Page 183 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:41 AM Page 166

166 Modern Analytical Chemistry

+
–
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Counting unknowns, we find that there are seven—[Ag ], [I ], [Ag(NH 3 ) 2 ],
–
+
+
[NH 3 ], [NH 4 ], [OH ], and [H 3 O ]. Four of the equations needed to solve this
problem are given by the equilibrium constant expressions
–
+
K sp =[ Ag ][ ] = .83 ´ 10 –17
I

+
[ Ag NH ) ]
(
3 2
2 b = = . 17 ´ 10 7
+
[ Ag ][ NH ] 2
3
+

–
[ NH ][ OH ] –5
4
K b = = .175 ´ 10
[ NH ]
3
+
–
K w =[ HO ][ OH ] = .100 ´ 10 –14
3
Three additional equations are needed. The first of these equations is a mass balance
for NH 3 .
+
+
= [NH 3 ] + [NH 4 ]+2 ´[Ag(NH 3 ) 2 ]
C NH 3
+
Note that in writing this mass balance equation, the concentration of Ag(NH 3 ) 2
+
must be multiplied by 2 since two moles of NH 3 occurs per mole of Ag(NH 3 ) 2 . The
second additional equation is a mass balance on iodide and silver. Since AgI is the
+
–
only source of I and Ag , every iodide in solution must have an associated silver
ion; thus
+
+
–
[I ] = [Ag ] + [Ag(NH 3 ) 2 ]
Finally, the last equation is a charge balance equation
+
–
+
+
+
–
[Ag ] + [Ag(NH 3 ) 2 ] + [NH 4 ]+[H 3 O ]=[I ] + [OH ]
Our problem looks challenging, but several assumptions greatly simplify the al-
+
gebra. First, since the formation of the Ag(NH 3 ) 2 complex is favorable, we will as-
sume that
+
+
[Ag ] << [Ag(NH 3 ) 2 ]
Second, since NH 3 is a base, we will assume that
–
+
[H 3 O ] << [OH ]
+
+
[NH 4 ] << [NH 3 ] + [Ag(NH 3 ) 2 ]
Finally, since K sp is significantly smaller than b 2 , it seems likely that the solubility of
AgI is small and
+
[Ag(NH 3 ) 2 ] << [NH 3 ]
Using these assumptions allows us to simplify several equations. The mass bal-
ance for NH 3 is now
= [NH 3 ]
C NH 3
–
and the mass balance for I is
+
–
[I ] = [Ag(NH 3 ) 2 ]
+
+
Simplifying the charge balance expression by dropping [H 3 O ] and [Ag ], and re-
–
+
placing [Ag(NH 3 ) 2 ] with [I ] gives