LAB. 4 Calcium lacta..

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Transcript LAB. 4 Calcium lacta..

Calcium lactate.
pentahydrate
NADA AL-THEYAB
calcium lactate
MF:C6H10CaO6 . 5H2O
Mwt: 308 Da
Chemical name : calcium 2-
hydroxylpropanoate pentahydrate
Uses of calcium lactate
 Use
in food as ( a baking powder)
 Given
medicinally as : antacid and also to
treat calcium deficiencies.
IDENTIFICATION
IDENTIFICATION OF CALCIUM
A. Flame test (ignition test)
 The
powdered tablet, when
moistened with hydrochloric acid ( use
beaker with spatula), and introduced on
platinum wire into flame, impart a
brick red colour to the flame
The flame is brick red
due topresence of Ca
B. Identification of calcium salt
1.
Make a solution of calcium salt (powdered calcium + water any
amount)
2.
Add 2 drops of methyl orange
3.
Neutralized with 6M ammonium hydroxide until the colour of the
solution change
4.
Add 3N HCl dropwise until the colour of the solution change again
5.
Add ammonium oxalate , white PPt. will form
6.
Divide the solution into two test tubes:
First test tube
Add 6N acetic acid
The ppt. is insoluble
Second test tube
add 6M HCL its dissolve the ppt.
ASSAY
COMPLEXOMETRIC TITRATION
Chelates: EDTA- the ultimate titrating agent
for metals

The most widely used chelating agent in titrations is ethylenediaminetetraacetic
acid(EDTA) the formula for EDTA is:

Each of two nitrogens and each of the four carboxyl groups contains a pair of
unshared electrons capable of complexing with metal ion. Thus, EDTA contains six
complexing groups. We will represents EDTA by the symbol H4Y.

It is the unprotonated ligand Y4- that forms complexes with metal ions, that is the
protons are displaced by the metal ion upon complexation.
Calcium-EDTA complexes
Chelates: EDTA- the ultimate
titrating agent for metals

EDTA most common complexmetric titrant

Forms 1:1 complexes with the majority of metals

Forms stable water soluble complexes.

Is a primary standard.

Is normally used like sodium salt to increase the solubility.

EDTA forms stable complexes with: Ba2+ , Ca2+ , Fe3+
,Fe2+ , Mg2+ , Mn2+ , Ni2+ , Hg2+ and others
Metal-EDTA Titration

A titration preformed by adding the chelating
agent(EDTA) to the sample (calcium metal ion).

Only some metal chelates are stable enough to
allow titrations in acid solution: others require
alkaline solution
Indicator Eriochrome Black T (E B T):

Is a complexometric indicator that is part of the
complexometric titrations

In its protonated form, Eriochrome Black T is blue. It turns
red when it forms a complex with calcium, magnesium,
or other metal ions.

When used as an indicator in an EDTA titration, the
characteristic blue end-point is reached when sufficient
EDTA is added and the metal ions bound to the indicator
are chelated by EDTA, leaving the free indicator
molecule.
EBT is blue in a
buffered solution
at pH 10. It turns
red when Ca2+
ions are added.
Complexometric titration (procedure)
1.
powdered 20 tablets and Weigh a quantity of powder
containing 0.3g of Ca-lactate pentahydrate or its equivalent(
needs a calculation )
2.
Dissolve it as completely as possible in 5 ml of water
3.
Titrate with 0.05M sodium edetate (add 15 ml)
4.
Add 8 ml of 5M sodium hydroxide (use ammonia buffer
instead)
5.
Add few grams of solochrome dark blue (use Eriochrome
Black T ( E.B.T ) instead)
6.
Continue titration until the colour change from pink to blue
(but our e.p red
violet
blue)
Calculation of calcium lactate
concentration using chemical factor
So, Titrant
Sample
1 moles of EDTADS
= 1 mole of Ca-lactate
1 moles of EDTADS
= 308 gm of Ca-lactate
1000 ml of 1M of EDTADS
= 308 gm of Ca-lactate
1 ml of 1/20 M EDTADS
= 0.308gm of Ca-lactate
1 ml of 0.05 M EDTADS
= 0.0308 / 20 gm of Ca-lactate
1ml of 0.05 M EDTADS
= 0.0154 gm of Ca-lactate
so, Equivalent factor (F) = 0.0154 gm/ml
Concentration of calcium lactate= E.P x F x f
% purity

Each ml of 0.05 M sodium edetate is equivalent to 0.0154gm of C6H10CaO6 .
5H2O

We dissolve a quantity of powder containing 0.3g of clcium lactate
so,
% purity = (X gm / 0.3) x 100
1ml of EDTADS = 0.0154 gm of ascorbic acid
e.p ml = X gm