I do not quiet understand what you said. Could you show your work.
What is the solubility of AgCl in 10 M NH3? The Kf of Ag(NH3)2+ from AgCl(s) in aqueous NH3 is 0.0031, which is for the balanced reaction shown below.
AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) + Cl-(aq)
Basically I said the definition for Kf is for
Ag^+ + 2NH3 ==> [Ag(NH3)2]^+ and that is something like 1.7E7. I don't think your Kf you quoted of 0.0031 is Kf; I think it is for
AgCl + 2NH3 ==> [Ag(NH3)2]^+ + Cl^- and that makes it Keq for that reaction and that is not the same as Kf which I showed above. The way you calculate Keq is Ksp*Kf (unless of course it is given in the problem) and if you use 1.8E-10 for Ksp and 1.7E7 for Kf you get 0.0031. As for showing my work, I'll be happy to check yours. If you had shown your work before I could have told you hours ago where the error is.
To find the solubility of AgCl in 10 M NH3, you need to use the concept of solubility product constant (Ksp) and the formation constant (Kf) of Ag(NH3)2+.
The solubility product constant (Ksp) relates the concentrations of the dissolved ions in a saturated solution. For a salt with the equation AgCl(s) ↔ Ag+(aq) + Cl-(aq), the Ksp expression is given by [Ag+][Cl-].
The formation constant (Kf) relates the concentration of the complex ion formed in a solution. In this case, the formation constant for the complex ion Ag(NH3)2+ is given as 0.0031, which represents the equilibrium constant for the reaction AgCl(s) + 2 NH3(aq) ↔ Ag(NH3)2+(aq) + Cl-(aq).
To determine the solubility of AgCl in 10 M NH3, you need to recognize that the Cl- and NH3 ions in the equation are spectator ions and will not affect the solubility of AgCl. Therefore, the equation can be simplified to:
AgCl(s) ↔ Ag(NH3)2+(aq)
Now, let's assume that the solubility of AgCl in 10 M NH3 is x. Since one AgCl molecule forms one Ag(NH3)2+ complex, the concentration of Ag(NH3)2+ in the saturated solution is also x.
Therefore, you can write the equation for the equilibrium constant as:
Kf = [Ag(NH3)2+]/[AgCl]
Substituting the given values:
0.0031 = x/x
Simplifying the equation:
0.0031 = 1
As you can see, this equation cannot be satisfied, which means that the solubility of AgCl in 10 M NH3 is essentially 0. This indicates that AgCl is insoluble in a solution with 10 M NH3.