outline
DESCRIPTION
Outline. EDTA Acid Base Properties a Y 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. - PowerPoint PPT PresentationTRANSCRIPT
Outline
EDTAEDTA Acid Base Properties Y nomenclature Conditional Formation Constants
EDTA TitrationEDTA Titration
EXAMPLE:
Calculate the conditional constant: =1.8 x 1010
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.
pCa at Equivalence
Equivalence Volume V2 = 25.0 mL
pCa at Pre-Equivalence Point
pCa at Post-Equivalence Point
pCa at Initial Point = 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)At 25.0 mL (Equivalence Point)
Ca2+ + Y4- -> CaY2-
Before 0.0025 moles 0.0025 moles
-
After - - 0.0025 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- CaY2-
I - - 0.0025 moles/V
C +x +x -x
E +x + x 0.0333 –x
]][[
][2
2'
CaEDTA
CaYK CaY
2' 0333.0
x
xK CaY
X = [Ca2+] = 1.4 x10-6
pX = p[Ca2+] = 5.866
0.0025moles/0.075 L0.0025moles/0.075 L
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+ + Y4- -> CaY2-
Before 0.0025 moles 0.0015 moles
-
After 0.0010 moles - 0.0015 moles
What can contribute to Ca2+ after reaction?
K’CaY = 1.8 x 1010
negligiblenegligible
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 Mp [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+ + Y4- -> CaY2-
Before 0.0025 moles 0.0028 moles
-
After - 0.0003 moles
0.0025 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-
I - 0.0003 moles/V 0.0025 moles/V
C +x +x -x
E +x 0.003846 + x 0.03205 –x
]][[
][2
2'
CaEDTA
CaYK CaY
))(003846.0(
03205.0'
xx
xK CaY
X = [Ca2+] = 4.6 x10-10
pX = p[Ca2+] = 9.334
0.078 L
Chapter 23
An Introduction toAn Introduction toAnalytical Analytical
SeparationsSeparations
Problems
Chapter 23 1, 15, 20 a and b, 27, 29, 30, 37, 44
Chapter 24 1, 3, 4, 5, 6 From 23 23-33,
What is Chromatography?
Parts of Column
column support stationary phase mobile phase
Types of Chromatography
1. Adsorption2. Partition
3. Ion Exchange4. Molecular Exclusion (gel-filtration)
5. Affinity chromatography
Section 23-3 A Plumber’s View of A Plumber’s View of ChromatographyChromatography
The chromatogram“Retention time”
“Relative retention time”“Relative Retention”
“Capacity Factor”
A chromatogramRetention 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 chromatogramAdjusted Retention time (t’
r) - is the retention time corrected for dead volume “the difference between tr and a non-retained
solute”
A chromatogramRelative Retention () -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.
1
2
'
'
r
r
t
t 1 '' 21 sottwhere rr
A chromatogramCapacity 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”.
m
mr
t
ttk
'
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”.
m
mr
t
ttk
'
42
42251'
m
mrbenzene t
ttk = 5.0
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”.
m
mr
t
ttk
'
42
42333'
m
mrtoulene t
ttk = 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.
1
2
'
'
r
r
t
t sec39.1
sec209
sec291
'
'
benzene
toulene
t
t
Efficiency of Separation
“Two factors”1) How far apart they are ()
2) Width of peaks
ResolutionResolution
Resolution
avw 2/1
r
av
r t589.0
w
tResolution
Example – measuring resolution
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?
av
r
w
tResolution
7
21
1.116)(13
407424Resolution
Why are bands broad?
Diffusion and flow related effects