Transcript EDTA Titrations

Outline

EDTA

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

Acid Base Properties
aY nomenclature
Conditional Formation Constants
EDTA 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.
Calculate the conditional constant: =1.8 x 1010
Equivalence Volume V2
= 25.0 mL
pCa at Initial Point
= 2.301
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.
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
Chapter 23
An Introduction to
Analytical Separations
Problems
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Chapter 23
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1, 15, 20 a and b, 27, 29, 30, 37, 44
Chapter 24
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1, 3, 4, 5, 6
From 23
23-33,
What is Chromatography?
Parts of Column
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column
support
stationary phase
mobile phase
Types of Chromatography
Adsorption
2.
Partition
3.
Ion Exchange
Molecular Exclusion (gel-filtration)
5.
Affinity chromatography
1.
4.
Section 23-3
A Plumber’s View of
Chromatography
The chromatogram
“Retention time”
“Relative retention time”
“Relative Retention”
“Capacity Factor”
A chromatogram
Retention time (tr) – the time required for a substance to pass from one
end of the column to the other.
Adjusted Retention time – is the retention time corrected for dead volume
“the difference between tr and a non-retained solute”
A chromatogram
Adjusted Retention time (t’r) - is the retention time corrected for dead
volume “the difference between tr and a non-retained solute”
A chromatogram
Relative Retention (a) -the ratio of adjusted retention times for any two
components. The greater the relative retention the greater the separation.
Used to help identify peaks when flow rate changes.
t 'r 2
a
t 'r1
where t 'r1  t 'r 2  so a  1
A chromatogram
Capacity Factor (k’) -”The longer a component is retained by the column,
the greater its capacity factor. To monitor performance of a column – one
should monitor the capacity factor, the number of plates, and peak
asymmetry”.
tr  tm
k'
tm
An Example
A mixture of benzene, toulene, and methane was injected into
a gas chromatograph. Methane gave a sharp peak in 42
sec, benzene was @ 251 sec and toulene eluted at 333
sec. Find the adjusted retention time (for each solute), the
capacity factor (for each solute) and the relative retention.
Adjusted retention time (t’r) = total time – tr (non retained component)
t’r(benzene) = 251 sec – 42 sec = 209 s
t’r (toulene) = 333-42 sec = 291 s
An Example
A mixture of benzene, toulene, and methane was injected into
a gas chromatograph. Methane gave a sharp peak in 42
sec, benzene was @ 251 sec and toulene eluted at 333
sec. Find the adjusted retention time (for each solute), the
capacity factor (for each solute) and the relative retention.
Capacity Factor (k’) -”The longer a component is retained by the column, the greater
its capacity factor. To monitor performance of a column – one should monitor the
capacity factor, the number of plates, and peak asymmetry”.
tr  tm
k'
tm
tr  t m 251 42
= 5.0
k 'benzene 

tm
42
An Example
A mixture of benzene, toulene, and methane was injected into
a gas chromatograph. Methane gave a sharp peak in 42
sec, benzene was @ 251 sec and toulene eluted at 333
sec. Find the adjusted retention time (for each solute), the
capacity factor (for each solute) and the relative retention.
Capacity Factor (k’) -”The longer a component is retained by the column, the greater
its capacity factor. To monitor performance of a column – one should monitor the
capacity factor, the number of plates, and peak asymmetry”.
tr  tm
k'
tm
tr  t m 333 42
k 'toulene 

tm
42
= 6.9
An Example
A mixture of benzene, toulene, and methane was injected into
a gas chromatograph. Methane gave a sharp peak in 42
sec, benzene was @ 251 sec and toulene eluted at 333
sec. Find the adjusted retention time (for each solute), the
capacity factor (for each solute) and the relative retention.
Relative Retention (a) -the ratio of adjusted retention times for any two components.
The greater the relative retention the greater the separation. Used to help identify
peaks when flow rate changes.
t 'r 2
a
t 'r1
t 'toulene 291sec
a

 1.39sec
t 'benzene 209sec
Efficiency of Separation
1)
“Two factors”
How far apart they are (a)
2)
Width of peaks
Resolution
Resolution
t r 0.589t r
Resolution

w av
w1/ 2 av
Example – measuring
resolution
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A peak with a retention time of 407 s has a width at the
base of 13 s. A neighboring peak is eluted at 424 sec with
a width of 16 sec. Are these two peaks well resolved?
t r
Resolution 
w av
424 407
Resolution 
 1.17
1
(13 16) 2
Why are bands broad?
Diffusion and flow related effects