4 practical class
ASSAY - VOLUMETRIC ANALYSIS
NAME: Alicia Vails Arrufat (F19038Z) SAMPLE: Sodium bromide
1) IDENTIFICATION REACTIONS OF IONS
CATIONS (describe briefly reactions):
a) Na+ + K2CO3    No precipitate is formed.
Na++ K[Sb(OH)6]    Na[Sb(OH)6] + K+ (dense, white precipitate).
b) Na+ + methoxyphenilacetic reagent    Voluminous, white, crystalline precipitate. Precipitate + NhV (d)    Precipitate dissolves completely
Dissolution + (NH4)2C03    No precipitate.
ANIONS (describe briefly reactions):
a) Br + Ag+     AgBr (curled, pale yellow precipitate).
AgBr + 2NH3    [Ag(NH3)2]+ Br (precipitate dissolves with difficulty).
b) 2Br+Pb02 + 2H+-» Br2 + PbO + H20
A violet color appears when dipping a filter paper with decolorized fuchsin solution R.
2) ASSAY: BACK TITRATION
Volumetric solutions: silver nitrate and ammonium thiocyanate.
Titre of volumetric solutions: 0,1M AgN03: 0.9998; 0,1M NH4SCN: 0.9897
Titration No.	m [g] (4 decimal places)	Consumption of VS [ml]	ASSAY
1.	1,9854	5,86	99,486
2.	2,0125	5,47	100,120
3.	2,0245	5,05	101,640
4.	2,0998	4,90	98,722
Average			100,005
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4 practical class
ASSAY - VOLUMETRIC ANALYSIS
CALCULATION PROCEDURE:
(VI x fl -V2 x f 2) x m x 100 X (%) =----—-
q
VI = 25ml f2 = 0,9897
fl = 0,9998 m = 10,29 mg
Titration n9 1
l,9854g in 100ml    0,19854g in 10ml    q = 0,19854 g = 198,54 mg V2 = 5,86ml
(25ml x 0,9998 - 5,86ml x 0,9897) x 10,29mg x 100
x (%) = ^-:- -J--:-2-= 99,486
198,54mg
Titration n9 2
2,0125g in 100ml    0,201254g in 10ml ^ q = 0,20125 g = 201,25 mg V2 = 5,47ml
(25ml x 0,9998 - 5,47ml x 0,9897) x 10,29mg x 100
x (%) = ^-:- „_ -}--:-s-= 100,120
201,25mg
Titration n^ 3
2,0245g in 100ml    0,20245g in 10ml    q = 0,20245 g = 202,45 mg V2 = 5,05ml
(25ml x 0,9998 - 5,05ml x 0,9897) x 10,29mg x 100
x (%) = ^-:- -}--:-2-= 101, 640
202,45mg
Titration n^ 4
2,0998g in 100ml    0,209984g in 10ml    q = 0,20998g = 209,98mg V2 = 4,90ml
(25ml x 0,9998 - 4,90ml x 0,9897) x 10,29mg x 100
x (%) =--'- 9no qq -"-:-"-= 98< 722
209,98mg
2
4 practical class
ASSAY - VOLUMETRIC ANALYSIS
Statistical calculation
R = Xmax - Xmin = 101,640 - 98,722 = 2,868 SD = kn XR = 0,4857 X 2,868 = 1, 3929
SD 1,3929 RSD = — X 100 =-- X 100 = 1, 3928
x 100,005
STATISTICAL EVALUATION:
Range: R= 2,868
Standard deviation (estimated from range): sd = 1,3929
Relative standard deviation: RSD = 1,3928
CONCLUSION (does your sample meet/not meet Ph. Eur): In the Pharmacopoeia we can see that for meeting the requirement we need a content of Sodium Bromide between 98,0 - 100,5 per cent. Our sample has a content of 100,005 so we can conclude that it meets the Pharmacopoeia requirements.
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