CGPET 2017 Solved Paper

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)=

y(0.4)2

x

Solving for x then x=1.15×10−3M