Cálculo de Eficiência de Extração

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90 Solutions Manual for Analytical Chemistry 2.1
 for two extractions of 50.0 mL of sample using 25.0 mL of organic 
solvent per extraction. Taking the square root of both sides 
. .
.. D 25 0 50 0
50 00 03162
# +
=
 and solving for D gives its minimum value of as 61.3. Because the 
analyte is a weak base, the distribution ratio’s value depends on the 
pH of the aqueous phase, with more basic pH levels favoring a larger 
value for D. From Practice Exercise 7.9, we know that
[ ]
[ ]
D K
K
OH
OH
aq
aq
b
D
=
+
-
-
.
[ ] ( . )
( . ) [ ]
61 3
1 0 10
5 00 10
OH
OH
aq
aq
3
2
#
#
=
+
- -
-
. [ ] ( . ) [ ].61 3 5 00 100 0613OH OHaq aq
2
#+ =
- -
. [ ].0 0613 438 7 OHaq=
-
 gives ][OHaq
- as 1.40×10–4, or a minimum pH of 10.15.
28. (a) To calculate the extraction eiciencies for HA and HB, we irst 
ind their respective distribution ratios at a pH of 7.00
[ ]
[ ]
. .
( . ) ( . )
.D
K
K
1 0 10 1 0 10
5 00 10 1 0 10
0 0500
H O
H O
aq
aq
7 3
2 7
HA
3 a,HA
D,HA 3
# #
# #
=
+
=
+
=+
+
- -
-
[ ]
[ ]
. .
( . ) ( . )
D
K
K
1 0 10 1 0 10
5 00 10 1 0 10
250
H O
H O
aq
aq
7
2 7
7H
3 a,H
D,H 3
B
B
B
# #
# #
=
+
=
+
=+
+
- -
-
 and then calculate the fraction of HA and HB that remain in the 
aqueous phase when the extraction is complete
. . .
. .Q DV V
V
0 0500 50 0 50 0
50 0 0 952,aq
org aq
aq
HA
#
=
+
=
+
=
. .
. .Q DV V
V
50 0 50 0
50 0 0 00398250,aq
org aq
aq
HB
#
=
+
=
+
=
 hus, the extraction eiciency for HA is 0.048 or 4.8% and for HB 
is 0.996 or 99.6%
 (b) he aqueous phase is enriched in the analyte, HA, with 95.2% of 
HA remaining unextracted.
 (c) he recovery for HA in the aqueous phase, RHA, is 0.952 or 95.2%; 
for HB, RHB is 0.00398 or 0.398%.
 (d) he separation factor, SHB,HA, is
.
. .S R
R
0 952
0 00398 4 18 10 3
HB,HA
HA
HB
#= = =
-
 (e) he error is
( )
( )
( )
E R
C
K C
R1
o
o
HA
HA
HA,HB HB
HB#= - +