Download Chapter 12 EDTA Titrations Coordination Number Geometries Ligands

Acids and Bases
Chapter 12
EDTA Titrations
• Definitions: Lewis – Electrons (acid: electron pair acceptor);
BrØnsted-Lowry (acid: proton donor)
Lewis acid-base concept in Metal-Chelate Complexes
Metal ions (electron pair acceptor) Lewis acid
Ligand (electron pair donor) Lewis base
Coordination Number
Complex Formation
Formation of coordinate bonds between Lewis Acids/Bases
K1
Ag ( aq ) + NH (aq)
+
• The atom of the ligand
that supplies the
nonbonding electrons
for the metal-ligand
bond is the donor atom.
• The number of these
atoms is the
coordination number.
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Ag ( NH ) ( aq ) + NH (aq)
3
Ag ( NH ) ( aq )
+
K2
+
3
1
[ Ag ( NH ) ]
[ Ag ][ NH ]
+
Ag ( NH ) ( aq )
3
+
K =
3
2
+
K =
3
2
+
[ Ag ( NH ) ]
[ Ag ( NH ) ][ NH ]
3
2
+
3
3
3
Formation constants (Kf) are the equilibrium constants for complex ion
formation. The overall, or cumulative, formation constants are
denoted i
Ag (aq) + 2 NH (aq)
+
Ag ( NH ) (aq)
Kf
+
3
3
2
+
K =
f
[ Ag ( NH ) ]
= β = K ⋅K
[ Ag ][ NH ]
3
2
+
2
2
1
2
3
Geometries
There are two common
geometries for metals with
a coordination number of
four:
Tetrahedral & Square planar
By far the mostencountered geometry,
when the coordination
number is six, is octahedral.
Ligands
• Monodentate ligand: binds to
a metal ion through only one
atom, e.g., CN• Multidentate ligand or
chelating ligand: has more
than one ligand donor atoms.
• In ethylenediamine,
NH2CH2CH2NH2 (i.e., en),
each N is a donor atom.
en is bidentate.
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Chelating Effect
A multidentate ligand to form more stable metal complexes
than those formed by similar monodentate ligand
Chelating Agents
• Porphyrins (tetradentate ligands, in
heme and chlorophyll)
• Adenosine triphosphate (ATP)
Some Metals Form 7 or 8
Coordinate Complexes
EDTA
Ethylenediaminetetraacetic acid (H4EDTA or H4Y)
Ethylenediaminetetraacetate anion (EDTA-4 or Y-4)
Ethylenediaminetetraacetate, mercifully abbreviated
EDTA, has six donor atoms.
EDTA is a primary standard material.
EDTA Complexes
• EDTA forms 1:1 complexes with most metals (Not with
Group 1A metals)
• EDTA complexes are usually stable water soluble
complexes with high formation constants
• Formation constant, Kf, (or stability constant):
M n + + Y 4− ↔ MY n −4
Kf =
[MY ]
[M ][Y ]
n−4
n+
4−
• Kf could have been defined for any form of EDTA
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pH affects EDTA titration
Acid-Base Properties of EDTA:
CH2CO2H
HO2CH2C
EDTA is a hexaprotic
system (H6Y2+) with
4 carboxylic acids
and 2 ammoniums
HNCH2CH2NH
CH2CO2H
HO2CH2C
H6Y2+
pK1 = 0.0
pK 2 = 1.5
pK 5 = 6.16
pK 6 = 10.24
pK 3 = 2.0
pK 4 = 2.66
Mn+ + Y4− ↔ MYn−4
pH:
Hn Yn−4
Fraction of EDTA in the form Y4αY =
4−
[H Y ]+ [H Y
4−
44−Y −
Y4-
[Y ]
] + [H Y ] + [H Y ] + [H Y ]+ [HY ]+ [Y ]
4−
αY =
ααY ==
pH Dependence of
6
2+
5
+
4
3
−
2
2−
3−
4−
[Y 4− ]
[Y 4− ]
=
[ EDTA] C EDTA
K1K2K3K4 K5K6
{[H ] +[H ] K +[H ] K K +[H ] K K K +[H ] K K K K +[H ]K K K K K + K K K K K K }
+ 6
+ 5
1
+ 4
1
2
+ 3
1
2
3
+ 2
1
2
3
4
+
1
2
3
4
5
1
2
3
4
5
6
Conditional Formation Constant
Fractional Composition Diagram for EDTA
αY =
4−
[Y ]
4-
[EDTA]
[Y ] = α [EDTA]
4-
Kf =
Y 4−
Conditional formation
constant:
[MY ]
[MY ]
[M ][Y ] = [M ]α [EDTA]
n−4
n+
n−4
4−
n+
Y
K f' = α Y 4- K f =
Fixing the pH by buffering, then α Y
At any fixed pH, find α Y
[ MY n − 4 ]
[ M n + ][Y 4 − ]
Kf =
M n + + Y 4− ⇔ MY n− 4
4−
4−
4-
[MY ]
[M ][EDTA ]
n−4
n+
is a constant.
