AgCl(s)==> Ag^+ + Cl^- k1 = 1.7E-10
Ag^+ + 2NH3 ==> Ag(NH3)2+ k2 = 1.6E7
....AgCl(s) + 2NH3 ==> Ag(NH3)2 + Cl^-
I..............2........0..........0
C.............-2x.......x..........x
E............2-2x.......x..........x
Substitute the E line into K (which is k1k2) and solve for x = (Cl^-) = AgCl.
Then convert M to grams AgCl per L then to 100 mL.
Post your work if you get stuck.
Calculate the mass of AgCl that will dissolve in 100ml of 2 M NH3 ksp AgCl = 1.7x10^-10 and Kf of Ag(NH3)2 = 1.6x10^7
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