Outline

EDTAEDTA Acid Base Properties Y nomenclature Conditional Formation Constants

EDTA TitrationEDTA Titration

EXAMPLE:

Calculate the conditional constant: =1.8 x 1010

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.

pCa at Equivalence

Equivalence Volume V2 = 25.0 mL

pCa at Pre-Equivalence Point

pCa at Post-Equivalence Point

pCa at Initial Point = 2.301

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)At 25.0 mL (Equivalence Point)

Ca2+ + Y4- -> CaY2-

Before 0.0025 moles 0.0025 moles

-

After - - 0.0025 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- CaY2-

I - - 0.0025 moles/V

C +x +x -x

E +x + x 0.0333 –x

]][[

][2

2'

CaEDTA

CaYK CaY

2' 0333.0

x

xK CaY

X = [Ca2+] = 1.4 x10-6

pX = p[Ca2+] = 5.866

0.0025moles/0.075 L0.0025moles/0.075 L

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+ + Y4- -> CaY2-

Before 0.0025 moles 0.0015 moles

-

After 0.0010 moles - 0.0015 moles

What can contribute to Ca2+ after reaction?

K’CaY = 1.8 x 1010

negligiblenegligible

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 Mp [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+ + Y4- -> CaY2-

Before 0.0025 moles 0.0028 moles

-

After - 0.0003 moles

0.0025 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-

I - 0.0003 moles/V 0.0025 moles/V

C +x +x -x

E +x 0.003846 + x 0.03205 –x

]][[

][2

2'

CaEDTA

CaYK CaY

))(003846.0(

03205.0'

xx

xK CaY

X = [Ca2+] = 4.6 x10-10

pX = p[Ca2+] = 9.334

0.078 L

Chapter 23

An Introduction toAn Introduction toAnalytical Analytical

SeparationsSeparations

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

1. Adsorption2. Partition

3. Ion Exchange4. Molecular Exclusion (gel-filtration)

5. Affinity chromatography

Section 23-3 A Plumber’s View of A Plumber’s View of ChromatographyChromatography

The chromatogram“Retention time”

“Relative retention time”“Relative Retention”

“Capacity Factor”

A chromatogramRetention 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 chromatogramAdjusted Retention time (t’

r) - is the retention time corrected for dead volume “the difference between tr and a non-retained

solute”

A chromatogramRelative 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.

1

2

'

'

r

r

t

t 1 '' 21 sottwhere rr

A chromatogramCapacity 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”.

m

mr

t

ttk

'

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”.

m

mr

t

ttk

'

42

42251'

m

mrbenzene t

ttk = 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 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”.

m

mr

t

ttk

'

42

42333'

m

mrtoulene t

ttk = 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.

1

2

'

'

r

r

t

t sec39.1

sec209

sec291

'

'

benzene

toulene

t

t

Efficiency of Separation

“Two factors”1) How far apart they are ()

2) Width of peaks

ResolutionResolution

Resolution

avw 2/1

r

av

r t589.0

w

tResolution

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?

av

r

w

tResolution

7

21

1.116)(13

407424Resolution

Why are bands broad?

Diffusion and flow related effects