Transcript Ca 2+ + Y 4 - Portal UniMAP

ERT207
Analytical Chemistry
Complexometric Titration
Pn Syazni Zainul Kamal
PPK Bioproses
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CO4: ABILITY TO DIFFERENTIATE
AND CALCULATE CONCENTRATION
OF ANALYTES OF VARIOUS
TITRIMETRIC METHODS (ACID-BASE,
COMPLEXATION, REDOX,
PRECIPITATION)
Types of Titrimetric Methods
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1.
2.
3.
4.
Classified into four groups based on type of
reaction involve;
Acid-base titrations
Complexometric titrations
Redox titrations
Precipitation titrations
Complexes
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Most metal ions (kation) can react with lone pair
electron from a molecule or anion to form covalent
bonds and produce coordination compound or
complexes.
Molecule or ion with at least 1 pair of unshared
electron can form covalent bond with metal ion =
ligands
The bonding between metal and ligand generally
involves formal donation of one or more of the ligand's
electron pairs
Eg of ligands = ammonia, cyanide ions, halide ions,
water (neutral/-ve charge mols or ions)
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Compound form between ligand and metal ion =
complexes or coordination compound (has charges or
neutral).
Examples of complex formation :
Ag+ + 2CNAg(CN)2Ag+ + 2NH3
Ag (NH3)2+
Metal ion
ligand
Complex/coordination compound
Metal ion – lewis acid (electron pair acceptor)
Ligand – lewis base (electron pair donor)
Coordination number – number of covalent bond formed
between metal ion and ligand
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A ligand that has one pair of unshared electron
such as NH3, is called unidentate.
Glycine (NH2CH2COOH) and ethylenediamine
(NH2CH2CH2NH2) which has two pairs of
unshared electron available for covalent
bonding, is called bidentate.
EDTA, has 6 pairs of unshared electrons =
hexadentate
EDTA Equilibrium
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Chelating agent - An organic agent that has
two or more group capable of complexing with
a metal ion (also called ligand)
Chelate – complex formed
Titration with chelating agent = chelometric
titration , a type of complexometric titration
Most widely used chelating agent in titration –
ethylenediaminetetraacetic acid (EDTA).
EDTA = Ethylenediaminetetraacetic acid
Hexadentate ligand
Has six bonding sites (the four carboxyl groups and
the two nitrogen providing six lone pairs electrons)
Tetraprotic acid (H4Y), can exist in many forms H3Y-,
H2Y2-, HY3- and Y4 only unprotonated ligand (Y4-) can complex with
metal ion
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Since EDTA is a tetraprotic acid, the dissociation
of EDTA as follows :
H4Y
H+ + H3Y-
H3Y-
H+ + H2Y2-
H2Y2-
H+ + HY3-
HY3-
H+ + Y4-
Ka1 = [H+][H3Y-] = 1.0 x10-2
[H4Y]
Ka2 = [H+][H2Y2-] = 2.1 x10-3
[H3Y-]
Ka3 = [H+][HY3-] = 6.9 x10-7
[H2Y2-]
Ka4 = [H+][Y4-] = 5.5 x10-11
[HY3-]
Complex formation constant of
EDTA
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EDTA can form complex with Ca2+ as the
following equilibrium :
Ca2+ + Y4-
CaY2-
(1)
The complex formation constant is :
Kf = KCaY2- = [CaY2-]
[Ca2+][Y4-]
(2)
Effect of pH on EDTA equilibria
If H+ concentration increases, equilibrium in equation 1 will shift to the left.
Chelating anion (Y4-) will react with H+. Dissociation of CaY2- in presence of acid
H+
CaY2
Ca2+ + Y4-
HY3-
H+
H+
H2Y2-
H3Y-
CH 4Y
2
CH4Y  [Ca ]
From the overall equilibrium
Ca2+ + H4Y
CaY2- + 4H+
CH4Y  [H 4 Y]  [H3Y ]  [H 2Y2 ]  [HY 3 ]  [Y 4 ]
H+
H4Y
Let us consider that CH4Y represent the total
uncomplexed EDTA
  
 
 

CH 4Y  Y 4  HY 3  H 2Y 2  H3Y   H 4Y 
If we substitute the values of [HY3-], [H2Y2-], [H3Y-] and [H4Y]
derived from the Ka values to this equation and divide each term
with [Y4-], we will get the following equation:CH 4Y
Y 
4

