Calculate the molar solubility of #\sf{AgI}# in a 3.0 M #\sf{NH_3}# solution, if the #\tt{K_(sp)}# is #\tt{1.5xx10^-16}# and the #\tt{K_f}# for #\tt{[Ag(NH_3)_2]^(+)}# is #\tt{1.5xx10^7}#?

#["I"^(-)]_(eq) = 1.42 xx 10^(-4) "M"#

Well, what I would do first is find the new equilibrium reaction that now occurs.

#"AgI"(s) rightleftharpoons cancel("Ag"^(+)(aq)) + "I"^(-)(aq)#, #K_(sp) = 1.5 xx 10^(-16)#
#ul(cancel("Ag"^(+)(aq)) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+)(aq))#, #K_f = 1.5 xx 10^7#
#"AgI"(s) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+) + "I"^(-)(aq)#

For this, the composite equilibrium constant is the product:

#beta = K_(sp)K_f#

#= 2.25 xx 10^(-9) = (["Ag"("NH"_3)_2^(+)]["I"^(-)])/(["NH"_3]^2)#

The #"3.0 M"# #"NH"_3# then serves as the initial concentration.

#"AgI"(s) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+) + "I"^(-)(aq)#
#"I"" "-" "" "3.0" "" "" "" "0" "" "" "" "" "0#
#"C"" "-" "-2x" "" "" "+x" "" "" "" "+x#
#"E"" "-" "3.0-2x" "" "" "x" "" "" "" "" "x#

Thus,

#2.25 xx 10^(-9) = x^2/(3.0 - 2x)^2#

And this is easily solvable without the quadratic formula.

#4.74 xx 10^(-5) = x/(3.0 - 2x)#

We can see #beta# is small, so the small x approximation applies:

#4.74 xx 10^(-5) ~~ x/3.0#

Therefore,

#color(blue)(x ~~ 1.42 xx 10^(-4) "M" -= ["I"^(-)]_(eq))#

That is then the molar solubility of #"I"^(-)#, and hence #"AgI"#, as they are #1:1#.