Transcript EDTA Titrations

What happens to the
absorbed energy?
s1
t1
Energy
so
EDTA Titrations
Outline


What is EDTA?
Metal-Chelate Complexes





Chelate Effect
EDTA




ATP4- with Mg2+
Fe(NTA)23Fe(DTPA)2-
Acid Base Properties
aY nomenclature
Conditional Formation Constants
EDTA Titration
Metal-Chelate Complexes
Lewis Acid/Base Chemistry
Monodentate
Multidentate and Chelates
Review:

What is a Lewis Acid? Examples?

And a Lewis Base? Examples?
Transition Metal with ligand
Central Metal ion is a Lewis Acid
Ligand – All ligands are Lewis Bases
Multidentate
Multidentate or chelating ligand
attaches to a metal ion through more
than one atom is said to be
multidentate, or a chelating ligand.
Examples?
ATP4- can also form complexes
with metals
Complex of Iron and NTA
Fe3+ + 2
Fe(NTA)23-
Medical Applications
The Thalassemia Story
The Chelate Effect
Question: Describe in your own
words, the “chelate effect”.
The Chelate Effect!
2+
H2
N
Cd(H2O)62+ + 2
H2N
NH2
K = B2 = 8 x 109
OH2
H2
N
+ 4 H2O
Cd
N
H2
OH2
N
H2
2+
H3CH2N
Cd(H2O)62+ + 4CH3NH2
K = B2 = 4 x 106
OH2
NH2CH3
+ 4 H2O
Cd
H3CH2N
OH2
NH2CH3
13-2 EDTA
“EDTA is by far, the most widely used chelator in
analytical chemistry. By direct titration or through
indirect series of reactions, virtually every element of
the periodic table can be measured with EDTA.”
- Daniel Harris
Acid/Base Properties
H
H
(H6Y2+)
Acid/Base Properties
pKa = 0.0
H
(H5Y+)
Acid/Base Properties
pKa = 0.0
pKa = 1.5
(H4Y)
Acid/Base Properties
pKa = 0.0
pKa = 2.0 -
pKa = 1.5
(H3Y-)
Acid/Base Properties
pKa = 0.0
pKa = 2.0 -
- pK = 2.7
a
pKa = 1.5
(H2Y-2)
Acid/Base Properties
pKa = 2.0 -
pKa = 6.16
pKa = 0.0
- pK = 2.7
a
pKa = 1.5
(HY-3)
Acid/Base Properties
pKa = 2.0 -
pKa = 10.24
pKa = 6.16
pKa = 0.0
- pK = 2.7
a
pKa = 1.5
(Y-4)
Fraction as Y4The fraction of EDTA in form Y4- is given
as a4Fraction of EDTA ion the form Y4-
[Y ]  aY 4 [EDTA]
4
Concentration
in the form Y4-
(13-3)
Total Concentration of EDTA
Fraction as Y4Equation 13-4 in text
aY 
4
K1K 2 K3 K 4 K 5 K 6
[ H  ]6  [ H  ]5 K1  [ H  ]4 K1K 2  [ H  ]3 K1K 2 K3  [ H  ]2 K1K 2 K3 K 4  [ H  ]1 K1K 2 K3 K 4 K5  K1K 2 K3 K 4 K5 K 6
Example
You make a solution of 0.10 M EDTA and you
buffer the pH to (a) 10.0. What is aY4- ? (b)
What is aY4- if the pH of the solution is
buffered to 11.0?
[Y ]  aY 4 [EDTA]
4
4
4
[Y ]  0.36(0.10M )
[Y ]  0.85(0.10M )
4
[Y ] pH 10.0  0.036M
4
[Y ] pH 11.0  0.085M
EDTA reactions with Metals
Silver – Ag+
Mercury - Hg2+
Iron (III) – Fe3+
EDTA
ethylenediaminetetraacetate anion
=> EDTA-4 => Y-4
+1 cation
Ag+ + Y-4 D AgY-3
EDTA
ethylenediaminetetraacetate anion
=> EDTA-4 => Y-4
+2 cation
Hg+2 + Y-4 D HgY-2
EDTA
ethylenediaminetetraacetate anion
=> EDTA-4 => Y-4
+3 cation
Fe+3 + Y-4 D FeY-1
EDTA
ethylenediaminetetraacetate anion
=> EDTA-4 => Y-4
+n ion
M+n + Y-4 D MY(n-4)+
EDTA
KMY
[MY(n-4)+]
= -------------[M][Y-4]
KMY
[MY(n-4)+]
= ------------------[M+n] * a4 * [EDTA]
[Y ]  aY 4 [EDTA]
4
Conditional formation constant!
K'MY = KMY
[MY(n-4)+]
x a4 = ------------------[M+n] [EDTA]
Example

