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Complex Ions

These problems are solved with the same method as all other equilibrium problems. In this case, the equilibrium reaction is the metal ion reacting with ligands creating the complex ion. The equilibrium constant values for this type of reaction, K_f, are usually quite large.

>> Example 1

What is the concentration of metal ion for a 0.13 M solution Ag(CN)₂^–? (K_f = 3.0 x 10²⁰)

Solution:

The reaction is

Ag⁺ + 2 CN^– Ag(CN)₂^–

Kf

=

[Ag(CN)2–] [Ag+][CN–]2

The ICE table

	Ag+	CN–	Ag(CN)2–
Initial	0	0	0.13
Change	+x	+2x	–x
Equilibrium	x	2x	0.13 – x

Using the K_f formula

3.0 x 1020

=

(0.13 – x) x(2x)

Assuming x is small (after all, K very much favors the complex ion),

3.0 x 1020

=

0.13 2x2

6.0 x 10²⁰x² = 0.13

x = 1.5 x 10^–11

Checking the assumption, x is small.

x = [Ag⁺] = 1.5 x 10^–11 M

>> Example 2

What is the concentration of metal ion in a solution of 0.17 M Pb(OH)₃^– at pH = 12.42? (K_f = 8 x 10¹³)

Solution:

The reaction is

Pb²⁺ + 3 OH^– Pb(OH)₃^–

Kf

=

[Pb(OH)3–] [Pb2+][OH–]3

=

8 x 1013

A pH of 12.42 implies a very basic solution. The concentration of OH^– in this solution is 0.026 M. The ICE table is

	Pb2+	OH–	Pb(OH)3–
Initial	0	0.026	0.17
Change	+x	+3x	–x
Equilibrium	x	0.026 + 3x	0.17–x

Using the K_f

8 x 1013

=

(0.17 – x) (x)(0.026 + 3x)

Assume 3x is small, and if 3x is, so is x.

8 x 1013

=

0.17 (x)(0.026)

2.08 x 10¹²x = 0.17

x = 8 x 10^–14

Well, x is small!

x = [Pb²⁺] = 8 x 10^–14 M

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