Outline EDTA Acid Base Properties aY 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. 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 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) Ca2+ Before After 0.0025 moles - + Y4- 0.0025 moles - -> CaY20.0025 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- D I - C +x E CaY2- K 'CaY [CaY 2 ] [ EDTA][Ca 2 ] 0.0025moles/0.075 0.0025 moles/V L - +x -x K 'CaY 0.0333 x x2 X = [Ca2+] = 1.4 x10-6 +x +x 0.0333 –x pX = p[Ca2+] = 5.866 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+ Before 0.0025 moles After 0.0010 moles + Y4- 0.0015 moles - -> CaY20.0015 moles What can contribute to Ca2+ after reaction? K’CaY = 1.8 x 1010 negligible 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 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+ Before After 0.0025 moles - + Y4- 0.0028 moles 0.0003 moles -> CaY20.0025 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- K 'CaY 0.078 L I C E - 0.0003 moles/V 0.0025 moles/V ' +x [CaY 2 ] [ EDTA][Ca 2 ] +x -x K CaY 0.03205 x (0.003846 x)(x) X = [Ca2+] = 4.6 x10-10 +x 0.003846 + x 0.03205 –x pX = p[Ca2+] = 9.334 Chapter 23 An Introduction to Analytical Separations 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 Adsorption 2. Partition 3. Ion Exchange Molecular Exclusion (gel-filtration) 5. Affinity chromatography 1. 4. Section 23-3 A Plumber’s View of Chromatography The chromatogram “Retention time” “Relative retention time” “Relative Retention” “Capacity Factor” 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” A chromatogram Adjusted Retention time (t’r) - is the retention time corrected for dead volume “the difference between tr and a non-retained solute” 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. t 'r 2 a t 'r1 where t 'r1 t 'r 2 so a 1 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”. tr tm k' tm 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”. tr tm k' tm tr t m 251 42 = 5.0 k 'benzene tm 42 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”. tr tm k' tm tr t m 333 42 k 'toulene tm 42 = 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. t 'r 2 a t 'r1 t 'toulene 291sec a 1.39sec t 'benzene 209sec Efficiency of Separation 1) “Two factors” How far apart they are (a) 2) Width of peaks Resolution Resolution t r 0.589t r Resolution w av w1/ 2 av 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? t r Resolution w av 424 407 Resolution 1.17 1 (13 16) 2 Why are bands broad? Diffusion and flow related effects