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Questions

Questions. From HW. 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.

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  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 (Y4-). 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 (Y4-) when titrated at pH 2.0. Express the concentration of iron in parts per million. Moles of EDTA = Moles of Fe2+ (0.01200 M)(0.01373L) = Moles of Fe2+ 0.00016476= Moles of Fe2+

  4. 2. • A 50.00-mL aliquot of a solution containing Iron (II) required 13.73 mL of 0.01200 M EDTA (Y4-) 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 Fe2+ at a pH of (a) 6.0, (b) 8.0, (c) 10.0. K’f = a Kf K’f = 5.6 x 10-3 (1.995 x 1014) K’f = 0.36(1.995 x 1014) K’f = 2.3 x 10-5 (1.995 x 1014) K’f = 4.589 x 109 K’f = 7.182 x 1013 K’f = 1.117 x 1012

  6. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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) Find equivalence Volume Moles Sr2+ = Moles EDTA At initial Point pSr = 2.000 (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 Sr2+ 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 0.0005000 moles 0.0002000 moles None 0.0003000 moles None 0.0002000 moles

  8. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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 0.0005000 moles 0.0004800 moles None 0.0000200 moles None 0.0004800 moles

  9. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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 0.0005000 moles 0.0004980 moles None 0.00000200 moles None 0.0004980 moles

  10. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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 SrY2- is source of Sr2+ 0.0005000 moles 0.0005000 moles None None None 0.0005000 moles

  11. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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 SrY2- is source of Sr2+ None None 0.0005000 moles/ 0.075 L +x +x -x 0.00666 –x +x +x K’ = 4.25 x 108 pSr = 5.40

  12. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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 SrY2- is source of Sr2+ None 0.000002/0.0751 L 0.0005000 moles/ 0.0751 L +x +x -x 0.006657 –x +x 2.666x10-5 +x K’ = 4.25 x 108 pSr = 6.2835

  13. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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 SrY2- is source of Sr2+ None 0.00002/0.076 L 0.0005000 moles/ 0.076 L +x +x -x 0.006578 –x +x 2.63x10-4 +x K’ = 4.25 x 108 pSr = 7.230

  14. 4. • Derive a titration curve for 50.00 mL of 0.01000 M Sr2+ 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 SrY2- is source of Sr2+ None 0.0001000/0.080 L 0.0005000 moles/ 0.080 L +x +x -x 0.00625 –x +x 0.00125+x K’ = 4.25 x 108 pSr = 7.929

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

  16. 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”

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

  18. 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.

  19. 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”.

  20. 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

  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. 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

  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”. = 6.9

  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. 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.

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

  25. Resolution

  26. Resolution

  27. 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?

  28. Data Analysis

  29. The Inlet

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

  31. Of particular concern in Gas Chromatography. Why?

  32. Diffusion is faster

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

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

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