Look at the two equations and the values of their

Look at the two equations and the values of their equilibrium constants:
Ag
+
+ Cl
–
AgCl(s) K = 1/K
sp
= 5.6 × 10
9
Ag
+
+ 2 Cl
–
[AgCl
2
]
–
K
f
= 2.5 × 10
5
The precipitation reaction is more product-favored, but less susceptible to changes in the concentration of the
chloride ion. In the dilute chloride solution, the silver ion reacts with just one chloride ion to make the solid,
and we see precipitation. In the presence of much higher chloride concentration, each silver ion must react with
two chloride ions, forming a complex ion, which is why we see no precipitation.
Complex Ion Formation
72 . Answer: see equations and expressions below
Strategy and Explanation: The charge of the reactant metal ion is determined by subtracting the Lewis base’s
charge(s), if any, from the complex ion charge.
(a) Ag
+
+ 2 CN
–
[Ag(CN)
2
]
–
K
f
=
[[Ag(CN)
2
]
"
]
[Ag
+
][CN
"
]
2
(b) Cd
2+
+ 4 NH
3
[Cd(NH
3
)
4
]
2+
K
f
=
[[Cd(NH
3
)
4
]
2 +
]
[Cd
2 +
][NH
3
]
4
73. Answer: see equations and expressions below
Strategy and Explanation: The charge of the reactant metal ion is determined by subtracting the Lewis base’s
charge(s), if any, from the complex ion charge.
(a) Co
3+
+ 6 Cl
–
[CoCl
6
]
3–
K
f
=
[[CoCl
6
]
3 "
]
[Co
3 +
][Cl
"
]
6
(b) Zn
2+
+ 4 OH
–
[Zn(OH)
4
]
2–
K
f
=
[Zn(OH)
4
]
2 "
]
[Zn
2 +
][OH
"
]
4
74 . Answer: 7.8 × 10
–3
mol or more
Strategy and Explanation: Adapt the method described in the solution to Question 70, and Problem-Solving
Example 17.13. The Na
2
S
2
O
3
salt provides a source of S
2
O
3
2–
, a Lewis base capable of forming a complex ion
with the silver ion. If all the solid is dissolved it must be have enough S
2
O
3
2–
to complex enough the silver ions
to prevent the precipitation of AgBr in the resulting Br
–
solution. First, we get the balanced equation for the
reaction from Problem-Solving Example 17.13 and determine the value of its equilibrium constant using Tables
17.2 and 17.3:
AgBr(s) + 2 S
2
O
3
2–
(aq) [Ag(S
2
O
3
)
2
]
3–
(aq) + Br
–
(aq) K =
[[Ag(S
2
O
3
)
2
]
3 "
][Br
"
]
[S
2
O
3
2 "
]
2
K = K
sp
× K
f
= [Ag
+
][Br
"
] #
[[Ag(S
2
O
3
)
2
]
3 "
]
[Ag
+
][S
2
O
3
2 "
]
2
=
[[Ag(S
2
O
3
)
2
]
3 "
][Br
"
]
[S
2
O
3
2 "
]
2
K = K
sp
× K
f
= (3.3 × 10
–13
) × (2.0 × 10
13
) = 6.6
Now, calculate the [Br
–
] and [[Ag(S
2
O
3
)
2
]
3–
], once all the AgBr has dissolved:

Chapter 17: Additional Aqueous Equilibria 805
[Br
–
] =
0.020 mol AgBr
1.0 L
"
1 mol Br
#
1 mol AgBr
= 0.020 M,
Similarly, [[Ag(S
2
O
3
)
2
]
3–
] = 0.020 M
Now, we can calculate the necessary [S
2
O
3
2–
]: 6.6 =
(0.020)(0.020)
[S
2
O
3
2 "
]
2
[S
2
O
3
2–
] = 7.8 × 10
–3
M
One mole of S
2
O
3
2–
is found in each mole of Na
2
S
2
O
3
, so we must add 7.8 × 10
–3
mol of Na
2
S
2
O
3
to this
1.0 L solution.
75. Answer: 1.2 × 10
–7
M
Strategy and Explanation: Take the product-favored reaction completely to products, then return it to
equilibrium using one variable, x.
Ag
+
(aq)
+ 2 NH
3
(aq)
[Ag(NH
3
)
2
]
+
(aq)
initial conc. (M) 0.063 c 0
change as forward reaction
occurs (M)
– 0.063 – 0.126 + 0.063
final conc. (M) 0 c – 0.126 0.063
change as reverse reaction
occurs (M)
+ x + 2x – x
equilibrium conc. (M) x 0.18 M = c – 0.126 + 2x 0.063 – x
K
f
=
[[Ag(NH
3
)
2
]
+
]
[Ag
+
][NH
3
]
2
= 1.6 × 10
7
1.6 × 10
7
=
(0.063 " x)
(x)(0.18)
2
Solve for x: (1.6 × 10
7
) × (x)(0.18)
2
= 0.063 – x
(5.2 × 10
5
)x = 0.063 – x
[1 + (5.2 × 10
5
)]x = 0.063
x = 1.2 × 10
–7
[[Ag(NH
3
)
2
]
+
] = 0.063 M – 1.2 × 10
–7
= 0.063 M
[Ag
+
] = 1.2 × 10
–7
M
76 . Answer: see equations below