1
4
H   H 
 1

Ka4
 2
K 3 K 4

H 
 3
K 2 K 3 K 4

H 
 4
K 1K 2 K 3 K 4
Where α4 is the fraction of the total EDTA exists as Y4- .
Y 4
4 
C H 4Y
Y 4   4CH 4Y
Conditional formation constant
The equation for the complex formation between
Ca and EDTA is :
Ca2+ + Y4CaY2(3)
Kf = KCaY2- = [CaY2-]
(4)
[Ca2+][Y4-]
α4 = [Y4-] , hence [Y4-] = α4 CH4Y
(5)
CH4Y
Replacing [Y4-] into equation (4) :
Kf = KCaY2- = [CaY2-]
[Ca2+] α4CH4Y
(6)
Kfα4 = [CaY2-] = K’f
[Ca2+]CH4Y
(7)
K’f is the conditional formation constant which depends on α4
therefore K’f depends on pH. We can use equation (7) to calculate the
value of equilibirum concentration for EDTA species at specific pH to
replace equation (4)
Metal-EDTA titration curves
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Titration is perform by adding the chelating
agent (EDTA) to the sample (metal).
Titration curve – plotting the changes in metal
ion concentration (pM) versus volume of
titrant (EDTA)
Example of complexometric titration is by
adding 0.100 M EDTA to 100 ml 0.100 M
Ca2+ solution buffered at pH 11
Ca2+ + Y4CaY2-
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Before titration started – only have Ca2+ solution.
pCa = - log [Ca2+]
Titration proceed – part of Ca2+ is reacted with EDTA to
form chelate. [Ca2+] gradually decrease.
pCa= -log [remaining Ca2+]
At equivalence point – have convert all Ca2+ to CaY2So pCa can be determined from the dissociation of
chelate at a given pH using Kf.
K’f = Kf α4 =
[CaY2-]
[Ca2+] CH4Y
Excess titrant added – pCa can be determined from the
dissociation of chelate at a given pH using Kf.
Exercise
Calculate pCa in 100 ml of a solution of 0.100
M Ca2+ at pH10 after addition of 0, 50, 100,
150 ml of 0.100 M EDTA. Kf for CaY2- is
5.0x1010 and α4 is 0.35.
K’f = Kf x α4
= 5.0x1010 x 0.35
= 1.75x1010
a) Addition of 0.00 ml EDTA
[Ca2+] = 0.100 M
pCa = - log 0.100
= 1.00
b) Addition of 50.00 ml EDTA
Initial mmol Ca2+ =100ml x 0.100 M =10 mmol
mmol EDTA added = 50ml x 0.100 M = 5 mmol
mmol Ca2+ left
= 5 mmol
[Ca2+] = 5 mmol
= 0.0333 M
(100+50)ml
pCa = - log 0.0333 = 1.48
c) Addition of 100 ml EDTA
Initial mmol Ca2+ =100ml x 0.100 M =10 mmol
mmol EDTA added =100ml x 0.100 M = 10 mmol
Equivalence point is reached. We have convert all Ca2+ to CaY2-. mmol CaY2- =
mmol initial Ca2+
[CaY2-] =
10 mmol
= 0.05 M
(100+100)ml
K’f = Kf α4 = [CaY2-]
[Ca2+] CH4Y
K’f = [CaY2-] = 1.75 x 1010
[Ca2+] CH4Y
0.05
= 1.75 x 1010
(x)(x)
x = 1.7 x 10-6 so pCa = - log 1.7x10-6 = 5.77
d) Addition of 150 ml EDTA
Initial mmol Ca2+ =100ml x 0.100 M =10 mmol
mmol EDTA added =150ml x 0.100 M =15 mmol
mmol EDTA excess
= 5 mmol
= 5 mmol = 0.02M [CaY2-] =
10
= 0.04M
(100+150)ml
(100+150)ml
K’f = [CaY2-] = 1.75 x 1010
[Ca2+] (0.02)
0.04
= 1.75 x 1010
(x)(0.02)
x = 1.14 x 10-10 so pCa = - log 1.14x10-10 = 9.94
CH
4Y