Subscribe Pharmaguideline on
1. Description
White or pale yellow powder or crystals.2. Solubility
Freely soluble in water forming an opalescent solution partly soluble in ethanol (95%).3. Identification
A. Physical characteristics
Procedure: Take about 0.25 g of sample in a glass-stoppered test tube and add 25 ml of water. When shaken, it produces plenty of foam.B. Color reaction
Reagent required0.1% w/v solution of methylene blue
1M sulphuric acid
Dichloromethane
Procedure: In clean conical flask take 0.1 ml of the solution prepared for identification A. Add 0.1 ml of a 0.1% w/v solution of methylene blue, mix and add 2 ml of 1M sulphuric acid, add 2 ml of dichloromethane and shake, the dichloromethane layer is intensely blue.
C. Reaction of sulfate
Reagent requiredBarium chloride solution
Ethanol AR
2M hydrochloric acid
Procedure: In a clean, test tube take about 10 mg of sample, mix with 10 ml of ethanol and heat to boiling on a water bath. Shake frequently. Filter immediately and evaporate ethanol. Dissolve the residue in 8 ml of water and 3 ml of 2M hydrochloric acid; evaporate the solution to half of its volume and cool. Filter the solution and to the filtrate add 1 ml of barium chloride solution. A white crystalline precipitate is produced.
4. Reaction characteristics of sodium
Limit: A yellow crystalline precipitate is produced.Reagent required
Magnesium Uranyl Acetate solution
Dilute Acetic acid
Procedure: Dissolve about 0.1 g of sample in 10 ml of water, acidify with dilute acetic acid and add a large excess of magnesium Uranyl acetate solution. A yellow crystalline precipitate is produced.
5. Alkalinity
Limit: Not more than 0.5 ml of 0.1M Hydrochloric acid is required to change the color of the solution.Reagent required
Phenol Red solution
0.1M hydrochloric acid
Procedure: In a clean and dried conical flask, take 1g of the sample, dissolve in 100 ml of CO2 free water and add 0.1 ml of phenol red solution. Not more than 0.5 ml of 0.1M HCl is required to change the color of the solution.
6. Non esterified alcohol
Limit: Not more than 4%Reagent required
Ethanol 95%
n-pentane AR
Anhydrous sodium sulphate AR
Sodium Chloride AR
Procedure: Weigh accurately about 10 g of sample (W) and dissolve in 100 ml of water. Add 100 ml of ethanol (95%) and extract the solution with three quantities, each of 50 ml of n-pentane, adding sodium chloride, if necessary, to promote separation of the two layers. Wash the combined organic layers with three quantities, each of 50 ml, of water. Dry the organic solution over anhydrous sodium sulphate, filter and collect the organic layer on a silica dish, which was previously dried at 110°C for two hours and weighed (W2). Evaporate the organic layer on a water-bath until the odor of pentane is no longer detectable. Heat the residue at 105°C for 15 minutes, cool in desiccators and weight (W3).
Calculation
W2 – W3 x 100
% Non-esterified alcohol = ----------------------
W
Where,
W = Weight of sample
W2 = Weight of empty crucible
W3 = Weight of crucible with residue
7. Sodium chloride and Sodium sulphate
Limit: Not more than a total of 8.0%Reagent required
2M nitric acid
Potassium chromate solution
0.1M Silver nitrate solution
0.05%w/v Dithizone in acetone
1M nitric acid
Dichloro acetic acid AR
Acetone AR
0.01M lead nitrate VS
Procedure
Sodium chloride
Weigh accurately about 5.0 g in a conical flask and dissolve in 50 ml of water, add 2M nitric acid dropwise until the solution is neutral to litmus paper, add 2 ml of potassium chromate solution and titrate with 0.1M silver nitrate. Each ml of 0.1M silver nitrate is equivalent to 0.005844g of NaCl.
Calculation
V x M x F x 100
% Sodium chloride = --------------------------
0.1 x W
Where,
V = Consumed volume of 0.1 Silver nitrate
M = Molarity of 0.1 M Silver nitrate
F = Factor
W= Weight of substance
Sodium Sulphate
Procedure: Weigh accurately about 0.5 g in a conical flask and dissolve in 20 ml of water, warm gently if necessary, and add 1 ml of a 0.05% w/v solution of Dithizone in acetone. If the solution is red, add 1M nitric acid dropwise, until it becomes bluish green. Add 2 ml of dichloroacetic acid solution and 80 ml of acetone and titrate with 0.01M lead nitrate until a permanent orange-red color is obtained. Each ml of 0.01M lead nitrate is equivalent to 0.001420g of Na2SO4.
Calculation
V x M x F x 100
% Sodium sulfate = --------------------------
0.1 x W
Where,
V = Consumed volume of 0.1 Lead nitrate
M = Molarity of 0.1 M Lead nitrate
F = Factor
W= Weight of substance
8. Assay
Limit: Not less than 85.0% sodium alkyl sulfateReagent required
Chloroform AR
Dilute sulphuric acid
Dimethyl yellow-oracet blue B solution
0.004M Benzothonium chloride VS
Procedure: Weigh accurately about 1.15g dissolve in sufficient water to produce 1000 ml, warm if necessary. Transfer 20.0 ml to a separator, add 15 ml of chloroform, 10 ml of dilute sulphuric acid and 1 ml of dimethyl yellow-oracet blue B solution and titrate with 0.004M Benzothonium chloride, shaking vigorously and allowing the layers to separate after each addition until the chloroform layer acquires a permanent clear green color. Each ml of 0.004M Benzothonium chloride is equivalent to 0.001154g of sodium alkyl Sulphate, calculated as C12H25NaO4S.
Calculation
V x M x F x 100
% Sodium chloride = -----------------------
0.1 x W
Where,
V = Consumed volume of 0.1 Lead nitrate
M = Molarity of 0.1 M Lead nitrate
F = Factor
W= Weight of substance
hello sir,
ReplyDeleteAssay of sodium lauryl sulphate is correct? here show molarity of lead nitrete but use benzothonium!
what's the factor of this calculation?
please help....