1. Questions From HW.

2. 1. The Zn in a 0.7556-g sample of foot powder was titrated with 21.27 mL of 0.01645 M EDTA (Y 4- ). Calculate the percent Zn in this sample. Moles of EDTA = Moles of Zn (0.01645 M)(0.02127L) = Moles of Zn 0.0003498915= Moles of Zn Convert to grams of Zn and compare to original value 0.0003498915 moles x 65.39 gram/mole = 0.022879 gram of Zn

3. 2. A 50.00-mL aliquot of a solution containing Iron (II) required 13.73 mL of 0.01200 M EDTA (Y 4- ) when titrated at pH 2.0. Express the concentration of iron in parts per million. Moles of EDTA = Moles of Fe 2+ (0.01200 M)(0.01373L) = Moles of Fe 2+ 0.00016476 = Moles of Fe 2+

4. 2. A 50.00-mL aliquot of a solution containing Iron (II) required 13.73 mL of 0.01200 M EDTA (Y 4- ) when titrated at pH 2.0. Express the concentration of iron in parts per million.

5. 13-5. Calculate the conditional constants for the formation of EDTA complex of Fe 2+ at a pH of (a) 6.0, (b) 8.0, (c) 10.0. K ’ f =  K f K ’ f =  x 10 -5 (1.995 x 10 14) K ’ f =  x 10 9 K ’ f =  x 10 -3 (1.995 x 10 14) K ’ f =  x 10 12 K ’ f =  (1.995 x 10 14) K ’ f =  x 10 13

6. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. At initial Point pSr = -log (0.0100) At initial Point pSr = 2.000 Find equivalence Volume Moles Sr 2+ = Moles EDTA (0.05000 L)x(0.01000M Sr2+) = 0.02000 M x Ve 25.0 mL = Ve

7. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Excess will determine pSr Sr 2+ + Y 4- ->SrY 2- Before ? After 0.0005000 moles0.0002000 molesNone 0.0003000 molesNone0.0002000 moles

8. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Excess will determine pSr Sr 2+ + Y 4- ->SrY 2- Before ? After 0.0005000 moles0.0004800 molesNone 0.0000200 molesNone0.0004800 moles

9. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Excess will determine pSr Sr 2+ + Y 4- ->SrY 2- Before ? After 0.0005000 moles0.0004980 molesNone 0.00000200 molesNone0.0004980 moles

10. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4- ->SrY 2- Before After 0.0005000 moles None 0.0005000 moles

11. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4-  SrY 2- I C E None 0.0005000 moles/ 0.075 L +x -x +x 0.00666 –x K ’ = 4.2 5 x 10 8 pSr = 5.40

12. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Post equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4-  SrY 2- I C E None0.000002/0.0751 L 0.0005000 moles/ 0.0751 L +x -x +x 2.666x10 -5 +x 0.00665 7 –x K ’ = 4.2 5 x 10 8 pSr = 6.2835

13. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Post equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4-  SrY 2- I C E None0.00002/0.076 L 0.0005000 moles/ 0.076 L +x -x +x 2.63x10 -4 +x 0.006578 –x K ’ = 4.2 5 x 10 8 pSr = 7.230

14. 4. Derive a titration curve for 50.00 mL of 0.01000 M Sr 2+ with 0.02000 M EDTA in a solution buffered to pH 11.0. Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, 30.00 mL of titrant. Post equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4-  SrY 2- I C E None0.0001000/0.080 L 0.0005000 moles/ 0.080 L +x -x +x 0.00125 +x 0.00625 –x K ’ = 4.2 5 x 10 8 pSr = 7.929

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

17. A chromatogram Retention time (t r ) – 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 t r and a non-retained solute”

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

19. A chromatogram Relative 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.

20. 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”.

21. 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 – t r (non retained component) t’ r (benzene) = 251 sec – 42 sec = 209 s t’ r (toulene) = 333-42 sec = 291 s

22. 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

23. 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

24. 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.

25. Efficiency of Separation “Two factors” 1) How far apart they are (  ) 2) Width of peaks

26. Resolution

27. Resolution

29. 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?

30. Data Analysis

32. The Inlet

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

38. Of particular concern in Gas Chromatography. Why?

39. Diffusion is faster

41. Gases from the headspace of a beer can!!

42. Packed column... Compare peak widths with your sample