Page 237 - Modern Analytical Chemistry

1400-CH07 9/8/99 4:04 PM Page 220

220 Modern Analytical Chemistry

] ]
[HA org tot [HA org
D = = - 7.27
]
[HA aq tot [HA aq ] +[A aq ]
Since the position of an acid–base equilibrium depends on the pH, the distribution
ratio must also be pH-dependent. To derive an equation for D showing this depen-
dency, we begin with the acid dissociation constant for HA.
-
aq A][
[ HO + aq ]
3
K a = 7.28
[ HA aq ]
–
Solving equation 7.28 for [A aq ]
[HA ]
K a aq
-
]
[A aq =
+
[HO aq ]
3

and substituting into equation 7.27 gives
]
[HA org
D =
+
] +( a ]/[H O aq ])
K [HA aq
[HA aq
3

Factoring [HA aq ] from the denominator
]
[HA org
D =
]{1 +( a 3 + ])}
[HA aq
K /[H O aq

and substituting equation 7.26
K D
D =
1 +( K a /[ HO + aq ])
3

gives, after simplifying, the sought-after relationship between the distribution ratio
and the pH of the aqueous solution
+
[
K D HO ]
aq
3
D = 7.29
[ HO + aq ] + K a
3
The value for D given by equation 7.29 can be used in equation 7.25 to determine
extraction efficiency.
7
EXAMPLE .16
–5
An acidic solute, HA, has an acid dissociation constant of 1.00 ´10 , and a
partition coefficient between water and benzene of 3.00. Calculate the
extraction efficiency when 50.00 mL of a 0.025 M aqueous solution of HA
buffered to a pH of 3.00, is extracted with 50.00 mL of benzene. Repeat for
cases in which the pH of the aqueous solution is buffered to 5.00 and 7.00.
SOLUTION
–3
+
When the pH is 3.00, the [H 3 O aq ] is 1.00 ´10 , and the distribution ratio for
the extraction is
3 00
(. )( . 1 00 ´10 -3 )
D = = . 297
. 1 00 ´10 -3 + . 1 00 ´ 10 -5