Download lecture 7 complexation

Complexation and EDTA
Chemistry 321, Summer 2014
Complexation allows the titration of metal ions
Chemical analysis scheme:
A (aq) +
analyte
Kf
R (aq)
reagent
direct analysis of A; need a relatively
clean sample with analyte having
inherent properties amenable to
quantitative analysis
AR (s)
neutral complex
precipitate (ppt)
Complexation allows the titration of metal ions
Chemical analysis scheme:
A (aq) +
analyte
Kf
R (aq)
reagent
AR (s)
neutral complex
precipitate (ppt)
indirect analysis of A
Neutral complex
• extractable (organic phase)
• precipitate (grav. analysis)
Charged complex
• should be highly colored to
facilitate selective monitoring
by absorbance spectrometry
Complexation allows the titration of metal ions
Chemical analysis scheme:
A (aq) +
analyte
R (aq)
reagent
Kf
AR (s)
neutral complex
precipitate (ppt)
[AR]
K f = [A] [R]
R often has acid/base properties so you should control the pH !
Complexation allows the titration of metal ions
Chemical analysis scheme:
A (aq) +
analyte
R (aq)
reagent
Kf
AR (s)
neutral complex
precipitate (ppt)
• Filtering to remove particulates from the sample may be required.
• If selectivity is not adequate (e.g., a sample with many components),
either A or AR species are often analyzed using chromatography.
Complexation allows the titration of metal ions
Organics are usually analyzed by extraction from the aqueous
phase (covered in latter part of course)
Metal ions form polydentate chelates with various organic
molecules (called chelating agents).
Neutral complex example (bidentate ligand)
Cu2+ + 2
8-hydroxyquinoline
(also known as
oxine)
+ 2 H+
In this case, the
copper ion is part
of two 5membered rings
Charged complex example (tridentate ligand)
Again, a 5membered
ring
1,10-phenanthroline
Note that the charge on the metal ion does not necessarily need
to equal the number of ligands (and note that the complex
retains the same charge as the ion).
Notes on complex stability
5- or 6-membered rings are thermodynamically stable
Extent of complexation is pH dependent
Number of ligands depends on the coordination number of
the metal ion
For quantitative analysis, as usual, a large Kf is desired
Monodentate chelating agents are not used for quantitative
analysis (no sharp endpoint)
Complexation with EDTA
EDTA = ethylene diamine tetra-acetic acid
It’s a polyprotic acid, which will be abbreviated “H4Y”.
Thus it has five different charge states; only the Y4– state
will complex metal cations.
Properties of the metal-EDTA complex
The chelation is “claw” type,
with all new rings created by
the chelation being 5membered. The coordination
of the metal ion itself is
octahedral, with four of the
ligands O– and the other two to
the lone pair on the N.
EDTA complexation equilibrium
Kf
MY|n–4|–
Mn+ + Y4–
n–4 –
[MY
]
K f = n+
[M ] [Y4– ]
These Kfs are listed in
Table C-4, page 807
One can standardize EDTA concentration using a metal
ion with a well-known Kf, then use the standardized
concentration to analyze a different metal ion.
A note of caution using EDTA
Since the concentration of Y4– is dependent on pH,
make sure there is sufficient EDTA in the solution in
the first place for the complex to form.
For instance, consider the equilibria of the calcium ion Ca2+
with EDTA (there are many typos in the text):
Kf
CaY2–
Ca2+ + Y4–
HY3–
H2Y2–
H3Y–
H4Y
A note of caution using EDTA
Since the concentration of Y4– is dependent on pH,
make sure there is sufficient EDTA in the solution in
the first place for the complex to form.
For instance, consider the equilibria of the calcium ion Ca2+
with EDTA (there are many typos in the text):
Kf
CaY2–
Ca2+ + Y4–
[CaY 2– ]
K f = [Ca 2+ ] [Y4– ]
HY3–
H2Y2–
H3Y–
H4Y
A note of caution using EDTA
Since the concentration of Y4– is dependent on pH,
make sure there is sufficient EDTA in the solution in
the first place for the complex to form.
For instance, consider the equilibria of the calcium ion Ca2+
with EDTA (there are many typos in the text):
Kf
CaY2–
Ca2+ + Y4–
2–
[CaY ]
K f = [Ca 2+ ] [Y4– ]
HY3–
H2Y2–
H3Y–
H4Y
Another way to express the
equilibrium is:
Ca2+ + H4Y
CaY2– + 4 H+
A note of caution using EDTA
Since the concentration of Y4– is dependent on pH,
make sure there is sufficient EDTA in the solution in
the first place for the complex to form.
For instance, consider the equilibria of the calcium ion Ca2+
with EDTA (there are many typos in the text):
Kf
CaY2–
Ca2+ + Y4–
2–
[CaY ]
K f = [Ca 2+ ] [Y4– ]
HY3–
H2Y2–
H3Y–
H4Y
Another way to express the
equilibrium is:
Ca2+ + H4Y
CaY2– + 4 H+
At increasing acidity (lower pH), H4Y will be favored
The alpha plot for EDTA complexation
The formal concentration of EDTA = CH4Y
= [Y4–] + [HY3–] + [H2Y2–] + [H3Y–] + [H4Y]
Note that complexed Y
(e.g., CaY2–) does not
count toward CH4Y
[Y 4– ]
so aY 4– = CH 4Y
and aY 4– ´ CH 4Y = [Y 4– ]
For complexation with EDTA, only the
fraction that is Y4– matters
Since complex
formation occurs
only with Y4–, the pH
of the solution
needs to be quite
alkaline.
