Solubility Equilibria Regarding #AgCl(s)# and Complex-Ion Formation #Ag(NH_3)_2^+# Question?

If in the end you manage to dissolve #"0.010 M"# of #"Cl"^(-)# in solution for your first equilibrium, and the two sequential reactions occurring are

#"AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl"^(-)(aq)# #" "bb((1))#

#K_(sp) = ["Ag"^(+)]["Cl"^(-)]#

#"Ag"^(+)(aq) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+)(aq)# #" "bb((2))#

#K_f = (["Ag"("NH"_3)_2^(+)])/(["Ag"^(+)]["NH"_3]^2)#,

then these add to be:

#"AgCl"(s) + 2"NH"_3(aq) -> "Cl"^(-)(aq) + "Ag"("NH"_3)_2^(+)(aq)#

and the net equilibrium constant #beta# would be the product.

#beta = K_(sp)K_f = 1.8 xx 10^(-10) cdot 1.0 xx 10^8 = 1.8 xx 10^(-2)#

#= cancel(["Ag"^(+)])["Cl"^(-)] cdot (["Ag"("NH"_3)_2^(+)])/(cancel(["Ag"^(+)])["NH"_3]^2)#

#= (["Ag"("NH"_3)_2^(+)]["Cl"^(-)])/(["NH"_3]^2)#

The ICE table for the net reaction could then look like:

#"AgCl"(s) + 2"NH"_3(aq) -> "Cl"^(-)(aq) + "Ag"("NH"_3)_2^(+)(aq)#

#"I"" "-" "" "" "["NH"_3]_i" "" ""0 M"" "" "" ""0 M"#
#"C"" "-" "" "-2s" "" "" "+s" "" "" "+s#
#"E"" "-" "["NH"_3]_i - 2s" ""0.010 M"" "" "" "s#

where #["NH"_3]_i# is the initial concentration of ammonia added to cause #"0.010 M"# of #"Cl"^(-)# to dissolve in solution from the initially added #"AgCl"(s)#.

Note that the #"0.010 M"# is after adding #"NH"_3(aq)#, *so it is assigned to the FINAL concentration, and not the initial.*

The initial, whatever it would have been without ammonia, would have been much less than the #"Cl"^(-)# that does dissociate from #"AgCl"(s)# from addition of #"NH"_3(aq)#.

At this point, you know #s#, the solubility, is #"0.010 M"# of #"Cl"^(-)#, so #2s = "0.020 M"# and you can then set up the mass action expression:

#1.8 xx 10^(-2) = ("0.010 M")^2/(["NH"_3]_i - "0.020 M")^2#

This is a perfect square, so:

#0.1342 = ("0.010 M")/(["NH"_3]_i - "0.020 M")#

Flipping the fraction gives:

#7.454 = (["NH"_3]_i - "0.020 M")/("0.010 M")#

And solving for the #"NH"_3# you added gives:

#color(blue)(["NH"_3]_i = "0.0946 M")#

And since this was in #"1.00 L"# solution, you added #"0.095 mols"# of #"NH"_3(aq)# to two sig figs.

#AgCl(s) rightleftharpoons Ag^(+) + Cl^(-)# with #K_(sp) = 1.8*10^-10#
#Ag^(+)(aq) + 2NH_3(aq) rightleftharpoons Ag(NH_3)_2^(+)(aq)# with #K_f = 1.0*10^8#

#AgCl(s) + 2NH_3(aq) rightleftharpoons Ag(NH_3)^+ + Cl^(-)# where,

#K_(sp) * K_f = ([Ag(NH_3)^+][Cl^-])/([NH_3]^2) approx 1.8*10^-2#

#1.8*10^-2 = (0.01x)/(2x)^2#
#0.072x^2-0.01x= 0#
#therefore x = 0.13#

You want to dissolve 0.01 mol of #"AgCl"#, so you will have 0.010 mol of #"Ag"("NH"_3)_2^"+"# and 0.010 mol of #"Cl"^"-"# at equilibrium. The initial amount of #"NH"_3# is unknown.

The equilibrium reactions and the ICE table will look like this:

#color(white)(mmmmmm)"AgCl(s)"color(white)(mmmmmmll) ⇌ "Ag"^"+""(aq)" +color(white)(m) "Cl"^"-""(aq)"; K_text(sp)#
#ul(color(white)(mmmmmm)"Ag"^"+""(aq)" + "2NH"_3"(aq)" ⇌ "Ag"("NH"_3)_2^"+" + "Cl"^"-""(aq)"; K_text(f))#
#color(blue)(color(white)(mmmmmm)"AgCl(s)" color(white)(l)+ "2NH"_3"(aq)" ⇌ "Ag"("NH"_3)_2^"+" + "Cl"^"-""(aq)"; K_text(oa) = K_text(sp)K_text(f))#

#"I/mol·L"^"-1":color(white)(mmmmmmmml)xcolor(white)(mmmmmml)0color(white)(mmmmm)0#
#"C/mol·L"^"-1":color(white)(mmmmmmll)"-0.020"color(white)(mmmll)"+0.010"color(white)(mm)"+0.010"#
#"E/mol·L"^"-1":color(white)(mmmmmm)x"-0.020"color(white)(mmmml)0.010color(white)(mmm)0.010#

#K_text(oa) = K_text(sp)K_text(f) = 1.8 × 10^"-10" × 1.0 × 10^8 = 0.018#

#K_text(oa) = ([ "Ag"("NH"_3)_2^"+"]["Cl"^"-"])/ (["NH"_3]^2) = 0.010^2/(x"-0.020")^2= 0.018#

#0.010/(x"-0.020") = 0.134#

#0.010 = 0.134(x"-0.020") = 0.134x - 0.00268#

#x = (0.010 + 0.00268)/0.134 = 0.01268/0.134= 0.095#

Check:

#0.010^2/("0.095 - 0.020")^2 = 0.00010/0.075^2 = 0.00010/0.00562 = 0.018#

It checks!