questions from hw.. 1. the zn in a 0.7556-g sample of foot powder was titrated with 21.27 ml of...
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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