Plotted in red in the
alpha plot (see eq. 912 and fig. 9-1) –
seems αY4– = 0 at or
below pH 8.
Ca2+
log αY4–
αY4– = 1.0 up here
A log alpha plot
4
6
8
10
12
14 pH
Ca2+
log αY4–
αY4– = 1.0 up here
At pH 8, αY4– = 0.0056 (not 0)
A log alpha plot
4
6
8
10
12
14 pH
Ca2+
log αY4–
αY4– = 1.0 up here
At pH 8, αY4– = 0.0056 (not 0)
Even minute amounts of Y4– can
produce complexation
4
6
8
10
12
14 pH
Solving for αY4–
To determine the mole fraction of Y4– (the complexing form) in
a solution, you can expand the fraction [Y4–]/CH4Y, as done for
other polyprotic acids previously. You end up with an
expression given in equation 9.12 (page 300) in the text:
[H + ] [H + ]2
[H + ]3
[H + ]4
= = 1 + + + + 4–
[Y ] aY 4–
K a1
K a1K a2 K a1K a2 K a3 K a1K a2 K a3K a4
CH 4Y
1
Where Ka1, Ka2, Ka3, and Ka4 are the stepwise acid dissociation
equilibrium constants. Note that the right side of the expression
only depends on hydrogen ion concentration (which is pH). The
limiting behavior of this equation conforms to the shape of the
curve given on the previous slide.
To determine the concentration of Ca2+
from titration with standardized EDTA
Replacing [Y4–] in the Kf expression:
2–
[CaY ]
K f = 2+
[Ca ] aY 4– CH 4Y
Strategy:
• Look up Kf for Ca2+/EDTA in Table C.4 ( = 5.01 × 1010)
• Calculate αY4– using equation 9.12
• Calculate [CaY2–] and CH4Y using ICE table
• Solve for [Ca2+]
Alternatively, the conditional
formation constant Kf ’ may be used
You can re-express the formation constant equilibrium
expression in such a way as to keep all the pH dependence out of
the right-hand side of the equation:
[CaY 2– ]
K f aY 4– = 2+
[Ca ] CH 4Y
Define Kf ’ as the conditional formation constant = Kf αY4–
Taking the logarithm of both sides: log Kf ’ = log Kf + log αY4–
Alternatively, the conditional
formation constant Kf ’ may be used
You can re-express the formation constant equilibrium
expression in such a way as to keep all the pH dependence out of
the right-hand side of the equation:
[CaY 2– ]
K f aY 4– = 2+
[Ca ] CH 4Y
Define Kf ’ as the conditional formation constant = Kf αY4–
Taking the logarithm of both sides: log Kf ’ = log Kf + log αY4–
pH dependent
constant for a given metal
Using the conditional formation constant
The goal is to quantitatively titrate 99.9% of a metal ion in a
solution using EDTA. To do this successively, you must predict
and control (using pH) the percent complexation.
Kf ’
Mn+
+
Y4–
MY|n–4|–
Initial
0
0
[MY]0
Change
+x
+x
–x
Equilibrium
x
x
[MY]0 –x
Using the conditional formation constant
In a quantitative titration, at the equivalence point, you can treat
the solution as if it were made from dissolving MY|n–4|– and then
letting that equilibrate into small amounts of Mn+ and various forms
of EDTA. In other words, [Mn+] = CH4Y = x (from ICE table)
[MY ]
[MY ]0 - x [MY ]0
K f ' = n+
= » 2
2
[M ] CH 4Y
x
x
(if x << [MY]0)
Using the conditional formation constant
In a quantitative titration, at the equivalence point, you can treat
the solution as if it were made from dissolving MY|n–4|– and then
letting that equilibrate into small amounts of Mn+ and various forms
of EDTA. In other words, [Mn+] = CH4Y = x (from ICE table)
[MY ]
[MY ]0 - x [MY ]0
K f ' = n+
= » 2
2
[M ] CH 4Y
x
x
(if x << [MY]0)
The quantitative goal of 99.9% of Mn+ titrated can be expressed:
free metal ion
x
-3
= ³ 1 ´ 10 EDTA - bound metal ion [MY ]0
0.1%
Estimating the minimum amount of metal
ion that can be titrated using EDTA
Rearranging the previous inequality:
Substituting into the Kf ’
expression:
-3
x ³ 10 [MY ]0
K f ' ³ (10
[MY ]0
-3
[MY ]0 )
2
Simplifying:
K f ' [MY ]0 ³ 106
Which, prior to titration
is equal to:
K f ' [M ]0 ³ 10
n+
6
Challenge problem
Consider a magnesium ion titration with EDTA. The initial concentration of the sample to be titrated [Mg2+] = 1.0 × 10–2 M. If
Kf for magnesium ion with EDTA is 6.2 × 108,
• can the magnesium ion be quantitatively (99.9%) titrated at
pH 10.00?
• can the magnesium ion be quantitatively (99.9%) titrated at
pH 8.00?
Show the calculation to support your answers.