This is what school teachers teach us (without explanation):

The net equation is, Ag^{+} + 2NH_{3} <-> Ag(NH_{3})_{2}^{+} + Cl^{-} K = K_{sp}K_{f} = 2.89*10^{-3}

**According to the net equation**, make x = [Cl^{-}] = [Ag(NH_{3})_{2}^{+}]

Solve: 2.89*10^{-3} = x^{2} / (1-2x)^{2}

And get x = 0.049M

this is correct. I'll try to explain this method to you. When dissolving AgCl in a solution containing 1 M NH3 we get the following 2 equilibria:

AgCl(s) <--> Ag

^{+} + Cl

^{-} : K

^{s} = 1,7 x 10

^{-10}Ag

^{+} + 2 NH

_{3} --> Ag(NH3)

_{2}^{+} : Ks = 1,7 x 10

^{7}So the total reaction occuring here is:

AgCl + 2 NH

_{3} <--> Ag(NH

_{3})

_{2}^{+} + Cl

^{-} : K

_{s } = ?

Now we must find this Ks-value for the toal reaction and to get this one, we need to combine these equilibria by using substitution:

K

_{s} = [Ag

^{+}][Cl

^{-}] = 1,7 x 10

^{-10}K

_{s} = [Ag(NH

_{3})

_{2}^{+}] / ( [Ag

^{+}][NH

_{3}]

^{2} ) = 1,7 x 107

The combined K

_{s}:

K

_{s} = [Ag(NH

_{3})

_{2}^{+}][Cl

^{-}] / ([Ag

^{+}][NH

_{3}]

^{2} )

Because 1,7 x 10

^{7} >> 1,7 x 10

^{-10} we can say, according the LeChatelier principle, All the Ag+ formed during solvation is complexed. So we don't know what [Ag

^{+}] is ( we know [Cl

^{-}] = [Ag(NH

_{3})

_{2}^{+}], as you correctly stated)

But from the The fact that from dissolving the complex K

_{s} = [Ag(NH

_{3})

_{2}^{+}] / ( [Ag

^{+}][NH

_{3}]

^{2} ), we can also say:

Ks * [Ag

^{+}] * [NH

_{3}]

^{2} [Ag(NH

_{3})

_{2}^{+}] --> [Ag

^{+}] = [Ag(NH

_{3})

_{2}^{+}] / (1,7 x 10

^{7} * [NH

_{3}]

^{2})

Let's put that in the other Ks, [Ag

^{+}][Cl

^{-}]:

1,7 x 10-10 = [Ag(NH

_{3})

_{2}^{+}] * [Cl

^{-}] / (1,7 x 10

^{7} * [NH

_{3}]

^{2})

Thus:

1,7 x 10

^{-10 } * 1,7 x 10

^{7} = [Ag(NH

_{3})

_{2}^{+}] * [Cl

^{-}] / [NH

_{3}]

^{2}) = 2,89 x 10

^{-3}You're with me so far ? We only used the solubility product of the complex ion and modified it so that we could put in the solubility product of AgCl.

Now let's assume we dissolve x mol AgCl in 1 L 1 M NH3. Then, as already said 1,7 x 10

^{7} >> 1,7 x 10

^{-10} and LeChatelier principle, [Ag

^{+}] = 0 and [Cl-] = [Ag(NH

_{3})

_{2}^{+}] = x.

Then we only need to see that x mol Ag

^{+} reacts with 2x mol NH

_{3}. We started with 1 mol / L NH

_{3}, so at equilibrium we have (1-2x) M NH

_{3}.

Let's plug that into our derived equation:

2,89 x 10

^{-3} = x

_{2} / (1-2x)

^{2}Taking the square root on both sides:

5,38 x 10

^{-2} = x / (1-2x)

--> 5,38 x 10

^{-2} - 1,08 x 10

^{-1}x = x

--> 5,38 x 10

^{-2} = 1,11x

--> x = [Cl

^{-}] = [Ag(NH

_{3})

_{2}^{+}] = 0,049 M

I hope this makes sense to you. I must say: this is very tough and rigorous maths for highschool chemistry. But i think you can just assume that when adding 2 equilibria, the equilibrium constant of the summed reaction is the product of the equilibrium constants of the separate reactions.