and evaluate Kf’
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EDTA Titration Curve
Example: Calculate the concentration of free Ca2+ in a
solution of 0.10 M CaY2- at pH 10 and pH 6. Kf for
CaY2- is 1010.65 (Table 12-2)
Ca
2+
+ EDTA ↔ CaY
1. Excess Mn+ left after each
addition of EDTA. Conc. of free
metal equal to conc. of unreacted
Mn+.
K = α Y 4− K f
2−
'
f
at pH = 10.00, K = α K = (0.30)(10 ) = 1.3 × 10
'
10.65
Y 4−
f
10
f
at pH = 6.00, K = α K = (1.8 ×10 )(10 ) = 8.0 × 10
−5
'
Y 4−
f
10.65
5
f
2. Equivalence point: [Mn+] =
[EDTA] Some free Mn+ generated
by MYn-4 ⇔ Mn+ + EDTA
Ca 2+ + EDTA ↔ CaY2−
K f' =
Conci
0
0
0.1
Concf
x
x
0.1 - x
[CaY ] 0.1 − x
[Ca ][EDTA ] = x
2−
2+
2
[
3. Excess EDTA. Virtually all
metal in MYn-4 form.
]
x = Ca 2+ = 2.7 ×10−6 M at pH = 10
= 3.5 ×10-4 M at pH = 6
EDTA Titration Curve
EDTA Titration Curve
1. Titration reaction:
'
M n+ + EDTA⇔ MYn−4 Kf =αY4− Kf
EXAMPLE: Derive a titration curve for the titration of 50.0
mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and
26.00 mL of 0.080 M EDTA.
Ca 2+ + EDTA → CaY 2 −
K f' = α Y 4− K f = 0.30 *1010.65 = 1.3 × 1010
2. Reaction completes at each
point in the titration if Kf’ is large.
At equivalence point, Vol. of EDTA = 25.0 mL
5.00 ml,
25.00 ml,
26.00 ml,
3. Plot pM (= -log[Mn+]) vs.
volume of EDTA added
EDTA Titration Curve
EXAMPLE: Derive a titration curve for the titration of 50.0
mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and
26.00 mL of 0.080 M EDTA.
Before the equivalence point
At the equivalence point
After the equivalence point
EDTA Titration Curve
EXAMPLE: Derive a titration curve for the titration of 50.0
mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and
26.00 mL of 0.080 M EDTA.
