Answer:
[tex]7.60x10^{-5}mol/L[/tex]
Explanation:
Hello, in this case, one can consider the following reaction:
[tex]AgI(s)+2NH_3(aq)<-->Ag(NH_3)_2(aq)^++I^-(aq)[/tex]
And the equilibrium constant for the whole process:
[tex]K=K_{sp}*K_{f}=1.7x10^7*8.5x10^{-17}=1.445x10^{-9}[/tex]
The equilibrium state turns out into:
[tex]K=\frac{[Ag(NH_3)_2^+][I^-]}{[NH_3]^2} \\K=\frac{x^2}{(2.0-2x)^2}\\\sqrt{K}= \sqrt{\frac{x^2}{(2.0-2x)^2}}\\2.0\sqrt{K}-2\sqrt{K}x-x=0\\7.60x10^{-5}-7.60x10^{-5}x-x=0\\x=\frac{7.60x10^{-5}}{1} =7.60x10^{-5}mol/L[/tex]
So that result accounts for the molar solubility of the AgI into the given solution.
