The document discusses the solubility of silver iodide (AgI) in the presence of ammonia. It provides the relevant equilibrium constants for the dissociation of AgI into Ag+ and I- ions (Ksp = 1.5 x 10^-16) and the formation of the Ag(NH3)2+ complex from Ag+ and NH3 (Kf = 1.6 x 10^7). It asks to write the net ionic equation for the increased solubility of AgI in ammonia and calculate the overall equilibrium constant by combining the individual constants.

1. Consider the insoluble compound silver iodide , AgI . The silver ion also forms a complex with
ammonia . Write a balanced net ionic equation to show why the solubility of AgI (s) increases in
the presence of ammonia and calculate the equilibrium constant for this reaction.
For Ag(NH 3 ) 2
+ , K f = 1.6×10 7 . Use the pull-down boxes to specify states such as (aq) or (s).
______ ( ) + _______ ( ) <---> ______ ( ) + _______ ( )
K= _________
Solution
AgI (s) ---------------> Ag+ (aq) + I- (aq) ; Ksp = 1.5 x 10^-16
Ag+ (aq) + 2 NH3 (aq) ------------> Ag(NH3)2+ ; Kf = 1.6 x 10^7
__________________________________________________________
AgI (s) + 2 NH3 (aq) ------------> Ag(NH3)2+ (aq) + I- (aq)
Keq = Kf x Ksp
= 1.6 x 10^7 x 1.5 x 10^-16
Keq = 2.4 x 10^-9
equilibrium constant Keq = 2.4 x 10^-9