Question

consider a solution made by mixing 500.0 mL of 4.0 M NH3 and 500.0 mL of 0.40 M AgNO3. Ag+ reacts with NH3 to form AgNH3+ and Ag(NH3)2^+. K1= 2.1x10^3; K2=8.2x10^3, what is the concentration of Ag(NH3)2^+ at equilibrium?

Answer 1

NH3 moles = 4 x 500/1000= 2

Ag+ moles = 0.4 x 500/1000= 0.2

Ag+ + 2NH3 -------------------------> Ag(NH3)2+

K= [Ag(NH3)2+]/[Ag+][NH3]^2 ------------------------------------------(1)

K1 = [Ag(NH3)+]/[Ag+][NH3] = 2.1 x 10^3 -----------------------------(2)

K2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3 ]= 8.2 x10^3 -----------------------(3)

K1 x K2 = [Ag(NH3)2+]/[Ag+][NH3]^2 = 2.1 x 10^3 x8.2 x10^3

[Ag(NH3)2+]/(0.2)(2)^2 = 1.7 x10^7

[Ag(NH3)2+] = 1.7 x10^7 x 0.2)(2)^2

[Ag(NH3)2+] = 1.36 10^7 M