At equivalence point, Vol. of EDTA = 25.0 mL
At equivalence point, Vol. of EDTA = 25.0 mL
5mL before the equivalence point
almost all the metal is in the form, CaY2-
[CaY ] = (0.040)
2-
[Ca ]
2+
25.0 - 5.0
50.0
=
(0.040)
= 0.0291 M
25.0
5.00 + 50.0
Fraction
Remaining
pCa
2+
= − log(0.0291) = 1.54
50.0
= 0.0267 M
25.00 + 50.0
Ca 2+ + EDTA ↔ CaY 2−
Conci
0
0
0.0267
Concf
x
x
0.0267 - x
K f' =
[CaY ] 0.0267 − x
[Ca ][EDTA] = x = 1.8 ×10
2−
2+
10
2
x = 1.2 ×10 −6 M
pCa 2+ = − log(1.2 × 10−6 ) = 5.91
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EDTA Titration Curve
EXAMPLE: Derive a titration curve for the titration of 50.0
mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and
26.00 mL of 0.080 M EDTA.
At equivalence point, Vol. of EDTA = 25.0 mL
26.00 mL
1.0 mL excess EDTA, after the equivalence point
[EDTA ] = (0.080)
[CaY ] = (0.040)
2-
K f' =
1.00
= 1.05 × 10 −3 M
50.0 + 26.00
50.0
= 2.63 ×10 −2 M
50.0 + 26.00
[CaY ]
2.63 ×10
[Ca ][EDTA] = [Ca ](1.05 ×10
2−
2+
−2
2+
−3
)
= 1.8 ×1010
Ca 2+ = 1.4 ×10 −9 M
pCa 2+ = 8.86
pH affects the titration of Ca2+ with EDTA
Auxiliary Complexing Agents
• A ligand that binds strongly enough to the metal to
prevent hydroxide precipitation, but weak enough to
be displaced by EDTA (e.g., ammonia, tartrate,
citrate, or trithanolamine)
Ammonia is a common auxiliary complex for transition metals
like zinc (p. 239)
K 'f' = α Zn 2+ α Y 4− K f
Kf’’ is the effective formation constant at a fixed concentration of
auxiliary complexing agent.
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Metal Ion Indicators
Metal Ion Indicators
• To detect the end point of EDTA titrations, we usually
use a metal ion indicator or an ion-selective electrode
(Ch. 15)
• Metal ion indicators change color when the metal ion is
bound to EDTA:
MgEbT + EDTA ↔ MgEDTA + EbT
(Red)
(Colorless ) (Blue)
– Eriochrome black T is an organic ion
• The indicator must bind less strongly than EDTA
EDTA Titration Techniques
• Direct titration: analyte is titrated with standard EDTA
with solution buffered at a pH where Kf’ is large
• Back titration: known excess of EDTA is added to
analyte. Excess EDTA is titrated with 2nd metal ion.
Example: 25.0 mL of an unknown Ni2+ solution was
treated with 25.00 mL of 0.05283 M Na2EDTA. The pH
of the solution was buffered to 5.5 and than back-titrated
with 17.61 mL of 0.02299 M Zn2+. What was the
unknown Ni2+ in M?
mol EDTA = (25.00 mL)(0.05283 M) = 1.32 mmol EDTA
Zn 2+ + Y 4- ↔ ZnY 2-
mol Zn 2+ = (17.61 mL)(0.02299 M) = 0.4049 mmol Zn 2+
Ni 2+ + Y 4- ↔ NiY 2mol Ni
2+
= 1.321 mmol EDTA - 0.4049 mmol Zn 2+ = 0.916 mmol
M Ni 2+ = (0.916 mmol)/(25.00 mL) = 0.0366 M
EDTA Titration Techniques
• Direct titration: analyte is titrated with standard EDTA with
solution buffered at a pH where Kf’ is large
• Back titration: known excess of EDTA is added to analyte.
Excess EDTA is titrated with 2nd metal ion.
• Displacement titration: For metals without a good indicator ion,
the analyte can be treated with excess Mg(EDTA)2-. The analyte
displaces Mg, and than Mg can be titrated with standard EDTA
• Indirect titration: Anions can be analyzed by precipitation with
excess metal ion and then titration of the metal in the dissolved
precipitate with EDTA.
• Masking agent: protects some componet of the analyte from
reaction with EDTA (render metal ions inactive without actually
removing them from solution). Demasking: releasing metal ion
from a masking agent.
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