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89Chapter 7 Collecting and Preparing Samples
 For (a) we need a D of
.
. ..
. .D 0 01
0 01 99 050 0
50 0 50 0
#
#
=
-
=
 and for (b) we need a D of
.
.
.
. . .
D
0 01
18 0
25 0
50 0 50 0 0 01
#
#
=
-
=
25. From equation 7.27, an extraction eiciency of 99.9%, requires that
. . .
.Q DV V
V
D0 001 50 0 50 0
50 0
aq
org aq
aq
#
= =
+
=
+
 for a single extraction of 50.0 mL of sample using 50.0 mL of organic 
solvent. Solving gives the minimum value of D as 999. Because the 
analyte is a weak acid, the distribution ratio’s value depends on the 
pH of the aqueous phase, with more acidic pH levels favoring a larger 
value for D. From equation 7.31, we know that
[ ]
[ ]
D
K
K
H O
H O
aq
aq
3 a
3D
=
+
+
+
[ ]
[ ]
( . )
( )
999
1 00 10
1200
H O
H O
aq
aq
5
3
3
#
=
+
+
+
-
[ ] . [ ]999 9 99 10 1200H O H Oaq aq
3
3 3#+ =
+ - +
. [ ]9 99 10 201 H Oaq
3
3# =
- +
 gives ][H Oaq3
+ as 4.97×10–5, or a maximum pH of 4.30.
26. For a pH of 7.00 ( ][H Oaq3
+ = 1.00×10–7), the distribution ratio, D, 
is
[ ]
[ ]
( . ) ( . )
( ) ( . )
.D
K
K
1 00 10 1 00 10
1200 1 00 10
11 9
H O
H O
aq
aq
7 5
7
3 a
D 3
# #
# #
=
+
=
+
=+
+
- -
-
 To ind the number of extractions, we make appropriate substitutions 
into equation 7.27 and solve for n
. . . .
.0 001 11 9 50 0 50 0
50 0 n
#
=
+
a k
( . ) .log logn0 001 0 0775= ^ h
.. n1 113 00=--
 inding that n is 2.7; thus, we need to complete at least three ex-
tractions to achieve an extraction eiciency of 99.9%.
27. From equation 7.27, an extraction eiciency of 99.9%, requires that
( ) . . .
.Q DV V
V
D0 001 25 0 50 0
50 0
aq
org aq
aq
2
2 2
#
= =
+
=
+
c am k