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75Chapter 6 Equilibrium Chemistry
 here are ive species whose concentrations deine this system (Ca2+, 
F–, HF, H3O+, and OH–), which means we need ive equations that 
relate the concentrations of these species to each other; these are the 
three equilibrium constant expressions
.K 3 9 10[Ca ][F ] 11
sp
2 2
#= =
+ - -
.K 6 8 10
[HF]
[H O ][F ] 4
a
3
#= =
+ -
-
.K 1 00 10[H O ][OH ] 14
w 3 #= =
+ - -
 a charge balance equation
2[Ca ] [H O ] [OH ] [F ]2
3+ = +
+ + - -
 and a mass balance equation
2 [Ca ] [HF] [F ]2
# = +
+ -
 To solve this system of ive equations, we make a guess for [Ca2+], and 
then use Ksp to calculate [F–], the mass balance equation to calculate 
[HF], Ka to calculate [H3O+], and Kw to calculate [OH–]. We eval-
uate each guess by rewriting the charge balance equation as an error 
function
error 2 [Ca ] [H O ] [OH ] [F ]2
3#= + - -
+ + - -
 searching for a [Ca2+] that gives an error suiciently close to zero. 
Successive iterations over a narrower range of concentrations for Ca2+ 
will lead you to a equilibrium molar solubility of 2.1×10-4 M.
 (b) To ind the solubility of AgCl we irst write down all relevant 
equilibrium reactions; these are
( ) ( ) ( )s aq aqAgCl Ag Cl? +
+ -
( ) ( ) ( )aq aq aqAg Cl AgCl?+
+ -
( ) ( ) ( )aq aq aqAgCl Cl AgCl2?+
- -
( ) ( ) ( )aq aq aqAgCl Cl AgCl2 3?+
- - -
( ) ( ) ( )aq aq aqAgCl Cl AgCl3 4?+
- - -
 here are six species whose concentrations deine this system (Ag+, 
Cl–, AgCl(aq), AgCl2
- , AgCl2
3
- , and AgCl3
4
- ), which means we need 
six equations that relate the concentrations of these species to each 
other; these are the ive equilibrium constant expressions
.K 1 8 10[Ag ][Cl ] 10
sp #= =
+ - -
.
( )
K
aq
5 01 10
[Ag ][Cl ]
[AgCl ]
3
1 #= =+ -
Be sure you understand why the concen-
tration of Ca
2+
 is multiplied by 2.