.

The supplied concentrated HCl was of 37% by weight & sp. gr. 1.19, then,

1 cc of HCl contains, (37/100) ×1.19 g of HCl = 0.4403 ml of HCl

0.4403 ml HCl needed for   1 cc

1           ,,    ,,         ,,       ,,   (1/0.4403),,

0.365      ,,    ,,         ,,       ,,    (0.365/0.4403),,

= 0.83 cc

So, 0.83 cc of hydrochloric acid was required to produce 1000 ml of 0.01 N solutions.

3.2.6 Standardization of 0.01N hydrochloric acid with sodium carbonate

The burette was filled with hydrochloric acid solution. 10 ml of sodium carbonate solution was taken in the conical flask and it was diluted with 10 ml of distilled water, 2 drops of methyl orange solution was added as an indicator and became red in color. Then it was titrated with hydrochloric acid till the color changed to orange.

Table 3.2.6.1: Standardization data chart for 0.01N hydrochloric acid

Strength of HCl in normality

V₁ = volume of sodium carbonate = 10 ml

S₁ – strength of sodium carbonate = 0.01 N

V₂ = volume of HCl = 10.50 ml

S₂ = strength of HCl =??

We know,

V₁ × S₁   = V2× S₂

Or, S₂ = (V₁ × S₁) / V₂ = (10×0.01 N) /10.50   = 0.0095 N

3.3 Methods

Two types of test were performed during the project work;

• Physical parameters test
• Chemical test (Spectrophotometric analysis for potency)

3.3.1 Physical parameters test

Following parameters were tested for capsules; including:

• Appearance (Color & Shape)
• Weight Variation
• Disintegration Time

3.3.1.1 Appearance Test

Physical appearance color and shape of all the capsules were identified visually. Following appearance were identified for Tetracycline HCl capsules:

Table 3.3.1.1: List of appearance of sample capsules

3.3.1.2 Weight Variation Test

The weight variation of the capsule can be measured by weighing each individual capsules and determining the percent difference from the intended amount. Guidelines in the USP 24 /NF19 Supplement 1 indicate that each capsule “shall be not less than 90% and not more than 110% of the theoretically calculated weight for each unit”. Mainly weight variation of drug from batch to batch of same company occurred by the given amount of any drug such as active drug or excipients. 20 capsules should be taken for this test.

• Ten capsules of each brand were taken
• Weighed out carefully on an electronic balance
• Calculated carefully

Specification:

Table 3.3.1.2: weight variation tolerance for capsule

For this study of weight variation test of Tetracycline HCl capsule percentages deviation was considered ±7.5%, as the average weight of tetracycline HCl capsule were found to be less or more than 250 mg.

Calculation of weight variation

To calculate the weight variation of capsules in a batch, it is required to determined the folloing data.

Weight of taken of capsule in that batch (W)

Average weight of capsule in that batch (W_A)

Highest weight of capsule in that batch (W_H)

Lowest weight of capsule in that batch (W_L)

W_H― W_A

So, % of variation for highest weight of capsule =                                  × 100

W_A

W_{L ―}W_A

And, % of variation for lowest weight of capsule =                                  × 100

W_A

3.3.1.3 Disintegration time test:

The disintegration test was performed to determine the time required for the capsules to disintegrate. The test is very important and necessary for all the capsules because the dissolution rate depends upon the time of disintegration which ultimately affects the rate of absorption of drugs.

5 samples were placed into a testing basket assembly which was introduced into a special holder and placed into a water bath. The water inside the water bath was heated to 37^oC. The disintegration baskets were moved smoothly up and down for 28-32 stokes per minute and for a distance of 55mm. The samples were disintegrated if no solid rest was left within the basket.

3.3.2 Potency test (active content)

3.3.2.1 For raw (Standard) materials

First, 100 mg of Tetracycline HCl was taken carefully by electric balance and dissolved in 100 ml volumetric flask with 0.01 M HCl solutions. Then, 1 ml solution was transferred to another 100 ml volumetric flask and 75 ml 0.01 M HCl was added to it. Again 5 ml of 5 M NaOH added and then the volume was made up to the mark with water.

A few minutes waited and measured the absorbance in a suitable spectrophotometer at 380 nm wave length.

3.3.2.2 For Capsule (Sample):

Weighted out accurately 100 mg of the powder by electric balance and dissolved in 100 ml volumetric flask with 0.01 M HCl solutions. Then, 1 ml solution was transferred to another 100 ml volumetric flask and 75 ml 0.01 M HCl was added to it. Again 5 ml of 5 M NaOH added and then the volume was made up to the mark with water.

A few minutes waited and measured the absorbance in a suitable spectrophotometer at 380 nm wave length.

Absorbance of the sample

Calculation:                                                      ×  100

                     Absorbance of the standard

= % of Tetracycline present in the sample

3.3.3 More Rules for Successful Spectrophotometric Analysis

3.3.3.1 Standard solution

About 50 mg of pure standard Tetracycline HCl weighted out accurately into a 250 ml volumetric flask. Added 100 ml of 0.1 N HCl well shaked and dissolved volume with sample solvent and then shaked. 5 ml solution was taken into 100 ml with same solvent hounded.

3.3.3.2 Sample solution

Weighted out accurately equivalent to about 50 mg of sample Tetracycline HCl into a 250 ml volumetric flask. Then followed the standard preparation.

The absorbance of standard and sample solution was determined maximum at 274 nm in a suitable Spectrophotometer using 0.1 N HCl acid as a blank.

AU                      WS

Calculation =                             ×                        × PS × W

AS                        WU

Where,

AU = Absorbance of Sample solution

AS= Absorbance of standard solution

WS= Weight of Standard in mg

WU= Sample taken in mg

PS= Potency of Standard sample in mg

W= Potency of

required sample in mg