Equação de Equilíbrio Químico

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78 Solutions Manual for Analytical Chemistry 2.1
 which we then solve for [F–]
] [ ]K C K [F H O ][Fa HF a 3- =
- + -
[ ] ]K C KH O ][F [Fa HF 3 a= +
+ - -
] [ K K C[F H O ]3 a a HF+ =
- +" ,
]
[ K
K C[F
H O ]3 a
a HFa
=
+
-
+
 Next, we solve Kw for [OH–]
[ ]
[ ]
KOH
H O3
w
=
-
+
 and then substitute this and the equation for [F–] into the charge 
balance equation
[ ] [
K
K
K C[H O ]
H O H O ]3
3
w
3 a
a HF
= +
+
+
+ +
 Rearranging this equation
[ ] [
K
K
K C 0[H O ]
H O H O ]3
3
w
3 a
a HFa
- -
+
=
+
+ +
 multiplying through by [H3O+]
[
[ ]
K
K
K C
0[H O ]
H O ]
H O2
3 w
3 a
a HF 3
- -
+
=
+
+
+
 multiplying through by [H3O+] + Ka
[ ]
[ ]
K K
K K K C 0
[H O ] [H O ] H O
H O
3 2
3 a 3 w 3
a w a HF 3
+ - -
- =
+ + +
+
 and gathering terms leaves us with the inal equation
( )[ ]K K C K K K 0[H O ] [H O ] H O3 2
3 a 3 a HF w 3 a w+ - + - =
+ + +