Let solubility by AgBr be x M. Thus, [Br−]=xM and [Ag+]=yM as Ag+ react with NH3 to form a examples. AgBr⇌Ag+(aq)+Br⊖(aq) ksp=[Ag+][Br−]=yx Since, the formation constant (Kt) of the complex is very high, assume that whole Ag+ formed is consumed Ag++2NH3→Ag(NH3)2+
x
0.4
0.4−2x
x
Ag(NH3)2+⇌Ag+(aq)+2NH3
x
0.4−2x
x−y
y
0.4−2x+2y
kd=
[Ag+][NH3]2
[Ag(NH3h2+]
=
y(0.4−2x+2y)2
x−y
x−y≈x since kd(1×10−8 is low ) and x<<0.4 kd(1×10−8)=