Transcript EDTA Titrations - Information Technology

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
0.022879g
%Zn 
x100%  3.0279 %
0.7556g
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+
0.00016476 mole 0.0032952 mole
Fe , M 

0.0500 L
1L
2
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.
0.0032952m ole 55.847g 1000m g  184 .0mg
x
x
L
1L
m ole
1g
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 = a Kf

14)
KKK’f’f’f=
=
=5.6
0.36
2.3xx(1.995
10
10-3-5(1.995
(1.995
x 1014)
xx10
1014)
912
KK’f’f =
= 7.182
4.589
1.117 xx 10
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)
Find equivalence Volume
Moles Sr2+ = Moles EDTA
At initial Point
pSr = 2.000
(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
?
0.0005000
moles
0.0002000 moles
After
0.0003000 moles
None
 0.0003000m oles 
pSr   log

 0.05000L  0.01000L 
None
0.0002000 moles
 2.3010
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
?
0.0005000
moles
0.0004800 moles
After
0.0000200 moles
None
None
0.0004800 moles
 0.00002000m oles 
pSr   log
  3.568
 0.05000L  0.02400L 
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
?
0.0005000
moles
0.0004980 moles
After
0.00000200 moles
None
None
0.0004980 moles
m oles 
 0.000002000
pSr   log
  4.5735
 0.05000L  0.02490L 
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
0.0005000 moles
0.0005000 moles
After
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-
I
None
None
C
+x
+x
E
+x
+x
 SrY20.0005000 moles/ 0.075 L
-x
0.00666 –x
2
[
SrY
]
[0.0066 x] x  3.94106
’
8
K = 4.25 x 10 
4
3 
[Y ][Sr ]
[ x][x]
pSr = 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+
I
None
C
+x
E
+x
+
Y4-
 SrY2-
0.000002/0.0751 L 0.0005000 moles/ 0.0751 L
+x
2.666x10-5 +x
-x
0.006657 –x
2
[
SrY
]
7
[0.006657 x]
’
8
x

5
.
205

10
K = 4.25 x 10 
4
3 
[Y ][Sr ] [ x][2.666105  x] pSr = 6.2835
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+
I
None
C
+x
E
+x
+
Y4-
0.00002/0.076 L
 SrY20.0005000 moles/ 0.076 L
+x
2.63x10-4 +x
-x
0.006578 –x
2
[
SrY
]
[0.006578 x]
’
8
K = 4.25 x 10 

[Y 4 ][Sr3 ] [ x][2.6310 4  x]
x  5.885108
pSr = 7.230
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+
I
None
C
+x
E
+x
+
Y4-
 SrY2-
0.0001000/0.080 L 0.0005000 moles/ 0.080 L
+x
0.00125 +x
-x
0.00625 –x
2
[
SrY
]
[0.00625 x]
’
8
K = 4.25 x 10 

[Y 4 ][Sr3 ] [ x][0.00125 x]
x  1.176x108
pSr = 7.929
9
8
pSr
7
6
5
4
3
2
1
0
0
10
20
Volume Titrant
30
40
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 0.589t 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
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