Transcript LAB. 3 ascorbic acid..

A N A LY S I S O F V I T- C ( A S C O R B I C
ACI D ) TAB L ETS
P R E PA R E D B Y N A D A A LT H E YA B
ASCORBIC ACID
• Chemical structure:
• MF: C6H8O6
• Mwt : 176 Da
• Chemical name:
1,2 dihydroxyethyl- 3,4 dihydroxyfuran2-one.
USES AND ACTION:
Ascorbic acid is a naturally occurring organic compound with
antioxidant properties
• It helps the body to absorb iron
• Helps wound to heal
• Helps red blood cell formation
• Helps to fight infections
IDENTIFICTION
OX I D - R E D R E AC T I O N S
OXID-RED REACTIONS
• Various methods are used to balance redox reactions, and we shall
use the half-reaction method, in this technique the reaction is
broke down into two parts the oxidizing part and reducing part
• In every redox reaction, an oxidizing agent reacts with a reducing
agent
• The oxidizing agent is reduced in the reaction while the reducing
agent is oxidized
OXID-RED REACTIONS
• The overall reaction can be broken down into these two half reaction
• To balance a reduction oxidation reaction each half reaction is first balanced , there must be a
net gain or loss of zero electrons in the overall reaction.
PROCEDURE:
1.
Grind Vit-C tablet and wight 0.25g
2.
Dissolve 0.25g of Vit-C powder in 5 ml dist. Water through strong shaking
3.
Filter
4.
Divide it into two equal volumes in two test tubes
 Add to the first tube 2,6-DCPIP dropwise
 Add to the second tube silver nitrate solution
Its Deep blue colour will be decolourized its
Blak PPt. from Ag-metal will be formed
THE MECHANISM 0F DCPIP
THE MECHANISM 0F SILVER NITRATE
ASSY (PURITY AND %
ASCORBIC ACID VI-C
TAB L ET
R E D O X T I T R AT I O N
STANDARD USED FOR REDOX REACTION
• Cerium (IV) is a powerful oxidizing agent, ( it hydrolyzed to form ceric
hydroxide if the solution is not acid)
• Sulfuric acid solutions of cerium (IV) are stable.
• Cerium (IV) solutions to be standardized are usually prepared from
ammonium cerium (IV) sulphate . 2H2O
TITRATION OF ASCORBIC ACID AGAINST
AMMONIUM CERIUM SULPHATE
Procedure:
1. Weight and powder 20 tablet
2. Dissolve a quantity of powder containing 0.15g of
ascorbic acid in a mixture of 30 ml of water and 20 ml of
1 M sulphuric acid
3. Titrate with 0.1M ammonium cerium (IV) sulphate
4. Using ferroin solution as indicator 8-10 drops
5. End point red
torques
(Ferroin is a sutible indicator for many cerate titrations)
CALCULATION OF ASCORBIC ACID CONCENTRATION
USING CHEMICAL FACTOR
Hint
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So, Titrant
Sample
2 moles of Ce[NH4(SO4)4]
= 1 mole of ascorbic acid
2 moles of Ce[NH4(SO4)4]
= 176.12 gm of ascorbic acid
moles of Ce[NH4(SO4)4]
= 176.12 /2 gm of ascorbic acid
1000 ml x 1M of Ce[NH4(SO4)4]
= 88.065 gm of ascorbic acid
1 ml of 1M Ce[NH4(SO4)4]
= 0.088 gm of ascorbic acid
1 ml of 1 / 10 M Ce[NH4(SO4)4]
= 0.088 / 10 gm of ascorbic acid
1ml of 0.1 M Ce[NH4(SO4)4]
= 0.0088 gm of ascorbic acid
so, Equivalent factor (F) = 0.045 gm/ml
Concentration of ascorbic acid = E.P x F x f
mole= M x V
mole= g wt/ F wt
CALCULATION OF ASCORBIC ACID CONCENTRATION
USING STOICHIOMETRIC CALCULATION:
mmol Ce +4
M x V (e.p ml)
= 2 mmole ascorbic acid
Hint
mmole= M x Vml
mmole= mg wt/ F wt
= 2 wt in mg of ascorbic acid / formula wt of ascorbic acid
(0.1 x e.p ml ) /2 = wt in mg of ascorbic acid / 176.12
wt of ascorbic acid =
mg
% PURITY
• We dissolve a quantity of powder containing 0.15g of ascorbic acid so,
% purity = (X gm / 0.15) x 100
1ml of Ce[NH4(so4)4] = 0.0088 gm of ascorbic acid
e.p ml = X gm