1. Outline EDTA EDTA Titration Acid Base Properties aY nomenclature
Conditional Formation Constants
EDTA Titration

3. Calculate the conditional constant: =1.8 x 1010
EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of M Ca+2 with M EDTA in a solution buffered to a constant pH of 10.0.
Calculate the conditional constant: =1.8 x 1010
Equivalence Volume V = 25.0 mL
pCa at Initial Point = 2.301
pCa at Equivalence
pCa at Pre-Equivalence Point
pCa at Post-Equivalence Point

4. EXAMPLE: At 25.0 mL (Equivalence Point) Ca2+ + Y4- -> CaY2- -
Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of M Ca+2 with M EDTA in a solution buffered to a constant pH of 10.0.
At 25.0 mL (Equivalence Point)
Ca2+
+ Y4-
->
CaY2-
Before
moles -
After
What can contribute to Ca2+ “after” reaction?

5. EXAMPLE: I - - 0.0025 moles/V C +x +x -x E +x + x 0.0333 –x
Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of M Ca+2 with M EDTA in a solution buffered to a constant pH of 10.0.
Ca Y4- D CaY2-
I moles/V
C +x x -x
E x x –x
0.0025moles/0.075 L
X = [Ca2+] = 1.4 x10-6
pX = p[Ca2+] = 5.866

6. Pre-Equivalence Point
Let’s try 15 mL

7. EXAMPLE: At 15.0 mL Ca2+ + Y4- -> CaY2- - K’CaY = 1.8 x 1010
Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of M Ca+2 with M EDTA in a solution buffered to a constant pH of 10.0.
At 15.0 mL
Ca2+
+ Y4-
->
CaY2-
Before
moles
moles -
After
moles
What can contribute to Ca2+ after reaction?
negligible
K’CaY = 1.8 x 1010

8. EXAMPLE:
Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of M Ca+2 with M EDTA in a solution buffered to a constant pH of 10.0.
At 15.0 mL
[Ca2+] = moles/0.065 L
[Ca2+] = M
p [Ca2+] = 1.812

9. Post Equivalence Point
Let’s Try 28 ml

10. EXAMPLE: At 28.0 mL Ca2+ + Y4- -> CaY2- -
Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of M Ca+2 with M EDTA in a solution buffered to a constant pH of 10.0.
At 28.0 mL
Ca2+
+ Y4-
->
CaY2-
Before
moles
moles -
After
moles
What can contribute to Ca2+ after titration?

11. EXAMPLE: I - 0.0003 moles/V 0.0025 moles/V C +x +x -x
Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of M Ca+2 with M EDTA in a solution buffered to a constant pH of 10.0.
Ca Y4-  CaY2-
I moles/V moles/V
C +x x -x
E x x –x
0.078 L
X = [Ca2+] = 4.6 x10-10
pX = p[Ca2+] = 9.334

13. An Introduction to Analytical Separations
Chapter 23
An Introduction to Analytical Separations

14. Problems Chapter 23 Chapter 24 1, 15, 20 a and b, 27, 29, 30, 37, 44
1, 3, 4, 5, 6
From ,

15. What is Chromatography?

17. Parts of Column
column
support
stationary phase
mobile phase

18. Types of Chromatography
Adsorption
Partition
Ion Exchange
Molecular Exclusion (gel-filtration)
Affinity chromatography

22. Section 23-3 A Plumber’s View of Chromatography
The chromatogram
“Retention time”
“Relative retention time”
“Relative Retention”
“Capacity Factor”

23. 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”

24. A chromatogram
Adjusted Retention time (t’r) - is the retention time corrected for dead volume “the difference between tr and a non-retained solute”

25. 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.

26. 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”.

27. t’r(benzene) = 251 sec – 42 sec = 209 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 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) = sec = 291 s

28. An Example
A mixture of benzene, toulene, and methane was injected into a gas chromatograph. Methane gave a sharp peak in 42 sec, benzene 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”.
= 5.0

29. An Example
A mixture of benzene, toulene, and methane was injected into a gas chromatograph. Methane gave a sharp peak in 42 sec, benzene 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”.
= 6.9

30. An Example
A mixture of benzene, toulene, and methane was injected into a gas chromatograph. Methane gave a sharp peak in 42 sec, benzene 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.

31. Efficiency of Separation
“Two factors”
How far apart they are (a)
Width of peaks

32. Resolution

33. Resolution

35. 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?

36. Diffusion and flow related effects
Why are bands broad?
Diffusion and flow related effects