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 (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

%1007556.0

022879.0% x

g

gZn %0279.3

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+

L

moleMFe

0500.0

00016476.0,2

L

mole

1

0032952.0

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.

g

mgx

mole

gx

L

mole

1

1000847.55

1

0032952.0L

mg0.184

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 = Kf

K’f = x 10-5 (1.995 x 1014)

K’f = x 109

K’f = x 10-3 (1.995 x 1014)

K’f = x 1012

K’f = (1.995 x 1014)

K’f = x 1013

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)

At initial Point pSr = 2.000

Find equivalence Volume

Moles Sr2+ = Moles EDTA

(0.05000 L)x(0.01000M Sr2+) = 0.02000 M x Ve

25.0 mL = Ve

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

Sr2+ + Y4- -> SrY2-

Before ? After

0.0005000 moles 0.0002000 moles None

0.0003000 moles None 0.0002000 moles

LL

molespSr

01000.005000.0

0003000.0log 3010.2

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

Sr2+ + Y4- -> SrY2-

Before ? After

0.0005000 moles 0.0004800 moles None

0.0000200 moles None 0.0004800 moles

LL

molespSr

02400.005000.0

00002000.0log 568.3

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

Sr2+ + Y4- -> SrY2-

Before ? After

0.0005000 moles 0.0004980 moles None

0.00000200 moles None 0.0004980 moles

LL

molespSr

02490.005000.0

000002000.0log 5735.4

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+

Sr2+ + Y4- -> SrY2-

Before After

0.0005000 moles 0.0005000 moles None

None None 0.0005000 moles

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+

Sr2+ + Y4- SrY2-

I C

E

None None 0.0005000 moles/ 0.075 L

+x +x -x

+x +x 0.00666 –x

K’ = 4.25 x 108

]][[

][34

2

SrY

SrY

]][[

]0066.0[

xx

x

61094.3 xpSr = 5.40

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+

Sr2+ + Y4- SrY2-

I C

E

None 0.000002/0.0751 L0.0005000 moles/ 0.0751 L

+x +x -x

+x 2.666x10-5 +x 0.006657 –x

K’ = 4.25 x 108

]][[

][34

2

SrY

SrY 710205.5 xpSr = 6.2835]10666.2][[

]006657.0[5 xx

x

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+

Sr2+ + Y4- SrY2-

I C

E

None 0.00002/0.076 L 0.0005000 moles/ 0.076 L

+x +x -x

+x 2.63x10-4

+x

0.006578 –x

K’ = 4.25 x 108

]][[

][34

2

SrY

SrY 810885.5 xpSr = 7.230]1063.2][[

]006578.0[4 xx

x

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+

Sr2+ + Y4- SrY2-

I C

E

None 0.0001000/0.080 L0.0005000 moles/ 0.080 L

+x +x -x

+x 0.00125

+x

0.00625 –x

K’ = 4.25 x 108

]][[

][34

2

SrY

SrY 810176.1 xxpSr = 7.929]00125.0][[

]00625.0[

xx

x

0

1

2

3

4

5

6

7

8

9

0 10 20 30 40

Volume Titrant

pS

r

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

Data Analysis

The Inlet

Why are bands broad?

Diffusion and flow related effects

Of particular concern in Gas Chromatography.Of particular concern in Gas Chromatography.Why?Why?

Diffusion is fasterDiffusion is faster

Gases from the headspace of a beer can!!

Packed column ... Compare peak widths with your sample