HW 9-UV visible spectroscopy Key

HW 9-UV visible spectroscopy Key -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW
9:
UV‐Visible
Spectroscopy
 (1)  Express
the
following
absorbances
as
%
transmiBance:
0.692,
0.038.
 0.20;
0.92
 (2)
Convert
the
following
transmiBance
data
to
absorbances:
15.8%,
0.085.
 0.74;
1.1
 (3)
A
soluSon
containing
6.23
ppm
KMnO4
had
a
transmiBance
of
0.195
in
a
1.00
cm
 cell
at
520
nm.
Calculate
the
molar
absorpSvity
(exSncSon
coefficient)
at
520
nm.
 A
=
0.71;
Molar
absorpSvity
=
0.11
ppm‐1cm‐1;
AlternaSvely,
ppm
could
be
converted
 to
molarity
to
give
an
absorpSvity
with
units
of
M‐1cm‐1.
 (4)
 A
 25.0
 mL
 aliquot
 of
 an
 aqueous
 quinine
 soluSon
 was
 diluted
 to
 50.0
 mL
 and
 found
to
have
an
absorbance
of
0.656
at
348
nm
when
measured
in
a
2.50
cm
 cell.
A
second
25.0
 mL
aliquot
was
mixed
with
10.00
 mL
of
a
soluSon
containing
 25.7
 ppm
of
quinine;
a[er
diluSon
to
50
 mL,
this
soluSon
had
an
absorbance
of
 0.976
(2.50
cm
cell).
Calculate
the
concentraSon
of
quinine
in
ppm
in
the
sample.
 The
 problem
 is
 idenScal
 to
 the
 one
 assigned
 with
 the
 calibraSon
 curve
 HW.
 It
 is
 a
 standard
addiSon
problem.
 The
unknown
concentraSon
is
5.2
ppm.
 (5)
The
acid‐base
indicator
HIn
undergoes
the
following
reacSon
in
dilute
aqueous
soluSon:

 

 
 
 
 
 
HIn

H+
+
In‐
 The
following
absorbance
data
were
obtained
for
a
5.00
x
10‐4
M
soluSon
of
HIn
in
0.1
M
NaOH
 and
0.1
M
HCl.
Measurements
were
made
at
wavelengths
of
485
nm
and
625
nm
with
1.00
 cm
cells.
 0.1
M
NaOH
A485
=
0.075;
A625
=
0.904
 0.1
M
HCl
A485
=
0.487;
A625
=
0.181
 In
the
 NaOH
soluSon
essenSally
all
of
the
indicator
is
present
as
In‐;
in
the
acidic
soluSon,
it
is
 essenSally
all
in
the
form
of
HIn.
 (a)  Calculate
molar
absorpSviSes
for
In‐
and
HIn
at
485
and
625
nm.
 In‐:
150
and
1808
M‐1
cm‐1
at
485
nm
and
625
nm
 Hin:
974
and
362
M‐1
cm‐1
at
485
nm
and
625
nm
 (b)
 Calculate
 the
 acid
 dissociaSon
 constant
 for
 the
 indicator
 if
 a
 pH
 5.00
 buffer
 containing
 a
 small
 amount
 of
 the
 indicator
 exhibits
 an
 absorbance
 of
 0.567
 at
 485
 nm
 and
 0.395
 and
 625
nm
(1.00
cm
cells).
 Solve
 for
 [In]
 and
 [HIn]
 by
 treaSng
 this
 as
 a
 2‐component
 Beer’s
 law
 problem.
 Then
 use
 the
 Henderson‐Haaselbalch
equaSon
to
solve
for
the
pKa.
 [In]
=
1.05
x
10‐4
M;
[HIn]
=
5.66
x
10‐4
M;
pKa
=
5.73
 (c)
 What
 is
 the
 pH
 of
 a
 soluSon
 containing
 a
 small
 amount
 of
 the
 indicator
 that
 exhibits
 an
 absorbance
of
0.492
at
485
nm
and
0.245
at
635
nm?

 Same
thing.
Find
the
concentraSons
and
use
the
HH
equaSon
with
the
pKa
determined
above.
 ...
View Full Document

This note was uploaded on 11/16/2011 for the course CHEM 4011 at Colorado.

Ask a homework question - tutors are online