Solubilidade e Constantes de Equilíbrio

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66 Solutions Manual for Analytical Chemistry 2.1
.
.
( ) ( )
x
x
x
x
s l2
0 010
0 010
0 0
2
2
2
I
C
E
CaF
—
—
—
H O
—
2HF H O
—
—
—
Ca2
2 3 2?+
+
+
+
+
+ +
 Substituting the equilibrium concentrations into the reaction’s equi-
librium constant expression
( . )
( ) ( )
( . )
.K
x x x
0 010
2
0 010
4 8 4 10
[H O ]
[Ca ][HF]
2
2
2
3
5
3
2
2 2
#= = = =+
+
-
 and solving gives x as 1.3×10–3. he molar solubility of CaF2 at a 
pH of 2.00 is the concentration of Ca2+, which is x or 1.3×10–3 M. 
Our solution here assumes that we can ignore the presence of F–; as 
a check on this assumption, we note that
]
[ ]
( . ) ( )
.
.
.
K 6 8 10 10
100 01
1 3
8 8[F
H O
[HF] 4 3
5
3
a,HF # #
#= = =
-
+
- -
-
 a concentration that is negligible when compared to the concentra-
tion of HF.
12. he solubility of Mg(OH)2 is determined by the following reaction
( ) ( ) ( )s aq aqMg(OH) Mg 2OH2
2
? +
+ -
 for which the equilibrium constant expression is
[ ] .K 7 1 10[Mg ] OHsp
2 122
#= =
+ - -
 he following table deines the equilibrium concentrations of Mg2+ 
and OH– in terms of their concentrations in the solution prior to 
adding CaF2 and the change in their concentrations due to the sol-
ubility reaction; note that the concentration of OH– is ixed by the 
bufer.
.
( ) ( ) ( )
.
x
x
s aq aq
0
1 0 10
1 0 10I
C
E
Mg(OH)
—
—
—
Mg 2OH
—
7
2
7
2 ?
#
#
+
+
-
-
+
-
 Substituting the equilibrium concentrations into the Ksp expression
( )( . ) .K x 1 0 10 107 1[Mg ][OH ] 7 2 12
sp
2 2
# #= = =
+ - - -
 and solving gives x as 710. he molar solubility of Mg(OH)2 at a pH 
of 7.00 is the concentration of Mg2+, which is x or 710 M; clearly, 
Mg(OH)2 is very soluble in a pH 7.00 bufer.
 If the solution is not bufered, then we have
To display this table within the available 
space, we identify the physical state only 
those species that are not present as aque-
ous ions or molecules.