Chapter 22  Transition Metals and Coordination Chemistry
Complex Ion Equilibria
Given K
f
and initial concentrations, find equilibrium concentration.
28.
0.40 moles of AgNO
3
and 2.5 moles of Na
2
S
2
O
3
(sodium thiosulfate) are dissolved in 1.00L of
aqueous solution.
A coordination complex forms.
Ag
+
(aq) + 2S
2
O
3
2
(aq)
[Ag(S
2
O
3
)
2
]
3
(aq)
K
f
= 1.7 x 10
13
Calculate [Ag
+
] in this solution.
A
.
8.1 x 10
15
M
B.
5.3 x 10
15
M
C.
1.7 x 10
13
M
D.
2.0 x 10
12
M
E.
1.7 x 10
13
M
Ag
+
(aq) + 2S
2
O
3
2
(aq)
[Ag(S
2
O
3
)
2
]
3
(aq)
K
f
= 1.7 x 10
13
Ag
+
(aq)
+
2S
2
O
3
2
(aq)
[Ag(S
2
O
3
)
2
]
3
(aq)
Initial 0.40 2.5 0
Change X 2X +X
Equilibrium 0.40X 2.52X +X
[Ag(S
2
O
3
)
2
]
3
/([Ag
+
][S
2
O
3
2
]
2
)= K
f
= 1.7 x 10
13
([X])/([0.40X] x ([2.52X]
2
)) = K
f
= 1.7 x 10
13
Quadratic equation.
Use the large K rule.
Bring the reaction to completion.
Ag
+
is the limiting reactant.
Ag
+
(aq)
+
2S
2
O
3
2
(aq)
[Ag(S
2
O
3
)
2
]
3
(aq)
Initial 0.40 2.5 0
Change 0.40 0.80 +0.40
Equilibrium 0
1.70
+0.40
[Ag
+
] = 0, but we need a finite value for the [Ag
+
].
Use the RighttoLeft Rule.
Ag
+
(aq)
+
2S
2
O
3
2
(aq)
[Ag(S
2
O
3
)
2
]
3
(aq)
New initial
0
1.70
0.40
Change +X +2X X
Equilibrium
+X
1.70 + 2X
0.40X
[Ag(S
2
O
3
)
2
]
3
/([Ag
+
][S
2
O
3
2
]
2
)
= K
f
= 1.7 x 10
13
[0.40X]/([X] x ([1.70 + 2X]
2
))
= 1.7 x 10
13
Quadratic equation.
Get around this by recognizing that due to a large K, there is much substance on the
right side and very little on the left side.
Therefore, the 2X is very small relative to the 1.70, and the –X is
very small relative to the 0.40.
That is, use the righttoleft rule.
[0.40]/([X][1.70]
2
)
= 1.7 x 10
13
X = 8.142 x 10
15
= [Ag
+
]
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
67
Complex Ion Formation Effect on Salt Solubility
Ligands can compete with counteranions for bonding with positive ions.
Adding a
potential ligand to the solution might thus affect solubility.
That is, if the ligand is strong
enough, it can attract a positive ion away from its electrostatic bond with a negatively
charged counterion, thus forming a new bond, and dissolving a precipitate.
Consider silver chloride, AgCl, as an example.
AgCl(s)
⇄
Ag
+
(aq)
+ Cl

(aq)
K
sp
= 1.6 x 10
10
If NH
3
is added to a solution containing silver ions, a complex ion, Ag(NH
3
)
2
+
, will form:
Ag
+
+ 2NH
3
⇄
Ag(NH
3
)
2
+
K
f
= 1.5 x 10
7
According to LeChatelier’s principle, as Ag
+
is removed (“
consumed
” by combining with
NH
3
), more AgCl will dissolve.
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
68
Chem 1622003 Quiz 3
Version A1
ET: Given K
sp
and K
f
, find K
c
.
Then, using K
c
and initial concentration, find solubility.
2mod. What is the solubility of AgI in water?
What is the solubility of AgI in a
3.00M KSCN solution?
K
sp
(AgI) = 1.5 x 10
16
;
K
f
([Ag(SCN)
4
]
3
) = 1.2 x 10
10
Solubility of AgI in H
2
O:
AgI(s)
Ag
+
(aq)
+
I

(aq)
Initial Y
0
0
Change X
+X
+X
Equilibrium YX
X
X
[Ag
+
][I

] = K
sp
[X][X] = 1.5 x 10
16
[X] = 1.2 x 10
8
M = AgI solubility in H
2
O
Solubility of AgI in KSCN:
AgI(s)
Ag
+
(aq) + I

(aq)
K
sp
= 1.5 x 10
16
Ag
+
(aq) + 4SCN

(aq)
Ag(SCN)
4
3
(aq)
K
f
= 1.2 x 10
10
AgI(s) + 4SCN

I

+ Ag(SCN)
4
3
K = 1.8 x 10
6
AgI(s)
+
4SCN

(aq)
I

(aq) +
Ag(SCN)
4
3
(aq)
Initial Y
3.00 0 0
Change X
4X
+X +X
Equilibrium YX
3.00  4X
+X
+X
([I

][Ag(SCN)
4
3
])/[SCN

]
4
= 1.8 x 10
6
([X][X])/[3.00  4X]
4
= 1.8 x 10
6
Quadratic equation; therefore, use small K rule.
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 pH, solubility equilibria lecture, equilibria lecture notes