Outline

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

3. 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

4. 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) What can contribute to Ca2+ “after” reaction?

5. 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-DCaY2- I - - 0.0025 moles/V C +x +x -x E +x + x 0.0333 –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: 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 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 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

9. Post Equivalence Point Let’s Try 28 ml

10. 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 What can contribute to Ca2+ after titration?

11. 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 0.078 L X = [Ca2+] = 4.6 x10-10 pX = p[Ca2+] = 9.334

13. Chapter 23 An Introduction toAnalytical Separations

14. 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,

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 23-3A 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. 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

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 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”. = 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 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”. = 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 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.

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. Why are bands broad? Diffusion and flow related effects