Question

5. The formation constant of the silver-ethylenediamine complex, Ag(NH₂CH₂ CH₂NH₂)⁺, is 5.0 x 10⁴. Calculate the concentration of Ag⁺ in equilibrium with a     0.10 M solution of the complex. (Assume no higher order complexes).

6. What would be the concentration of Ag⁺ in (problem 5) if the solution contained also 0.10 M ethylene diamine (NH₂CH₂ CH₂NH₂).  

Answer 1

Ag^+ + en <==> Ag(en)^+

Kf = [Ag(en)^+]/[Ag+][en]

	Ag^+	en	Ag(en)^+
initial	0	0	0.1
change	+x	+x	-x
equilibrium	x	x	0.1-x

Kf = 0.1-x/x^2

5*10^4 = 0.1-x/x^2

or, 5*10^4x^2 +x-0.1 = 0

or, x =0.0014 M

[Ag+] in equilibrium is 0.0014 M

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	Ag^+	en	Ag(en)^+
initial	0	0.1	0.1
change	+x	+x	-x
equilibrium	x	0.1+x	0.1-x

Kf = 0.1-x/x(0.1+x)

5*10^4 = 0.1-x/(0.1x+x^2)

or, 5*10^4x^2 +5*10^3x-0.1 +x= 0

or, 5*10^4x^2 + 5x*10^3 -0.1 = 0

or, x =1.99*10^-5 M

[Ag+] in equilibrium is 1.99*10^-5 M

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