Questions From HW.

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

2. A mL aliquot of a solution containing Iron (II) required mL of 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+ ( M)( L) = Moles of Fe = Moles of Fe 2+

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

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) K ’ f =  K f K ’ f =  x (1.995 x 10 14) K ’ f =  x 10 9 K ’ f =  x (1.995 x 10 14) K ’ f =  x K ’ f =  (1.995 x 10 14) K ’ f =  x 10 13

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

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

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

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

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

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

4. Derive a titration curve for mL of M Sr 2+ with M EDTA in a solution buffered to pH Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, mL of titrant. Post equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4-  SrY 2- I C E None / L moles/ L +x -x +x 2.666x x –x K ’ = x 10 8 pSr =

4. Derive a titration curve for mL of M Sr 2+ with M EDTA in a solution buffered to pH Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, mL of titrant. Post equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4-  SrY 2- I C E None /0.076 L moles/ L +x -x +x 2.63x x –x K ’ = x 10 8 pSr = 7.230

4. Derive a titration curve for mL of M Sr 2+ with M EDTA in a solution buffered to pH Calculate pSr values after the addition of 0.00, 10.00, 24.00, 24.90, 25.00, 25.10, 26.00, mL of titrant. Post equivalence - EQUILIBRIUM OF SrY 2- is source of Sr 2+ Sr 2+ + Y 4-  SrY 2- I C E None /0.080 L moles/ L +x -x +x x –x K ’ = x 10 8 pSr = 7.929

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”

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”

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”

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.

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”.

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 – t r (non retained component) t’ r (benzene) = 251 sec – 42 sec = 209 s t’ r (toulene) = 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 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

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

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.

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

Resolution

Resolution

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?

Data Analysis

The Inlet

Why are bands broad? Diffusion and flow related effects

Of particular concern in Gas Chromatography. Why?

Diffusion is faster

Gases from the headspace of a beer can!!

Packed column... Compare peak widths with your sample