Question

Calculate The concentration of Ag⁺  present in solution at equilibrium when equilibrium concentration of [NH₃] =0.20 M. The initial Ag(NH₃)₂⁺ concentration is 0.010M. The K_f of Ag(NH₃)₂⁺ is 1.7 x 10⁷.

Ag⁺(aq)+ 2NH₃(aq)--- >Ag(NH₃)₂⁺(aq)

Answer 1

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

initial     0 M     0 M                   0.01 M

change      +x      +x                     -x

equili      +x      0.2 M                 0.01-x

Kf = [ Ag(NH3)2+]/[Ag+][NH3]^2

(1.7*10^7) = (0.01-x)/(x*0.2^2)

x = 1.47*10^-8 M

[Ag+] at equilibrium = x = 1.47*10^-8 M