Question

Calculate the equilibrium concentration of Ag^+(aq ) in a solution that is initially 0.140 M AgNO3 and 0.850 M NH3, and in which the following reaction takes place: Ag^+(aq)+2NH3​(aq)<---->Ag(NH3)2^+(aq) (Kf = 1.70x10^7)

Answer 1

Given,

Concentration of AgNO₃ solution = 0.140 M

Concentration of NH₃ = 0.850 M

If the Volume of the solution is 1.0 L, then

Moles of Ag⁺ ion from AgNO₃ = 0.140 moles

Moles of NH₃ = 0.850 moles

Now,

Ag⁺(aq) + 2NH₃(aq) Ag(NH₃)₂⁺

	Ag+(aq)	2NH3(aq)	Ag(NH3)2+
I(Moles)	0.140	0.850	0
C(Moles)	-0.140	-0.280	+0.140
E(Moles)	0	0.57	0.140

Now, the concentrations are,

[Ag(NH₃)₂⁺] = 0.140 mol /1 L = 0.140 M

[2NH₃] = 0.57 mol /1 L =0.57 M

Drawing an ICE chart,

Ag(NH₃)₂⁺ Ag⁺(aq) + 2NH₃(aq)

	Ag(NH3)2+	Ag+(aq)	2NH3(aq)
I(M)	0.140	0	0.57
C(M)	-x	+x	+2x
E(M)	0.140-x	+x	0.57+2x

The formation constant is,

K_f = [Ag(NH₃)₂⁺] / [Ag⁺] [NH₃]² = 1.7 x 10⁷

Writing the dissociation constant expression,

[Ag⁺] [NH₃]² / [Ag(NH₃)₂⁺] = 1/ K_f = K_d

Substituting the known values,

[x] [ 0.57+2x]² / [0.140-x] = 1/1.7 x 10⁷

x (0.57)² / (0.140) = 5.88 x 10^-8 ---------------[ 0.57-2x] 0.57 and 0.140-x 0.140 Since, x <<<<0.57 and 0.140

x = 2.5 x 10^-8

[Ag⁺] = x =  2.5 x 10^-8 M

Thus, the silver ion concentration is 2.5 x 10^-8 M at equilibrium.