Calculate the concentration of Ni2+ in a
solution that was prepared by mixing
50.0 mL of 0.0300 M Ni2+ with 50.0 mL
of 0.0500 M EDTA. The solution was
buffered to pH of 3.00.
Two Parts
1. Reaction
2. Then equilibrium is established
EDTA Titrations
Figure 1310
Theoretical
titration
curves
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
Calculate the conditional constant:
Equivalence Volume
pCa at Initial Point
pCa at Equivalence
pCa at Pre-Equivalence Point
pCa at Post-Equivalence Point
Example
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA
in a solution buffered to a constant pH of 10.0.
[CaY-2]
K'CaY = KCaY * a4 = ---------------[Ca+2] * [EDTA]
where aY4- = 0.36
K'CaY =
at pH = 10.0
KCaY * a4 = 4.9 x 1010 * 0.36 = 1.8
KCaY = 4.9 x 1010
x 1010
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
Equivalence Volume
1 Mole of EDTA = 1 Mole of Metal
M1V1 = M2V2 (Careful of Stoichiometry)
50.0 mL (0.0500 M) = 0.1000 M (V2)
V2 = 25.0 mL
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
K'CaY = 1.8 x 1010
0.00 mL EDTA added
Initial Point
pCa = - log[Ca+2]
= - log(0.00500 M) = 2.301
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
At 25.0 mL (Equivalence Point)
Ca2+
Before
After
0.0025 moles
-
+
Y4-
0.0025 moles
-
-> CaY20.0025 moles
What can contribute to Ca2+ “after” reaction?
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
Ca2+ + Y4- D
I
-
C
+x
E
CaY2-
K 'CaY
[CaY 2 ]

[ EDTA][Ca 2 ]
0.0025moles/0.075
0.0025 moles/V L
-
+x
-x
K 'CaY 
0.0333  x
x2
X = [Ca2+] = 1.4 x10-6
+x
+x
0.0333 –x
pX = p[Ca2+] = 5.866
Pre-Equivalence Point
Let’s try 15 mL
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
At 15.0 mL
Ca2+
Before
0.0025 moles
After
0.0010 moles
+
Y4-
0.0015 moles
-
-> CaY20.0015 moles
What can contribute to Ca2+ after reaction?
K’CaY = 1.8 x 1010
negligible
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
At 15.0 mL
[Ca2+] = 0.0010 moles/0.065 L
[Ca2+] = 0.015384 M
p [Ca2+] = 1.812
Post Equivalence Point
Let’s Try 28 ml
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
At 28.0 mL
Ca2+
Before
After
0.0025 moles
-
+
Y4-
0.0028 moles
0.0003 moles
-> CaY20.0025 moles
What can contribute to Ca2+ after titration?
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the
titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a
solution buffered to a constant pH of 10.0.
Ca2+ + Y4-  CaY2-
K 'CaY
0.078 L
I
C
E
-
0.0003 moles/V 0.0025 moles/V
'
+x
[CaY 2 ]

[ EDTA][Ca 2 ]
+x
-x
K CaY
0.03205 x

(0.003846 x)(x)
X = [Ca2+] = 4.6 x10-10
+x
0.003846 + x
0.03205 –x
pX = p[Ca2+] = 9.334
Experimental Considerations
EDTA Titration Techniques
Erichrome Black T
MgIn + EDTA
(red)
a
MgEDTA + In
(colorless) (blue)
Figure 13-13
Guide to EDTA
titrations, light
color, pH range
for quantitative
analysis, dark
area where
ammonia must
be present